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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 3, , The equation of the graph is y = 2x + 1. This represents a line with slope 2 and, y−intercept 1. With these informations, we can sketch the portion of the graph of, f . The expression 2x + 1 is defined for all real numbers, and hence the domain of, f is the set of all real numbers, which we denote by R. The graph shows that the, range is also R., Example 1.2. Sketch the graph of g(x) = x2 and find the domain and range., , AP, P, , ., , Solution. The equation of the graph is y = x2 ., , Y, , x, , −2, , −1, , 0, , 1, , 2, , y, , 4, , 1, , 0, , 1, , 4, , 8, 6, 4, , CO, R, , 2, −2 −1 0, , y = x2, X, 1 2, , From the tabular column we can plot the points (−2, 4), (−1, 1), (0, 0), (1, 1) and (2, 4), and join them to produce the graph, which represents a parabola. The domain of, g is R. The range of g consists of all values of g(x), that is, all numbers of the form, , U, , x2 . But x2 ≥ 0 for all numbers x and any positive number y is a square. Hence, the, range of g is {y : y ≥ 0} = [0, ∞]., , ST, , The graph of a function is a curve in the xy−plane. But whether all curves in, the xy−plane represent the graph of some function. This can be answered by the, vertical line test., , The vertical line test. A curve in the xy−plane is the graph of a function of, , x if and only if no vertical line intersects the curve more than once., Consider the following graphs., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 4, , Engineering Mathematics - I, , Y, , Y, x=a, , x=a, , (a, b), , (a, b), , a, , 0, , X, , 0, , a, , X, , AP, P, , ., , (a, c), , In the first graph, each vertical line x = a intersects the curve only once, at, (a, b), then exactly one function value is defined by f (a) = b. But in the second, graph, the line x = a intersects the curve twice at (a, b) and (a, c), that is, the, function assigns two different values to a. Hence, the first curve represents the, , CO, R, , graph of a function while the second graph is not., As an illustration consider the parabola x = y2 − 4. The curve that represents, this equation is give below., , This curve does not represent the, , Y, , graph of a function of x because, we can, see that every vertical line intersects, , U, , x = y2 − 4, , the parabola twice. This parabola is the, , union of the graphs of two functions of, , (−4, 0), , ST, , 0, , X, , x. Rewriting the equation as y2 = x + 4, √, we get y = ± x + 2., , Hence, the upper and lower halves of the parabola are the graphs of the, √, √, functions f (x) = x + 4 and g(x) = − x + 4. The separate graphs are given below., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , Y, , 5, , y=, , √, , Y, x+4, (−4, 0), , (−4, 0), , 0, , √, y=− x+4, , AP, P, , ., , X, , 0, , X, , If we reverse the roles of x and y, then we get the equation x = h(y) = y2 − 4, definitely define x as a function of y with y as the independent variables and x as, the dependent variable and in this case, the parabola represent the graph of the, function h., , CO, R, , Piecewise difined functions., , , , , , 1 − x, , Example 1.3. Consider the function f defined by f (x) = , , , , x2, For all x ≤ −1., , -3, , -2, , y=1−x, , 4, , 3, , 6, 5, , 2, , 4, , For all x > −1., , ST, , ., , , if x > −1, , Y, , -1, , U, , x, , , if x ≤ −1, , 3, , x, , -1, , 0, , 1, , 2, , 3, , 2, , y = x2, , 1, , 0, , 1, , 4, , 9, , 1, , The combined graph of the function f is, , as follows, , −4 −3 −2 −1 0, , 1, , 2, , 3, , 4, , X, , Notice that the point (−1, 1) in y = x2 is not included and there is a discontinuity, between the graph for all x ≤ 1 (i.e.y = 1 − x) and the graph for all x > 1 (i.e. y = x2 )., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 6, , Engineering Mathematics - I, , Example 1.4. Consider the absolute value function f (x) = |x|, which is defined as, , , , , , if x ≥ 0, x, , . The graph of f (x) consists of two branches namely, follows f (x) = , , , , −x , if x < 0, y = x for all x ≥ 0 and y = −x for all x < 0., , ., , x, y, , 0, , 1, , 2, , 3, , 0, , 1, , 2, , 3, , AP, P, , The combined graph of f is given below., y = x (x ≥ 0), , Y, , 3 y = |x|, 2, , x, , y = −x (x < 0), -3, , -2, , -1, , 0, , 3, , 2, , 1, , 0, , −4 −3 −2 −1 0, , 1, , CO, R, , y, , 1, , 3, , 2, , 4, , Notice that the point (0, 0) is not included in the graph of y = −x (x < 0)., , , , , , x, , , , , , , Example 1.5. Consider the function f (x) = , 2−x, , , , , , , , , 0, , U, 1, , 0, , 1, , ST, , y, , 0, , , if 1 < x ≤ 2 ., , , if x > 2, y = 0 (x > 2), , y = 2 − x (1 ≤ x ≤ 2), , y = x (0 ≤ x ≤ 1), x, , , if 0 ≤ x ≤ 1, , x, , 1, , 2, , x, , 2, , 3, , 4, , 5, , y, , 1, , 0, , y, , 0, , 0, , 0, , 0, , 1 2, , 3, , 4, , The graph of f is given below., Y, 1, 0, , 5, , X, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , X
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 7, , Example 1.6. Consider the greatest integer function defined by f (x) = [x]., By definition [x]= Largest integer that is less than or equal to x, = max{m ∈ Z; m ≤ x}., Example. [2.7] = 2, [−1.3] = −2, [4] = 4, The graph of the function is as follows, , ., , corresponding points., 5, 4, 3, 2, 1, , The Examples 1.3 to 1.6 indicate that, there is jump from one value to the, next. Hence, the above functions are, called piecewise functions. The function, X, , discussed in Example 1.6 is called step, , CO, R, , −2 −1 0 1 2 3 4 5, −1, −2, symmetry, , AP, P, , The open dots are not included at the, Y, , function., , Even function. If a function f satisfies f (−x) = f (x) for every number x in its, , U, , domain, then f is called an even function., , Example. f (x) = x2 is even,, , Y, , ST, , since f (−x) = (−x)2 = x2 = f (x)., Geometrically, tha graph of an even, , function is symmetric about the y−axis., Consider the curve of y = x2 − 4., , f (−x), , f (x), , −x, , 0, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , x, , X
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 8, , Engineering Mathematics - I, , Odd function. If f statifies f (−x) = − f (x) for every number x in the domain, then, f is called an odd function., Example. Consider f (x) = x3, f (−x) = (−x)3 = −x3 = − f (x)., , ∴ f (x) = x3 is an odd function., , Y, , − f (x), , AP, P, , ., , f (x), , −x, , x, , 0, , X, , The graph of the odd function is symmetric about the origin., , Example 1.7. Determine the nature of the following functions., , Solution., , (ii) g(x) = 1 + x4 ., , (iii) h(x) = x + x2 ., , CO, R, , (i) f (x) = x5 − x., , (i) f (x) = x5 − x., , f (−x) = (−x)5 − (−x) = −x5 + x = −(x5 − x) = − f (x)., , ∴ f (x) is odd., (ii) g(x) = 1 + x4 ., , g(−x) = 1 + (−x)4 = 1 + x4 = g(x)., , U, , ∴ g(x) is even., , (iii) h(x) = x + x2 ., , ST, , h(−x) = (−x) + (−x)2 = −x + x2 ., , h(−x) , h(x) and h(−x) , −h(x), ∴ h(x) is neither even nor odd., , Increasing and decreasing functions. A function f is called increasing on an, , interval I, if f (x1 ) < f (x2 ) whenever x1 < x2 in I., It is called decreasing on I, if f (x1 ) > f (x2 ) whenever x1 < x2 in I., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 9, , 1.2 Limit of a function, Definition. Intuitive definition of a limit. Suppose f (x) is defined. When x is, near the number a, then we write lim f (x) = L. This means that the limit of f (x),, x→a, , as x approaches a equals L., lim f (x) = L can also be represented as f (x) → L as x → a., , x→a, , ., , AP, P, , One sided limits. The left hand limit of f (x) as x approaches a is written as, lim f (x) = L which means that the limit of f (x) as x approaches a from the left, , x→a−, , is equal to L if we can make the values of f (x) arbitrarily close to L by taking x, sufficiently close to a with x less than a., , Similarly, if we require that x greater than a, we get the right hand limit of, f (x) as x approaches a is equal to L and we write lim f (x) = L., x→a+, , Result. lim f (x) = L if and only if lim f (x) = L and lim f (x) = L., Infinite limits., , x→a−, , x→a+, , CO, R, , x→a, , (i) Let f ba a function defined on both sides of a, except possibly at a itself. Then, lim f (x) = ∞ means that the value of f (x) can be made arbitrarily large by taking x, , x→a, , sufficiently close to a, but not equal to a., , (ii) Let f be a function defined on both sides of a, expect possibly at a itself. Then, , U, , lim f (x) = −∞ means that the values of f (x) can be made arbitrarily large negative, , x→a, , ST, , by taking x sufficiently close to a, but not equal to a., 1, Example. lim 2 = ∞., x→0 x, !, 1, lim − 2 = −∞., x→0, x, Asymptote. The vertical line x = a is called a vertical asymptote of the curve, , y = f (x) if atleast one of the following statements is true:, lim f (x) = ∞, lim f (x) = ∞, lim f (x) = ∞., , x→a, , x→a−, , x→a+, , lim f (x) = −∞, lim f (x) = −∞, lim f (x) = −∞., x→a, x→a−, x→a+, 2x, Example. (i) For the function f (x) =, , x = 3 is a vertical asymptote., x−3, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 10, , Engineering Mathematics - I, , π, is a vertical asymptote., 2, Laws of limits. Suppose that c is a constant and the limits lim f (x) and lim g(x), (ii) For the curve f (x) = tan x, x =, , x→a, , x→a, , exist, then the following laws are true., , , 1. lim f (x) + g(x) = lim f (x)+ lim g(x)., x→a, , x→a, , x→a, , , 2. lim f (x) − g(x) = lim f (x) − lim g(x)., x→a, , x→a, , x→a, , , , 4. lim f (x)g(x) = lim f (x) × lim g(x)., x→a, , x→a, , x→a, , # lim f (x), f (x), . if lim g(x) , 0., = x→a, 5. lim, lim g(x), x→a, x→a g(x), x→a, n, , , 6. lim f (x) n = lim f (x) , where n is a positive integer., ", , x→a, , x→a, , 7. lim c = c., x→a, , 8. lim x = a., x→a, , CO, R, , ., , x→a, , , , 3. lim c f (x) = c lim f (x)., , AP, P, , x→a, , , , 9. lim xn = an , where n is a positive integer, x→a, , 10. lim, , x→a, , 11. lim, , p, n, , x=, , √n, a, where n is a positive integer and a > 0., , f (x) =, , q, n, , lim f (x)., , x→a, , U, , x→a, , √n, , Direct substitution property. If f is a polynomial or a rational function and a, is in the domain of f , then lim f (x) = f (a)., x→a, , ST, , Theorem. If f (x) ≤ g(x), when x is near a and the limits of f and g both exist as x, approaches a, then lim f (x) ≤ lim g(x)., x→a, , x→a, , If f (x) ≤ g(x) ≤ h(x), when x is near a and, , The Squeeze Theorem., , lim f (x) = lim h(x) = L, then lim g(x) = L., , x→a, , x→a, , x→a, , The above theorem is sometimes called Sandwich Theorem or the Pinching, Theorem., , ✎, , ☞, , ✍, , ✌, , Worked Examples, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 12, , Engineering Mathematics - I, (3 + x)2 − 9, ., x→0, x, , Example 1.12. Evaluate lim, , (3 + x)2 − 9, (3 + x − 3)(3 + x + 3), = lim, x→0, x→0, x, x, x(6 + x), = lim, x→0, x, , Solution. lim, , = lim (6 + x) = 6., x→0, , AP, P, , x2 + 9 − 3, ., x→0, x2, Solution. On rationalizing the numerator we get, √, √, √, x2 + 9 − 3, x2 + 9 − 3, x2 + 9 + 3, =, lim, ×, lim, √, x→0, x→0, x2, x2, x2 + 9 + 3, 2, x +9−9, = lim √, x→0 x2 x2 + 9 + 3, x2, = lim √, x→0 x2 x2 + 9 + 3, 1, 1, 1, = lim √, = ., =, x→0, x2 + 9 + 3 3 + 3 6, , Example 1.13. Find lim, , CO, R, , ., , √, , U, , Example 1.14. Show that lim |x| = 0., x→0, , , , , , ifx ≥ 0, , x, ., Solution. We know that |x| = , , , , −x , ifx < 0, Since |x| = −x for x < 0, we have lim |x| = lim −x = 0., x→0−, , x→0−, , Also |x| = x for x > 0, we have lim |x| = lim x = 0., x→0+, , x→0+, , ST, , ∴ lim |x| = lim |x| = 0., x→0−, , x→0+, , Hence, lim |x| = 0., x→0, , |x|, Example 1.15. Prove that lim, does not exist., x→0 x, , , , , , ifx ≥ 0, , x, Solution. We know that |x| = , ., , , , −x , ifx < 0, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 13, , |x|, x, = lim = 1., x→0+ x, x→0+ x, −x, |x|, = lim, = −1., lim, x→0− x, x→0− x, |x|, |x|, |x|, Since lim, , lim, , lim, does not exist., x→0+ x, x→0− x x→0 x, √, , , , x − 4 , ifx > 4, , , determine whether lim f (x) exists., Example 1.16. If f (x) = , , , x→4, , 8 − 2x , ifx < 4, √, Solution. We have f (x) = x − 4 for x > 4., √, ∴ lim f (x) = lim x − 4 = 0., , ., , x→4+, , x→4+, , Also, f (x) = 8 − 2x for x < 4., ∴ lim f (x) = lim (8 − 2x) = 0., x→4−, , x→4−, , AP, P, , Now, lim, , Since lim f (x) = lim f (x) = 0, we have lim f (x) exists and lim f (x) = 0., x→4+, , x→4−, , x→4, , x→4, , Example 1.17. Show that for the greatest integer function [x], lim [x] does not, , CO, R, , x→3, , exist., , Solution. By definition [x] = 3 for 3 ≤ x < 4., ∴ lim [x] = lim 3 = 3., x→3+, , x→3+, , Also [x] = 2 for 2 ≤ x < 3., ∴ lim [x] = lim 2 = 2., x→3−, , x→3−, , Since lim [x] , lim [x], lim [x] does not exist., x→3−, , x→3, , U, , x→3+, , ST, , 1.3 Continuity, , Definition. A function f is continuous at a number a, if lim f (x) = f (a)., x→a, , f is discontinuous at a if it is not continuous at a., , Note. For proving a function f to be continuous, we have to prove the following., 1. f (a) must be defined., 2. lim f (x) exists., x→a, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 14, , Engineering Mathematics - I, 3. lim f (x) = f (a)., x→a, , x2 − x − 2, ., x−2, Here, f (x) is not defined at x = 2., , Example. Consider f (x) =, , (i) f (2) is defined and f (2) = 1., (ii) lim f (x) = lim, x→2, , x→2, , , ifx , 2, , ., , , ifx = 2, , AP, P, , ., , ∴ f is not continuous at x = 2., , , x2 −x−2, , , , x−2, If we define f (x) as f (x) = , , , , 1, Then, , (x − 2)(x + 1), x2 − x − 2, = lim, = lim (x + 1) = 3, x→2, x→2, x−2, x−2, , (iii) Since f (2) = 1 , lim f (x), f (x) is not continuous at x = 2., x→2, , Definition. A function f is said to be continuous from the right at a number a if, , CO, R, , lim f (x) = f (a), and f is said to be continuous from the left at a if lim f (x) = f (a)., , x→a+, , x→a−, , Example. Consider the greatest integer function f (x) = [x]. We notice that, lim f (x) = lim [x] = n = f (n), , x→n+, , x→n+, , lim f (x) = lim [x] = n − 1 , f (n)., , x→n−, , x→n−, , ∴ f (x) = [x] is continuous from the right but not continuous from the, left., , U, , Definition. A function f is continuous on an interval if it is continuous at every, number in the interval., , ST, , Theorem. If f and g are continuous at a and c is a constant, then the following, functions are also continuous at a., 1. f + g, , 2. f − g, , 3. c f, , 4. f g, , 5., , f, g, , g(a) , 0., , Theorem., , (a) Any polynomial is continuous everywhere. i.e., it iscontinuous on R = (−∞, ∞)., (b) Any rational function is continuous wherever it is defined, i.e., it is continuous, , on its domain., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 15, , Theorem. The following types of functions are continuous at every number in, their domains., 1. Polynomials., 2. Rational functions., , ., , 4. Trigonometric functions., 5. Inverse trigonometric functions., 6. Exponential functions., 7. Logarithmic functions., , AP, P, , 3. Root functions., , x→a, , CO, R, , Theorem. If f is continuous at b, and lim g(x) = b, then lim f (g(x)) = f (b)., x→a, , x→a, i.e., lim f (g(x)) = f lim g(x) ., x→a, , Theorem. If g is continuous at a and f is continuous at g(a), then the composite, function f og given by ( f og)(x) = f (g(x)) is continuous at a., The intermediate value theorem. Suppose that f is continuous on the closed, interval [a, b] and let N be any number between f (a) and f (b), where f (a) , f (b)., , U, , Then there exists a number c in (a, b) such that f (c) = N., Result. The intermediate value theorem is useful in the location of the roots, , ST, , of the given equation., , Example 1.18. Find the domain where the fnction f is continuous. Also find the, , numbers, at, , , , , , 1 + x2, , , , , , , f (x) = , 2−x, , , , , , , , , (x − 2)2 x, , which, , the, , function, , f, , is, , discontinous, , where, , , if x ≤ 0, , , if 0 < x ≤ 2 ., , [A.U. Dec. 2015], , , if x > 2, Solution. The function f changes its value at x = 0, and x = 2., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 16, , Engineering Mathematics - I, , lim f (x) = lim (1 + x2 ) = 1., , x→0−, , x→0−, , lim f (x) = lim (2 − x) = 2., , x→0+, , x→0+, , Since, lim f (x) = lim f (x), f is not continuous at x = 0., x→0−, , x→0+, , lim f (x) = lim (2 − x) = 0., , x→2−, , x→2−, , lim f (x) = lim (x − 2)2 = 0., , x→2+, , x→2+, , Since, lim f (x) = lim f (x) = 0 = f (2), f (x) is continuous at x = 2., x→2−, , AP, P, , ., , x→2+, , ∴ The only number at which the function is discontinuous is at x = 0., The domain of continuity of f is {(−∞, 0) ∪ (0, ∞)., The graph of f is given below, , The combined graph of the function f is, For all x ≤ 0., , as follows, , y = f (x) = 1 + x2 ., -3, , y, , 10, , For all 0 < x ≤ 2., , CO, R, , x, , Y, , -2, , -1, , 0, , 10, , 5, , 2, , 1, , 9, 8, , 1 + x2, , y = f (x) = 2 − x., 0, , y, , 2, , 6, , 1, , 2, , 5, , 1, , 0, , 4, , U, , x, , 7, , 3, , y = f (x) = (x − 2)2 ., , 2, , ST, , For all x > 2., , x, , 2, , 3, , 4, , y, , 0, , 1, , 4, , 1, −4 −3 −2 −1 0, , (x − 2)2, , 2−x, 1, , 2, , 3, , 4, , X, , Notice that the point (−1, 1) in y = x2 is not included and there is a discontinuity, , between the graph for all x ≤ 1 (i.e.y = 1 − x) and the graph for all x > 1 (i.e. y = x2 )., Example 1.19. Show that there is a root of the equation 4x3 − 6x2 + 3x − 2 between, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 17, , 1 and 2., Solution.Let f (x) = 4x3 − 6x2 + 3x − 2., f (1) = 4(1)3 − 6(1)2 + 3(1) − 2 = 4 − 6 + 3 − 2 = −1 < 0., f (2) = 4(2)3 − 6(2)2 + 3(2) − 2, , AP, P, , ., , = 4 × 8 − 6 × 4 + 3 × 2 − 2 = 32 − 24 + 6 − 2 = 12 > 0., The function changes its sign between 1 and 2., , Y, , Since f (x) is a polynomial, which is, continuous in its domain the graph of, y = f (x) must cross the x−axis atleast, , CO, R, , at one point say x = c between x = 1, , f (1), , c, , and x = 2 such that f (c) = 0, which is, the root of the equation f (x) = 0. This, , 1, , 2, , X, , f (2), , proves that the given equation f (x) = 0, has a root between 1 and 2., , U, , Example 1.20. Prove that the equation x3 − 15x + 1 = 0 has atmost one real root, in the interval [−2, 2]., , [A.U.Nov.2016], , ST, , Solution. Let f (x) = x3 − 15x + 1, f (−2) = −8 + 30 + 1 = 23 = +ve., f (−1) = −1 + 15 + 1 = 15 = +ve., , + + +, , f (0) = 1 = +ve., , f (1) = 1 − 15 + 1 = −13 = −ve., f (2) = 8 − 30 + 1 = −21 = −ve., , −2 −1 0, , 1 2, − −, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 18, , Engineering Mathematics - I, Hence, f (0) > 0 > f (1)., f changes sign between 0 and 1., ∴ By intermediate value theorem, there is a number c between 1 and 2, such that f (c) = 0., ∴ The given equation has atleast one root c in the interval (0, 1)., Since f (x) changes sign in the interval [−2, 2] only once between 0 and, 1, y = f (x) crosses the x axis only once., , AP, P, , ., , i.e., f (x) attains 0 only once in [−2, 2], , Hence, there is atmost one real root in the interval [−2, 2]., , 1.4 Derivatives and differentiation rules, , CO, R, , Definition. The tangent line to the curve y = f (x) at the point P(a, f (a)) is the line, f (x) − f (a), ., through P with slope m = lim, x→a, x−a, , ✎, , ☞, , ✍, , ✌, , Worked Examples, , ST, , U, , Definition. The derivative of a function f at a number a, denoted by f ′ (a) is, f (a + h) − f (a), defined as f ′ (a) = lim, , if the limit exists., h→0, h, Note. Let x = a+h. As h → 0, x → a. The equivalent definition for the derivative, f (x) − f (a), ., is f ′ (a) = lim, x→a, x−a, , Example 1.21. Find an equation to the tangent line to the parabola y = x2 at the, point P(1, 1)., , Solution. Given a = 1, f (a) = 1, f (x) = x2 ., f (x) − f (1), x→1, x−1, (x − 1)(x + 1), x2 − 1, = lim, = lim (x + 1) = 1 + 1 = 2., = lim, x→1, x→1, x→1 x − 1, x−1, , Slope m = lim, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 19, , Equation of the tangent line through (1, 1) is, y − y1 = m(x − x1 )., y − 1 = 2(x − 1), y = 2x − 2 + 1, , ., , AP, P, , y = 2x − 1., Example 1.22. Find an equation to the tangent line to the hyperbola y =, point (3, 1)., Solution. Given a = 3, f (a) = f (3) = 1., , 3, at the, x, , f (a + h) − f (a), h→0, h, f (3 + h) − f (3), = lim, h→0, h, 3, −1, = lim 3+h, h→0, h, 3−3−h, = lim, h→0 h(3 + h), −h, −1, −1, = lim, = lim, =, ., h→0 h(3 + h), h→0 (3 + h), 3, , CO, R, , We have m = lim, , ST, , U, , ∴ The equation of the tangent line is, , y − y1 = m(x − x1 )., y−1=, , −1, (x − 3), 3, , 3y − 3 = −x + 3, x + 3y − 3 − 3 = 0, x + 3y − 6 = 0., , Example 1.23. Find the derivative of the function f (x) = x2 − 8x + 9 at a number, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 20, , Engineering Mathematics - I, , a. Find also the equation the tangent line at the point (3, −6)., f (a + h) − f (a), h→0, h, 2, (a + h) − 8(a + h) + 9 − (a2 − 8a + 9), = lim, h→0, h, 2, 2, a + h + 2ah − 8a − 8h + 9 − a2 + 8a − 9, = lim, h→0, h, h2 + 2ah − 8h, = lim, h→0, h, , ., , = lim (h + 2a − 8) = 2a − 8., h→0, , AP, P, , Solution. f ′ (a) = lim, , We have m = f ′ (a) = f ′ (3) = 2 × 3 − 8 = −2., ∴ The equation of the tangent line is, , y − y1 = m(x − x1 )., , CO, R, , y + 6 = −2(x − 3), = −2x + 6, , 2x + y = 0., , Derivative of a function. Let f (x) be a given function. The derivative of f (x) at, f (x + h) − f (x), ., any variable point x is defined by f ′ (x) = lim, h→0, h, , U, , Example 1.24. If f (x) = x3 − x, find a formula for f ′ (x)., Solution.Given f (x) = x3 − x., , f (x + h) − f (x), h→0, h, (x + h)3 − (x + h) − (x3 − x), = lim, h→0, h, 3, 2, x + 3x h + 3xh2 + h3 − x − h − x3 + x, = lim, h→0, h, 2, 2, 3, 3x h + 3xh + h − h, = lim, h→0, h, , ST, , f ′ (x) = lim, , = lim 3x2 + 3xh + h2 − 1 = 3x2 − 1., h→0, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 21, , Example 1.25. If f (x) =, Solution.Given f (x) =, , √, , √, , x, find the derivative of f . State the domain of f ′ ., , x., , f (x + h) − f (x), h→0, h, √, √, x+h− x, = lim, h→0, h, √, √, √, √, x+h− x, x+h+ x, × √, = lim, √, h→0, h, x+h+ x, x+h−x, = lim √, √ , h→0 h, x+h+ x, , ., , h, = lim √, √ , h→0 h, x+h+ x, = lim √, , 1, 1, √ = √ ., √ = √, x+ x 2 x, x+h+ x, 1, , CO, R, , h→0, , AP, P, , f ′ (x) = lim, , From this we notice that f ′ (x) exists if x > 0., ∴ The domain of f ′ is (0, ∞)., , U, , Example 1.26. Where is the function f (x) = |x| differentiable., [A.U. Nov. 2015], , , , , if x ≥ 0, , x, ., Solution. We have that |x| = , , , , −x if x < 0, Consider x > 0. Then |x| = x. We choose h small enough such that x + h > 0., ∴ |x + h| = x + h., , ST, , ∴ When x > 0,, , f (x + h) − f (x), h→0, h, |x + h| − |x|, = lim, h→0, h, x+h−x, h, = lim, = lim = lim 1 = 1., h→0, h→0, h, h h→0, , f ′ (x) = lim, , ∴ f is differentiable for x > 0., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 22, , Engineering Mathematics - I, Consider x = 0., f (0 + h) − f (x), h, |0 + h| − |x|, = lim, h→0, h, |h|, [if it exists], = lim, h→0 h, , f ′ (0) = lim, , h→0, , ∴ f is not differentiable at x = 0., Consider x < 0., In this case |x| = −x., , AP, P, , ., , Let us find the left and right limits., h, |h|, = lim = lim 1 = 1., lim, h→0+ h, h→0+, h→0+ h, −h, |h|, = lim, = lim −1 = −1., lim, h→0− h, h→0, h→0− h, Since the two limits are different, f ′ (0) does not exist., , CO, R, , Choose h small so that x + h < 0., ∴ |x + h| = −(x + h), ∴ For x < 0,, , f (x + h) − f (x), h→0, h, −(x + h) − (−x), = lim, h→0, h, −h, −x − h + x, = lim, = lim −1 = −1., = lim, h→0 h, h→0, h→0, h, , U, , f ′ (x) = lim, , ST, , ∴ f is differentiable for x < 0., , Combining all the three cases we obtain that f is differentiable at all x except 0., , , , , ifx > 0, , 1, ′, ., ∴ f (x) = , , , , −1 ifx < 0, Theorem. If f is differentiable at a, then f is continuous at a., , Proof. Given that f is differentiable at a., f (x) − f (a), f ′ (x) = lim, exits., x→a, x−a, , (1), , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 23, , f (x) − f (a), (x − a), x−a, Taking limits as x → a both sides we get, , Now, f (x) − f (a) =, , f (x) − f (a), (x − a), x→a, x−a, f (x) − f (a), = lim, × lim (x − a), x→a, x→a, x−a, , lim f (x) − f (a) = lim, , x→a, , = f ′ (a) × 0, , [by (1)], , lim f (x) − f (a) = 0, , ., , (2), , AP, P, , x→a, , , , Also lim f (x) = lim f (a) + ( f (x) − f (a)), x→a, , x→a, , = lim + lim ( f (x) − f (a)), x→a, , x→a, , [by (1)], , = f (a) + 0, lim f (x) = f (a), , x→a, , CO, R, , ∴ f (x) is continuous at a., Result. The converse of the above theorem is false., i.e., If a function is continuous at a need not be differentiable at a., For example consider the function f (x) = |x|., lim f (x) = lim |x| = 0 = f (0)., , x→a, , x→a, , But by example 1.26, f (x) = |x| is not differentiable at 0., , U, , Higher derivatives. If a function f is differentiable, then its derivative f ′ is also, a function. Hence, f ′ may have a derivative of its own denoted by ( f ′ )′ = f ′′ . f ′′ is, , ST, , called !the second derivative of f . The second derivative of y = f (x) is denoted by, d2 y, d dy, = 2 . In a similar way we can have the third derivative of f denoted by, dx dx, dx, !, d3 y, d d2 y, =, . If this process is, ( f ′′ )′ = f ′′′ . If y = f (x) then we have y′′′ = f ′′′ (x) =, dx dx2, dx3, continued we can get in general, the nth derivative of f denoted by f (n) is obtained, dn y, by differentiating f n times. Generally if y = f (x) then we write y(n) = f (n) (x) = n ., dx, Rules on differentiation, 1. The, , sum, , rule., , If, , f, , and, , g, , are, , both, , differentiable,, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , then
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 24, , Engineering Mathematics - I, , d , , d , d , f (x) + g(x) =, f (x) +, g(x) ., dx, dx, dx, Proof. Let F(x) = f (x) + g(x). Then,, F(x + h) − F(x), h, f (x + h) + g(x + h) − [ f (x) + g(x)], = lim, h→0, h, [ f (x + h) − f (x)] + [g(x + h) − g(x)], = lim, h→0, h, [ f (x + h) − f (x)], = lim, + lim [g(x + h) − g(x)]h, h→0, h→0, h, , F ′ (x) = lim, , ., , = f ′ (x) + g′ (x)., , AP, P, , h→0, , CO, R, , 2. Constant multiple rule. If c is a constant and f is a differentiable function,, , , d , d , c f (x) = c, f (x) ., then, dx, dx, Proof. Let F(x) = c f (x). Then,, F(x + h) − F(x), h→0, h, c f (x + h) − c f (x), = lim, h→0, h, c[ f (x + h) − f (x)], = lim, h→0, h, [ f (x + h) − f (x)], = c lim, h→0, h, , , d, = c f ′ (x) = c, f (x) ., dx, , U, , F ′ (x) = lim, , ST, , 3. The difference rule., If f and g are both differentiable, then, , d , , d , d , f (x) − g(x) =, f (x) −, g(x) ., dx, dx, dx, Proof. We have f (x) − g(x) = f (x) + (−)g(x)., , , d , d , f (x) − g(x) =, f (x) + (−1)g(x), dx, dx, d , , d , f (x) +, (−1)g(x), =, dx, dx, , , d , d , =, f (x) + (−1), g(x), dx, dx, , [by sum rule], [by constant multiple rule], , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 25, , =, , d , , d , f (x) −, g(x) ., dx, dx, , 4. The product rule., If f and g are both differentiable,, , , , d , d , d , f (x) × g(x) = f (x), g(x) + g(x), f (x) ., dx, dx, dx, , then, , f (x + h)g(x + h) − f (x)g(x), h→0, h, f (x + h)g(x + h) − f (x + h)g(x) + f (x + h)g(x) − f (x)g(x), = lim, h→0, h, f (x + h){g(x + h) − g(x)} + g(x){ f (x + h) − f (x)}, = lim, h→0, h, (, ), g(x + h) − g(x), f (x + h) − f (x), = lim f (x + h), + g(x), h→0, h, h, f (x + h) − f (x), g(x + h) − g(x), + lim g(x) lim, = lim f (x + h) lim, h→0, h→0, h→0, h→0, h, h, , Proof. ( f (x)g(x))′ = lim, , AP, P, , ., , CO, R, , ( f (x)g(x))′ = f (x)g′ (x) + g(x) f ′ (x)., 5. Quotient rule., If f d and g, ", #, d, g(x) dx, f (x) − f (x) dx, g(x), d f (x), =, ., 2, dx g(x), [g(x)], !′, , both, , differentiable,, , f (x+h) f (x) , , , , , , , g(x+h) − g(x) , = lim , , , , , , h→0 , h, , ), (, f (x + h)g(x) − g(x + h) f (x), = lim, h→0, hg(x + h)g(x), (, ), f (x + h)g(x) − f (x)g(x) + f (x)g(x) − g(x + h) f (x), = lim, h→0, hg(x + h)g(x), (, ), g(x)[ f (x + h) − f (x)] − f (x)[g(x + h) − g(x)], = lim, h→0, hg(x + h)g(x), , , f (x+h)− f (x), , , , , − f (x) g(x+h)−g(x), , g(x), h, h, = lim , , , , , h→0 , g(x + h)g(x), , ST, , U, , Proof., , f (x), g(x), , are, , lim g(x), , =, , h→0, , f (x + h) − f (x), g(x + h) − g(x), − lim f (x), h→0, h, h, lim g(x + h)g(x), h→0, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , then
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 26, , Engineering Mathematics - I, g(x + h) − g(x), f (x + h) − f (x), − lim f (x) lim, h→0, h→0, h→0, h→0, h, h, =, g(x)g(x), g(x) f ′ (x) − f (x)g′ (x), =, ., [g(x)]2, lim g(x) lim, , Derivatives of simple functions, , ., , d, (c) = 0., dx, , Proof. Let f (x) = c., f (x + h) − f (x), h→0, h, c−c, 0, = lim, = lim = 0., h→0 h, h→0 h, , f ′ (x) = lim, , AP, P, , 1. If c is a constant, prove that, , 2. If n is a positive integer, then prove that, , CO, R, , Proof. Let f (x) = xn ., , d n, (x ) = nxn−1 ., dx, , f (x + h) − f (x), h→0, h, (x + h)n − xn, = lim, h→0, h, n, n, x + C1 xn−1 h + nC2 xn−2 h2 + nC3 xn−3 h3 + · · · + hn − xn, = lim, h→0, h, n(n−1) n−2 2, n(n−1)(n−2) n−3 3, n−1, nx h + 2 x h +, x h + · · · + hn, 6, = lim, h→0, h, n(n − 1) n−2 1 n(n − 1)(n − 2) n−3 2, n−1, x h +, x h + · · · + hn−1, = lim nx +, h→0, 2, 6, n(n − 1) n−2 1, n(n − 1)(n − 2) n−3 2, = lim nxn−1 + lim, x h + lim, x h + · · · + lim hn−1, h→0, h→0, h→0, h→0, 2, 6, , ST, , U, , f ′ (x) = lim, , = nxn−1 + 0 + 0 + · · · + 0 = nxn−1 ., , General rule. If n ia any real number, then, , d n, (x ) = nxn−1 ., dx, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 27, , 3. Derivative of e x . Prove that, , d x, (e ) = e x ., dx, , eh − 1, = 1., h→0, h, , Definition of the number e. e is a number such that lim, Proof. Let f (x) = e x ., f (x + h) − f (x), h→0, h, e x+h − e x, = lim, h→0, h, x, h, e e − ex, = lim, h→0, h, !, h, x e −1, = lim e, h→0, h, !, eh − 1, x, = lim e × lim, = ex × 1 = ex ., h→0, h→0, h, , ., , AP, P, , f ′ (x) = lim, , CO, R, , Two important limits, sin θ, = 1., θ→0 θ, , 1. Prove that lim, , D, , Proof. Consider a circle with center, O and radius 1 unit., , Let arc(AB), , U, , subtends an angle θ at the centre, π, and assume that 0 < θ < . Draw, 2, BC perpendicular to OA. Let the, , B, E, θ, 0 C, , A, , ST, , tangents at A and B meet at E., Produce AE and OB to intersect at D., , Now we have arcAB = θ., , From ∆OAB, sin θ =, , BC, OB, , =, , BC, 1 ., , ∴ |BC| = sin θ., From the figure we have, |BC| < |AB| < arcAB., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 29, , Proof. Let f (x) = sin(x), f (x + h) − f (x), h→0, h, sin(x + h) − sin(x), = lim, h→0, h, sin(x) cos(h) + cos(x) sin(h) − sin(x), = lim, h→0, h, sin(x)(cos(h) − 1) + cos(x) sin(h), = lim, h→0, h, !, sin(x)(cos(h) − 1) cos(x) sin(h), = lim, +, h→0, h, h, !, sin(h), (cos(h) − 1), + lim cos(x), = lim sin(x), h→0, h→0, h, h, !, sin(h), (cos(h) − 1), = sin(x) lim, + cos(x) lim, h→0, h→0, h, h, , f ′ (x) = lim, , AP, P, , ., , = sin(x) × 0 + cos(x) × 1 = 0 + cos(x) = cos(x)., , CO, R, , d, (cos x) = − sin x., dx, Proof. Let f (x) = cos(x), , 2. Prove that, , f (x + h) − f (x), h→0, h, cos(x + h) − cos(x), = lim, h→0, h, cos(x) cos(h) − sin(x) sin(h) − cos(x), = lim, h→0, h, cos(x)(cos(h) − 1) − sin(x) sin(h), = lim, h→0, h, !, cos(x)(cos(h) − 1) sin(x) sin(h), = lim, −, h→0, h, h, !, sin(h), (cos(h) − 1), = lim cos(x), − lim sin(x), h→0, h→0, h, h, , ST, , U, , f ′ (x) = lim, , = cos(x) × 0 − sin(x) × 1 = − sin(x)., , f ′ (x) = − sin x., 3. Prove that, , d, (tan x) = sec2 x., dx, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 30, , Engineering Mathematics - I, Proof. Let f (x) = tan(x), !, d, d, (sin x) − sin x dx, (cos x), cos x dx, d, d sin x, =, (tan x) =, 2, dx, dx cos x, cos x, cos x × cos x − sin x × (− sin x), =, cos2 x, 2, 1, cos2 x + sin x, =, = sec2 x., =, cos2 x, cos2 x, , AP, P, , ., , d, (cosecx) = −cosecx cot x., dx, !, d, d, (1) − 1 × dx, (sin x), sin x dx, d, d, 1, Proof., (cosecx) =, =, dx, dx sin x, sin2 x, sin x × 0 − 1 × (cos x), =, sin2 x, −1, cos x, − cos x, =, ×, = −cosecx cot x., =, 2, sin x sin x, sin x, , 4. Prove that, , U, , CO, R, , d, (sec x) = sec x tan x., dx, !, d, d, (1) − 1 × dx, (cos x), cos x dx, d, d, 1, Proof., (sec x) =, =, 2, dx, dx cos x, cos x, cos x × 0 − 1 × (− sin x), =, cos2 x, sin x, sin x, 1, =, =, ×, = sec x tan x., 2, cos x cos x cos x, , 5. Prove that, , d, (cot x) = −cosec2 x., dx, d, d, d, d cos x sin x dx (cos x) − cos x × dx (sin x), Proof., (cot x) =, =, dx, dx sin x, sin2 x, sin x × (− sin x) − cos x × (cos x), =, sin2 x, − sin2 x − cos2 x −(sin2 x + cos2 x), 1, =, =, = − 2 = −cosec2 x., 2, 2, sin x, sin x, sin x, , ST, , 6. Prove that, , The chain rule. If g is differentiable at x and f is differentiable at g(x), then the, composite function F = f og defined by F(x) = f (g(x)) is differentiable at x and F ′ is, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 31, , given by the product F ′ (x) = f ′ (g(x)) × g′ (x)., Proof. Let y = F(x) = f (g(x)), Let u = g(x) ⇒ y = f (u), Let ∆x be a small change in x, then ∆u is the corresponding change in u., ∆u = g(x + ∆x) − g(x)., Then the corresponding change in y is given by, , AP, P, , Now,, , ∆y, dy, = lim, dx ∆x→0 ∆x, ∆y ∆u, = lim, ×, ∆x→0 ∆u, ∆x, ∆y, ∆u, = lim, × lim, ∆x→0 ∆u, ∆x→0 ∆x, ∆y, ∆u, [∆u → 0 as ∆x → 0], = lim, × lim, ∆x→0 ∆u, ∆x→0 ∆x, dy du, =, ×, du dx, d, d, ( f (u)) ×, (g(x)) = f ′ (u) × g′ (x), =, du, dx, , CO, R, , ., , ∆y = f (u + ∆u) − f (u), , d, [F(x)] = f ′ (g(x)) × g′ (x), dx, , U, , F ′ (x) = f ′ (g(x)) × g′ (x), , Example 1.27. Evaluate limπ, , ST, , x→ 2, , ✎, , ☞, , ✍, , ✌, , Worked Examples, , 1 + cos 2x, ., (π − 2x)2, , [A.U.Jan. 2015], , 1 + cos 2x, 2 cos2 x, =, lim, x→ 2 (π − 2x)2, x→ π2 (π − 2x)2, π, Let, −x=y, 2, π, As x → , y → 0, 2, 1 + cos 2x, 2 cos2 x, = limπ , ∴ limπ, 2, x→ 2, x→ 2 (π − 2x)2, 4 π−x, , Solution. limπ, , 2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 33, , Applying the product rule we obtain, , dy, d 2, =, x sin x, dx dx, d, d, = x2 (sin x) + sin x (x2 ), dx, dx, = x2 × (cos x) + sin x × (2x), = x2 cos x + 2x sin x., , =, =, =, , d, d, (sec x) − sec x dx, (1 + tan x), (1 + tan x) dx, , (1 + tan x)2, , , , (1 + tan x)(sec x tan x) − sec x 0 + sec2 x, (1 + tan x)2, sec x tan x + sec x tan2 x − sec3 x, (1 + tan x)2, sec x(tan x + tan2 x − sec2 x), (1 + tan x)2, sec x(tan x − (sec2 x − tan2 x)), (1 + tan x)2, sec x(tan x − 1), (1 + tan x)2, sec x(tan x − 1), ., (1 + tan x)2, , U, , =, , AP, P, , f ′ (x) =, , CO, R, , ., , dy, sec x, , find, . For what values of x, does the graph of, 1 + tan x, dx, f , has a horizontal tangent., sec x, Solution. Given f (x) =, ., 1 + tan x, By the quotient rule, we have, Example 1.31. If f (x) =, , =, , ST, , f ′ (x) =, , The points where the graph of f has a horizontal tangent are given by f ′ (x) = 0, i.e.,, , sec x(tan x − 1), =0, (1 + tan x)2, , ⇒ sec x(tan x − 1) = 0, But sec x , 0, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 34, , Engineering Mathematics - I, , ∴ tan x − 1 = 0, , π, i.e., tan x = 1 ⇒ x = nπ + , n is an integer., 4, , Example 1.32. Find the points on the curve y = x4 − 6x2 + 4 where the tangent line, is horizontal., , ., , i.e.,4x3 − 6 × 2x = 0, 4x3 − 12x = 0, 4x(x2 − 3) = 0, , dy, = 0., dx, , AP, P, , Solution. Horizontal tangent occur at points where, , √, i.e.,x = 0 or x2 − 3 = 0 ⇒ x2 = 3 ⇒ x = ± 3., When x = 0, y = 4., , U, , CO, R, , The point is (0, 4)., √ 4, √ 2, √, When x = 3, y =, 3 − 6 3 + 4 = 9 − 18 + 4 = −5., √, The point is ( 3, −5)., √ 4, √ 2, √, When x = − 3, y = − 3 − 6 − 3 + 4 = 9 − 18 + 4 = −5., √, The point is (− 3, −5)., √, √, The given curve has horizontal tangents at the points (0, 4), ( 3, −5) and (− 3, −5)., Example 1.33. At what point on the cure y = e x , is the tangent line parallel to the, line y = 2x., , ST, , Solution. Given y = e x ⇒ y′ = e x ., , slope of the tangent at x = e x ., , Since the tangent line is parallel to the line y = 2x,, Slope of the tangent = 2., i.e., e x = 2, , ⇒, , x = log 2., , When x = log 2, y = elog 2 = 2., ∴ The required point is (log 2, 2)., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 35, , Example 1.34. If f (x) = xe x find f (n) (x)., Solution. Given f (x) = xe x ., , d, xe x, dx, d x, d, =x, e + e x (x) = xe x + e x .1 = e x (x + 1)., dx, dx, , d, (x + 1)e x, f ′′ (x) =, dx, d x, d, = (x + 1), e + e x (x + 1), dx, dx, , ., , AP, P, , f ′ (x) =, , = (x + 1)e x + e x .1 = e x (x + 1 + 1) = e x (x + 2)., , d, (x + 2)e x, dx, d x, d, = (x + 2), e + e x (x + 2), dx, dx, , f ′′′ (x) =, , CO, R, , = (x + 2)e x + e x .1 = e x (x + 2 + 1) = e x (x + 3)., Applying this process successively n times we get, , f (n) (x) = e x (x + n)., , Example 1.35. If y =, , x2 + x − 2, ., x3 + 6, d, d, (x2 + x − 2) − (x2 + x − 2) dx, (x3 + 6), (x3 + 6) dx, , U, , Solution. Given y =, , dy, x2 + x − 2, , find, ., 3, dx, x +6, , dy, =, dx, , ST, , (x3 + 6)2, (x3 + 6)(2x + 1) − (x2 + x − 2)(3x2 ), =, (x3 + 6)2, 4, 3, 2x + x + 12x + 6 − 3x4 − 3x3 + 6x2, =, (x3 + 6)2, −x4 − 2x3 + 6x2 + 12x + 6, =, ., (x3 + 6)2, , Example 1.36. Find the equation of the tangent line to the curve y =, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , ex, at the, 1 + x2
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 39, , 1.5 Implicit differentiation, logarithmic differentiation and, Inverse trigonometric function., , ✍, , Example 1.40. If x2 + y2 = 25, find, , AP, P, , ., , Whenever an equation containing two variables x and y, it is not always, dy, directly. In such, possible to express y interns of x and it is not easy to find, dx, case, we differentiate both sides with respect to x and then solve the resultant, dy, equation, for, ., dx, ☞, ✎, Worked Examples, ✌, , dy, . Find also the equation of the tangent to the, dx, , circle at the point (3, 4)., Solution. Given x2 + y2 = 25., , Differentiating both sides w.r.to x we get, , CO, R, , dy, =0, dx, dy, x+y, =0, dx, dy, y, = −x, dx, , 2x + 2y, , ⇒, , dy −x, =, ., dx, y, , dy −3, =, ., dx, 4, −3, Slope of the tangent at (3, 4) = m =, ., 4, Equation of the tangent is, , U, , At (3, 4), , −3, (x − 3), 4, , ST, , y−4=, , 4y − 16 = −3x + 9, 3x + 4y = 16 + 9, , 3x + 4y = 25., , dy, . Find the tangent to the Folium of Descartes, dx, x3 + y3 = 6xy at the point (3, 3). At what point in the first quadrant is the tangent, , Example 1.41. If x3 +y3 = 6xy, find, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 41, , x3 = 0 or x3 − 16 = 0, x3 = 0 or x3 = 16, But x , 0, 4, , ∴ x3 = 16 = 24 ⇒ x = 2 3 ., 4 2, 4, 8, 5, 8, when x = 2 3 , y = 12 2 3 = 12 2 3 = 2 3 −1 = 2 3 ., , 4, , 5, , ., , AP, P, , ∴ In the first quadrant, the tangent line is horizontal at (2 3 , 2 3 )., , dy, if sin(x + y) = y2 cos x., dx, Solution. Given sin(x + y) = y2 cos x., , Example 1.42. Find, , Differentiating both sides w.r.to x we get, , U, , CO, R, , !, dy, dy, = y2 (− sin x) + cos x × 2y, cos(x + y) 1 +, dx, dx, dy, dy, cos(x + y) + cos(x + y), = −y2 sin x + 2y cos x, dx, dx, dy, dy, 2, = −y sin x − cos(x + y), cos(x + y) − 2y cos x, dx, dx, , , dy, (cos(x + y) − 2y cos x) = − y2 sin x + cos(x + y), dx, , , 2, dy − y sin x + cos(x + y), =, (cos(x + y) − 2y cos x), dx, y2 sin x + cos(x + y), =, ., 2y cos x − cos(x + y), , ST, , Example 1.43. If sin y = x sin(a + y) find, , dy, ., dx, , Solution. Given sin y = x sin(a + y), sin y, ∴x=, sin(a + y), Differentiating with respect to y we get, sin(a), dx sin(a + y) cos y − sin y cos(a + y) sin(a + y − y), =, =, =, 2, 2, dy, sin (a + y), sin (a + y), sin2 (a + y), dy sin2 (a + y), ∴, =, ., dx, sin(a), , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 42, , Engineering Mathematics - I, , √, p, dy, 1, =−, ., Example 1.44. If x 1 + y + y 1 + x = 0, prove that, dx √ (1 + x)2, √, p, p, Solution. Given x 1 + y + y 1 + x = 0 ⇒ x 1 + y = −y 1 + x, squaring both sides we get, , x2 (1 + y) = y2 (1 + x), , x2 + x2 y − y2 − y2 x = 0, , ., , x2 − y2 + x2 y − y2 x = 0, (x − y)(x + y) + xy(x − y) = 0, (x − y)(x + y + xy) = 0, ∴ x − y = 0 or x + y + xy = 0, , CO, R, , i.e.,x = y or x + y + xy = 0, , AP, P, , x2 + x2 y = y2 + y2 x, , But x , y., , (1), , ∴ x + y + xy = 0, , ST, , U, , Differentiating both sides w.r.to x we get, !, dy, dy, 1+, + x +y =0, dx, dx, dy, dy, +x +y=0, 1+, dx, dx, dy, (1 + x) = −y − 1, dx, dy −(y + 1), =, ., dx, (1 + x), But form (1) we have y(1 + x) = −x ⇒ y =, x, , −x, ., 1+x, , 1 − 1+x, dy, 1, 1+x−x, =−, ., =−, =−, 2, dx, 1+x, (1 + x), (1 + x)2, r, q, √, dy, Example 1.45. If y = x + x + x + · · · ∞, find, ., dx, r, q, √, Solution. Given y = x + x + x + · · · ∞., ∴, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 45, , dy, =1, sec2 y, dx, dy, 1, 1, 1, =, =, =, ., 2, 2, dx sec y 1 + tan y 1 + x2, dy, ., dx, Solution. Given y = cot−1 x., , 4. If y = cot−1 x, find, , cot y = x., , dy, ., dx, Solution. Given y = sec−1 x., , 5. If y = sec−1 x, find, , CO, R, , sec y = x., , AP, P, , ., , Differentiating w.r.to x, dy, =1, −cosec2 y, dx, −1, dy, −1, −1, =, =, =, ., 2, 2, dx cosec y 1 + cot y 1 + x2, , Differentiating w.r.to x, dy, sec y tan y, =1, dx, dy, 1, 1, 1, 1, =, = p, = √, = p, ., dx sec y tan y x tan2 y x sec2 y − 1 x x2 − 1, , dy, ., dx, Solution. Given y = cosec−1 x., , U, , 6. If y = cosec−1 x, find, , cosecy = x., , ST, , Differentiating w.r.to x, dy, −cosecy cot y, =1, dx, dy, −1, −1, −1, −1, = √, =, = p, = p, ., 2, 2, dx cosecy cot y x cot y x cosec y − 1 x x2 − 1, ✎, , ☞, , ✍, , ✌, , Worked Examples, , 1, Example 1.48. Find the derivatives of (i) −1, sin x, , , 1, −1 −1, Solution. (i) Let y =, = sin x ., sin−1 x, , (ii) x tan−1, , √ , x., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 46, , , −2, 1, dy, 1, = (−1) sin−1 x, =, ., √, , √, 2, dx, 1 − x2, 1 − x2, sin−1 x, √ , (ii) Let y = x tan−1 x ., √, , , dy, 1, x, 1, −1 √ x ., −1 √ x .1 =, +, tan, +, tan, = (x), √ 2 × 2√, 2(1+x), x, dx, 1 + ( x), √ , Example 1.49. Differentiate w.r.to x (i) sin−1 x, (ii) (1 + x2 ) tan−1 x, cos x , ., (iii) sec−1 (x4 ), (iv) tan−1, 1 + sin x, √, Solution. (i) Let y = sin−1 ( x)., 1, 1, 1, 1, dy, = q, √ 2 × 2 √ x = 2 √ x √1 − x = 2 √ x(1 − x) ., dx, 1− x, (ii) Let (1 + x2 ) tan−1 x., 1, dy, = (1 + x2 ) ×, + tan−1 x × 2x = 1 + 2x tan−1 x., dx, 1 + x2, , AP, P, , ., , Engineering Mathematics - I, , CO, R, , (iii) Let sec−1 (x4 )., 1, 4, dy, = q, ., × 4x3 = √, 8−1, 2, dx, x, x, 4, 4, x, x −1, , cos x , ., 1 + sin, x , , , , sin π2 − x , −1 , , ., , = tan , π, 1 + cos 2 − x, , , , sin 2 π4 − 2x , , ., = tan−1 , π, x , 1 + cos 2 4 − 2, , , , , , 2 sin π4 − 2x cos π4 − 2x , ., , , = tan−1 , , x, π, 2, 2 cos 4 − 2, π x , = tan−1 tan, −, ., 4, 2, π x, −, y=, 4 2, dy −1, =, ., dx, 2, !, 2x, dy, −1, ., Example 1.50. If y = tan, find, 2, dx, 1−x, , ST, , U, , (iv) Let y = tan−1, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 48, , Engineering Mathematics - I, , dy, = e x log b log b = b x log b., dx, dy, (ii) If y = logb x find, ., dx, Solution. Given y = logb x, ∴ by = x., , ✍, , AP, P, , ., , Differentiating w.r.to x, dy, by log b, = 1., dx, dy, = y1 = 1 ., dx b log b x log b ✎, ☞, Worked Examples, ✌, , CO, R, , Example 1.53.! Differntiate (i) log(x3 + 1) (ii) log(sin x) (iii) log(2 + sin x), p, x+1, (v) log |x| (vi) log(x)., (iv) log √, x+2, Solution. (i) Let y = log(x3 + 1)., , dy, 3x2, 1, 3x2 = 3, ., = 3, dx x + 1, x +1, (ii) Let y = log(sin x)., , 1, dy, =, cos x = cot x., dx sin x, , U, , (iii) Let y = log10 (2 + sin x)., , ST, , dy, 1, cos x, =, cos x =, ., dx (2 + sin x) log 10, (2 + sin x) log 10, !, x+1, ., (iv) Let y = log √, x+2, , √, = log (x + 1) − log x + 2 ., = log (x + 1) − log (x + 2)1/2 ., , 1, log (x + 2) ., 2, dy, 1, 1, =, −, ., dx (x + 1) 2(x + 2), = log (x + 1) −, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 49, , , , , , , if x > 0, log x, , (v) Let y = log |x| = , ., , , , log(−x) , if x < 0, , , , , 1, 1, , , , ,, if, x, >, 0, , , dy , x , if x > 0, x, =, ., =, , , , dx , , , 1 , if x < 0, 1 (−1) , if x < 0 , −x, x, , dy 1, = for allx , 0., dx x, p, , (vi) Let y = log x = log x 1/2 ., , , dy 1, 1, 1, log x −1/2 =, =, ., p, dx 2, x 2x log x, 3 √, dy, x 4 x2 + 1, ., , find, Example 1.54. If y =, 5, dx, (3x, + 2), √, 3, x 4 x2 + 1, Solution. Given y =, (3x + 2)5, Taking log on both sides we get, , CO, R, , ., , AP, P, , ∴, , p, , , log y = log x3/4 + log x2 + 1 − log(3x + 2)5, =, , 3, 1, log x + log(x2 + 1) − 5 log(3x + 2)., 4, 2, , U, , Differentiating w.r.to x we get, 1, 3, x, 1, 15, 1 dy 3 1 1, = × + × 2, ×3=, + 2, 2x − 5, −, y dx 4 x 2 x + 1, 3x + 2√, 4x x + 1 3x + 2, ", #, ", #, 3, 2, 4, x, x x +1 3, x, 3, dy, 15, 15, =y, + 2, =, +, ., ∴, −, −, dx, 4x x + 1 3x + 2, (3x + 2)5 4x x2 + 1 3x + 2, , ST, , Example 1.55. If y = x, , √, , √, , x, , Solution. Given y = x x ., √, log y = x log x., , find, , dy, ., dx, , Differentiating w.r.t x, , 1, 1 dy √ 1, = x + log x √, y dx, x, 2 x, 1, 1, = √ + √ log x, x 2 x, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 50, , Engineering Mathematics - I, 1, 1, = √ 1 + log x, 2, x, , 1, = √ 2 + log x, 2 x, , !, , √, , 1, dy, x x, =y √, = √, ., dx, 2 x 2 + log x, 2 x 2 + log x, , , d, (sin x)cos x ., dx, Solution. Let y = (sin x)cos x ., , [A.U. Jan. 2015], , ., , AP, P, , Example 1.56. Find, , Taking logarithms on both sides we get, log y = log (sin x)cos x = cos x log (sin x)., Differentiating w.r.to x., , CO, R, , 1, 1 dy, = cos x, cos x + log(sin x)(− sin x), y dx, sin x, cos2 x, − sin x log(sin x), =, sin, " 2, " x2, #, #, dy, cos x, cos x cos x, =y, − sin x log(sin x) = (sin x), − sin x log(sin x) ., dx, sin x, sin x, Example 1.57. If y = (sin x) x , find, Solution. Let y = (sin x) x ., , dy, ., dx, , Taking log both sides we get, , U, , log y = log (sin x) x = x log sin x., Differentiating w.r.t. x., , ST, , 1, 1 dy, =x, cos x + log sin x.1, y dx, sin x, = x cot x + log sin x., , , , , , dy, = y x cot x + log sin x = (sin x) x x cot x + log sin x ., dx, , Example 1.58. If y = x, Solution. Let y = x x, , xx, , xx, , ···∞, , ∞, x···, , , find, , dy, ., dx, , ⇒ y = xy ., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 54, , Engineering Mathematics - I, , (ix) cos(xy) = 1 + sin y, , (vii) tan−1 (x2 y) = x + xy2, (viii) sin(xy) = cos(x + y), , (x) x sin y + y sin x = 1, , ., , 3. If y = (sin x)sin x, , ∞, sin x···, , , find, , dy, ., dx, , dy, ., dx, , AP, P, , 2. If (sin x)cos y = (sin y)cos x , find, , 4. If x1+y + y1+x = a, where a is a constant, find, , dy, ., dx, , 5. Find the derivatives of the following using logarithmic differentiation., , (i) (x2 + 2)2 (x4 + 4)4, , q, , (x−1), (x4 +1), , (iii) x x, , (v) (cos x) x, , CO, R, , (ii), , (iv) xsin x, , 1, , (vi) (tan x) x, , 6. Find the equation of the tangent line to the curve y sin 2x = x cos 2y at ( π2 , π4 )., 7. Find the equation of the tangent line to the hyperbola x2 − xy − y2 = 1 at the, point (2, 1)., , ST, , U, , 8. Find the equation to the tangent line to the Cardioid x2 + y2 = (2x2 + 2y2 − x)2, , , at 0, 21 ., , 9. Find the derivative of the following functions., , (i) tanh, , √, , √, , (v) cosh−1 ( x)., , x., , (ii) sinh(log(x))., (iii) ecosh 3x ., (iv) x coth, , √, 1 + x2 ., , (vi) x sinh−1, , , x, 3, , −, , √, 9 + x2 ., , (vii) coth−1 (sec x)., (viii) tanh−1 (sin x)., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 55, , 10. At what point of the curve y = cosh x does the tangent have slope 1., Definition. Let c be a number in the domain D of a function f . Then f (c) is the, absolute maximum value of f on D if f (c) ≥ f (x) for all x in D. f (c) is the absolute, minimum value of f on D if f (c) ≤ x for all x in D., Maximum and minimum values of f are called extreme values of f ., Definition. The number f (c) is a local maximum value of f if f (c) ≥ f (x) when x, Increasing and decreasing test, , AP, P, , ., , is near c. f (c) is the local minimum value of f if f (c) ≤ f (x) when x is near c., , (i) If f ′ (x) > 0, on an interval, then f is increasing on that interval., , (ii) If f ′ (x) < 0, on an interval, then f is decreasing on that interval., , then f ′ (c) = 0., , CO, R, , Fermat’s theorem. If f has a local maximum or minimum at c, and f ′ (c) exits,, Critical number. A critical number of a function f is a number c in the domain, of f such that either f ′ (c) = 0 or f ′ (c) does not exist., Result. If f has a local maximum or minimum at c, then c is a Critical number, of f ., , First derivative test., , U, , function f ., , Suppose that c is a critical number of a continuous, , ST, , (i) If f ′ changes from positive to negative at c, then f has a local maximum at c., (ii) If f ′ changes from negative to positive at c, then f has a local minimum at c., (iii) If f ′ is positive to the left and right of c or negative to the left and right of c,, , then f has no local maximum or minimum at c., , Second derivative test. Suppose f ′′ is continuous near c., (i) If f ′ (c) = 0 and f ′′ (c) > 0, then f has a local minimum at c., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 56, , Engineering Mathematics - I, , (ii) If f ′ (c) = 0 and f ′′ (c) < 0, then f has a local maximum at c., , Concavity. If the graph of f lies above all of its tangents on an interval I, then it, is called concave upward on I. If the graph of f lies below all of its tangents on I,, it is called concave downward on I., Concavity test, , AP, P, , ., , (i) If f ′′ (x) > 0, for all x in I, then the graph of f is concave upward on I., , (ii) If f ′′ (x) < 0, for all x in I, then the graph of f is concave downward on I., , Point of inflexion. A point P on a curve y = f (x) is called a point of inflexion, if f is continuous there and the curve changes from concave upward to concave, , CO, R, , downward or from concave downward to concave upward at P., Note. Points of inflection occurs at points where f ′′ (x) = 0., ✎, , ☞, , ✍, , ✌, , Worked Examples, , Example 1.59. Find the intervals on which the function f (x) = sin x + cos x is, increasing or decreasing., , [A.U.Jan.2015], , U, , downwards., , Also find the curve is concave upwards or concave, , Solution. f (x) = sin x + cos x., f ′ (x) = cos x − sin x., , ST, , f ′′ (x) = − sin x − cos x., , At critical points f ′ (x) = 0, i.e., cos x − sin x = 0., i.e., sin x = cos x., i.e., tan x = 1., π 5π, ∴x= ,, ., 4 4, Let us evaluate the intervals at which f decrease or increases., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 57, Interval, π, 0<x<, 4, π, 5π, <x<, 4, 4, 5π, < x < 2π, 4, , +, +, , f, π, increases in (0, ), 4, π 5π, decreases in ( , ), 4 4, 5π, increases in ( , 2π), 4, , π, Since f ′ changes from positive to negative at x = ,, 4, π, f has a maximum at x = ., 4, √, π, π, π, 1, 1, Maximum value of f = f ( ) = sin + cos = √ + √ = 2., 4, 4, 4, 2, 2, 5π, ′, ., Also f changes from negative to positive at x =, 4, √, 1, 5π, 5π, 5π, 1, Minimum value of f = f ( ) = sin, + cos, = − √ − √ = − 2., 4, 4, 4, 2, 2, Let us evaluate the intervals at which the curve is concave upwards or, , AP, P, , ., , f ′ (x), , concave downwards., , CO, R, , Making f ′′ = 0 we get, − sin x − cos x = 0., , ST, , U, , i.e., sin x + cos x = 0., !, , π, 3π, +x, (or) cos x = cos, −x, cos x = − sin x = cos, 2, π2 , π, Taking cos x = cos, + x implies x = + x, 2, 2, π, i.e., = 0 which is absurd., 2, !, 3π, 3π, Taking cos x = cos, − x implies x =, −x, 2, 2, 3π, 3π, ⇒x=, i.e., 2x =, 2, 4, !, !, 3π, 3π, , 2π, ,, The interval to be considered are 0,, 4, 4, Interval, 3π, 0<x<, 4, 3π, < x < 2π, 4, , f ′ (x), , f, , -, , concave downwards, , +, , concave upwards, , Since f ′′ changes sign from negative to positive at x =, , 3π, 3π, ,x=, is a, 4, 4, , point of inflexion., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 58, , Engineering Mathematics - I, , Example 1.60. Find the intervals on which the function f (x) = x4 − 2x2 + 3 is, increasing or decreasing. Also find the local maximum and minimum values of f, by using first derivative test., , [A.U.Nov.2016], , Solution. f (x) = x4 − 2x2 + 3., , f ′ (x) = 4x3 − 4x = 4x(x2 − 1) = 4x(x − 1)(x + 1)., , Critical numbers occur at points where f ′ (x) = 0, , AP, P, , x = −1, x = 0, x = 1., , The critical numbers are x = −1, x = 0, x = 1., , Let us evaluate the intervals at which f decreases or increases., Interval, , 4x, , x−1, , x+1, , f ′ (x), , f, , x < −1, , -, , -, , -, , -, , decreases in (−∞, −1), , −1 < x < 0, , -, , -, , +, , +, , increases in (−1, 0), , 0<x<1, , +, , -, , +, , -, , decreases in (0, 1), , x>1, , +, , +, , +, , +, , increases in (1, ∞), , CO, R, , ., , i.e., 4x(x − 1)(x + 1) = 0., , Since f ′ changes from negative to positive at x = −1,, f has a local minimum at x = −1., , Minimum value of f = f (−1) = 1 − 2 + 3 = 1., , Also f ′ changes from positive to negative at x = 0., , U, , ∴ f has a maximum at x = 0., , Maximum value of f = f (0) = 3., , ST, , Again f ′ changes from negative to positive at x = 1., , ∴ f has a local minimum at x = 1., , Minimum value of f = 1 − 2 + 3 = 2., , Example 1.61. Find the intervals on which the function f (x) = 3x4 − 4x3 − 12x2 + 5, , is increasing and decreasing. Also find the local maximum and minimum values, of f by using first derivative test., , [A.U.Nov.2016], , Solution. f (x) = 3x4 − 4x3 − 12x2 + 5., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 59, , f ′ (x) = 12x3 − 12x2 − 24x = 12x(x2 − x − 2) = 12x(x − 2)(x + 1)., , Critical numbers occur at points where f ′ (x) = 0, i.e., 12x(x − 2)(x + 1) = 0., x = −1, x = 0, x = 2., The critical numbers are x = −1, x = 0, x = 2., , Let us evaluate the intervals at which f decreases or increases., 12x, , x−2, , x+1, , f ′ (x), , x < −1, , -, , -, , -, , -, , decreases in (−∞, −1), , −1 < x < 0, , -, , -, , +, , +, , increases in (−1, 0), , 0<x<2, , +, , -, , +, , -, , decreases in (0, 2), , x>2, , +, , +, , +, , +, , increases in (2, ∞), , f, , AP, P, , ., , Interval, , Since f ′ changes from negative to positive at x = −1,, f has a local minimum at x = −1., , CO, R, , Minimum value of f = f (−1) = 3 + 4 − 12 + 5 = 20., , Also f ′ changes from positive to negative at x = 0., , ∴ f has a local maximum at x = 0., Maximum value of f = f (0) = 5., , Again f ′ changes from negative to positive at x = 2., ∴ f has a minimum at x = 2., , U, , Minimum value of f = 3 × 16 − 4 × 8 − 12 × 4 + 5 = 48 − 32 − 48 + 5 = −27., , Example 1.62. Find the local maximum and minimum values of the function, , ST, , f (x) = x + 2 sin x, 0 ≤ x ≤ 2π., , Solution. f (x) = x + 2 sin x., f ′ (x) = 1 + 2 cos x., , f ′′ (x) = −2 sin x., , Local maximum and minimum occur at points where f ′ (x) = 0., i.e., 1 + 2 cos x = 0., 2 cos x = −1., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 60, , Engineering Mathematics - I, cos x =, , −1, 2 ., , cos x is negative in the II and III quadrants., x=, , 2π, 3, , and x =, , 4π, 3 ., , First derivative test, 4π 4π, 2π 2π, ,, <x<, ,, < x < 2π., 3 3, 3 3, Let us evaluate the intervals at which f decreases and increases., , The intervals to be considered are 0 < x <, , ., , f ′ (x) = 1 + 2 cos x, +, +, , f, , AP, P, , Interval, 2π, 0<x<, 3, 4π, 2π, <x<, 3, 3, 4π, < x < 2π, 3, , increasing, , decreasing, increasing, 2π, 3 ,, , Since f ′ changes from positive to negative at, f has a local maximum at x =, 2π, 3, , + 2 sin 2π, 3 =, , 2π, 3, , +2, , CO, R, , ∴ Maximum value of f =, , 2π, 3 ., , √, , 3, 2, , =, , Since f ′ changes from negative to positive at x =, f has a minimum at x =, , ∴ Minimum value of f =, , 4π, 3 ., 4π, 3 +, , 2 sin 4π, 3 =, , 4π, 3, , Second derivative test, When x =, , 2π, 3 ,, , f ′′ (x) = −2 sin 2π, 3 = −2, , U, , ∴ f has a maximum at x =, Maximum value of f =, , 2π, 3, , √, , √, 3., , √, , √, 3 > 0., , =, , 4π, 3, , √, 3., , −, , √, 3., , ST, , 4π, 3 ,, , √ , − 3, 2, , √, = − 3 < 0., , − 3, f ′′ (x) = −2 sin 4π, 3 = −2 2 =, 4π, ., ∴ f has a minimum at x =, 3, 4π √, − 3., ∴ Minimum value of f =, 3, , When x =, , , , 3, 2, , 2π, 3 ., , +, , +2, , 2π, 3 +, 4π, 3 ,, , Example 1.63. Find the local maxima and minima and the points of inflexion for, the function f (x) = x4 − 4x3 ., , Solution. f (x) = x4 − 4x3 ., , f ′ (x) = 4x3 − 12x2 ., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 61, , f ′′ (x) = 12x2 − 24x., , Local maximum and minimum occur at points where f ′ (x) = 0, i.e., 4x3 − 12x2 = 0., , 4x2 (x − 3) = 0., x = 0, x = 3., , When x = 0, f ′′ (x) = 0., ∴ f (x) does not have either mxaimum or minimum., , ∴ f has a local minimum at x = 3., , AP, P, , ., , When x = 3, f ′′ (x) = 12 × 9 − 24 × 3 = 108 − 72 = 36 > 0, Minimum value of f = 81 − 4 × 27 = 81 − 108 = −27., , At the points of inflexion, f ′′ (x) = 0, , 12x2 − 24x = 0 ⇒ 12x(x − 2) = 0 ⇒ x = 0, x = 2., When x = 0, f (0) = 0., , CO, R, , ∴ (0, 0) is a point of inflexion., , When x = 2, f (2) = 24 − 4 × 23 = 16 − 32 = −16., Another point of inflexion is (2, −16)., , Example 1.64. Find the maxima and minima of the function f (x) = x2/3 (6 − x)1/3 ., , Solution. f (x) = x2/3 (6 − x)1/3 ., , ST, , U, , 2, 1, f ′ (x) = x2/3 (6 − x)−2/3 (−1) + (6 − x)1/3 x−1/3 ., 3, 3, ", #, 2/3, 1/3, 1, −x, 2(6 − x), =, ., +, 3 (6 − x)2/3, x1/3, ", #, 1 −x + 2(6 − x), =, 3 x1/3 (6 − x)2/3, ", #, 1 −x + 12 − 2x, =, 3 x1/3 (6 − x)2/3, ", # ", #, 12 − 3x, 1, 4−x, =, = 1/3, ., 3 x1/3 (6 − x)2/3, x (6 − x)2/3, Now, f ′ (x) = 0 when x = 4 and f ′ (x) does not exist when x = 0 and x = 6., ∴ The critical numbers are 0, 4, 6., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 62, , Engineering Mathematics - I, Let us evaluate the intervals at which f increases or decreases., Interval, , 4−x, , x1/3, , (6 − x)2/3, , f ′ (x), , f, , −∞ < x < 0, , +, , -, , +, , -, , decreases on (−∞, 0), , 0<x<4, , +, , +, , +, , +, , increases on (0, 4), , 4<x<6, , -, , +, , +, , -, , decreases on (4, 6), , x>6, , -, , +, , +, , -, , decreases on (6, ∞), , ., , ∴ Minimum value of f = f (0) = 0., , AP, P, , Since f ′ changes from negative to positive at x = 0, f has a local minimum at x = 0., Also f ′ changes from positive to negative at x = 4., ∴ f has a local maximum at x = 0., , ∴ Maximum value of f = f (4) = 42/3 (6 − 4)1/3 = (22 )2/3 (21/3 ) = 24/3 21/3 = 25/3 ., Again f ′ does not changes sign at x = 6., , CO, R, , ∴ f has no maximum or minimum at x = 6., , Example 1.65. Using first derivative test, examine for maximum and minimum, of the function f (x) = x3 − 3x + 3, x ∈ R., , Solution. f (x) = x3 − 3x + 3., , f ′ (x) = 3x2 − 3 = 3(x2 − 1) = 3(x − 1)(x + 1)., , Critical numbers occur at points where f ′ (x) = 0., i.e., 3(x − 1)(x + 1) = 0., , U, , i.e., x = −1, x = 1., , The critical numbers are x = −1, 1., , ST, , Let us evaluate the intervals at which f decreases or increases., Interval, , x−1, , x+1, , f ′ (x), , f, , x < −1, , -, , -, , +, , increases in (−∞, −1), , −1 < x < 1, , -, , +, , -, , decreases in (−1, 1), , x>1, , +, , +, , -, , increases in (1, ∞), , Since f ′ changes from positive to negative at x = −1., f has a local maximum at x = −1., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 63, , ∴ Maximum value of f = f (−1) = (−1)3 − 3(−1) + 3 = −1 + 3 + 3 = 5., Also f ′ changes from negative to positive at x = 1., , ∴ f has a local minimum at x = 1., ∴ Minimum value of f = f (1) = 13 − 1 × 3 + 3 = 1 − 3 + 3 = 1., Example 1.66. Using first derivative test examine the maximum and minimum, of f (x) = sin2 x, 0 < x < π., , AP, P, , ., , Solution. f (x) = sin2 x., f ′ (x) = 2 sin x cos x = sin 2x., , Critical numbers occur at points where f ′ (x) = 0., i.e., sin 2x = 0., ∴ 2x = 0, or 2x = π., ∴ x = 0, or x = π2 ., , The critical numbers are x = 0, x = π2 ., , CO, R, , Let us evaluate the intervals at which f decreases or increases., Interval, , 0<x<, x>, , sin 2x, , f ′ (x), , f, , +, , +, , increases in (0, π2 ), , -, , -, , decreases in ( π2 , π), , π, 2, , π, 2, , Since f ′ changes from positive to negative at x = π2 ., ∴ f has a maximum at x = π2 ., , Example, , U, , ∴ Maximum value of f = f ( π2 ) = sin2, 1.67. Find, , the, , maximum, , , π, 2, , and, , = 1., minimum, , of, , the, , ST, , f (x) = 2x3 − 3x2 − 36x + 10., , Solution. f (x) = 2x3 − 3x2 − 36x + 10., f ′ (x) = 6x2 − 6x − 36., , f ′′ (x) = 12x − 6., , Critical pints occur at f ′ (x) = 0., i.e., 6x2 − 6x − 36 = 0., , ∴ x2 − x − 6 = 0., , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , function
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 64, , Engineering Mathematics - I, ∴ (x − 3)(x + 2) = 0., ∴ x = −2, x = 3., The critical points are at x = −2, x = 3., , At x = −2, f ′′ (x) = 12 × (−2) − 6 = −24 − 6 = −30 = −ve, ∴ f is maximum at x = −2., Maximum value of, f = 2(−2)3 − 3(−2)2 − 36(−2) + 10 = −16 − 12 + 72 + 10 = 54., , ., , ∴ f has minimum at x = 3., ∴ Minimum value of, , AP, P, , At x = 3, f ′′ (x) = 12 × 3 − 6 = 36 − 6 = 30 = +ve., , f = f (3) = 2 × 33 − 3 × 32 − 36 × 3 + 10 = 54 − 27 − 108 + 10 = −71., , Example 1.68. Find the maxima and minima of the function f (x) = sin x(1 + cos x),, 0 ≤ x ≤ 2π., , CO, R, , Solution. Given f (x) = sin x(1 + cos x), , f ′ (x) = sin x(0−sin x)+(1+cos x)(cos x) = − sin2 x+cos x+cos2 x = cos x+cos 2x., f ′′ (x) = − sin x − 2 sin 2x., , At critical points f ′ (x) = 0., , cos x + cos 2x = 0, , U, , cos x + 2 cos2 x − 1 = 0, , ST, , i.e., 2 cos2 x + cos x − 1 = 0, √, √, −1 ± 1 − 4 × 2 × (−1) −1 ± 1 + 8 −1 ± 3, =, =, cos x =, 2×2, 4, 4, −1 + 3 1, −1 − 3, cos x =, =, or cos x =, = −1., 4, 2, 4, π, 5π, 1, gives x = and, ., 2, 3, 3, cos x = −1 gives x = π., √, √, √, π, When x = , f ′′ (x) = − sin π3 − 2 sin 2π, = − 23 − 2 23 = − 3 2 3 < 0., 3, 3, π, ∴ f has a maximum at x = ., 3, cos x =, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential calculus, , 65, , √, √, , , 3, π, π, Maximum value of f = f, (1 + 21 ) = 3 4 3 ., = sin 3 1 + cos 3 =, √ , √ √2 √, √, 5π ′′, − 3, 10π, 5π, When x =, , f (x) = − sin 3 − 2 sin 3 = − 2 − 2 − 23 = 23 + 2 23 = 3 2 3 > 0., 3, 5π, ∴ f has a minimum at x =, ., 3, √, √, √, , − 3, , 5π, π, 5π, (1 + 21 ) = − 2 3 × 23 = −34 3 ., ∴ Minimum value of f = f 3 = sin 3 1 + cos 5 3 =, 2, When x = π, f ′′ (x) = − sin π − 2 sin 2π = 0., , π, 3, , ∴ Minimum value =, , √, 3 3, 4√ ., −3 3, 4 ., , AP, P, , ∴ Maximum value =, , ST, , U, , CO, R, , ., , ∴ f has neiher maximum nor minimum at x = π., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , ., , 2.1 Limits and Continuity, , AP, P, , 2 Function of Several Variables, , Limit. Let f (x, y) be a function of two independent variables x and y. f (x, y) is said, to have the limit ℓ as (x, y) tends to (x0 , y0 ), if corresponding to any positive number, ǫ, however small, there is a positive number η such that, , CO, R, , | f (x, y) − ℓ| < ǫ where 0 < d < η,, , where d is the distance between (x, y) and (x0 , y0 ) given by d2 = (x − x0 )2 + (y − y0 )2 ., Note. If the limit exists, then it can be written as lim f (x, y) = ℓ., x→x0, y→y0, , Continuity. f (x, y) is said to be continuous at (x0 , y0 ) if, to any positive number ǫ,, however small, there corresponds a positive number η such that, , U, , | f (x, y) − f (x0 , y0 )| < ǫ where 0 < d < η,, , ST, , where d is the distance between (x, y) and (x0 , y0 ) given above., Result 1. The above definition is equivalent to lim f (x, y) = f (x0 , y0 )., x→x0, y→y0, , Result 2. Usually the limit is the same irrespective of the path along which, the point (x, y) approaches (x0 , y0 )., (, ), , , i.e., lim lim f (x, y) = lim lim f (x, y) ., x→x0, , y→y0, , y→y0 x→x0, , But this is not true always., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , ., , 2.1 Limits and Continuity, , AP, P, , 2 Function of Several Variables, , Limit. Let f (x, y) be a function of two independent variables x and y. f (x, y) is said, to have the limit ℓ as (x, y) tends to (x0 , y0 ), if corresponding to any positive number, ǫ, however small, there is a positive number η such that, , CO, R, , | f (x, y) − ℓ| < ǫ where 0 < d < η,, , where d is the distance between (x, y) and (x0 , y0 ) given by d2 = (x − x0 )2 + (y − y0 )2 ., Note. If the limit exists, then it can be written as lim f (x, y) = ℓ., x→x0, y→y0, , Continuity. f (x, y) is said to be continuous at (x0 , y0 ) if, to any positive number ǫ,, however small, there corresponds a positive number η such that, , U, , | f (x, y) − f (x0 , y0 )| < ǫ where 0 < d < η,, , ST, , where d is the distance between (x, y) and (x0 , y0 ) given above., Result 1. The above definition is equivalent to lim f (x, y) = f (x0 , y0 )., x→x0, y→y0, , Result 2. Usually the limit is the same irrespective of the path along which, the point (x, y) approaches (x0 , y0 )., (, ), , , i.e., lim lim f (x, y) = lim lim f (x, y) ., x→x0, , y→y0, , y→y0 x→x0, , But this is not true always., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 68, , Engineering Mathematics - I, , Consider the following example., x − 2y, ., Let f (x, y) =, x + 2y, As (x, y) → (0, 0) along the line y = mx, we have, lim, , which is (different for, different slopes., ) lineswith, x, x − 2y, =1, = lim, Also lim lim, x→0 y→0 x + 2y, ) x→0 x !, (, −2y, x − 2y, = lim, = −1., and lim lim, y→0 2y, y→0 x→0 x + 2y, Hence, as (x, y) approaches (0, 0) along different paths, f (x, y) approaches different, limits., , CO, R, , ., , lim, , AP, P, , (x,y)→(0,0), , x − 2y, (x,y)→(0,0) x + 2y, x − 2mx, = lim, x→0 x + 2mx, x(1 − 2m), = lim, x→0 x(1 + 2m), 1 − 2m, ,, =, 1 + 2m, , f (x, y) =, , Hence, the two repeated limits are not equal and hence f (x, y) is, , discontinuous at the origin., , Note. If a function is continuous at every point of the domain a ≤ x ≤ a′ ,, b ≤ y ≤ b′ , it is said to be continuous in that domain., ✎, , ☞, , ✍, , ✌, , U, , Worked Examples, , Example 2.1. Find lim, , ST, , x→1, y→2, , 2x2 y, , x2 + y2 + 1, , Solution., , Example 2.2. Find lim, , x→∞, y→2, , lim, , x→1, y→2, , ., , 2×1×2 4 2, 2x2 y, =, = = ., x2 + y2 + 1 1 + 4 + 1 6 3, , xy + 1, ., x2 + 2y2, , , y + 1x, x y + 1x, 2, xy + 1, =, lim, =, = 0., Solution. lim 2, =, lim, , , , , , , 2, y, y, 2, 2, x→∞, x→∞ x + 2y, x→∞ 2, ∞, y→2 x 1 + 2 x, y→2, y→2 x 1 + 2 x, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, (, ), , , x−y, Example 2.3. If f (x, y) =, , show that lim lim f (x, y) , lim lim f (x, y) ., x→0 y→0, y→0 x→0, 2x + y, Solution., ), (, (, ), x−y, lim lim f (x, y) = lim lim, x→0 y→0 2x + y, x→0 y→0, x 1, = lim, = ., x→0 2x, 2, (, ), , , x−y, lim lim f (x, y) = lim lim, y→0 x→0, y→0 x→0 2x + y, !, −y, = −1., = lim, y→0 y, , AP, P, , ., , 69, , ∴ The two limits are not equal., , x3 y2 + x2 y3 − 3, , show that lim f (x, y) = lim f (x, y)., x→0, y→0, 2 − xy, y→0, x→0, !, Solution. lim f (x, y) = lim lim f (x, y), , Example 2.4. If f (x, y) =, , x→0 y→0, , CO, R, , x→0, y→0, , x3 y2 + x2 y3 − 3, = lim lim, x→0 y→0, 2 − xy, !, −3, −3, =, = lim, ., x→0 2, , 2, lim f (x, y) = lim lim f (x, y), , !, , y→0 x→0, , y→0, x→0, , !, , ST, , U, , x3 y2 + x2 y3 − 3, = lim lim, y→0 x→0, 2 − xy, !, −3, −3, = lim, =, ., y→0 2, 2, −3, ∴ lim f (x, y) = lim f (x, y) =, ., x→0, y→0, 2, y→0, , x→0, , 2.2 Partial Derivatives, We know that, if y is a continuous function of the independent variable x, and, ∆y, if, ∆y the increment in y corresponding to an increment ∆x in x, then lim, ∆x→0 ∆x, exists, is called the differential coefficient or the derivative of y with respect to x., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 70, , Engineering Mathematics - I, , Now let us consider a function of several independent variables., Let u be a function of two independent variables x and y and let ∆u be the, , AP, P, , ., , increment in u corresponding to an increment in x, y remaining constant. If u is a, ∆u, if exists is called the partial derivative, continuous function of x, the limit lim, ∆x→0 ∆x, ∂u, of u with respect to x, and is represented by, or u x ., ∂x, ∂u, u(x + ∆x, y) − u(x, y), Hence,, = lim, ., ∂x ∆x→0, ∆x, In the same way, if u is a continuous function of y, the increment ∆u in u,, corresponding to an increment ∆y in y, x remaining constant, then the limit, u(x, y + ∆y) − u(x, y), if exists is called the partial derivative of u with respect to, lim, ∆y→0, ∆y, ∂u, y and is represented by, or uy ., ∂y, It can be easily seen that the change in the value of u corresponding to a, change in the value of x will not in general be the same as the change in the value, , CO, R, , of u corresponding to an equal change in the value of y. For example, the area A of, a rectangular box is a function of the length x and the breadth y, being given by, the relation A = xy. When x alone changes,, , ∆A = (x + ∆x)y − xy = y∆x., , When y alone changes, ∆A = x(y + ∆y) − xy = x∆y., , ST, , U, , Though ∆x = ∆y, the values of ∆A will not be equal unless x = y. i.e., unless the, ∂A ∂A, ,, ., box is in the form of a square. Hence, we can see that, in general, ∂x, ∂y, Likewise, if u is a continuous function of several independent variables, x, y, z, . . . , we can define the partial derivatives of u with respect to each of these, , independent variables. As an example, let us consider a function with three, independent variables x, y and z., , ∂u, ∂u ∂u, ,, and, ∂x ∂y, ∂z, treating x, y and z respectively alone as variables can be obtained. We can also, ∂2 u ∂2 u ∂2 u ∂2 u ∂2 u ∂2 u, find the higher order derivatives 2 , 2 , 2 ,, ,, ,, , . . .., ∂x ∂y ∂z ∂x∂y ∂y∂x ∂x∂z, Let u = f (x, y, z). The first partial derivative of u denoted by, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 71, , They are evaluated as follows., ∂ ∂u ∂2 u ∂ ∂u , ∂2 u, ∂2 u ∂ ∂u , =, =, =, ,, and, ., ∂z ∂z, ∂x2 ∂x ∂x ∂y2 ∂y ∂y, ∂z2, , ., , and, , ∂ ∂u , ∂2 u, =, ∂x∂y ∂x ∂y, ∂2 u, ∂ ∂u , =, ., ∂y∂x ∂y ∂x, , AP, P, , Also,, , ∂2 u, ∂2 u, =, ., ∂x∂y ∂y∂x, The third and higher orders of the partial derivatives can be obtained similarly., , Generally,, , ✎, , Worked Examples, , ✍, , ☞, , ✌, , CO, R, , ∂V, Example 2.5. If V = πr2 h where r and h are independent variables, find, and, ∂r, ∂V, ., ∂h, Solution. Given: V = πr2 h., , U, , Differentiating partially w.r.t. r,, ∂V, = πh × 2r = 2πrh., ∂r, Differentiating partially w.r.t. h,, ∂V, = πr2 × 1 = πr2 ., ∂h, , Example 2.6. If u = log(e x + ey ), show that, , ST, , Solution. Given: u = log(e x + ey )., ex, ∂u, = x, ., ∂x e +y ey, e, ∂u, =, ., ∂y e x + ey, x, ∂u ∂u e + ey, +, =, = 1., ∂x ∂y e x + ey, , ∂u ∂u, +, = 1., ∂x ∂y, , Example 2.7. If z = tan−1, , x, ∂z, ∂z, , prove that x + y = 0., y, ∂x, ∂y, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 72, , Engineering Mathematics - I, , Solution. Given: z = tan−1, , x, ., y, , 1 1, y2 1, y, ∂z, ., =, =, = 2, 2, 2, 2, ∂x 1 + x2 y x + y y x + y2, y, ∂z, xy, ., = 2, ∂x x + y2, ∂z, 1 x, xy2, x, =, −, =, −, =− 2, 2, 2, 2, 2, 2, ∂y, y, y (x + y ), x + y2, x, 1+ 2, y, xy, ∂z, y =− 2, ∂y, x + y2, ∂z, xy, xy, ∂z, − 2, = 0., x +y = 2, 2, ∂x, ∂y x + y, x + y2, , ., , AP, P, , x, , Example 2.8. If u = (x − y)4 + (y − z)4 + (z − x)4 , find the value of, , CO, R, , Solution. u = (x − y)4 + (y − z)4 + (z − x)4 ., , ∂u ∂u ∂u, +, + ., ∂x ∂y ∂z, , ∂u, = 4(x − y)3 + 4(z − x)3 (−1) = 4(x − y)3 − 4(z − x)3 ., ∂x, , U, , ∂u, = 4(y − z)3 − 4(x − y)3 ., ∂y, ∂u, = 4(z − x)3 − 4(y − z)3 ., ∂z, ∂u ∂u ∂u, +, +, = 0., ∂x ∂y ∂z, , ST, , Example 2.9. If u = log(x3 + y3 + z3 − 3xyz), prove that, Solution. u = log(x3 + y3 + z3 − 3xyz)., , ∂u ∂u ∂u, 3, +, +, =, ., ∂x ∂y ∂z, x+y+z, , ∂u, 3x2 − 3yz, ., = 3, ∂x x + y3 + z3 − 3xyz, ∂u, 3y2 − 3xz, = 3, ., ∂y x + y3 + z3 − 3xyz, ∂u, 3z2 − 3yx, ., = 3, ∂z, x + y3 + z3 − 3xyz, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 78, , Engineering Mathematics - I, , u xyx, , = yxy−1 log x + xy−1 = xy−1 (y log x + 1), y, + (y log x + 1)(y − 1)xy−2, = xy−1, x, = xy−2 y + xy−2 (y − 1)(y log x + 1), , = xy−2 (y + y2 log x + y − y log x − 1), = xy−2 (2y − 1 + y log x(y − 1)), , AP, P, , y, ∂2 u, ∂2 u, ∂2 u, , find the value of x2 2 + 2xy, + y2 2 ., Example 2.18. If u = (x − y) f, x, ∂x∂y, ∂x, ∂y, [May 2001], y, Solution. u = (x − y) f, x, , CO, R, , ., , From (1) and (2) we have u xxy = u xyx ., , (2), , (1), , ST, , U, , y −y , y, ∂u, = (x − y) f ′, ., +, f, ∂x, x x2, x, −y , , y, ∂2 u, ′′ y, ′ y, · y(−2)x−3, =, −(x, −, y), ·, f, −, (x, −, y), f, x x2 , x, ∂x2, x2 , y, y, y −y, − 2 · f′, ., · 1 + f′, x, x x2, x, y y y y, y2 ′′ y , y, ′ y, = (x − y) 4 f, + 2 3 (x − y) f, − 2 f′, − 2 f′, x, x, x, x, x, x, x, x, , , , , , , 2, 2, y, y, 2y, y, ∂ u y, + (x − y) f ′, − 2y f ′, ., x2 2 = 2 (x − y) f ′′, x, x, x, x, ∂x, x, , , , , y 1, y, ∂u, · +f, (−1), = (x − y) f ′, ∂y, x x, x, , y, y ′ y, = 1−, f, −f, ., x, x, x, 1, −1 !, y ′′ y 1, ∂2 u , ′ y, ′ y, f, ·, · ., −, f, +, f, =, 1, −, x, x x, x, x, x x, ∂y2, (x − y) ′′ y 2 ′ y , =, − f, f, x, x, x, x2, , , 2, 2, 2, 2y ′ y , y, 2∂ u, ′′ y, −, ., y, f, = 2 (x − y) f, x, x, x, ∂y2, x, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , (2)
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 79, , ∂u, w.r.t. x partially we get, ∂y, !, , −y , , y ′′ y −y , −1, ∂2 u, ′ y, ′ y, = 1−, f, (−y), −, f, +, f, ∂x∂y, x, x x2, x, x x2, x2, , , , , , , y, y, (x − y)y ′′ y, y, y, + 2 f′, + 2 f′, =−, f, 3, x, x, x, x, x, x, (x − y) · y ′′ y 2y ′ y , =−, + 2f, ., f, x, x, x3, x, 4y2 y , −2y2, ∂2 u, ′′ y, = 2 (x − y) f, +, f′, ., 2xy, ∂x∂y, x, x, x, x, ∂2 u, ∂2 u, ∂2 u, (1) + (2) + (3) ⇒ x2 2 + 2xy, + y2 2 = 0., ∂x∂y, ∂x, ∂y, , Differentiating, , AP, P, , ., , (3), , Homogeneous function. A function f (x, y) is said to be a homogeneous function, in x and y of degree n if f (tx, ty) = tn f (x, y) for any positive t., Example, , CO, R, , (i) f (x, y) = x2 + y2 + 2xy is a homogeneous function of degree 3 in x and y., , (ii) If tan−1 yx = u then tan u is a homogeneous function of degree 0., , Note. If u = f (x, y) is a homogeneous function of degree n in x and y, then u can, y, be written as u = xn F ., x, 3, 3, x +y, ., Example. Let u =, x+y, t3 x3 + t3 y3 t3 (x3 + y3 ), x3 + y3, =, = t2, = t2 u., tx + ty, t(x + y), x+y, , U, , u(tx, ty) =, , ST, , u is a homogeneous function of degree 2., , y3 , x3 1 + 3, x, Now u = , y, x 1+, x, y 3, 1 + x , 2, =x, y, 1+, x, = x2 F, , y, x, , where F, , y, x, , 1+, =, , y 3, , x, y ., 1+, x, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 80, , Engineering Mathematics - I, , ∴ u is a homogeneous function of degree 2 in x and y., , 2.3 Euler’s Theorem, , ., , AP, P, , Statement. If f (x, y) is a homogeneous function of degree n in x and y having, ∂f, ∂f, +y, = nf., continuous partial derivatives, then x, ∂x, ∂y, Proof., Given that f (x, y) is a homogeneous function of degree n in x and y., Hence f (x, y) can be written as, , y, ., x, y, y −y , = nxn−1 F, + xn F ′, x, x x2, y, y, = nxn−1 F, − xn−2 yF ′ ., x, x, , , n ′ y 1, n−1 ′ y, =x F, =x F, ., x x, x, y, y, y, = nxn F, − xn−1 yF ′, + xn−1 yF ′, x, x, x, y, = nxn F, x, , f (x, y) = xn F, , CO, R, , ∂f, ∂x, ∂f, ∂y, ∂f, ∂f, x, +y, ∂x, ∂y, , ∂f, ∂f, +y, = n f (x, y)., ∂x, ∂y, , U, , x, , If f (x1 , x2 , . . . , xn ) is a homogeneous function of degree n in, ∂f, ∂f, ∂f, x1 , x2 , . . . , xn , then x1, + x2, + · · · + xn, = nf., ∂x1, ∂x2, ∂xn, Euler’s theorem on higher partial derivatives, , ST, , Extension., , Statement. If f (x, y) is a homogeneous function of degree n in x and y then,, ∂2 f, ∂2 f, ∂2 f, + y2 2 = n(n − 1) f ., x2 2 + 2xy, ∂x∂y, ∂x, ∂y, Proof. Given, f (x, y) is a homogeneous function of degree n in x and y., , ∴ By Euler’s theorem we have, x, , ∂f, ∂f, +y, = n f., ∂x, ∂y, , (1), , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 81, , Differentiating this partially w.r.t. x we obtain, x, , =⇒ x, , ∂2 f, ∂f, ∂2 f ∂ f, +, y, =n ., +, ∂x∂y, ∂x, ∂x2 ∂x, , ∂2 f, ∂f ∂f, ∂f, ∂2 f, =n, −, = (n − 1) ., +, y, ∂x∂y, ∂x ∂x, ∂x, ∂x2, , Multiplying both sides by x we get, ∂2 f, ∂f, ∂2 f, = (n − 1)x ., +, xy, 2, ∂x∂y, ∂x, ∂x, , Differentiating (1) partially w.r.t. y we get, x, , (2), , ∂2 f, ∂2 f ∂ f, ∂f, +y 2 +, =n, ∂y∂x, ∂y, ∂y, ∂y, , ∂2 f, ∂f ∂f, ∂2 f, ∂f, +y 2 =n, −, = (n − 1) ., ∂y∂x, ∂y ∂y, ∂y, ∂y, , CO, R, , x, , AP, P, , ., , x2, , Since the partial derivatives of f are continuous we have, , ∴x, , ∂2 f, ∂2 f, =, ., ∂x∂y ∂y∂x, , ∂2 f, ∂2 f, ∂f, + y 2 = (n − 1) ., ∂x∂y, ∂y, ∂y, , U, , Multiplying both sides by y we get, , ∂2 f, ∂f, ∂2 f, + y2 2 = (n − 1)y ., ∂x∂y, ∂y, ∂y, , ST, , xy, , (1) + (2) =⇒ x2, , (2), , 2, ∂f, ∂f , ∂2 f, ∂2 f, 2∂ f, +, y, +, y, +, 2xy, =, (n, −, 1), x, ∂x∂y, ∂x, ∂y, ∂x2, ∂y2, , = (n − 1)n f, , [By Euler’s theorem], , = n(n − 1) f., ✎, , ☞, , ✍, , ✌, , Worked Examples, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 83, , ∴ By Euler’s theorem,, ∂, ∂ u, (e ) + y (eu ) = 2eu ., ∂x, ∂y, ∂u, ∂u, =⇒ xeu + yeu, = 2eu, ∂x, ∂y, ∂u, ∂u , eu x + y, = 2eu ., ∂x, ∂y, ∂u, ∂u, = 2., x +y, ∂x, ∂y, ∂u, ∂u, ∂u, y z, Example 2.21. If u = + , find the value of x + y + z ., x x, ∂x, ∂y, ∂z, y z, Solution. Given: u = + ., x x, y z, ty tz, +, = t0 +, = t0 u., u(tx, ty, tz) =, tx tx, x x, , ., , AP, P, , x, , =⇒ u is a homogeneous function in x, y, z of degree 0., By Euler’s theorem,, , ∂u, ∂u, ∂u, +y +z, = nu = 0.u = 0., ∂x, ∂y, ∂z, ∂u, ∂u, ∂u, x y z, = 0., Example 2.22. If u = + + , then show that x + y + z, y z x, ∂x, ∂y, ∂z, x y z, Solution. u(x, y, z) = + + ., y z x, tx ty tz, u(tx, ty, tz) =, + +, ty tz tx, tx y z, =, + +, t y z x, , U, , CO, R, , x, , = t0 u(x, y, z)., , ST, , u is a homogeneous function of degree 0 in x, y, z., ∂u, ∂u, ∂u, ∴ By Euler’s theorem, x + y + z, = 0.u = 0., ∂x, ∂y, ∂z, , Example 2.23. If sin u =, Solution. sin u(x, y) =, , x2 y2, ∂u, ∂u, , show that x + y, = 3 tan u., x+y, ∂x, ∂y, , x2 y2, ., x+y, , sin u(tx, ty) =, , t2 x2 t2 y2 t4 (x2 y2 ), =, = t3 sin u., tx + ty, t(x + y), , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , [Jun 2012]
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 84, , Engineering Mathematics - I, , sin u is a homogeneous function of degree 3 in x and y., By Euler’s theorem,, ∂, ∂, (sin u) + y (sin u) = 3 sin u., ∂x, ∂y, ∂u, ∂u, x cos u + y cos u, = 3 sin u, ∂x, ∂y, ∂u, ∂u , = 3 sin u, cos u x + y, ∂x, ∂y, ∂u, ∂u 3 sin u 3 sin u, x +y, =, =, = 3 tan u., ∂x, ∂y, cos u, cos u, x y z, ∂u, ∂u, ∂u, Example 2.24. If u = f , , , prove that x + y + z, = 0., y z x, ∂x, ∂y, ∂z, Solution., , ., , AP, P, , x, , x y z, , ,, y z x, tx ty tz , u(tx, ty, tz) = f, , ,, ty tz tx, x y z, = f , ,, y z x, , CO, R, , u(x, y, z) = f, , = u(x, y, z) = t0 u(x, y, z), , u is a homogeneous function of degree 0 in x, y, z., ∂u, ∂u, ∂u, ∴ By Euler’s theorem, x + y + z, = 0.u = 0., ∂x, ∂y, ∂z, x3 + y3 , , U, , Example 2.25. If u = tan−1, , show that x, , ∂u, ∂u, +y, = sin 2u., ∂x, ∂y, , ST, , x−y, x3 + y3, ., Solution. tan u =, x−y, tan u is a homogeneous function of degree 2 in x and y., By Euler’s theorem, , ∂, ∂, (tan u) + y (tan u) = 2 tan u, ∂x, ∂y, ∂u, ∂u, sec2 ux + sec2 uy, = 2 tan u, ∂x, ∂y, ∂u 2 tan u 2 sin u, ∂u, =, cos2 u, =, x +y, ∂x, ∂y, cos u, sec2 u, , x, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 85, , = sin 2u., x+y , ∂u 1, ∂u, = tan u., Example 2.26. If u = sin−1 √, √ prove that x + y, ∂x, ∂y 2, x+ y, x+y , x+y, Solution. u = sin−1 √, √ ⇒ sin u = √, √ ., x+ y, x+ y, sin u is a homogeneous function of degree 21 in x and y., , [Jun 2009], , By Euler’s theorem,, , AP, P, , CO, R, , ., , ∂, 1, ∂, (sin u) + y (sin u) = sin u., ∂x, ∂y, 2, ∂u, ∂u 1, cos ux + cos uy, = sin u, ∂x, ∂y 2, ∂u 1, ∂u, = tan u., x +y, ∂x, ∂y 2, , , ∂u −1, ∂u, x + y , −1 , =, cot u. [Dec 2011], Example 2.27. If u = cos √, √ , prove that x + y, ∂x, ∂y, 2, , x+ y, x + y , Solution. u = cos−1 √, √ ., x + y, , x + y , cos u = √, √ ., x+ y, cos u is a homogeneous function of degree 21 in x and y., x, , ∴ By Euler’s theorem, , ∂, ∂, (cos u) + y (cos u) =, ∂x, ∂y, ∂u, ∂u, =, x(− sin u) + y(− sin u), ∂x, ∂y, !, ∂u, ∂, − sin u x + y, =, ∂x, ∂y, ∂u, ∂, ∴ x +y =, ∂x, ∂y, , ST, , U, , x, , 1, · cos u., 2, 1, cos u., 2, 1, cos u., 2, −1 cos u −1, =, cot u., 2 sin u, 2, , y, x, + tan−1, ., Example 2.28. Verify Euler’s theorem for the function u = sin−1, y, x, x, y, Solution. Given: u = sin−1, + tan−1, ., y, x, tx , ty , u(tx, ty) = sin−1, + tan−1, ty, tx, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 86, , Engineering Mathematics - I, , x, y , = t0 sin−1, + tan−1, y, x, , = t0 u(x, y)., =⇒ u is a homogeneous function of degree 0 in x and y., ∂u, ∂u, ∴ By Euler’s theorem, x + y, = 0., ∂x, ∂y, Verification., , ., , x2 y, 1, − 2 2, 2, y2 − x2 y x (x + y ), y, 1, − 2, = p, ., 2, 2, x + y2, y −x, = p, , y, , x, xy, ∂u, = p, − 2, 2, 2, ∂x, x, + y2, y −x, −x , 1, 1 1, ∂u, = s, +, 2, ∂y, y2 x, x2 y, 1, +, 1− 2, x2, y, x, −xy, + 2, = p, 2, y2 y2 − x2 x + y, x, x, =− p, + 2, 2, 2, x, +, y2, y y −x, , (1), , ST, , U, , CO, R, , x, , AP, P, , 1, 1, 1 −y , ∂u, = s, +, ∂x, y, y2 x2, x2, 1, +, 1− 2, x2, y, , y, , ∂u, x, xy, =−p, + 2, 2, 2, ∂y, x, + y2, y −x, , (1) + (2) =⇒ = x, , (2), , ∂u, ∂u, +y, = 0., ∂x, ∂y, , Hence, Euler’s theorem is verified., , 1, 1, Example 2.29. Verify Euler’s theorem for the function u = x 2 + y 2 (xn + yn )., , , Solution. It is easy to verify that u is a homogeneous function of degree n + 12 in, , x and y., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 87, , Hence, by Euler’s theorem we have, x, Verification, , (1), , AP, P, , 1 1 , 1, 1, ∂u, = x 2 + y 2 nxn−1 + (xn + yn ) x− 2, ∂x, 2, 1, 1, 1, ∂u, 1, = n x 2 + y 2 xn + (xn + yn )x 2 ., x, ∂x, 2, , Since the function is symmetric in x and y we have, , 1, 1, 1, ∂u, 1, = n x 2 + y 2 yn + (xn + yn )y 2 ., ∂y, 2, 1, 1, 1, 1, 1, ∂u, ∂u, = n x 2 + y 2 (xn + yn ) + (xn + yn ) x 2 + y 2, x +y, ∂x, ∂y, 2, 1, 1 1, = n+, x 2 + y 2 (xn + yn ), 2, 1, = n + u,, 2, , y, , CO, R, , ., , ∂u , 1, ∂u, +y, = n + u., ∂x, ∂y, 2, , which verifies Euler’s theorem., Example 2.30. If u = x2 tan−1, , y, x, , − y2 tan−1, , x, y, , , find the value of, , 2, ∂2 u, ∂2 u, 2∂ u, +, 2xy, ., +, y, ∂x∂y, ∂x2, ∂y2, Solution. u is a homogeneous function of degree 2 in x and y., ∂2 u, ∂2 u, ∂2 u, + y2 2 = 2(2 − 1)u = 2u [n = 2]., By Euler’s theorem, x2 2 + 2xy, ∂x∂y, ∂x, ∂y, , U, , x2, , ST, , y, y, ∂2 z, ∂2 z, ∂2 z, + g , show that x2 2 + 2xy, + y2 2 = 0., Example 2.31. If z = x f, x, x, ∂x∂y, ∂x, ∂y, y, y, Solution. Let u = x f, and v = g ., x, x, ∴ z = u + v., u is a homogeneous function of degree 1 in x and y., ∴ By Euler’s theorem,, x2, , 2, ∂2 u, ∂2 u, 2∂ u, +, 2xy, = n(n − 1)u = 1(1 − 1)u = 0., +, y, ∂x∂y, ∂x2, ∂y2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , (1)
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 88, , Engineering Mathematics - I, , v is a homogeneous function of degree 0 in x and y., By Euler’s theorem,, x2, , 2, ∂2 v, ∂2 v, 2∂ v, +, y, +, 2xy, = 0(0 − 1)v = 0., ∂x∂y, ∂x2, ∂y2, , (1) + (2) ⇒ x2, , (2), , 2, ∂2 (u + v), ∂2 (u + v), 2 ∂ (u + v), +, y, +, 2xy, =0, ∂x∂y, ∂x2, ∂y2, , 2, ∂2 z, ∂2 z, 2∂ z, +, y, +, 2xy, = 0., ∂x∂y, ∂x2, ∂y2, !, 3, 3, −1 x + y, , prove that, Example 2.32. If u = tan, x−y, ∂2 u, ∂2 u, ∂2 u, + y2 2 = 2 sin u cos 3u., x2 2 + 2xy, ∂x∂y, ∂x, !∂y, 3 + y3, x, ., Solution. u = tan−1, x−y, !, x3 + y3, tan u =, x−y, tan u is an homogeneous function of degree 2 in x and y., , CO, R, , ., , AP, P, , x2, , ST, , U, , By Euler’s theorem,, ∂, ∂, x (tan u) + y (tan u) = 2 tan u., ∂x, ∂y, ∂u, ∂u, = 2 tan u., x · sec2 u + y · sec2 u, ∂x, ∂y, !, ∂u, ∂u, 2, = 2 tan u., sec u x + y, ∂x, ∂y, ∂u, ∂u 2 tan u, sin u, x +y, =, · cos2 u, =2·, 2, ∂x, ∂y, cos u, sec u, , x, , = 2 sin u cos u, , ∂u, ∂u, +y, = sin 2u., ∂x, ∂y, , (1), , Differentiating (1) partially w.r.t. x we get, ∂2 u ∂u, ∂2 u, ∂u, x· 2 +, +y, = 2 cos 2u ., ∂x, ∂x∂y, ∂x, ∂x, Multiplying by x we get, x2, , ∂2 u, ∂2 u, ∂u, ∂u, +, xy, = 2 cos 2u · x ., +, x, 2, ∂x, ∂x∂y, ∂x, ∂x, , (2), , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 89, , Differentiating (1) partially w.r.t. y we get, x, , ∂2 u ∂u, ∂u, ∂2 u, +y 2 +, = 2 cos 2u ., ∂y∂x, ∂y, ∂y, ∂y, , Multiplying w.r.t. y we get, xy, , (3), , (2) + (3) ⇒, !, 2, 2, ∂2 u, ∂u, ∂u, ∂u, ∂u, 2∂ u, 2∂ u, + 2xy, +x +y, x, +y, = 2 cos 2u x + y, ∂x∂y, ∂x, ∂y, ∂x, ∂y, ∂x2, ∂y2, !, 2, 2, 2, ∂u, ∂ u, ∂ u, ∂ u, ∂u, x2 2 + 2xy, + y2 2 = (2 cos 2u − 1) x + y, ∂x∂y, ∂x, ∂y, ∂x, ∂y, , AP, P, , ., , ∂2 u, ∂u, ∂2 u, ∂u, + y2 2 + y, = 2 cos 2u · y ., ∂y∂x, ∂y, ∂y, ∂y, , = (2 cos 2u − 1) sin 2u, , = sin 2u{2(2 cos2 u − 1) − 1}, = sin 2u{4 cos2 u − 3}, , CO, R, , = 2 sin u cos u(4 cos2 u − 3), = 2 sin u(4 cos3 u − 3 cos u), = 2 sin u · cos 3u., , U, , 2.4 Total Derivative - Implicit functions, We know that if y = f (u) and u = φ(x), then, , dy dy du, =, . We shall now extend this, dx du dx, , ST, , result to function of several variables., Let z = f (x, y) possess continuous partial derivatives and let x, y be differentiable, , functions of an independent variable t. Let ∆x, ∆y and ∆z be increments in x, y and, z respectively, corresponding to an increment ∆t in t. Then the total increment in z, , is given by, , ∆z = f (x + ∆x, y + ∆y) − f (x, y), = f (x + ∆x, y + ∆y) − f (x, y + ∆y) + f (x, y + ∆y) − f (x, y)., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 90, , Engineering Mathematics - I, , By mean value theorem, f (x + ∆x, y + ∆y) − f (x, y + ∆y) = f x (x + θ1 x, y + ∆y)∆x and, f (x, y + ∆y) − f (x, y) = fy (x, y + θ2 ∆y)∆y, 0 < θ1 , θ2 < 1., ∆z = f x (x + θ1 x, y + ∆y)∆x + fy (x, y + θ2 ∆y)∆y, , (1), , where θ1 and θ2 are positive proper fractions., , ., , ∆x, ∆y, ∆z, = f x (x + θ1 x, y + ∆y), + fy (x, y + θ2 ∆y) ., ∆t, ∆t, ∆t, , AP, P, , ∴, , Taking limits as ∆t → 0, we get the total derivative, , dx, dy, dz, = f x (x, y) + fy (x, y) ., dt, dt, dt, i.e.,, , dz ∂z dx ∂z dy, =, +, ., dt ∂x dt ∂y dt, , (2), , In the same way if u = f (x1 , x2 , . . . , xn ) possesses continuous partial derivatives, , CO, R, , with respect to each of the independent variables and if x1 , x2 , . . . , xn are, differentiable functions of an independent variable t, then the total derivative of u, with respect to t is given by, , ∂u dx1, ∂u dx2, ∂u dxn, du, =, +, + ··· +, ., dt, ∂x1 dt, ∂x2 dt, ∂xn dt, , (3), , U, , Note. (2) expresses the rate of change of z w.r.t. t along the curve, x = x(t), y = y(t)., , ST, , Composite functions. Let z = f (u, v) and u and v are themselves functions of the, independent variables x, y so that u = φ(x, y) and v = ψ(x, y)., ∂z, , we consider y as a constant so that u and v may be supposed to be, To find, ∂x, functions of x only. Hence, by the above result we have, ∂z ∂u ∂z ∂v, ∂z ∂z ∂u ∂z ∂v, ∂z, =, +, . Similarly, =, +, ., ∂x ∂u ∂x ∂v ∂x, ∂y ∂u ∂y ∂v ∂y, Now, in (2), if we replace t by x then y is a function of x we get, dz, ∂z ∂z dy, =, +, ., dx ∂x ∂y dx, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 96, , Engineering Mathematics - I, , Example 2.42. If z = f (x, y) where x = eu cos v, y = eu sin v, show that, ∂z, ∂z, ∂z, y + x = e2u ., ∂u, ∂v, ∂y, Solution. z is a composite function of u and v., , [Jan 2000], , ∂z ∂x ∂z ∂y ∂z u, ∂z, ∂z, =, +, =, e cos v + eu sin v., ∂u ∂x ∂u ∂y ∂u ∂x, ∂y, ∂z ∂z ∂x ∂z ∂y ∂z, ∂z, =, +, =, (−eu sin v) + eu cos v., ∂v ∂x ∂v ∂y ∂v ∂x, ∂y, ∂z, ∂z u, ∂z u, ∂z, ∂z, y, = y e cos v + y e sin v = eu sin veu cos v + eu sin veu sin v, ∂u, ∂x, ∂y, ∂x, ∂y, ∂z, ∂z, = e2u sin v cos v + e2u sin2 v ., ∂x, ∂y, ∂z, ∂z, ∂z, x = −eu sin veu cos v + eu cos veu cos v, ∂v, ∂x, ∂y, ∂z, ∂z, = −e2u sin v cos v + e2u cos2 v, ∂x, ∂y, , AP, P, , ., , ∂z, ∂z, ∂z, + x = e2u ., ∂u, ∂v, ∂y, y − x z − x, ∂u, ∂u, ∂u, ,, prove that x2 + y2 + z2, = 0., Example 2.43. If u = f, xy, zx, ∂x, ∂y, ∂z, y − x z − x, ,, Solution. Given: u = f, xy, zx, z−x, 1 1, 1 1, y−x, ,s=, ,r = − , s = − ., Let r =, xy, zx, x y, x z, ∴ u = f (r, s)., , CO, R, , Adding we get y, , Now, , U, , ∴ u is a function of r and s and r and s are functions of x, y and z., , ∂u ∂r ∂u ∂s ∂u −1 ∂u −1 , +, =, +, ∂r ∂x ∂s ∂x ∂r x2, ∂s x2, ∂u ∂u, =− − ., ∂r ∂s, ∂u ∂r ∂u ∂s ∂u −1 ∂u, + (0), =, +, =, ∂r ∂y ∂s ∂y ∂r y2, ∂s, ∂u, =, ., ∂r, ∂u ∂r ∂u ∂s ∂u, ∂u 1 , =, +, =, (0) +, ∂r ∂z ∂s ∂z ∂r, ∂s z2, =, , ST, , ∂u, ∂x, ∂u, ∴ x2, ∂x, ∂u, ∂y, ∂u, y2, ∂y, ∂u, ∂z, , (1), , (2), , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 98, , Engineering Mathematics - I, , Solution., , AP, P, , CO, R, , ., , ∂z ∂x ∂z ∂y ∂z, ∂z, ∂z, =, +, =, cos α +, sin α., ∂u ∂x ∂u ∂y ∂u ∂x, ∂y, ∂2 z, ∂ ∂z ∂y, ∂ ∂z ∂x, +, =, ∂u2 ∂x ∂u ∂u ∂y ∂u ∂u, , , ∂ ∂z, ∂z, ∂ ∂z, ∂z, =, cos α +, sin α cos α +, cos α +, sin α sin α, ∂x ∂x, ∂y, ∂y ∂x, ∂y, 2, 2, 2, ∂ z, ∂ z, ∂ z, ∂2 z, = 2 cos2 α +, sin α cos α +, cos α sin α + 2 sin2 α., ∂x∂y, ∂y∂x, ∂x, ∂y, ∂z ∂z ∂x ∂z ∂y ∂z, ∂z, =, +, =, (− sin α) +, cos α., ∂v ∂x ∂v ∂y ∂v ∂x, ∂y, ∂ ∂z ∂x, ∂2 z, ∂ ∂z ∂y, =, +, ∂v2 ∂x ∂v ∂v ∂y ∂v ∂v, , , ∂ ∂z, ∂z, ∂ ∂z, ∂z, =, −, sin α +, cos α (− sin α) +, −, sin α +, cos α cos α., ∂x, ∂x, ∂y, ∂y, ∂x, ∂y, 2, 2, 2, 2, ∂ z, ∂ z, ∂ z, ∂ z, sin α cos α −, cos α sin α + 2 cos2 α., = 2 sin2 α −, ∂x∂y, ∂y∂x, ∂x, ∂y, (1) + (2) ⇒, , (1), , (2), , ∂2 z ∂2 z ∂2 z, ∂2 z, 2, 2, +, =, (sin, α, +, cos, α), +, (sin2 α + cos2 α), ∂u2 ∂v2 ∂x2, ∂y2, =, , ∂2 z ∂2 z, +, ., ∂x2 ∂y2, , ST, , U, , u, u, Example 2.46. If F is, # of x and y and if x = e sin v, y = e cos v, prove, " 2a function, 2, 2, 2, ∂ F ∂ F, ∂ F ∂ F, + 2, [Jan 2013], that 2 + 2 = e−2u, ∂x, ∂y, ∂u2, ∂v, ∂F ∂F ∂x ∂F ∂y, Solution., =, ·, +, ·, ∂u, ∂x ∂u ∂y ∂u, ∂F u, ∂F u, =, · e sin v +, · e cos v., ∂x, ∂y, !, !, ∂ ∂F ∂x, ∂ ∂F ∂y, ∂2 F, =, +, 2, ∂x ∂u ∂u ∂y ∂u ∂u, ∂u, ", ", #, #, ∂ ∂F u, ∂F u, ∂ ∂F u, ∂F u, ∂x, ∂y, =, · e sin v +, · e cos v, +, · e sin v +, · e cos v, ∂x ∂x, ∂y, ∂u ∂y ∂x, ∂y, ∂u, " 2, #, " 2, #, ∂2 F u, ∂2 F u, ∂ F u, ∂ F u, u, · e sin v +, · e cos v +, =, · e cos v · e sin v +, · e sin v · eu cos v, ∂x∂y, ∂y∂x, ∂x2, ∂y2, ∂2 F 2u, ∂2 F 2u, ∂2 F, ∂2 F, · e sin v cos v +, · e sin v cos v + 2 · e2u cos2 v (1), = 2 · e2u sin2 v +, ∂x∂y, ∂y∂x, ∂x, ∂y, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , AP, P, , ∂F ∂F ∂x ∂F ∂y, =, ·, +, · ., ∂v, ∂x ∂v ∂y ∂v, ∂F u, ∂F u, · e cos v −, · e sin v., =, ∂x, ∂y, !, !, ∂2 F, ∂ ∂F ∂y, ∂ ∂F ∂x, +, =, ∂x ∂v ∂v ∂y ∂v ∂v, ∂v2, ", #, ", #, ∂F u, ∂ ∂F u, ∂F u, ∂ ∂F u, u, · e cos v −, · e sin v e cos v +, · e cos v −, · e sin v (−eu sin v), =, ∂x ∂x, ∂y, ∂y ∂x, ∂y, ∂2 F 2u, ∂2 F, ∂2 F, ∂2 F 2u, · e sin v cos v −, · e sin v cos v + 2 · e2u sin2 v. (2), = 2 · e2u cos2 v −, ∂x∂y, ∂y∂x, ∂x, ∂y, ∂2 F ∂2 F ∂2 F 2u ∂2 F 2u, (1) + (2) ⇒ 2 + 2 = 2 e + 2 · e, ∂u, ∂v, ∂x, ∂y, " 2, #, ∂2 F, 2u ∂ F, =e, + 2, ∂x2, ∂y, " 2, #, 2, 2, 2, ∂ F ∂ F, ∂ F ∂ F, e−2u, + 2 = 2 + 2., 2, ∂u, ∂v, ∂x, ∂y, , CO, R, , ., , 99, , Example 2.47. Transform the equation z xx + 2z xy + zyy = 0 by changing the, independent variables using u = x − y and v = x + y., , [Jun 2012], , Solution. Consider z as a function of u and v and u and v are functions of x and y., , ST, , U, , ∂z ∂u ∂z ∂v, ∂z, =, ·, +, ·, ∂x ∂u ∂x ∂v ∂x, ∂z, ∂z, =, ·1+, ·1, ∂u, ∂v, ∂z ∂z, =, + ., ∂u ∂v!, !, ∂ ∂z ∂u, ∂ ∂z ∂v, ∂2 z, =, ·, ·, +, ∂x2 ∂u ∂x ∂x ∂v ∂x ∂x, !, !, ∂ ∂z ∂z, ∂ ∂z ∂z, ·1+, ·1, =, +, +, ∂u ∂u ∂v, ∂v ∂u ∂v, ∂2 z, ∂2 z, ∂2 z, ∂2 z, = 2+, +, + 2, ∂u∂v ∂v∂u ∂v, ∂u, , i.e.,z xx = zuu + zuv + zvu + zvv, , (1), , ∂z ∂z ∂u ∂z ∂v, =, ·, +, ·, ∂y ∂u ∂y ∂v ∂y, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 100, , Engineering Mathematics - I, ∂z, ∂z, (−1) +, ·1, ∂u, ∂v, ∂z ∂z, =− + ., ∂u !∂v, !, ∂2 z, ∂ ∂z ∂u, ∂ ∂z ∂v, =, +, ∂y2 ∂u ∂y ∂y ∂v ∂y ∂y, !, !, ∂z ∂z, ∂, ∂z ∂z, ∂, − +, (−1) +, − +, ·1, =, ∂u ∂u ∂v, ∂v ∂u ∂v, ∂2 z, ∂2 z, ∂2 z, ∂2 z, = 2−, −, + 2., ∂u∂v ∂v∂u ∂v, ∂u, i.e., zyy = zuu − zuv − zvu + zvv, !, ∂ ∂z, ∂2 z, =, ∂x∂y ∂x ∂y, !, !, ∂ ∂z ∂v, ∂ ∂z ∂u, +, =, ∂u ∂y ∂x ∂v ∂y ∂x, !, !, ∂z ∂z, ∂, ∂z ∂z, ∂, − +, ·1+, − +, ·1, =, ∂u ∂u ∂v, ∂v ∂u ∂v, ∂2 z, ∂2 z, ∂2 z, ∂2 z, =− 2 +, −, + 2., ∂u∂v ∂v∂u ∂v, ∂u, , (2), , i.e., z xy = −zuu + zuv − zvu + zvv, , (3), , (1) + (2) + 2 × (3) ⇒, , CO, R, , ., , AP, P, , =, , z xx + 2z xy + zyy = 2zuv − 2zvu + 4zvv, , U, , 0 = 2zuv − 2zvu + 4zvv ., , ∴ zuv − zvu + 2zvv = 0, , ST, , which is the required equation., , Example 2.48. If z = f (x, y)! where x = u2 − v2 and y = 2uv, prove that, 2, ∂2 z, ∂2 z ∂2 z, 2, 2 ∂ z, +, =, 4(u, +, v, ), +, ., [Jan 2012, Jan 2010], ∂u2 ∂v2, ∂x2 ∂y2, Solution. z is a function of x and y and x and y are functions of u and v., ∴ z is a composite function of u and v., ∂z, ∂z ∂x ∂z ∂y, =, ·, +, ·, ∂u ∂x ∂u ∂y ∂u, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 103, , 2.5 Jacobian and its Properties, Definition. If u and v are continuous functions of two independent variables x, and y having first order partial derivatives, then the Jacobian determinant or, the Jacobian of u and v is defined by, ∂u, ∂y, ∂v, ∂y, , ∂u, ∂x, ∂v, ∂x, , ., , AP, P, , ., , u, v , ∂(u, v), or J, or J =, ∂(x, y), x, y, , If u, v, w are continuous functions of three independent variables x, y, z having first, order partial derivatives, then the Jacobian of u, v, w w.r.t. x, y, z is defined as, ∂u, ∂y, ∂v, ∂y, ∂w, ∂y, , ∂u, ∂z, ∂v, ∂z, ∂w, ∂z, , ., , CO, R, , ∂(u, v, w), =, ∂(x, y, z), , ∂u, ∂x, ∂v, ∂x, ∂w, ∂x, , Properties of Jacobians, , Property I. If u and v are functions of r and s where r and s are functions of x, y, then, , ∂(u, v) ∂(u, v) ∂(r, s), =, ., ∂(x, y) ∂(r, s) ∂(x, y), , ST, , U, , Proof. Since u and v are functions of x and y, we have, ux =, , ∂u ∂u ∂r ∂u ∂s, =, +, = ur r x + u s s x ., ∂x ∂r ∂x ∂s ∂x, , uy =, , ∂u ∂u ∂r ∂u ∂s, =, +, = ur ry + u s sy ., ∂y ∂r ∂y ∂s ∂y, , vx =, , ∂v ∂v ∂r ∂v ∂s, =, +, = vr r x + v s s x ., ∂x ∂r ∂x ∂s ∂x, , vy =, , ∂v ∂v ∂r ∂v ∂s, =, +, = vr ry + v s sy ., ∂y ∂r ∂y ∂s ∂y, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 104, , Engineering Mathematics - I, , ur u s r x ry, ∂(u, v) ∂(r, s), =, ∂(r, s) ∂(x, y), vr v s s x sy, =, , =, , ., , =, =, , ur u s r x, , sx, , v s ry, , sy, , vr, , ur r x + u s s x ur ry + u s sy, vr r x + v s s x, , vr ry + v s sy, , u x uy, vx, , AP, P, , Now, , vy, , ∂(u, v), ., ∂(x, y), , CO, R, , Property II. If J1 is the Jacobian of u, v with respect to x, y and J2 is the Jacobian, ∂(u, v) ∂(x, y), •, = 1., of x, y w.r.t. u, v then J1 J2 = 1. i.e.,, ∂(x, y) ∂(u, v), Proof. u is a function of x and y., Differentiating partially with respect to u and v we get, ∂u ∂x ∂u ∂y, +, = u x xu + uy yu ., ∂x ∂u ∂y ∂u, , 0=, , ∂u ∂x ∂u ∂y, +, = u x xv + uy yv ., ∂x ∂v ∂y ∂v, , U, , 1=, , ST, , v is a function of x and y., , Differentiating with respect to u and v partially we get, , 0=, , ∂v ∂x ∂v ∂y, +, = v x xu + vy yv ., ∂x ∂u ∂y ∂v, , 1=, , ∂v ∂x ∂v ∂y, +, = v x xv + vy yv ., ∂x ∂v ∂y ∂v, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , u x uy xu xv, ∂(u, v) ∂(x, y), =, ∂(x, y) ∂(u, v), v x vy yu yv, =, , =, , ., , =, , u x uy xu yu, vy xv yv, , vx, , u x xu + uy yv u x xv + uy yv, v x xu + vy yu, 1 0, 0 1, , J1 J2 = 1., , v x xv + vy yv, , AP, P, , Now, , 105, , Property III. If the functions u, v, w of three independent variables x, y, z are not, independent, then the Jacobian of u, v, w with respect to x, y, z vanishes., , CO, R, , Solution. Given: u, v and w are not independent variables., =⇒ There exists a relation F(u, v, w) = 0., , Differentiating this w.r.t x, y and z we get, , ∂F ∂u ∂F ∂v ∂F ∂w, +, +, =0, ∂u ∂x ∂v ∂x ∂w ∂x, , U, , ∂F ∂u ∂F ∂v ∂F ∂w, +, +, =0, ∂u ∂y ∂v ∂y ∂w ∂y, , ∂F ∂u ∂F ∂v ∂F ∂w, +, +, = 0., ∂u ∂z ∂v ∂z ∂w ∂z, , ∂F, ∂F ∂F, ,, and, from the above three equations we get, ∂u ∂v, ∂w, , ST, Eliminating, , ∂u, ∂x, ∂u, ∂y, ∂u, ∂z, , =⇒, , ∂v, ∂x, ∂v, ∂y, ∂v, ∂z, , ∂w, ∂x, ∂w, ∂y, ∂w, ∂z, , = 0., , ∂(u, v, w), = 0., ∂(x, y, z), , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 108, , Engineering Mathematics - I, ∂x, ∂x, = v,, = u., ∂u, ∂v, u+v, ., y=, u−v, , ∂y u − v − (u + v) u − v − u − v, −2v, =, =, =, ., 2, 2, ∂u, (u − v), (u − v), (u − v)2, , ., , AP, P, , 2u, ∂y (u − v).1 − (u + v)(−1), =, ., =, 2, ∂v, (u − v), (u − v)2, v, u, −2v, 2u, (u − v)2 (u − v)2, 2uv, 4uv, 2uv, +, =, =, 2, 2, (u − v), (u − v), (u − v)2, ∂(u, v) (u − v)2, =, ., ∂(x, y), 4uv, , ∂(x, y), =, ∂(u, v), , ∂x, ∂u, ∂y, ∂u, , ∂x, ∂v, ∂y, ∂v, , =, , J1 J2 = 1., , CO, R, , Example 2.58. If x = u(1 − v), y = uv then compute J1 and J2 and prove that, Solution. We have, , J1 =, , ∂(u, v), ∂(x, y), , J2 =, ∂(u, v), ∂(x, y), , x = u − uv, y = uv., , ∂x, ∂u, ∂y, ∂u, , ∂x, ∂v, ∂y, ∂v, , 1 − v −u, , = u − uv + uv = u., v, u, We shall express u and v interms of x and y., , U, , =, , ST, , J1 =, , x = u − uv = u − y ⇒ x + y = u., y = uv ⇒ v =, , y, y, =, ., u x+y, , 1, we get J2 = ., u, 1, Now J1 J2 = u = 1., u, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 112, , Engineering Mathematics - I, , derivatives in a neighbourhood of (a, b) then, , ., , Put x = a + h, y = b + k then h = x − a, k = y − b., ∴ The Taylor’s series can be written as, , AP, P, , ∂, ∂, f (a, b), f (a + h, b + k) = f (a, b) + h + k, ∂x, ∂y, 1 ∂, ∂ 2, 1 ∂, ∂ 3, +, h +k, f (a, b) +, h +k, f (a, b) + · · · +, 2! ∂x, ∂y, 3! ∂x, ∂y, , , = f (a, b) + h f x (a, b) + k fy (a, b), , 1 2, +, h f xx (a, b) + 2hk f xy (a, b) + k2 fyy (a, b), 2!, 1 3, h f xxx (a, b) + 3h2 k f xxy (a, b) + 3hk2 f xyy (a, b), +, 3!, , + k3 fyyy (a, b) + · · · ., , CO, R, , , , f (x, y) = f (a, b) + (x − a) f x (a, b) + (y − b) fy (a, b), , 1, +, (x − a)2 f xx (a, b) + 2(x − a)(y − b) f xy (a, b) + (y − b)2 fyy (a, b), 2!, 1, (x − a)3 f xxx (a, b) + 3(x − a)2 (y − b) f xxy (a, b), +, 3!, , + 3(x − a)(y − b)2 f xyy (a, b) + (y − b)3 fyyy (a, b) + · · · ., , U, , This is known as the Taylor’s expansion of f (x, y) in the neighbourhood of (a, b) or, about the point (a, b)., , ST, , Put a = 0, b = 0. we get, , , , f (x, y) = f (0, 0) + x f x (0, 0) + y fy (0, 0), , 1 2, x f xx (0, 0) + 2xy f xy (0, 0) + y2 fyy (0, 0), +, 2!, , 1 3, x f xxx (0, 0) + 3x2 y f xxy (0, 0) + 3xy2 f xyy (0, 0) + y3 fyyy (0, 0) + · · · ., +, 3!, , This, , is, , called, , Maclaurin’s, , series, , ✎, , for, , f (x, y), , Worked Examples, , ✍, , in, , ☞, , powers, , of, , ✌, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , x, , and, , y.
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 113, , Example 2.63. Expand e x sin y in powers of x and y as far as the terms of third, [Jun 2013], , Solution., , f (0, 0) = 0, , f x (x, y) = e x sin y, , f x (0, 0) = 0, , fy (x, y) = e x cos y, , fy (0, 0) = 1, , f xx (x, y) = e x sin y, , f xx (0, 0) = 0, , f xy (x, y) = e x cos y, , f xy (0, 0) = 1, , fyy (x, y) = −e x sin y, , fyy (0, 0) = 0, , f xxx (x, y) = e x sin y, , f xxx (0, 0) = 0, , f xxy (x, y) = e x cos y, , f xxy (0, 0) = 1, , f xyy (x, y) = −e x sin y, , f xyy (0, 0) = 0, , CO, R, , ., , f (x, y) = e x sin y, , AP, P, , degree., , fyyy (x, y) = −e x cos y, , fyyy (0, 0) = −1., , ST, , U, , , 1 2, Now f (x, y) = f (0, 0) + x f x (0, 0) + y fy (0, 0) +, x f xx (0, 0) + 2xy f xy (0, 0) + y2 fyy (0, 0), 2!, , 1 3, x f xxx (0, 0) + 3x2 y f xxy (0, 0) + 3xy2 f xyy (0, 0) + y3 fyyy (0, 0) + · · ·, +, 3!, , 1, = 0 + x.0 + y.1 + x2 .0 + 2xy.1 + y2 .0, 2, , 1 3, 2, x .0 + 3x y.1 + 3xy2 .0 + y3 (−1) + · · ·, 6, x2 y y3, = y + xy +, −, + ··· ., 2, 6, , Example 2.64. Expand e x loge (1 + y) in powers of x and y upto terms of third, degree., , [Jan 2014, Dec 2011, Jan 2003], , Solution., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 114, , Engineering Mathematics - I, , fyyy =, , 2e x, (1 + y)3, , f (x, y) = e x loge (1 + y), f (0, 0) = 0, , f x (x, y) = e x loge (1 + y), , ., , ex, 1+y, x, , f xx = e log(1 + y), ex, f xy =, 1+y, −e x, fyy =, (1 + y)2, f xxx = e x log(1 + y), , CO, R, , ex, 1+y, −e x, f xyy =, (1 + y)2, By Maclaurin’s series we have,, f xxy =, , f x (0, 0) = 0, fy (0, 0) = 1, , AP, P, , fy (x, y) =, , f xx (0, 0) = 0, , f xy (0, 0) = 1, , fyy (0, 0) = −1, , f xxx (0, 0) = 0, , f xxy (0, 0) = 1, , f xyy (0, 0) = −1, fyyy (0, 0) = 2, , ST, , U, , f (x, y) = f (0, 0) + x f x (0, 0) + y fy (0, 0), , 1 2, +, x f xx (0, 0) + 2xy f xy (0, 0) + y2 fyy (0, 0), 2!, 1 3, x f xxx (0, 0) + 3x2 y f xxy (0, 0), +, 3!, , + 3xy2 f xyy (0, 0) + y3 fyyy (0, 0) + · · · ., , 1, e x loge (1 + y) = 0 + x.0 + y.1 + (x2 .0 + 2xy.1 + y2 (−1))+, 2, 1 3, [x · 0 + 3x2 y · 1 + 3xy2 (−1) + y3 (2)] + . . . ., 6, y2 x2 y xy2 y3, +, −, +, − ···, = y + xy −, 2, 2, 2, 3, , Example 2.65. Expand x2 y + 3y − 2 in powers of x − 1 and y + 2 upto third degree, terms., , [Jun 2012], , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , Solution. f (x, y) = x2 y + 3y − 2., , 115, , f (1, −2) = 1 × (−2) + 3(−2) − 2, = −2 − 6 − 2 = −10., , f x (x, y) = 2xy., , f x (1, −2) = −4., , 2, , fy (x, y) = x + 3., , fy (1, −2) = 4., , f xx (x, y) = 2y., , fyy (x, y) = 0., f xxx (x, y) = 0., f xxy (x, y) = 2., f xyy (x, y) = 0., , CO, R, , fyyy (x, y) = 0., , f xy (1, −2) = 2., , AP, P, , ., , f xy (x, y) = 2x., , f xx (1, −2) = −4., fyy (1, −2) = 0., , f xxx (1, −2) = 0., , f xxy (1, −2) = 2., , f xyy (1, −2) = 0., , fyyy (1, −2) = 0., , By Taylor’s theorem we have,, , ST, , U, , f (x, y) = f (a, b) + (x − a) f x (a, b) + (y − b) fy (a, b), i, 1 h, +, (x − a)2 f xx (a, b) + 2(x − a)(y − b) f xy (a, b) + (y − b)2 fyy (a, b), 2!, 1 h, (x − a)3 f xxx (a, b) + 3(x − a)2 (y − b) f xxy (a, b), +, 3!, i, +3(x − a)(y − b)2 f xyy (a, b) + (y − b)3 fyyy (a, b) + · · ·, , x2 y + 3y − 2 = f (1, −2) + (x − 1) f x (1, −2) + (y + 2) fy (1, −2), i, 1 h, +, (x − 1)2 f xx (1, −2) + 2(x − 1)(y + 2) f xy (1, −2) + (y + 2)2 fyy (1, −2), 2!, 1 h, (x − 1)3 f xxx (1, −2) + 3(x − 1)2 (y + 2) f xxy (1, −2), +, 3!, i, +3(x − 1)(y + 2)2 f xyy (1, −2) + (y + 2)3 fyyy (1, −2) + · · ·, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 116, , Engineering Mathematics - I, i, 1h, = −10 − 4(x − 1) + 4(y + 2) + −4(x − 1)2 + 2(x − 1)(y + 2), 2, i, 1h, + 6(x − 1)2 (y + 2) + · · ·, 6, , = −10 − 4(x − 1) + 4(y + 2) − 2(x − 1)2 + (x − 1)(y + 2) + (x − 1)2 (y + 2) + · · ·, Example 2.66. Find the Taylor’s series expansion of x2 y2 + 2x2 y + 3y2 in powers of, (x + 2) and y − 1 upto third degree terms., Solution., f (x, y) = x2 y2 + 2x2 y + 3y2 ., f x (x, y) = 2xy2 + 4xy., fy (x, y) = 2x2 y + 2x2 + 6y., f xx (x, y) = 2y2 + 4y., , f (−2, 1) = 4 + 8 + 3 = 15., , f x (−2, 1) = −4 − 8 = −12., , fy (−2, 1) = 8 + 8 + 6 = 22., , f xx (−2, 1) = 2 + 4 = 6., , f xy (−2, 1) = −8 − 8 = −16., , CO, R, , f xy (x, y) = 4xy + 4x., , AP, P, , ., , [Jan 2012, Jun 2010, Jan 2010], , fyy (x, y) = 2x2 + 6., f xxx (x, y) = 0., , f xxy (x, y) = 4y + 4., f xyy (x, y) = 4x., , f xxx (−2, 1) = 0., , f xxy (−2, 1) = 4 + 4 = 8., , f xyy (−2, 1) = −8., , fyyy (−2, 1) = 0., , ST, , U, , fyyy (x, y) = 0., , fyy (−2, 1) = 8 + 6 = 14., , By Taylor’s theorem we have,, f (x, y) = f (a, b) + (x − a) f x (a, b) + (y − b) fy (a, b), i, 1 h, +, (x − a)2 f xx (a, b) + 2(x − a)(y − b) f xy (a, b) + (y − b)2 fyy (a, b), 2!, 1 h, (x − a)3 f xxx (a, b) + 3(x − a)2 (y − b) f xxy (a, b), +, 3!, i, +3(x − a)(y − b)2 f xyy (a, b) + (y − b)3 fyyy (a, b) + · · ·, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 117, , x2 y2 + 2x2 y + 3y2 = 15 − 12(x + 2) + 22(y − 1), i, 1h, + 6(x + 2)2 − 2 × 16(x + 2)(y − 1) + 14(y − 1)2, 2, i, 1h, + 24(x + 2)2 (y − 1) − 24(x + 2)(y − 1)2 + · · ·, 6, , = 15 − 12(x + 2) + 22(y − 1) + 3(x + 2)2 − 16(x + 2)(y − 1) + 7(y − 1)2, + 4(x + 2)2 (y − 1) − 4(x + 2)(y − 1)2 + · · ·, , x3 + y3 + xy2 in powers of (x − 1) and (y − 2)., Solution., f (x, y) = x3 + y3 + xy2 ., f x (x, y) = 3x2 + y2 ., , [May 2011], , f (1, 2) = 1 + 8 + 4 = 13., , f x (1, 2) = 3 + 4 = 7., , fy (1, 2) = 12 + 4 = 16., , CO, R, , fy (x, y) = 3y2 + 2xy., , AP, P, , ., , Example 2.67. Use Taylor’s formula to expand the function defined by f (x, y) =, , f xx (x, y) = 6x., , f xx (1, 2) = 6., , f xy (x, y) = 2y., , f xy (1, 2) = 4., , fyy (1, 2) = 12 + 2 = 14., , f xxx (1, 2) = 6., , f xxy (x, y) = 0., , f xxy (1, 2) = 0., , f xyy (x, y) = 2., , f xyy (1, 2) = 2., , fyyy (x, y) = 6., , fyyy (1, 2) = 6., , ST, , f xxx (x, y) = 6., , U, , fyy (x, y) = 6y + 2x., , By Taylor’s theorem we have,, f (x, y) = f (a, b) + (x − a) f x (a, b) + (y − b) fy (a, b), i, 1 h, +, (x − a)2 f xx (a, b) + 2(x − a)(y − b) f xy (a, b) + (y − b)2 fyy (a, b), 2!, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 118, , Engineering Mathematics - I, , 1 h, (x − a)3 f xxx (a, b) + 3(x − a)2 (y − b) f xxy (a, b), 3!, i, +3(x − a)(y − b)2 f xyy (a, b) + (y − b)3 fyyy (a, b) + · · ·, i, 1h, x3 + y3 + xy2 = 13 + 7(x − 1) + 16(y − 2) + 6(x − 1)2 + 8(x − 1)(y − 2) + 14(y − 2)2, 2, i, 1h, + 6(x − 1)3 + 6(x − 1)(y − 2)2 + 6(y − 2)3 + · · ·, 6, +, , = 13 + 7(x − 1) + 16(y − 2) + 3(x − 1)2 + 4(x − 1)(y − 2) + 7(y − 2)2, , AP, P, , ., , + (x − 1)3 + (x − 1)(y − 2)2 + (y − 2)3 + · · ·, , Example 2.68. Expand e−x log y as a Taylor’s series in powers of x and y − 1 upto, third degree terms., , [Jun 2011], , Solution., f (x, y) = e−x log y., , CO, R, , f x (x, y) = −e−x log y., , f (0, 1) = 0., , e−x, ., y, , fy (x, y) =, , f xx (x, y) = e−x log y., , f x (0, 1) = 0., , fy (0, 1) = 1., , f xx (0, 1) = 0., , e−x, ., y, , f xy (0, 1) = −1., , fyy (x, y) = −, , e−x, ., y2, , fyy (0, 1) = −1., , U, , f xy (x, y) = −, , f xxx (x, y) = −e−x log y., , ST, , e−x, , f xxx (0, 1) = 0., , ., , f xxy (0, 1) = 1., , f xyy (x, y) =, , e−x, ., y2, , f xyy (0, 1) = 1., , fyyy (x, y) =, , 2e−x, ., y3, , fyyy (0, 1) = 2., , f xxy (x, y) =, , y, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 119, , By Taylor’s theorem we have,, , 1, 1, [−2x(y − 1) − (y − 1)2 ] + [3x2 (y − 1) + 3x(y − 1)2 + 2(y − 1)3 ] + · · ·, 2!, 3!, y, Example 2.69. Expand f (x, y) = tan−1, about (1, 1) upto the second degree terms., x, Hence compute f (1.1, 0.9) approximately., [Jan 2005], y, Solution. Given, f (x, y) = tan−1, x, (a, b) = (1, 1), a = 1, b = 1, 1 π, =, f (1, 1) = tan−1, 1, 4, 1 −y , x2 y, −y, −1, fx =, =, −, = 2, f x (1, 1) =, 2, 2, 2 + y2 )x2, 2, y, 2, x, (x, x, +, y, 1 + x2, 1 1, 1, x, x2, fy =, fy (1, 1) =, = 2, = 2, 2, 2 )x, 2, y x, 2, (x, +, y, x, +, y, 1 + x2, 2xy, 2 1, f xx = −y(−1)(x2 + y2 )−2 2x = 2, f xx (1, 1) = =, 2, 2, 4 2, (x + y ), 2, 2, 2, 2, x + y − x2x, y −x, f xy =, = 2, f xy (1, 1) = 0, 2, 2, 2, (x + y ), (x + y2 )2, −1, −2xy, fyy (1, 1) =, fyy = x(−1)(x2 + y2 )−2 2y = 2, 2, 2, 2, (x + y ), , AP, P, , e−x log y = y − 1 +, , ST, , U, , CO, R, , ., , f (x, y) = f (a, b) + (x − a) f x (a, b) + (y − b) fy (a, b), i, 1 h, (x − a)2 f xx (a, b) + 2(x − a)(y − b) f xy (a, b) + (y − b)2 fyy (a, b), +, 2!, 1 h, (x − a)3 f xxx (a, b) + 3(x − a)2 (y − b) f xxy (a, b), +, 3!, i, +3(x − a)(y − b)2 f xyy (a, b) + (y − b)3 fyyy (a, b) + · · ·, , By Taylor’s theorem we have, f (x, y) = f (a, b) + (x − a) f x (a, b) + (y − b) fy (a, b), , 1, (x − a)2 f xx (a, b) + 2(x − a)(y − b) f xy (a, b) + (y − b)2 fyy (a, b) + · · ·, +, 2!, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 120, , Engineering Mathematics - I, tan−1, , = f (1, 1) + (x − 1) f x (1, 1) + (y − 1) fy (1, 1), , 1, (x − 1)2 f xx (1, 1) + 2(x − 1)(y − 1) f xy (1, 1) + (y − 1)2 fyy (1, 1), +, 2!, −1 , π, 1 1, 1, = + (x − 1), + (y − 1) + (x − 1)2 + 2(x − 1)(y − 1)0, 4, 2, 2 2, 2, , 2 (−1), + ···, + (y − 1), 2, , π 1, 11, 1, = − (x − 1 − y + 1) +, (x − 1)2 − (y − 1)2 + · · ·, 4 2, 2 2, 2, π 1, 1 2, 2, = − (x − y) + (x − y − 2x + 2y) + · · ·, 4 2, 4, y π 1, 1, 1, 1, tan−1, = − (x − 1) + (y − 1) + (x − 1)2 − (y − 1)2 + · · · ., x, 4 2, 2, 4, 4, x, , AP, P, , ., , y, , π 1, 1, 1, 1, f (1.1, 0.9) = − (0.1) + (−0.1) + (0.1)2 − (−0.1)2 approximately, 4 2, 2, 4, 4, π, = − 0.1 = 0.685 approximately., 4, , Solution., , CO, R, , , π, Example 2.70. Find the Taylor’s series expansion of e x sin y at the point − 1,, 4, upto third degree terms., [Jan 2009], , f (x, y) = e x sin y, , U, , f x (x, y) = e x sin y, , fy (x, y) = e x cos y, , ST, , f xx = e x sin y, , f xy = e x cos y, , fyx = e x cos y, fyy = −e x sin y, , f xxx = e x sin y, , π, 1, = √, 4, e 2, π, 1, f x − 1,, = √, 4, e 2, π, 1, fy − 1,, = √, 4, e 2, π, 1, f xx − 1,, = √, 4, e 2, π, 1, f xy − 1,, = √, 4, e 2, π, 1, fyx − 1,, = √, 4, e 2, π, −1, fyy − 1,, = √, 4, e 2, π, 1, f xxx − 1,, = √, 4, e 2, , f − 1,, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, π, 1, = √, 4, e 2, π, −1, − 1,, = √, 4, e 2, π, −1, − 1,, = √, 4, e 2, , π, −1, − 1,, = √, 4, e 2, , f xxy − 1,, , f xyy = −e x sin y, , f xyy, , fyyx = −e x sin y, , fyyx, , fyyy = −e x sin y, , fyyy, , ., By Taylor’s theorem we have, , AP, P, , f xxy = e x cos y, , 121, , ST, , U, , CO, R, , f (x, y) = f (a, b) + (x − a) f x (a, b) + (y − b) fy (a, b), , 1, (x − a)2 f xx (a, b) + 2(x − a)(y − b) f xy (a, b) + (y − b)2 fyy (a, b), +, 2!, 1, +, (x − a)3 f xxx (a, b) + 3(x − a)2 (y − b) f xxy (a, b), 3!, , + 3(x − a)(y − b)2 f xyy (a, b) + (y − b)3 fyyy (a, b) + · · · ., , π, π, π, π, e x sin y = f − 1, + (x + 1) f x − 1, + (y − ) fy − 1,, 4, 4, 4, 4, π, π, π, 1, 2, (x + 1) f xx − 1, + 2(x + 1)(y − ) f xy − 1,, +, 2!, 4, 4, 4, π 2, π , + (y − ) fyy − 1,, 4, 4, 1, π, π, π, 3, + (x + 1) f xxx − 1, + 3(x + 1)2 (y − ) f xxy − 1,, 6, 4, 4, 4, π 3, π, π , π 2, + ··· ., + 3(x + 1)(y − ) f xyy − 1, + (y − ) fyyy − 1,, 4, 4, 4, 4, 1, 1, 1, π, 1, = √ + √ (x + 1) + √ y −, + √ (x + 1)2, 4, e 2 e 2, e 2, 2 2e, π, 1, 1, π, 1, + √ (x + 1) y −, − √ y − 2 + √ (x + 1)3, 4, 4, 2e, 2 2e, 6 2e, √, √, π, 2, 2, π, π, 1, (x + 1)2 y −, (x + 1) y − 2 − √ y − 3 + · · · ., −, +, e, 4, e, 4, 4, 6 2e, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 122, , Engineering Mathematics - I, , π, Example 2.71. Expand e x cos y near the point 1,, by Taylor’s series as far as, 4, quadratic terms., [Jan 1996], Solution., , f x (x, y) = e x cos y, , ., , fy (x, y) = −e x sin y, f xx = e x cos y, f xy = −e x sin y, , CO, R, , fyy = −e x cos y, , e, π, = √, 4, 2, e, π, π, = e cos = √, f x 1,, 4, 4, 2, e, π, =−√, fy 1,, 4, 2, e, π, f xx 1,, = √, 4, 2, π −e, f xy 1,, = √, 4, 2, π −e, fyy 1,, = √, 4, 2, , f 1,, , AP, P, , f (x, y) = e x cos y, , By Taylor’s theorem we have, , ST, , U, , f (x, y) = f (a, b) + (x − a) f x (a, b) + (y − b) fy (a, b), , 1, +, (x − a)2 f xx (a, b) + 2(x − a)(y − b) f xy (a, b) + (y − b)2 fyy (a, b), 2!, π, π, π, π, x, e cos y = f 1, + (x − 1) f x 1, + (y − ) fy 1,, 4, 4, 4, 4, , π, π, π, 1, (x − 1)2 f xx 1, + 2(x − 1)(y − ) f xy 1,, +, 2!, 4, 4, 4, π 2, π , + (y − ) fyy 1,, + ···, 4, 4, e, e, π (−e), = √ + (x − 1) √ + y −, √, 4, 2, 2, 2, π (−e), π 2 (−e) , 1, e, + (x − 1)2 √ + 2(x − 1) y −, √ + y−, √ + ···, 2, 4, 4, 2, 2, 2, !, e, π 1, π, 1, π, = √ 1 + (x − 1) − y −, + (x − 1)2 − y − (x − 1) − y − 2 + · · · ., 4, 2, 4, 2, 4, 2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 123, , 2.7 Maxima and Minima for functions of two variables, Definition. Let f (x, y) be a continuous function defined in a closed and bounded, domain D of the xy plane and let (a, b) be an interior point of D., (i) f (a, b) is said to be a local maximum value of f (x, y) at the point (a, b) if there, exists a neighborhood N of (a, b) such that f (x, y) < f (a, b) for all points (x, y) in, N., , other than (a, b)., f (x, y), , AP, P, , ., , (ii) f (a, b) is said to be a local minimum if f (x, y) > f (a, b) for all points (x, y) in N, , f (x, y), , f (a, b), , f (a, b), , X, , Y, , (x, y) (a, b) (x, y), , CO, R, , (x, y) (a, b) (x, y), , X, , Y, , Local maximum or local minimum values are called extreme values., Stationary point of f (x, y), , A point (a, b) satisfying f x = 0 and fy = 0 is called a stationary point of f (x, y)., , U, , Necessary conditions for Maximum or minimum, If f (a, b) is an extreme value of f (x, y) at (a, b), then (a, b) is a stationary point of, f (x, y) if f x and fy exist at (a, b) and f x (a, b) = 0, fy (a, b) = 0., , ST, , Sufficient conditions for extreme values of f (x, y), Let (a, b) be a stationary point of the differentiable function f (x, y)., i.e., f x (a, b) = 0, fy = (a, b) = 0., Let us define f xx (a, b) = r, f xy (a, b) = s, fyy (a, b) = t., (i) If rt − s2 > 0 and r < 0, then f (a, b) is a maximum value., (ii) If rt − s2 > 0 and r > 0 then f (a, b) is a minimum value., , (iii) If rt − s2 < 0, then f (a, b) is not an extreme value but (a, b) is a saddle point of, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 124, , Engineering Mathematics - I, , f (x, y)., (iv) If rt − s2 = 0, then no conclusion is possible and further investigation is, required., Working rule to find maxima and minima of f (x, y), ∂f, ∂f, and fy =, and solve for f x = 0 and fy = 0 as simultaneous, step (1). Find f x =, ∂x, ∂y, equations in x and y., , AP, P, , ., , Let (a, b), (a1 , b1 ), . . . be the solutions which are stationary points of f (x, y)., ∂2 f, ∂2 f, ∂2 f, ,t = 2 ., step (2). Find r = 2 , s =, ∂x∂y, ∂x, ∂y, step (3). Evaluate r, s, t at each stationary point., At the point (a, b) if, , (i) rt − s2 > 0 and r < 0 then f (a, b) is a maximum value of f (x, y)., , (ii) rt − s2 > 0 and r > 0 then f (a, b) is a minimum value of f (x, y)., (iii) rt − s2 < 0 then (a, b) is called a saddle point., , Critical Point, , CO, R, , (iv) rt − s2 = 0, no conclusion can be made, further investigation is required., , A point (a, b) is a critical point of f (x, y) if f x = 0 and fy = 0 at (a, b) or f x and fy do, not exist at (a, b)., , Maxima or Minima occur at a critical point., ✎, , ☞, , ✍, , ✌, , Worked Examples, , [ Jun 2013, Jan 2012, May 2011, Jun 2010], , Given f (x, y) = x3 + y3 − 12x − 3y + 20., , ST, , Solution., , U, , Example 2.72. Examine f (x, y) = x3 + y3 − 12x − 3y + 20 for its extreme values., , f x = 3x2 − 12, , fy = 3y2 − 3, , r = f xx = 6x, , s = f xy = 0, , t = fyy = 6y, , For stationary points, solve f x = 0 and fy = 0., f x = 0 ⇒ 3x2 − 12 = 0 ⇒ x2 − 4 = 0 ⇒ x2 = 4 ⇒ x = ±2., , fy = 0 ⇒ 3y2 − 3 = 0 ⇒ y2 = 1 ⇒ y = ±1., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 125, , Stationary points are (2, 1), (2, −1), (−2, 1) and (−2, −1)., , rt − s2 = 6x6y − 0 = 36xy., , At (2, 1), rt − s2 = 36 × 2 × 1 = 72 > 0., , At (2, 1), r = 6(2) = 12 > 0., ∴ (2, 1) is a minimum point., Minimum value is f (2, 1) = 8 + 1 − 24 − 3 + 20 = 29 − 27 = 2., ∴ (2, −1) is a saddle point., , At (−2, 1), rt − s2 = 36 × (−2) × 1 = −72 < 0., ∴ (−2, 1) is a saddle point., At (−2, −1), rt − s2 = 36 × −2 × −1 = 72 > 0., r = 6(−2) = −12 < 0., ∴ (−2, −1) is a maximum point., , AP, P, , ., , At (2, −1), rt − s2 = 36 × 2 × −1 = −72 < 0., , CO, R, , Maximum value f (−2, −1) = −8 − 1 + 24 + 3 + 20 = 47 − 9 = 38., Example 2.73. Examine f (x, y) = x3 + y3 − 3axy for maximum and minimum, values., Solution., , [Jan 1999], , Given f (x, y) = x3 + y3 − 3axy., f x = 3x2 − 3ay, , U, , fy = 3y2 − 3ax, , ST, , r = f xx = 6x, s = f xy = −3a t = fyy = 6y., , For stationary points, solve f x = 0 and fy = 0., x2, f x = 0 ⇒ 3x2 − 3ay = 0 ⇒ ay = x2 ⇒ y = ., a, x4, 2, fy = 0 ⇒ 3y − 3ax = 0 ⇒ 2 − ax = 0., a, , x3, 3, 3, x 2 − a = 0 ⇒ x(x − a ) = 0 ⇒ x = 0 or x = a., a, x=0⇒y=0, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 127, , fy = 0 ⇒ x3 y(2 − 2x − 3y) = 0., x = 0, y = 0, 2x + 3y = 2., (1), , 2x + 3y = 2, , (2), , 1, we get 2x = 1 ⇒ x = ., 2, 1, 1, When x = , (1) ⇒ 2 + 3y = 3 ⇒ 3y = 1 ⇒ y = ., 2, 3, 1 1, ∴ The stationary points are (0, 0), , ., 2 3, At (0, 0), rt − s2 = 0.0 − 0 = 0., , AP, P, , ., , Solving 4x + 3y = 3, , ST, , U, , CO, R, , We can not say maximum or minimum. Further investigation is required., 1 1, At , ,, 2 3, 1, 1 1, 1, 1, − 12 × − 6 × × + 6 ×, r=, 9, 4, 2 3, 2, 1, 1, 1, =, − 3 − 1 + 3 = (−1) − ., 9, 9, 9, 1, 1, 11 1 1 1, 1, t =2 −2 −6, = − − =− ., 8, 16, 83 4 8 4, 8, 11, 11, 1 1 2 1 1 1 2, 2, =, s = 6, −8, −9, − −, 43, 83, 49, 2 3 4, 1 1 2 1 2, 1, =, −, = −, =, ., 4 3, 12, 144, 1 1 , 1, −, −, rt − s2 = −, 9, 8, 144, 1, 2−1, 1, −, =, > 0 and r < 0., =, 72 144, 144, 1 1, ∴ ,, is a maximum point., 2 3, 1 1 1 1, 1 1, 1 6 − 3 − 2, 1, 1, Maximum value is f ,, =, 1− −, =, =, =, ., 2 3, 89, 2 3, 72, 6, 72 × 6 432, , Example 2.75. Find the extreme values of the function f (x, y) = x4 + y2 + x2 y., Solution. Given: f (x, y) = x4 + y2 + x2 y., f x = 4x3 + 2xy., , r = f xx = 12x2 + 2y., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 129, , f x = 2x − y − 2., fy = −x + 2y + 1., r = f xx = 2., s = f xy = −1., t = fyy = 2., For stationary points, solve f x = 0, fy = 0, , AP, P, , ., , 2x − y − 2 = 0., , (1), , −x + 2y + 1 = 0., , (2), , (1) ⇒ y = 2x − 2., , (2) ⇒ −x + 2(2x − 2) + 1 = 0, , CO, R, , −x + 4x − 4 + 1 = 0, 3x − 3 = 0, 3x = 3, , U, , x = 1., , ∴ y = 2 − 2 = 0., , ST, , The stationary point is (1, 0)., rt − s2 = 4 + 1 = 5 > 0 and, , r = 2 > 0., , ∴ (1, 0) is a minimum point., Minimum value of f = 1 − 2 = −1., , Example 2.77. Find the extreme values of the function f (x, y) = x3 +y3 −3x−12y+20., [Jan 2012], Solution. f (x, y) = x3 + y3 − 3x − 12y + 20., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 130, , Engineering Mathematics - I, f x = 3x2 − 3., fy = 3y2 − 12., r = f xx = 6x., s = f xy = 0., , ., , 3x2 − 3 = 0., 3x2 = 3., , AP, P, , t = fyy = 6y., For stationary points, solve f x = 0, fy = 0., 3y2 − 12 = 0., 3y2 = 12., y2 = 4., , x = ±1., , y = ±2., , CO, R, , x2 = 1., , The stationary points are (1, 2), (−1, 2), (1, −2), (−1, −2)., At (1, 2), rt − s2 = 36xy = 36 × 1 × 2 = 72 > 0., r = 6 > 0., , ∴ (1, 2) is a minimum point., , Minimum value of f = 1 + 8 − 3 − 24 + 20 = 2., , U, , At (−1, 2), rt − s2 = 36xy = 36 × (−1) × 2 = −72 < 0., ∴ (−1, 2) is a saddle point., , At (1, −2), rt − s2 = 36xy = 36 × 1 × (−2) = −72 < 0., , ST, , ∴ (1, −2) is a saddle point., , At (−1, −2), rt − s2 = 36xy = 72 > 0., r = 6 × (−1) = −6 < 0., , ∴ (−1, −1) is a maximum point., , Maximum value of f = −1 − 8 + 3 + 24 + 20 = 38., Maxima = 38., Minima = 2., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 132, , Engineering Mathematics - I, , r = 36 × 3 × 4 − 12 × 9 × 4 − 6 × 3 × 8, = 432 − 432 − 144, = −144 < 0., t = 12 × 9 − 2 × 81 − 6 × 27 × 2, = 108 − 162 − 324, , ., , AP, P, , = −378., s = 34 × 9 × 2 − 8 × 27 × 2 − 9 × 9 × 4, = 612 − 432 − 324, = −144., rt − s2 = (−144)(−378) − (−144)2, = 54432 − 20736, , = 33696 > 0, > 0 and r < 0, (3, 2) is a maximum point., , CO, R, , Since rt −, , s2, , ∴ Maximum value of f = 27 × 4(6 − 3 − 2) = 108., , Example 2.79. Examine for minimum and maximum values sin x + sin y + sin(x + y)., Solution. We have f (x, y) = sin x + sin y + sin(x + y)., , U, , f x = cos x + cos(x + y), , fy = cos y + cos(x + y)., , r = f xx = − sin x − sin(x + y), , ST, , s = f xy = − sin(x + y), t = fyy = − sin y − sin(x + y)., , For stationary points, solve f x = 0 and fy = 0., i.e., cos x + cos(x + y) = 0., , (1), , cos y + cos(x + y) = 0., , (2), , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 133, , (1) − (2) =⇒ cos x − cos y = 0., i.e., cos x = cos y, =⇒x = y, Now (1) =⇒ cos x + cos 2x = 0, i.e., cos 2x = − cos x., , ., , AP, P, , cos 2x = cos(π − x), =⇒2x = π − x., i.e., 3x = π, π, x= ., 3, , CO, R, , π, π, When x = , y = ., 3, π π 3, ∴ ,, is a stationary point., 3 3 , π π, At ,, 3 3, , ST, , U, , √, √, √, π, 3, 2π − 3, r = − sin − sin, =, −, = − 3 < 0., 3, 3, 2 √ 2√, √, √, 3, 3, 3, 2π, =−, ,t = −, −, = − 3., s = − sin, 3, 2, 2, 2, 3 9, 2, rt − s = 3 − = > 0., 4 4, π π, Since rt − s2 > 0 and r < 0, f (x, y) has a maximum value at , ., 3√ 3 √, √, √, π π, 3, 3, 3 3 3, π, π, 2π, ∴ Maximum value = f ,, = sin + sin + sin, =, +, +, =, ., 3 3, 3, 3, 3, 2, 2, 2, 2, Example, , 2.80., , Find, , the, , maximum, , and, , minimum, , of, , [Jan 1997], , sin x sin y sin(x + y), 0 < x, y < π., Solution. Given f (x, y) = sin x sin y sin(x + y)., f x = sin y[sin x cos(x + y) + sin(x + y) cos x] = sin y sin(2x + y)., r = f xx = 2 sin y cos(2x + y)., , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , values
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , ∴ Another stationary point is, π π, At , ., 3 3, , 2π 2π , ,, ., 3 3, , AP, P, , √, π, 3(−1), r = 2 sin cos π = 2, < 0., 3, 2, √, , 3, 4π, π, π, s = sin, = sin π +, = − sin = −, ., 3, 3, 3, 2, √, √, π, 2 3(−1), t = 2 sin cos π =, = − 3., 3, 2, √, − 3 2, √, √, 3 9, = 3 − = > 0., rt − s2 = (− 3)(− 3) −, 2, 4 4, , π π, ,, is a maximum point., 3 3, √, π π, π, π, 2π, 3, Maximum value = f ,, = sin . sin sin, =, 3 √ 3 √ 3, 3, 2, √3 √, 3 3 3 3 3, =, =, ., 2 2 2, 8, 2π 2π , At, ,, 3 3, √, √, 3, 2π, 6π, r = 2 sin, cos, =2, .1 = 3 > 0., 3, 3, 2, 6π √, 2π, cos, = 3., t = 2 sin, 3, 3, √, , 3, π, 8π, =, = sin 3π −, ., s = sin, 3, 3, 2, √, 3 2, √ √, 3 9, = 3 − = > 0., rt − s2 = 3 3 −, 2, 4 4, 2π 2π , ,, is a minimum point., ∴, 3 3, 2π 2π , 2π, 2π, 4π, = sin, Minimum value = f, ,, sin, sin, =, 3 3, 3, 3, 3, ∴, , √, , , 3, π, sin π −, 2, 3, , ST, , U, , CO, R, , ., , 135, , √, √ √, √, 3 3, 3 −3 3, −, =, ., 2 2, 2, 8, , Example 2.81. In a plane triangle, find the maximum value of cos A cos B cos C., [Jan 2000], , Solution. The angles of the ∆le ABC satisfy 0 < A, B, C < π and A + B + C = π,, , =⇒ C = π − (A + B). Replacing C, we get, f (A, B) = cos A cos B cos(π − (A + B)) = − cos A cos B cos(A + B),, , 0 < A, B < π., , fA = − cos B[cos A(− sin(A + B)) + cos(A + B)(− sin A)] = cos B sin(2A + B)., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, Finally 2A + B = 2π and A + 2B = 2π ⇒ A − B = 0 ⇒ A = B., 2π, 2π, ,B =, ., 3A = 2π ⇒ A =, 3, 3, π, 2π 2π, +, = 4 > π not possible., A+B=, 3, 3, 3, π π, ∴ The only stationary point is , ., 3 3, π π, At , ,, 3 3, π, 1, r = 2 cos cos π = 2 (−1) < 0., 3, 2, π, t = 2 cos cos π = −1., 3, , 4π , π, 1, = cos π +, =− ., s = cos, 3, 3, 2, −1 2, 1 3, 2, rt − s = (−1)(−1) −, = 1 − = > 0., 2, 4 4, π π, ∴ ,, is a maximum point., 3 3, π π, π, π, π 1, = cos cos cos = ., ∴ Maximum value of f is f ,, 3 3, 3, 3, 3 8, , AP, P, , ., , 137, , CO, R, , Example 2.82. A flat circular plate is heated so that the temperature at any, point (x, y) is U(x, y) = x2 + 2y2 − x. Find the coldest point on the plate., Solution. Given U = x2 + 2y2 − x., , U x = 2x − 1, , U, , r = U xx = 2., , s = U xy = 0., , Uy = 4y., t = Uyy = 4., , ST, , rt − s2 = 8 > 0 and r = 2 > 0., , ∴ All points are minimum points., , At minimum, U x = 0 and Uy = 0., 1, U x = 0 ⇒ 2x − 1 = 0 ⇒ x = ., 2, Uy = 0 ⇒ 4y = 0 ⇒ y = 0., 1 , ∴ The minimum point is , 0 ., 2, 1 , ∴ The coldest point on the plate is , 0 ., 2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , [Jan 2005]
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 138, , Engineering Mathematics - I, , 2.8 Constrained Maxima and Minima - Lagrange’s Method, Lagrange’s Method, Let f (x, y, z) be the function for which the extreme values are to be found subject to, the condition, (1), , φ(x, y, z) = 0., , Construct the auxiliary function F(x, y, z) = f (x, y, z) + λφ(x, y, z), where λ is an, multiplier., , AP, P, , ., , undetermined parameter independent of x, y, z which is called the Lagrange’s, Any relative extremum of f (x, y, z) subject to (1) must occur at a, , stationary point of F(x, y, z)., , ∂F, ∂F, ∂F, ∂F, = 0,, = 0,, = 0,, = 0., ∂x, ∂y, ∂z, ∂λ, ⇒ f x + λφ x = 0, fy + λφy = 0, fz + λφz = 0, φ(x, y, z) = 0., fy, fx, fz, =, =, = −λ and φ(x, y, z) = 0., φ x φy φz, Solving these equations we can find the values of x, y, z which are stationary, , CO, R, , The stationary points of F are given by, , points of F and the values of f at these points give the maximum and minimum, values of f (x, y, z)., , ✎, , ☞, , ✍, , ✌, , Worked Examples, , Example 2.83. Find the maximum value of xm yn z p subject to x + y + z = a., [Jan 2009], , U, , Solution. Let f = xm ym z p ., , (1), , φ = x + y + z − a = 0., , ST, , We have to maximise f subject to (1)., , Let F = f + λφ where λ is the Lagrange’s multiplier., F = xm yn z p + λ(x + y + z − a)., , F x = mxm−1 yn z p + λ, Fy = nxm yn−1 z p + λ, Fz = pxm yn z p−1 + λ., , To find the stationary points, solve F x = 0, Fy = 0, Fz = 0, φ = 0., F x = 0 ⇒ mxm−1 yn z p = −λ., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 142, , Engineering Mathematics - I, , √ 1, 1, 1, ∴ y = ± ,z = ± 2 = ± √ ., 2, 2, 2, , 1, 1, 1, The stationary points are given by x = ± , y = ± , z = ± √ ., 2, 2, 2, To have maximum value, we must have xy positive., 1 1 1 1 1 −1 −1 −1 1 −1 −1 −1 , ,, , √ ,, ,, , √ ., ∴ The points are , , √ , , , √ ,, 2 2, 2 2, 2 2, 2 2 2, 2, 2, 2, 111, 0, ∴ Maximum T = 400, = 50 C., 222, , AP, P, , ., , Example 2.87. Find the shortest and longest distance from the point (1, 2, −1) to, the sphere x2 + y2 + z2 = 24 using Lagrange’s method of constrained maxima and, , minima., , [Jun 2011, Jan 2002], , Solution. Let P(x, y, z) be any point on the sphere., Let A be the point (1, 2, −1)., AP =, , q, , (x − 1)2 + (y − 2)2 + (z + 1)2 ., , CO, R, , Let f (x, y, z) = (x − 1)2 + (y − 2)2 + (z + 1)2 ., , (1), , AP is minimum or maximum if f is minimum or maximum., ∴ The problem is now reduced to minimise or maximise f subject to, φ(x, y, z) = x2 + y2 + z2 − 24 = 0., , Consider the auxiliary function, , where, , U, , F = f + λφ = (x − 1)2 + (y − 2)2 + (z + 1)2 + λ(x2 + y2 + z2 − 24), Lagrange’s multiplier., , F x = 2(x − 1) + 2λx, Fy = 2(y − 2) + 2λy, Fz = 2(z + 1) + 2λz., , ST, , To find the stationary points solve, F x = 0, Fy = 0, Fz = 0, φ = 0., , 1, x−1, =1− ., x, x, y−2, 2, Fy = 0 ⇒ 2(y − 2) + 2λy = 0 ⇒ y − 2 = −λy ⇒ −λ =, =1− ., y, y, 1, z+1, =1+ ., Fz = 0 ⇒ 2(z + 1) + 2λz = 0 ⇒ z + 1 = −λz ⇒ −λ =, z, z, 2, 1, 1, ∴1− =1− =1+ ., x, y, z, F x = 0 ⇒ 2(x − 1) + 2λx = 0 ⇒ x − 1 = −λx ⇒ −λ =, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , λ, , is, , the
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Function of Several Variables, , 143, , 1, 2, 1 2, = 1 − we get = ⇒ y = 2x., x, y, x y, 1, 1, Taking 1 − = 1 + we get z = −x., x, z, 2, 1, −y, Taking 1 − = 1 + we get z =, = −x., y, z, 2, We have x2 + y2 + z2 = 24 ⇒ x2 + 4x2 + x2 = 24 ⇒ 6x2 = 24 ⇒ x2 = 4 ⇒ x = ±2., Taking 1 −, , When x = 2, y = 4, z = −2, the first point is (2, 4, −2). Let this point be P1 ., , AP, P, , ., , When x = −2, y = −4, z = 2, the second point is (−2, −4, 2). Let this point be P2 ., √, √, √, √, √, P1 A = 1 + 4 + 1 = 6, P2 A = 9 + 36 + 9 = 54 = 3 6, √, ∴ Shortest distance = 6, √, Longest distance = 3 6., , Example 2.88. A rectangular box open at the top is to have a volume of 32cc., Find the dimensions of the box which requires least amount of material for its, construction., , [Jun 2012, Dec 2011, Jun 2010, Jan 2005], , CO, R, , Solution. Let the dimensions of the box be Length = x, Breadth = y, height = z., Given: Volume = 32cc., , (1), , =⇒ xyz = 32, x, y, z > 0., , We have to minimize the amount of material used for the construction of the box., , U, , Let S be the surface area of the box whose top is open, (2), , ∴ S = xy + 2xz + 2yz, , By Lagrange’s method, , ST, , Let F = s + λφ = xy + 2yz + 2xz + λ(xyz − 32) where λ is the Lagrange’s multiplier., F x = y + 2z + λyz, Fy = x + 2z + λxz, Fz = 2y + 2x + λxy., , To find the stationary points, solve, F x = 0, Fy = 0, Fz = 0, φ = 0, F x = 0 ⇒ y + 2z + λyz = 0, =⇒ y + 2z = −λyz, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , ., , 3.1 The definite integral, Integration as the limit of a sum, , CO, R, , y = f (x), , AP, P, , 3 Integral Calculus, , n rectangles, , f (xk ), , ·········, , x1 = a x2 x3, , ∆x, , xn+1 = b, , U, , Consider the graph of the positive function y = f (x). Let us find the area, , ST, , under the curve y = f (x),between the x−axis and the ordinates x = a and x = b., b−a, . Let the x− co ordinates, Divide this area into n rectangles of equal width ∆x =, n, at the left hand side of the rectangles be x1 = a, x2 , x3 , . . . xn+1 = b. Consider a typical, rectangle, the kth one with height f (xk ). The area of this rectangle is f (xk )∆x. The, sum of all the areas of the n rectangles is, f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x., In summation convention form, this can be written as, , n, P, , f (xk )∆x. This, , k=1, , gives an estimate of the area under the curve but it is not exact. To improve the, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 150, , Engineering Mathematics - I, , estimate we must take a large number of very thin rectangles., , This can be, , achieved by allowing n → ∞ and making ∆x → 0., n, P, f (xk )∆x., ∴ Area under the curve = lim, n=∞ k=1, , The lower and upper limits on the sum correspond to the first and the, , last rectangle where x = a and x = b respectively. Hence the above limit can be, b, P, f (x)∆x. Since the number of rectangles increase without bound,, written as lim, ∆x→0 x=a, , we drop the subsript k from xk and write f (x) which is the value of f at a typical, , a, , ∴, , Rb, , n, P, , f (x)dx = lim, , n=∞ k=1, , a, , f (xk )∆x = lim, , AP, P, , ., , value of x. If this can actually be found, it is called the definite integral of f (x), Rb, from x = a and x = b and it is written as f (x)dx., b, P, , ∆x→0 x=a, , f (x)∆x., , Note. If we approximate each strip by a rectangle that has the same base as the, strip and whose height is the same as the right edge of the strip; (i.e., xk is taken, , CO, R, , at the right end points of the kth rectangle and f (xk ) as the height ) then also the, above result will be achieved., , Definition of a definite integral. If f is a function defined for a ≤ x ≤ b, we, b−a, . We let, divide the interval [a, b] into n subintervals of equal width ∆x =, n, x0 (= a), x1 , x2 , · · · xn (= b) be the end points of these subintervals and if we choose, x1∗ , x2∗ , · · · xn∗ as any sample ponits in these subintervals, so that xi∗ lies in the ith, , a, , U, , subinterval [xi−1 , xi ]. Then the definite integral of f from a to b is defined as, n, Rb, P, f (xi∗ )∆x, provided that the limit exists and gives the same value, f (x)dx = lim, n=∞ i=1, , ST, , for all possible choices of sample points., , If it does exist, we say that f is, , integrable on [a, b]., , Result (1)., , Rb, a, , f (x)dx = lim, , n, P, , n=∞ i=1, , integer N such that, , Rb, a, , f (xi∗ )∆x can also be written as, for ∈> 0, there is an, , f (x)dx −, , every choice of xi∗ in [xi−1 , xi ]., , n, P, , i=1, , f (xi∗ )∆x <∈ for every integer n > N and for, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 151, , Result (2). The sum, , n, P, , i=1, , f (xi∗ )∆x is called the Reimann sum., , Theorem 1. If f is continuos on [a, b], (or) if f has only a finite number of jump, Rb, discontinuities, then f is integrable on [a, b].[i.e., f (x)dx exists]., a, , Theorem 2.If f is integrable on [, b], then, , Rb, , f (x)dx = lim, , n=∞ i=1, , a, , ✎, , Worked Examples, , ✍, , f (xi )∆x where ∆x =, , ☞, , AP, P, , ., , b−a, and xi = a + i∆x., n, , n, P, , Example 3.1. Using the area property evaluate, , R1, , ✌, , x2 dx. Show that the sum of the, , 0, , CO, R, , 1, areas of the upper approximating rectangles approaches ., 3, Solution. Consider the function y = f (x) = x2 . Divide the area under the curve, 1−0 1, = ., y = x2 between x = 0 and x = 1 into 4 rectangles of equal width ∆x =, 4, 4, 1, The width of each rectangle is ., ", # " 4# ", #, ", #, 1, 1 1, 1 3, 3, The subintervals are 0, , , , ,, and , 1 ., 4, 4 2, 2 4, 4, The height of the rectangles is approximated to the value of the, 1 1 3, ordinates at x = 0, , & ., 4 2 4, , ST, , U, , !, !, !, 1, 1, 3, ∴ Area of the required portion = f (0)∆x + f, ∆x + f, ∆x + f, ∆x., 4, 2, 4, !2, !2, !2, 1 1, 1 1, 3 1, 2 1, =0 . +, . +, . +, ., 4, 4 4, 2 4, 4 4, ", #, 1, 1, 9, 1, 0+, + +, =, 4, 16 4 16, ", #, 1 1+4+9, =, 4, 16, 1 14, 7, = ×, = ., 4 16 32, If the height of the rectangles is approximated to the value of the, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 152, , Engineering Mathematics - I, , ordinates at the right edge of each rectangle then,, !, !, !, 1, 3, 1, ∆x + f, ∆x + f, ∆x + f (1)∆x., Area = f, 4, 2, 4, !2, !2, !2, 1 1, 1 1, 3 1, 1, =, . +, . +, . + 12 ·, 4 4, 2 4, 4 4, 4, #, ", ", #, 1, 9, 1 1 + 4 + 9 + 16, 1 1, 1 30 15, + +, +1 =, = ., =, = ×, 4 16 4 16, 4, 16, 4 16 32, , AP, P, , U, , CO, R, , ., , Now, let us divide the area, with width, " into, # "n rectangles, #, " each of their, #, 1−0 1, 1, 1 2, n−1 n, ∆x =, = . The subintervals are 0, , , , · · · ,, , =1 ., n, n, n, n n, n n, The heights of the rectangle are the values of the function f (x) = x2 at, 1 2 3, n, the points , , · · · ., n n n, n, !2, !2, !2, n 2 1, 2 1, 3 1, 1 1, . +, . +, . + ··· +, ., ∴ Area =, n n, n n, n n, n n, #, ", n2, 1 12 22 32, +, +, +, ·, ·, ·, +, =, n n2 n2 n2, n2, h, i, 1, = 3 12 + 22 + 32 + · · · + n2, n ", #, 1 n(n + 1)(2n + 1), (n + 1)(2n + 1), = 3, =, 6, n, 6n2, (n + 1)(2n + 1), lim (Area) = lim, n=∞, n=∞, 6n2, 2, n (1 + 1n )(2 + 1n ) 2 1, = = ., = lim, n=∞, 6 3, 6n2, Properties of definite integrals., , ST, , The following properties can be easily proved using Riemann sums., , (i), , Ra, b, , (ii), , Ra, , f (x)dx = −, , Rb, , f (x)dx., , a, , f (x)dx = 0., , a, , (iii), , Rb, a, , cdx = c(b − a),where c is any constant., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, Rb, , [ f (x) + g(x)]dx =, , Rb, , c f (x)dx = c, , Rb, , [ f (x) − g(x)]dx =, , a, , (vi), , a, , ., , f (x)dx +, , Rb, , Rb, , g(x)dx., , a, , a, , a, , (v), , Rb, , f (x)dx,where c is any constant., , a, , (vii) If a < c < b then, , Rb, a, , Rb, , f (x)dx −, , f (x)dx =, , a, , Rb, , g(x)dx., , a, , Rc, , f (x)dx +, , a, , Rb, c, , Comparison properties of the integral., (i) If f (x) ≥ 0 for a ≤ x ≤ b, then, , Rb, , f (x)dx., , AP, P, , (iv), , 153, , f (x)dx ≥ 0., , a, , Rb, , (ii) If f (x) ≥ g(x) for a ≤ x ≤ b, then, , f (x)dx ≥, , g(x)dx., , a, , CO, R, , a, , Rb, , (iii) If m ≤ f (x) ≤ M for a ≤ x ≤ b, then m(b − a) ≤, Proof. Given m ≤ f (x) ≤ M, , Rb, a, , f (x)dx ≤ M(b − a)., , Integrating between a and b we get, Rb, Rb, Rb, mdx ≤ f (x)dx ≤ Mdx, Rb, , U, , m dx ≤, , ST, , a, , a, , a, , a, , Rb, , f (x)dx ≤ M, , a, , m(b − a) ≤, , Rb, a, , Rb, , dx, , a, , f (x)dx ≤ M(b − a)., , The first fundamental theorem of calculus, Statement. If f is continuous on [a, b], then the function g defined by, g(x) =, , Rx, a, , f (t)dt, a ≤ x ≤ b, , is continuous on [a, b] and is differentiable on (a, b) and g′ (x) = f (x)., Proof.Choose x and x + b ∈ (a, b)., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 154, , Engineering Mathematics - I, , Now, g(x + h) − g(x) =, =, , =, , Zx+h, , f (t)dt −, , a, , Zx, , Zx, , f (t)dt +, , Zx+h, , a, Zx+h, , y, f (t)dt, , a, , f (t)dt −, , x, , Zx, , f (t)dt, , m, , M, , a, , f (t)dt., , 0, , x, , u, , v x+h, , ., , AP, P, , x, , R, g(x + h) − g(x) 1 x+h, =, f (t)dt., (1), h, h x, Let us assume that h > 0. Since f is continuous on [x, x + h], by the, , Hence, for h , 0, we have, , extreme value theorem, there exists numbers u and v in [x, x + h] such that, f (u) = m and f (v) = M, where m and M are the absolute minimum and maximum, have, , CO, R, , values of f on [x, x + h]. By the comparison property of the definite integral we, mh ≤, , x+h, R, x, , f (u)h ≤, , f (t)dt ≤ Mh., , x+h, R, x, , f (t)dt ≤ f (v)h., , Since h > 0, dividing throughout by h we get, x+h, R, f (t)dt ≤ f (v)., f (u) ≤ h1, , (2), , x, , U, , (2) can be proved in a similar way for h < 0., , Now, allowing h → 0 and since u and v lie in [x, x + h], we have u → x, , ST, , and v → x. Since f is continuous at x,we have, ∴ lim f (u) = lim f (u) = f (x), h→0, , u→x, , and lim f (v) = lim f (v) = f (x)., h→0, , v→x, , By (2) we have, 1, lim f (u) ≤ lim, h→0, h→0 h, , Zx+h, , f (t)dt ≤ lim f (v), h→0, , x, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , x
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 155, g(x + h) − g(x), ≤ f (x) [by (1)], h→0, h, , f (x) ≤ lim, , i.e., f (x) ≤ g′ (x) ≤ f (x)., By Squeeze Theorem we get, g′ (x) = f (x)., , (3), , Since every differentiable function is continuous,, , AP, P, , If x = a or x = b, from (3) we get the one sided limits at a and b., This shows that g is continuous on [a, b]., , Rx √, Example 3.2. Find the derivative of the function g(x) =, 1 + t2 dt., 0, √, Solution. Since f (t) = 1 + t2 is continuous on [0, x], by part 1 of the Fundamental, √, theorem of calculus we get g′ (x) = 1 + x2, 4, , , d Rx, sec tdt., Example 3.3. Find, dx 1, 4, Solution. Let u = x, x4, , u, , Z, , Z, , d , d , , sec tdt, ∴, sec tdt =, , dx , dx , 1, 1, , u, , Z, d , du, [By Chain rule], =, sec tdt ·, du, dx, , ST, , U, , CO, R, , ., , we obtain g is also continuous on (a, b)., , 1, , = sec u · 4x3, , [by FTC 1], , = sec x4 · 4x3 ., , Second Fundamental theorem of calculus, Rb, Statement.If f is continuous on [a, b], then f (x)dx = F(b) − F(a) where F is any, a, , antiderivative of f . [i.e, a function such that F ′ = f ]., Rx, Proof. Let g(x) = f (t)dt., a, , By the First Fundamental theorem of calculus we have g′ (x) = f (x). i.e,, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 156, , Engineering Mathematics - I, , g is an antiderivative of f ., If F is any other antiderivative of f on [a, b], we know that F and g differ, by a constant., ∴ F(x) = g(x) + c for a < x < b., Since both F and g are continuous on [a, b] the above relation holds when x = a and, x = b., , ., , x→a+, , Also lim g(x) = g(a), x→a+, , & lim F(x) = F(b)., x→b−, , AP, P, , i.e lim F(x) = F(a), , & lim g(x) = g(b)., x→b−, , Hence we obtain F(x) = g(x) + c for all x in [a, b], Ra, When x = a, we have g(a) = f (t)dt = 0., a, , (1), , Using (1) for x = b and x = a we get, , F(b) − F(a) = [g(b) + c] − [g(a) + c], , CO, R, , = g(b) − g(a), = g(b) =, , Zb, , f (t)dt., , a, , Result. The First and Second Fundamental Theorem of Calculus are combined, together is called the Fundamental Theorem of Calculus., Statement of Fundamental Theorem of Calculus., , U, , Suppose f is continuous on [a, b], Rx, (i) If g(x) = f (t)dt, then g′ (x) = f (x)., a, , Rb, , ST, , (ii) f (x)dx = F(b) − F(a), where F is any antiderivative of f . (i.e., F ′ = f .), a, , Note (1). The Fundamental theorem of calculus says that differentiation and, integration are inverse processors., Note (2). By part(ii) of the Fundamental theorem of calculus we have, Rb, f (x)dx = F(b) − F(a) where F ′ = f ., a, , i.e, , Rb, a, , F ′ (x)dx = F(b) − F(a)., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 157, , The above result is sometimes called as the Net Change Theorem., Net Change Theorem, Statement. The integral of a rate of change is the net change., Rb, i.e F ′ (x)dx = F(b) − F(a)., a, ✎, , ☞, , ✍, , ✌, , Worked Examples, , ., , 0, , (x2 − 2x)dx by using Reimann sum by taking right end, , AP, P, , Example 3.4. Evaluate, , R3, , points as the sample points. Hence verify it by using fundamental theorem of, calculus., Solution. Let f (x) = x2 − 2x., , follows., , CO, R, , Divide the interval [0, 3] into n sub intevals with equal width, 3−0 3, ∆x =, = ., n, n, ", # ", # ", #, ", #, 3, 3 6, 6 9, 3n − 3 3n, The intervals are 0, , , , , , · · ·, ,, ., n, n n, n n, n, n, The sample points xi∗ are the right end points of the sub intervals., 3n, 3 6 9, ∴ xi∗ = , , , · · · ., n n n, n, The corresponding f (xi∗ ) = xi∗ 2 − 2xi∗ are calculated and tabulated as, , 3, n, , xi∗, , 9, n2, , −, , U, , f (xi∗ ), , 6, n, , 6, n, , 36, n2, , −, , 9, n, , 12, n, , 81, n2, , −, , 18, n, , ···, ···, , 3n, n, 9n2, n2, , −, , 6n, n, , By Reimann’s sum we have,, n, X, , ST, , Z3, , (x2 − 2x)dx = lim, , n=∞, , 0, , f (xi∗ )∆x, , i=1, , !, !, !, ! #, 9, 36 12 3, 81 18 3, 9n2 6n 3, 6 3, +, +, +, ·, ·, ·, +, −, −, −, −, n=∞ n2, n n, n n, n n, n n, n2, n2, n2, " 3, !, !, !, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, = lim, · 1 − 2 · 2 + 3 · 22 − 2 · 4 + 3 · 32 − 2 · 6, 3, n=∞ n, n, n, n, n, n, !#, 3, 32, 3, 2, + · · · + 3 · n − 2 · 2n, n, n, , = lim, , ", , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 158, , Engineering Mathematics - I, , 32, 33 2, 2, 2, 2, (2 + 4 + 6 + · · · + 2n), 1, +, 2, +, 3, +, ·, ·, ·, +, n, −, lim, n=∞ n2, n=∞ n3, h n(n + 1) i, 27 h n(n + 1)(2n + 1) i, 9, = lim 3, − lim 2 · 2, n=∞ n, n=∞ n, 6, 2, 1, 1, 2, 3, h n (1 + 1n ) i 27, h n (1 + n )(2 + n ) i, 27, − 9 lim, =, lim, × 2 − 9 = 0., =, n=∞, 6 n=∞, 6, n3, n2, , = lim, , ., , 0, , Example 3.5. Evaluate the integral, , AP, P, , By the Fundamental Theorem of Calculus we have, !3, Z3, x3, x2, 27, 2, (x − 2x)dx =, −2·, =, − 9 = 9 − 9 = 0., 3, 2 0, 3, , R1 √, 1 − x2 dx interms of areas., 0, , √, R1 √, 1 − x2 dx represents the area under the curve y = 1 − x2 between, Solution., 0, , Now, y =, , p, 1 − x2, , y2 = 1 − x2, 2, , 2, , y, , CO, R, , the x−axis and the lines x = 0 and x = 1., , 0, , 1x, , x + y = 1., , U, , This is a circle with centre at the origin and radius = 1 unit. Since we, , ST, , need the area between x = 0 and x = 1, the required area is the area of the portion, π, 1, of the circle that lies in the first quadrant = · π × 12 = ., 4, 4, R1 √, π, 1 − x2 dx = ., ∴, 4, 0, R3, Example 3.6. Evaluate the integral (x − 1)dx interms of areas., 0, , Solution., , R3, 0, , (x − 1)dx represents the area under the curve y = (x − 1) between the, , x−axis and the lines x = 0 and x = 3., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , ., , 3.1 The definite integral, Integration as the limit of a sum, , CO, R, , y = f (x), , AP, P, , 3 Integral Calculus, , n rectangles, , f (xk ), , ·········, , x1 = a x2 x3, , ∆x, , xn+1 = b, , U, , Consider the graph of the positive function y = f (x). Let us find the area, , ST, , under the curve y = f (x),between the x−axis and the ordinates x = a and x = b., b−a, . Let the x− co ordinates, Divide this area into n rectangles of equal width ∆x =, n, at the left hand side of the rectangles be x1 = a, x2 , x3 , . . . xn+1 = b. Consider a typical, rectangle, the kth one with height f (xk ). The area of this rectangle is f (xk )∆x. The, sum of all the areas of the n rectangles is, f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x., In summation convention form, this can be written as, , n, P, , f (xk )∆x. This, , k=1, , gives an estimate of the area under the curve but it is not exact. To improve the, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 150, , Engineering Mathematics - I, , estimate we must take a large number of very thin rectangles., , This can be, , achieved by allowing n → ∞ and making ∆x → 0., n, P, f (xk )∆x., ∴ Area under the curve = lim, n=∞ k=1, , The lower and upper limits on the sum correspond to the first and the, , last rectangle where x = a and x = b respectively. Hence the above limit can be, b, P, f (x)∆x. Since the number of rectangles increase without bound,, written as lim, ∆x→0 x=a, , we drop the subsript k from xk and write f (x) which is the value of f at a typical, , a, , ∴, , Rb, , n, P, , f (x)dx = lim, , n=∞ k=1, , a, , f (xk )∆x = lim, , AP, P, , ., , value of x. If this can actually be found, it is called the definite integral of f (x), Rb, from x = a and x = b and it is written as f (x)dx., b, P, , ∆x→0 x=a, , f (x)∆x., , Note. If we approximate each strip by a rectangle that has the same base as the, strip and whose height is the same as the right edge of the strip; (i.e., xk is taken, , CO, R, , at the right end points of the kth rectangle and f (xk ) as the height ) then also the, above result will be achieved., , Definition of a definite integral. If f is a function defined for a ≤ x ≤ b, we, b−a, . We let, divide the interval [a, b] into n subintervals of equal width ∆x =, n, x0 (= a), x1 , x2 , · · · xn (= b) be the end points of these subintervals and if we choose, x1∗ , x2∗ , · · · xn∗ as any sample ponits in these subintervals, so that xi∗ lies in the ith, , a, , U, , subinterval [xi−1 , xi ]. Then the definite integral of f from a to b is defined as, n, Rb, P, f (xi∗ )∆x, provided that the limit exists and gives the same value, f (x)dx = lim, n=∞ i=1, , ST, , for all possible choices of sample points., , If it does exist, we say that f is, , integrable on [a, b]., , Result (1)., , Rb, a, , f (x)dx = lim, , n, P, , n=∞ i=1, , integer N such that, , Rb, a, , f (xi∗ )∆x can also be written as, for ∈> 0, there is an, , f (x)dx −, , every choice of xi∗ in [xi−1 , xi ]., , n, P, , i=1, , f (xi∗ )∆x <∈ for every integer n > N and for, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 151, , Result (2). The sum, , n, P, , i=1, , f (xi∗ )∆x is called the Reimann sum., , Theorem 1. If f is continuos on [a, b], (or) if f has only a finite number of jump, Rb, discontinuities, then f is integrable on [a, b].[i.e., f (x)dx exists]., a, , Theorem 2.If f is integrable on [, b], then, , Rb, , f (x)dx = lim, , n=∞ i=1, , a, , ✎, , Worked Examples, , ✍, , f (xi )∆x where ∆x =, , ☞, , AP, P, , ., , b−a, and xi = a + i∆x., n, , n, P, , Example 3.1. Using the area property evaluate, , R1, , ✌, , x2 dx. Show that the sum of the, , 0, , CO, R, , 1, areas of the upper approximating rectangles approaches ., 3, Solution. Consider the function y = f (x) = x2 . Divide the area under the curve, 1−0 1, = ., y = x2 between x = 0 and x = 1 into 4 rectangles of equal width ∆x =, 4, 4, 1, The width of each rectangle is ., ", # " 4# ", #, ", #, 1, 1 1, 1 3, 3, The subintervals are 0, , , , ,, and , 1 ., 4, 4 2, 2 4, 4, The height of the rectangles is approximated to the value of the, 1 1 3, ordinates at x = 0, , & ., 4 2 4, , ST, , U, , !, !, !, 1, 1, 3, ∴ Area of the required portion = f (0)∆x + f, ∆x + f, ∆x + f, ∆x., 4, 2, 4, !2, !2, !2, 1 1, 1 1, 3 1, 2 1, =0 . +, . +, . +, ., 4, 4 4, 2 4, 4 4, ", #, 1, 1, 9, 1, 0+, + +, =, 4, 16 4 16, ", #, 1 1+4+9, =, 4, 16, 1 14, 7, = ×, = ., 4 16 32, If the height of the rectangles is approximated to the value of the, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 152, , Engineering Mathematics - I, , ordinates at the right edge of each rectangle then,, !, !, !, 1, 3, 1, ∆x + f, ∆x + f, ∆x + f (1)∆x., Area = f, 4, 2, 4, !2, !2, !2, 1 1, 1 1, 3 1, 1, =, . +, . +, . + 12 ·, 4 4, 2 4, 4 4, 4, #, ", ", #, 1, 9, 1 1 + 4 + 9 + 16, 1 1, 1 30 15, + +, +1 =, = ., =, = ×, 4 16 4 16, 4, 16, 4 16 32, , AP, P, , U, , CO, R, , ., , Now, let us divide the area, with width, " into, # "n rectangles, #, " each of their, #, 1−0 1, 1, 1 2, n−1 n, ∆x =, = . The subintervals are 0, , , , · · · ,, , =1 ., n, n, n, n n, n n, The heights of the rectangle are the values of the function f (x) = x2 at, 1 2 3, n, the points , , · · · ., n n n, n, !2, !2, !2, n 2 1, 2 1, 3 1, 1 1, . +, . +, . + ··· +, ., ∴ Area =, n n, n n, n n, n n, #, ", n2, 1 12 22 32, +, +, +, ·, ·, ·, +, =, n n2 n2 n2, n2, h, i, 1, = 3 12 + 22 + 32 + · · · + n2, n ", #, 1 n(n + 1)(2n + 1), (n + 1)(2n + 1), = 3, =, 6, n, 6n2, (n + 1)(2n + 1), lim (Area) = lim, n=∞, n=∞, 6n2, 2, n (1 + 1n )(2 + 1n ) 2 1, = = ., = lim, n=∞, 6 3, 6n2, Properties of definite integrals., , ST, , The following properties can be easily proved using Riemann sums., , (i), , Ra, b, , (ii), , Ra, , f (x)dx = −, , Rb, , f (x)dx., , a, , f (x)dx = 0., , a, , (iii), , Rb, a, , cdx = c(b − a),where c is any constant., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, Rb, , [ f (x) + g(x)]dx =, , Rb, , c f (x)dx = c, , Rb, , [ f (x) − g(x)]dx =, , a, , (vi), , a, , ., , f (x)dx +, , Rb, , Rb, , g(x)dx., , a, , a, , a, , (v), , Rb, , f (x)dx,where c is any constant., , a, , (vii) If a < c < b then, , Rb, a, , Rb, , f (x)dx −, , f (x)dx =, , a, , Rb, , g(x)dx., , a, , Rc, , f (x)dx +, , a, , Rb, c, , Comparison properties of the integral., (i) If f (x) ≥ 0 for a ≤ x ≤ b, then, , Rb, , f (x)dx., , AP, P, , (iv), , 153, , f (x)dx ≥ 0., , a, , Rb, , (ii) If f (x) ≥ g(x) for a ≤ x ≤ b, then, , f (x)dx ≥, , g(x)dx., , a, , CO, R, , a, , Rb, , (iii) If m ≤ f (x) ≤ M for a ≤ x ≤ b, then m(b − a) ≤, Proof. Given m ≤ f (x) ≤ M, , Rb, a, , f (x)dx ≤ M(b − a)., , Integrating between a and b we get, Rb, Rb, Rb, mdx ≤ f (x)dx ≤ Mdx, Rb, , U, , m dx ≤, , ST, , a, , a, , a, , a, , Rb, , f (x)dx ≤ M, , a, , m(b − a) ≤, , Rb, a, , Rb, , dx, , a, , f (x)dx ≤ M(b − a)., , The first fundamental theorem of calculus, Statement. If f is continuous on [a, b], then the function g defined by, g(x) =, , Rx, a, , f (t)dt, a ≤ x ≤ b, , is continuous on [a, b] and is differentiable on (a, b) and g′ (x) = f (x)., Proof.Choose x and x + b ∈ (a, b)., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 154, , Engineering Mathematics - I, , Now, g(x + h) − g(x) =, =, , =, , Zx+h, , f (t)dt −, , a, , Zx, , Zx, , f (t)dt +, , Zx+h, , a, Zx+h, , y, f (t)dt, , a, , f (t)dt −, , x, , Zx, , f (t)dt, , m, , M, , a, , f (t)dt., , 0, , x, , u, , v x+h, , ., , AP, P, , x, , R, g(x + h) − g(x) 1 x+h, =, f (t)dt., (1), h, h x, Let us assume that h > 0. Since f is continuous on [x, x + h], by the, , Hence, for h , 0, we have, , extreme value theorem, there exists numbers u and v in [x, x + h] such that, f (u) = m and f (v) = M, where m and M are the absolute minimum and maximum, have, , CO, R, , values of f on [x, x + h]. By the comparison property of the definite integral we, mh ≤, , x+h, R, x, , f (u)h ≤, , f (t)dt ≤ Mh., , x+h, R, x, , f (t)dt ≤ f (v)h., , Since h > 0, dividing throughout by h we get, x+h, R, f (t)dt ≤ f (v)., f (u) ≤ h1, , (2), , x, , U, , (2) can be proved in a similar way for h < 0., , Now, allowing h → 0 and since u and v lie in [x, x + h], we have u → x, , ST, , and v → x. Since f is continuous at x,we have, ∴ lim f (u) = lim f (u) = f (x), h→0, , u→x, , and lim f (v) = lim f (v) = f (x)., h→0, , v→x, , By (2) we have, 1, lim f (u) ≤ lim, h→0, h→0 h, , Zx+h, , f (t)dt ≤ lim f (v), h→0, , x, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , x
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 155, g(x + h) − g(x), ≤ f (x) [by (1)], h→0, h, , f (x) ≤ lim, , i.e., f (x) ≤ g′ (x) ≤ f (x)., By Squeeze Theorem we get, g′ (x) = f (x)., , (3), , Since every differentiable function is continuous,, , AP, P, , If x = a or x = b, from (3) we get the one sided limits at a and b., This shows that g is continuous on [a, b]., , Rx √, Example 3.2. Find the derivative of the function g(x) =, 1 + t2 dt., 0, √, Solution. Since f (t) = 1 + t2 is continuous on [0, x], by part 1 of the Fundamental, √, theorem of calculus we get g′ (x) = 1 + x2, 4, , , d Rx, sec tdt., Example 3.3. Find, dx 1, 4, Solution. Let u = x, x4, , u, , Z, , Z, , d , d , , sec tdt, ∴, sec tdt =, , dx , dx , 1, 1, , u, , Z, d , du, [By Chain rule], =, sec tdt ·, du, dx, , ST, , U, , CO, R, , ., , we obtain g is also continuous on (a, b)., , 1, , = sec u · 4x3, , [by FTC 1], , = sec x4 · 4x3 ., , Second Fundamental theorem of calculus, Rb, Statement.If f is continuous on [a, b], then f (x)dx = F(b) − F(a) where F is any, a, , antiderivative of f . [i.e, a function such that F ′ = f ]., Rx, Proof. Let g(x) = f (t)dt., a, , By the First Fundamental theorem of calculus we have g′ (x) = f (x). i.e,, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 156, , Engineering Mathematics - I, , g is an antiderivative of f ., If F is any other antiderivative of f on [a, b], we know that F and g differ, by a constant., ∴ F(x) = g(x) + c for a < x < b., Since both F and g are continuous on [a, b] the above relation holds when x = a and, x = b., , ., , x→a+, , Also lim g(x) = g(a), x→a+, , & lim F(x) = F(b)., x→b−, , AP, P, , i.e lim F(x) = F(a), , & lim g(x) = g(b)., x→b−, , Hence we obtain F(x) = g(x) + c for all x in [a, b], Ra, When x = a, we have g(a) = f (t)dt = 0., a, , (1), , Using (1) for x = b and x = a we get, , F(b) − F(a) = [g(b) + c] − [g(a) + c], , CO, R, , = g(b) − g(a), = g(b) =, , Zb, , f (t)dt., , a, , Result. The First and Second Fundamental Theorem of Calculus are combined, together is called the Fundamental Theorem of Calculus., Statement of Fundamental Theorem of Calculus., , U, , Suppose f is continuous on [a, b], Rx, (i) If g(x) = f (t)dt, then g′ (x) = f (x)., a, , Rb, , ST, , (ii) f (x)dx = F(b) − F(a), where F is any antiderivative of f . (i.e., F ′ = f .), a, , Note (1). The Fundamental theorem of calculus says that differentiation and, integration are inverse processors., Note (2). By part(ii) of the Fundamental theorem of calculus we have, Rb, f (x)dx = F(b) − F(a) where F ′ = f ., a, , i.e, , Rb, a, , F ′ (x)dx = F(b) − F(a)., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 157, , The above result is sometimes called as the Net Change Theorem., Net Change Theorem, Statement. The integral of a rate of change is the net change., Rb, i.e F ′ (x)dx = F(b) − F(a)., a, ✎, , ☞, , ✍, , ✌, , Worked Examples, , ., , 0, , (x2 − 2x)dx by using Reimann sum by taking right end, , AP, P, , Example 3.4. Evaluate, , R3, , points as the sample points. Hence verify it by using fundamental theorem of, calculus., Solution. Let f (x) = x2 − 2x., , follows., , CO, R, , Divide the interval [0, 3] into n sub intevals with equal width, 3−0 3, ∆x =, = ., n, n, ", # ", # ", #, ", #, 3, 3 6, 6 9, 3n − 3 3n, The intervals are 0, , , , , , · · ·, ,, ., n, n n, n n, n, n, The sample points xi∗ are the right end points of the sub intervals., 3n, 3 6 9, ∴ xi∗ = , , , · · · ., n n n, n, The corresponding f (xi∗ ) = xi∗ 2 − 2xi∗ are calculated and tabulated as, , 3, n, , xi∗, , 9, n2, , −, , U, , f (xi∗ ), , 6, n, , 6, n, , 36, n2, , −, , 9, n, , 12, n, , 81, n2, , −, , 18, n, , ···, ···, , 3n, n, 9n2, n2, , −, , 6n, n, , By Reimann’s sum we have,, n, X, , ST, , Z3, , (x2 − 2x)dx = lim, , n=∞, , 0, , f (xi∗ )∆x, , i=1, , !, !, !, ! #, 9, 36 12 3, 81 18 3, 9n2 6n 3, 6 3, +, +, +, ·, ·, ·, +, −, −, −, −, n=∞ n2, n n, n n, n n, n n, n2, n2, n2, " 3, !, !, !, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, = lim, · 1 − 2 · 2 + 3 · 22 − 2 · 4 + 3 · 32 − 2 · 6, 3, n=∞ n, n, n, n, n, n, !#, 3, 32, 3, 2, + · · · + 3 · n − 2 · 2n, n, n, , = lim, , ", , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 158, , Engineering Mathematics - I, , 32, 33 2, 2, 2, 2, (2 + 4 + 6 + · · · + 2n), 1, +, 2, +, 3, +, ·, ·, ·, +, n, −, lim, n=∞ n2, n=∞ n3, h n(n + 1) i, 27 h n(n + 1)(2n + 1) i, 9, = lim 3, − lim 2 · 2, n=∞ n, n=∞ n, 6, 2, 1, 1, 2, 3, h n (1 + 1n ) i 27, h n (1 + n )(2 + n ) i, 27, − 9 lim, =, lim, × 2 − 9 = 0., =, n=∞, 6 n=∞, 6, n3, n2, , = lim, , ., , 0, , Example 3.5. Evaluate the integral, , AP, P, , By the Fundamental Theorem of Calculus we have, !3, Z3, x3, x2, 27, 2, (x − 2x)dx =, −2·, =, − 9 = 9 − 9 = 0., 3, 2 0, 3, , R1 √, 1 − x2 dx interms of areas., 0, , √, R1 √, 1 − x2 dx represents the area under the curve y = 1 − x2 between, Solution., 0, , Now, y =, , p, 1 − x2, , y2 = 1 − x2, 2, , 2, , y, , CO, R, , the x−axis and the lines x = 0 and x = 1., , 0, , 1x, , x + y = 1., , U, , This is a circle with centre at the origin and radius = 1 unit. Since we, , ST, , need the area between x = 0 and x = 1, the required area is the area of the portion, π, 1, of the circle that lies in the first quadrant = · π × 12 = ., 4, 4, R1 √, π, 1 − x2 dx = ., ∴, 4, 0, R3, Example 3.6. Evaluate the integral (x − 1)dx interms of areas., 0, , Solution., , R3, 0, , (x − 1)dx represents the area under the curve y = (x − 1) between the, , x−axis and the lines x = 0 and x = 3., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 159, y, , The required area = A1 − A2 ., A1 is a triangle with base = 2units, and height = 2units., , , Put x = 3 in y = x − 1., , x, , A2, , x=0, , 1, × 2 × 2 = 2., 2, A2 is a triangle with base = 1 unit and height = 1unit., 1, 1, ∴ Area of A2 = × 1 × 1 = ., 2, 2, R3, 1 3, (x − 1)dx = 2 − = ., 2 2, 0, ∴ Area of A1 =, , Example 3.7. Prove that, , Rb, , xdx =, , Solution., , Rb, a, , b2 − a2, by interpretting interms of areas., 2, [A.U Nov 2016], , CO, R, , a, , AP, P, , ., , x=3, , A1, , xdx represents the area under the curve y = x between the x−axis, , and the lines x = a and x = b., , ST, , U, , Required area = Area of ∆OCD− Area of ∆OAB, 1, 1, = × OD × CD − × OB × AB, 2, 2, 1, 1, = b·b− a·a, 2, 2, b2 a2 b2 − a2, =, −, =, 2, 2, 2, Zb, b2 − a2, ., ∴, xdx =, 2, , y, C, A, , b, , a, 0, , a, , b−a, B, , Db, , x, , a, , Example 3.8. Express lim, , n, P, , n→∞ i=1, , (xi3 + xi sin xi )∆x as an integral on the interval [0, π]., , Solution. According to Riemann sum,the limit, Rπ, n, P, lim (xi3 + xi sin xi )∆x = (x3 + x sin x)dx., n→∞ i=1, , 0, , Example 3.9. Evaluate the Riemann sum for f (x) = x3 − 6x, taking the sample, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 160, , Engineering Mathematics - I, , R3, points to be the right end points and a = 0, b = 3, n = 6. Also Evaluate (x3 − 6x)dx., 0, , Solution. Divide the interval [0, 3] into 6 subintervals with equal width, i h, i h, i h, i h, i h, i, h, Hence, the intervals are 0, 21 , 21 , 1 , 1, 23 , 23 , 2 , 2, 52 , 25 , 3, 1, 3, 5, The right end points are , 1, , 2, , 3., 2, 2, 2, The value of f (x) at these points are as follows., , 1, 2, , 1, 8, , 1, , 1, , 3, 2, , 27, 8, , 2, , −6x, , f (xk ), , f (xk )∆x, , AP, P, , x3, , -3, , - 23, 8, , - 23, 16, , −6, , −5, , - 52, , −9, , - 45, 8, , - 45, 16, , 8, , -12, , -4, , -2, , 5, 2, , 125, 8, , −15, , 5, 8, , 3, , 27, , −18, , 9, , 5, 16, 9, 2, , CO, R, , ., , x(k), , 3−0 1, = ., 6, 2, , Reimann Sum =, , n, X, , f (xk )∆xk, , k=1, , 23 5 45, 5, 9, − −, −2+, +, 16 2 16, 16 2, −23 − 40 − 45 − 32 + 5 + 72, =, 16, −140 + 77 −63, =, =, ., 16, 16, , U, , =−, , R3, Let us evaluate (x3 − 6x)dx., 0, , ST, , Divide the interval [0, 3] into n subintervals with equal width, 3−0 3, = ., ∆x =, n, n, i h, i h, i, h, i, h, 3n, Hence the intervals are 0, 3n , n3 , n6 , 6n , n9 , · · · 3n−3, n , n, , The sample points xi∗ are the right end points of the sub intervals, ∴ xi∗ = n3 , 6n , n9 , · · · 3n, n., , The corresponding f (xi∗ ) = xi3 − 6xi are calculated and tabulated as, follows., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 161, 3, n, , xi∗, f (xi∗ ), , 33, n3, , 6, n, , −6·, , 63, n3, , 3, n, , −6·, , 9, n, 6, n, , 93, n3, , −6·, , 9, n, , ···, ···, , 3n, n, 33 n3, n3, , −6·, , 3n, n, , By Reimann sum, Z3, n, X, (x3 − 6x)dx = lim, f (xi∗ )∆x, n=∞, , 0, , i=1, , !, !, ! #, 33, 63, 33 n3, 3 3, 6 3, 3n 3, + 3 −6·, + ··· +, = lim, −6·, −6·, n=∞ n3, n n, n n, n n, n, n3, " 4, #, 2, , 3, 3, 3, 3, 3, 3, = lim 4 1 + 2 + 3 + · · · + n − 6 · 2 (1 + 2 + 3 + · · · + n), n=∞ n, n, , !, !, 2, 34 n(n + 1) 2, 3 n(n + 1) , −6· 2, = lim 4, , n=∞ n, 2, 2, n, " 4 2, #, 3 n (n + 1)2, 32 n(n + 1), = lim 4, −6· 2, n=∞ n, 4, 2, n, , !2, !, , 1, 1 , 4, 2, 4 n 1 +, n 1 + , 3, n, n , 32, , = lim 4 ·, −6· 2, n=∞ n, , 4, 2, n, , , , CO, R, , ., , AP, P, , ", , , !2, !, , 81, 1, 1, − 27 · 1 + , = lim · 1 +, n=∞ 4, n, n, 81, 81 − 108, 27, =, − 27 =, =− ., 4, 4, 4, , Fundamental, , theorem, , calculus gives the relation between, R, antiderivatives and integrals. The notation f (x)dx is traditionally used for an, Rb, antiderivative of f and is called an indefinite integral. Thus f (x)dx = F(x), , ST, , The, , U, , Indefinite Integrals., , of, , a, , means F ′ (x) = f (x)., R, d, [F(x)] = f (x) then f (x)dx = F(x), dx, Z, d, In other words if, [F(x)] = f (x), then, f (x)dx = F(x)., dx, d, Also we have, [F(x) + c] = f (x) then, dx, R, f (x)dx = F(x) + c,where c is called the constant of integration., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 162, , Engineering Mathematics - I, , With this notation, we can derive the basic results in integration, using the, basic results in differentiation., Basic results., d n, (x ) = nxn−1 ., dx, !, d xn+1, = xn ., (or), dx n + 1, R, xn+1, + c., ∴ xn dx =, n+1, R 1, d, 1, 2., (log x) = ⇒, dx = log x + c., dx, x, x, R, d x, (e ) = e x ⇒ e x dx = e x + c., 3., dx, , 5., , R, , b x dx =, , bx, + c., log b, , R, , sin xdx = − cos x + c., , R, , sec2 xdx = tan x + c., , R, , sec x tan xdx = sec x + c., , CO, R, , 4., , R, , cos xdx = sin x + c., , R, , cosec2 xdx = − cot x + c., , R, , cosecx cot xdx = −cosecx + c., , 12., , R, , 1, dx = tan−1 x + c., 1 + x2, , 13., , R, , 6., 7., 8., 9., , ST, , 10., , U, , ., , AP, P, , 1. We have, , 11., , 14., , R, , R, , 1, dx = sin−1 x + c., √, 1 − x2, , x, , √, , 1, x2, , −1, , dx = sec−1 x + c., , sinh xdx = cos hx + c., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, 15., 16., 17., , R, , 163, , cosh xdx = sin hx + c., , R, , √, , R, , √, , 1, x2 − 1, 1, x2, , dx = cos h−1 x + c or log(x +, , √, , dx = sin h−1 x + c or log(x +, , √, , x2 − 1) + c., , x2 + 1) + c., , +1, Basic properties of indefinite integrals., , ., , properties are true., R, R, R, (i) [ f (x) + g(x)]dx = f (x)dx + g(x)dx + c., (ii), , (iii), , R, , R, , [ f (x) − g(x)]dx =, k f (x)dx = k, , R, , R, , f (x)dx −, , f (x)dx + c., , R, , g(x)dx + c., , ✎, , AP, P, , If f (x) and g(x) are functions of x and k is a constant, then the following, , Worked Examples, , Z, , ✍, , ☞, , ✌, , CO, R, , Example 3.10. Evaluate (10x4 − 2 sec2 x)dx., Z, Z, Z, 4, 2, 4, Solution., (10x − 2 sec x)dx = 10 x dx − 2 sec2 xdx + c, Example 3.11. Evaluate, , Z2, 1, , !, Z2, Z2, Z2, 1, 1, 2, x−2 dx, x dx − 3 x 2 dx +, x − 3 x + 2 dx =, x, 2, , √, , U, , Solution., , Z2, , ST, , 1, , x5, −12!tan x + c = 2x5 − 2 tan x + c., = 10√·, 5, 2, x − 3 x + 2 dx., x, , 1, , 1, , 1, , 3 2, !2, x 2 , x−1, x, , , =, − 3 3 +, 3 1, −1 1, 2, !, 3 2, , =, , =, =, =, , 1, , !2, i, 1 3, 1, 2h 3, 3, 2, [2 − 1 ] − 3 × 2 − 1 −, 3, 3, x 1, √, 1, 1, [8 − 1] − 2[2 2 − 1] − [ − 1], 3, 2, √, 1, 7, −4 2+2+, 3, 2, √, √, √, 21 + 12 + 3, 36, −4 2=, − 4 2 = 6 − 4 2., 6, 6, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 166, , Engineering Mathematics - I, Let t = 2x + 3, , Now,, , 2, , (2x + 3) dx =, , dt 1, =, t ·, 2, 2, 2, , t2 dt =, , 1 t3, 1, · + c = · (2x + 3)3 + c., 2 3, 6, , AP, P, , Standard results., Z, 1. Evaluate (ax + b)n dx., , Z, , Solution. Let ax + b = t., , dt, adx = dt ⇒ dx = ., a, Z, Z, (ax + b)n+1, dt 1 tn+1, +c=, + c., (ax + b)n dx =, tn = ·, a, a n+1, a(n + 1), , 3., 4., 5., , ST, , 6., , CO, R, , 2., , In a similar way, the following results can be easily derived., Z, 1, log(ax + b), dx =, + c., ax + b, a, Z, eax+b, eax+b dx =, + c., a, Z, cos(ax + b), + c., sin(ax + b)dx = −, a, Z, sin(ax + b), + c., cos(ax + b)dx =, a, Z, tan(ax + b), + c., sec2 (ax + b)dx =, a, Z, cosec(ax + b), + c., cosec2 (ax + b)dx = −, a, Z, sec(ax + b), sec(ax + b) tan(ax + b)dx =, + c., a, Z, cos(ax + b), + c., cosec(ax + b) cot(ax + b)dx = −, a, , U, , ., , Z, , dt = 2dx, dt, ∴ dx =, 2Z, , 7., , 8., , 9., , ✎, , ☞, , ✍, , ✌, , Worked Examples, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 170, , Engineering Mathematics - I, Z, Z, 3, p, √, 3, 1, u2, 2, 2, 2x 1 + x dx =, udu =, u 2 du = 3 + c = (1 + x2 ) 2 + c., 3, 2, , Example 3.24. Evaluate, Solution. Let u = 1 − 4x3 ., , ., , Z, , p, x2 1 − 4x3 dx., , du = −4 × 3x2 dx., du, x2 dx = −, 12, !, Z, Z, p, √ −du, 2, 3, ∴, x 1 − 4x dx =, u, 12, Z, 1, 1, u 2 du, =−, 12, 3, , 1 u2, =−, +c, 12 32, , 3, 3, 1 2, 1, (1 − 4x3 ) 2 + c = − (1 − 4x2 ) 2 + c., 12 3, 18, , CO, R, , =−, , AP, P, , ∴, , Z, , Z, , ST, , U, , 1, Example 3.25. Evaluate, dx., x )(1 + e−x ), (1, +, e, Z, Z, 1, 1, dx, Solution., dx =, x, −x, 1, (1 + e ) (1 + e ), (1 + e x )(1 + x ), e, Z, ex, =, dx, (1 + e x )(1 + e x ), Z, ex, dx, =, 1 + ex = t, x )2, (1, +, e, Z, dt, e x dx = dt, =, 2, t, Z, 1, 1, t−1, +c=− +c=−, + c., =, t−2 dt =, −1, t, 1 + ex, , Example 3.26. Evaluate, , Z, , x3 cos(x4 + 2)dx., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 173, #, ", 1 2, 2, 4, 2 27, 2 23, 2 52, =, (1 + x ) + (1 + x ) − (1 + x ) + c, 2 7, 3, 5, 7, 3, 5, 1, 1, 2, = (1 + x2 ) 2 + (1 + x2 ) 2 − (1 + x2 ) 2 + c., 7, 3, 5, , x, , ex, x, 2, , e −1, , dx., , dx, = dt, 2, Z, Z, x x, e2 e2, ex, dx =, dx, ∴, x, x, e2 − 1, e2 − 1, Z, x, x, e2, =, · e 2 dx, x, Z e2 − 1, 1+t, 2dt, =, t, !, Z, 1, + 1 dt, =2, t, "Z, Z #, 1, =2, dt + dt + c, t, x, , e2 ·, , CO, R, , ., , Solution. Let e 2 = t., , Z, , AP, P, , Example 3.30. Evaluate, , U, , = 2[log t + t] + c, x, , x, = 2[log e 2 − 1) + e 2 − 1 ] + c., , ST, , Z, 1, Example 3.31. Evaluate, dx., Z 1 + tan x, Z, 1, 1, dx =, dx, Solution., sin x, 1 + tan x, 1+, cos x, Z, Z, cos x, 1, 2 cos x, =, dx =, dx, sin x + cos x, 2, sin x + cos x, Z, sin x + cos x + cos x − sin x, 1, dx, =, 2, sin x + cos x, !, Z, cos x − sin x, 1, 1+, =, dx, 2, sin x + cos x, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 174, , Engineering Mathematics - I, , Z, Z, 1, 1, cos x − sin x, dx +, dx, 2, sin x + cos x, Z 2, 1, dt, 1, = x+, 2, 2, t, 1, 1, = x + log(t) + c, 2, 2, x 1, = + log(sin x + cos x)) + c., 2 2, Z, tan x, Example 3.32. Evaluate, dx., sec x + cos x, sin x, Z, Z, tan x, cos x, dx, dx =, Solution., 1, sec x + cos x, + cos x, Z cos x, sin x, =, dx, 2, Z 1 + cos x, −dt, =, 1 + t2, =, , dt = (cos x − sin x)dx, , AP, P, , ., , t = sin x + cos x, , CO, R, , = − tan−1 (t) + c, , = − tan−1 (cos x) + c, , t = cos x, , dt = − sin xdx, , −dt = sin xdx, , Z4 √, 2x + 1dx., Example 3.33. Evaluate, 0, , Solution. Let 2x + 1 = t., , U, , 2dx = dt ⇒ dx =, , When x = 0, t = 1., , dt, ., 2, , When x = 4, t = 9., , ST, , Z4 √, Z9, √ dt, ∴, 2x + 1dx =, t, 2, 0, , 1, , 1, =, 2, , Z9, , 1, , t 2 dt, , 1, , 3 9, 1 2 3, 1 t 2 , 1, 1, 26, = 3 = × [9 2 − 1] = [27 − 1] = × 26 = ., 2 2, 2 3, 3, 3, 3, 1, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 175, , Example 3.34. Evaluate, , Z2, 1, , 1, dx., (3 − 5x)2, , Solution. Let 3 − 5x = t., −5dx = dt ⇒ dx = −, , When x = 0, t = −2., , When x = 2, t = −7., !, Z2, Z−7, dt, 1, 1, ∴, =, −, 5, (3 − 5x)2, t2, 1, , −2, , 1, =−, 5, , Z−7, , AP, P, , ., , dt, ., 5, , t−2 dt, , −2, , CO, R, , !−7, 1 t−1, =−, 5 −1 −2, !−7, !, ", #, 1 1, 1 −2 + 7, 1 5, 1, 1 −1 1, =, +, =, = ·, = ., =, 5 t −2 5 7, 2, 5, 14, 5 14 14, , Example 3.35. Evaluate, , Ze, , log x, dx., x, , 1, , Solution. Let t = log x., , U, , 1, dt = dx., x, When x = 1, t = log 1 = 0., , ST, , When x = e, t = log e = 1., !1, Ze, Z1, 1, 1, t2, log x, tdt =, ∴, dx =, = −0= ., x, 2 0 2, 2, 1, , 0, , Integrals of symmetric functions, Theorem. Suppose f is continuous on [−a, a], Ra, Ra, (i) If f is even, then f (x)dx = 2 f (x)dx., (ii) If f is odd, then, , −a, Ra, , 0, , f (x)dx = 0., , −a, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 177, , By the properties of definite integrals, ∴, , Z2, , f (x)dx = 2, , Z2, , f (x)dx, , 0, , −2, , =2, , Z2, , (x6 + 1)dx, , ., , 0, , 0, , AP, P, , 0, , 2, , Z2 , Z, , , = 2 x6 dx + dx, , , , , , !, " 7, #, , x7 2, 2, 2 , , −0+2−0, + (x)0 = 2, = 2 , 7 0, 7, ", #, #, ", 128, 2 × 142 284, 128 + 14, =2, =, +2 =2, =, ., 7, 7, 7, 7, tan x, dx., 1 + x2 + x4, , CO, R, , Example 3.37. Evaluate, , Z1, , −1, , tan x, ., Solution. Let f (x) =, 1 + x2 + x4, tan(−x), f (−x) =, 1 + (−x)2 + (−x)4, − tan x, =, = − f (x)., 1 + x2 + x4, ∴ f (x) is odd., , ST, , U, , By the properties of definite integrals, Z1, f (x)dx = 0, i.e,, , Z1, , −1, , −1, , tan x, dx = 0., 1 + x2 + x4, , 3.4 Integration of rational functions by Partial Fractions., Techniques of resolving a given fraction into partial fractions., p(x), where p(x) and q(x) are, A rational function is generally of the form, q(x), , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 178, , Engineering Mathematics - I, , algebraic expressions., , AP, P, , ., , If the degree of p(x) is less than the degree of q(x) then the fraction, p(x), is called a proper fraction., q(x), p(x), If the degree of p(x) ≥ degree of q(x), then the given fraction, is, q(x), improper., p(x), into partial fractions, the following, While resolving a fraction, q(x), technique can be adopted., p(x), be proper and if q(x) is expressed as linear factors., q(x), 1+x, , then we can express, For example, if the given fraction is, (x − 1)(x + 2)(x − 3), the fraction in the following way, 1+x, A, B, C, =, +, +, ., (x − 1)(x + 2)(x − 3) x − 1 x + 2 x − 3, , Case (i) Let, , Case (iii) Let, , p(x), be proper and q(x) contains nonfactorisable second degree, q(x), , U, , factors., , CO, R, , p(x), be proper and q(x) contains repeated linear factors., q(x), x2 + 2x − 3, Consider the following example, .This must be written as, (x − 1)3 (2x + 3), B, C, D, A, x2 + 2x − 3, +, ., +, +, =, 3, 2, 3, x, −, 1, 2x, +3, (x − 1) (2x + 3), (x − 1), (x − 1), , Case (ii) Let, , Consider the following example, , x2 + 3x + 2, . This can be resolved, (x − 1)2 (x2 + 4)(x2 + 9), , ST, , into partial fractions as follows., x2 + 3x + 2, B, Cx + D Ex + F, A, +, + 2, =, + 2, ., 2, 2, 2, 2, (x − 1) (x + 4)(x + 9) x − 1 (x − 1), x +4, x +9, , Methodology for improper fractions., , Case (i) If degree of p(x) = degree of q(x)., x2 + x + 1, ., Example 1. Consider, (x − 1)(x − 2), This must be resolved as, x2 + x + 1, B, C, = A+, +, ., (x − 1)(x − 2), x−1 x−2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 179, , ", , AP, P, , CO, R, , ., , p(x), = A constant + the usual methods adopted for proper fractions. ], q(x), Cx + D, x3 + 2x + 4, B, + 2, Example 2., = A+, ., 2, x, −, 1, (x − 1)(x + 4), x +9, Case (ii) If the degree of p(x) is one more than that of q(x), then, p(x), = A First degree expression + usual method adopted in proper functions., q(x), x4 + x + 4, C, Dx + E, Example 1., = Ax + B +, ., + 2, 2, x−1, (x − 1)(x + 2), x +2, x3 + 2x + 3, C, D, Example 2., = Ax + B +, ., +, 2, x + 2 (x + 2)2, (x + 2), Case (iii) If the degree of p(x) is 2 more than that of q(x), then, p(x), = A second degree expression + usual method adopted in proper functions., q(x), E, F, x7 + 2x + 4, D, Gx + H, +, +, Example 1., = Ax2 + Bx + C +, ., + 2, 2, 2, 2, x + 2 x − 1 (x − 1), (x + 2)(x − 1) (x + 1), x +1, This procedure can be extended for any improper fraction depending on the nature, p(x), ., of the fraction, q(x), Standard results., 1, dx., + a2, Solution. Let x = a tan θ., , 1. Evaluate, , R, , x2, , ST, , U, , dx = a sec2 θdθ., Z, Z, 1, 1, · a sec2 θdθ, dx, =, ∴, 2 tan2 θ + a2, x2 + a2, a, Z, 1, =, · a sec2 θdθ, a2 (tan2 θ + 1), Z, 1, 1, =, · sec2 dθ, 2, a, Z sec θ, 1, dθ, =, a, 1, 1 −1 x , = θ + c = tan, + c., a, a, a, R, 1, 2. Evaluate, dx., 2, x − a2, Solution. Let, , x2, , 1, 1, A, B, =, =, +, 2, (x − a)(x + a) x − a x + a, −a, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 180, , Engineering Mathematics - I, , =, , A(x + a) + B(x − a), (x − a)(x + a), , ∴ 1 = A(x + a) + B(x − a)., Put x = a., 1 = A · 2a ⇒ A =, , 1, ., 2a, , Put x = −a., , AP, P, , CO, R, , ., , 1, ., 2a ", #, −1, 1, 1, 1, 1, 1, 2a, 2a, +, =, −, ∴ 2, =, x − a x + a 2a x − a x + a, x − a2, !, Z, Z, 1, 1, 1, 1, −, dx, ∴, dx, =, 2a x − a x + a, x2 − a2, !, Z, Z, 1, 1, 1, =, dx −, dx, 2a, x−a, x+a, x − a, , 1 , 1, =, + c., log, log(x − a) − log(x + a) + c =, 2a, 2a, x+a, 1 = B · (−2a) ⇒ B = −, , 1, dx., − x2, 1, 1, A, B, A(a + x) + B(a − x), =, Solution. Let 2, =, +, =, 2, (a − x)(a + x) a − x a + x, (a − x)(a + x), a −x, , 3. Evaluate, , R, , a2, , ST, , U, , ∴ 1 = A(a + x) + B(a − x)., Put x = a., 1, 1 = A · 2a ⇒ A = ., 2a, Put x = −a., 1, 1 = B · (2a) ⇒ B = ., 2a, , ", #, 1, 1, 1, 1, 1, 1, 2a, 2a, +, =, +, ∴ 2, =, a − x2 a − x a + x 2a a − x a + x, !, Z, Z, 1, 1, 1, 1, ∴, dx, +, dx =, 2a a − x a + x, a2 − x2, !, Z, Z, 1, 1, 1, =, dx +, dx, 2a, a−x, a+x, #, ", 1 log(a − x), + log(a + x) + c, =, 2a, −1, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 181, , , 1 , log(a + x) − log(a − x) + c, 2a, a + x, 1, log, + c., =, 2a, a−x, Z, 1, dx, Type 1. Evaluation of integrals of the form, 2, ax + bx + c, Method. Express the denominator to any one of the forms x2 − a2 , x2 + a2 or a2 − x2, =, , and then apply the correct formula., , ☞, , ✍, , ✌, , Z, , AP, P, , Worked Examples, , 1, Example 3.38. Evaluate, dx., 2, xZ + 8x − 7, Z, 1, 1, Solution., dx =, dx, 2, 2, x + 8x − 7, (x + 4) − 7 − 16, Z, 1, =, dx, (x + 4)2 − 23, Z, 1, dx, =, √, 2, (x + 4) − ( 23)2, √ , , 1, x + 4 − 23 , = √ log , √ + c., 2 23, x + 4 + 23, Z, 1, dx., Example 3.39. Evaluate, 2, Z, Z1 + x − x, 1, 1, Solution., dx =, dx, 2, 1+x−x, 1 − (x2 − x), Z, 1, dx, , =, 1 − (x − 21 )2 − 41, Z, 1, =, dx, 1 − (x − 21 )2 + 14, Z, 1, dx, =, 1 2, 5, 4 − (x − 2 ), Z, 1, =, dx, √, 5 2, ( 2 ) − (x − 21 )2, √5, , 1, , +, x, −, , 1, 2, 2, + c, = √ log √, 5, 1, −, x, +, 2 25, 2, 2, , √, 5 + 2x − 1 , 1, = √ log √, + c., 5, 5 − 2x + 1, , ST, , U, , CO, R, , ., , ✎, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 182, , AP, P, , Z, 1, Example 3.40. Evaluate, dx., 2, 2x, Z −x+5, Z, 1, 1, ! dx, dx =, Solution., 2x2 − x + 5, x 5, 2 x2 − +, 2 2, Z, 1, 1, dx, =, !, 2, 2, 1, 5, 1, + −, x−, 4, 2 16, Z, 1, 1, =, dx, !2, 2, 1, 40 − 1, x−, +, 4, 16, Z, 1, 1, dx, =, !2, 2, 39, 1, +, x−, 4, 16, Z, 1, 1, =, √ 2 dx, !, 2, 2, 39 , 1, , x−, + , 4, 4, , , !, x − 1 , 4x − 1, 1 1, 2, 4, −1 , −1, +c, = √ tan √ + c = √ tan, √, 39, 2 39, 39, 39, , CO, R, , ., , Engineering Mathematics - I, , 4, , 4, , Type 2. Evaluation of integrals using partial fractions., , Z, , U, , x+5, dx., x2 + x − 2, x+5, A, B, x+5, =, =, +, ., Solution. Let 2, (x, +, 2)(x, −, 1), x, +, 2, x, −, 1, x +x−2, A(x − 1) + B(x + 2), =, (x + 2)(x − 1), , ST, , Example 3.41. Evaluate, , ∴ x + 5 = A(x − 1) + B(x + 2)., Put x = 1., 6 = 3B ⇒ B = 2., , Put x = −2., 3 = −3A ⇒ A = −1., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 184, , Engineering Mathematics - I, sec2 xdx = dt., , ., , Z, 1, sec2 x, dt, dx =, 2, 2, tan x + 3 tan x + 2, Z t + 3t + 2, 1, =, dt, (t + 1)(t + 2), Z, (t + 2) − (t + 1), =, dt, (t + 1)(t + 2), !, Z, 1, 1, −, dt, =, (t + 1) t + 2, Z, Z, 1, 1, =, dt −, dt + c, (t + 1), t+2, , AP, P, , ∴, , Z, , CO, R, , = log(t + 1) − log(t + 2) + c, !, t+1, +c, = log, t+2, !, tan x + 1, + c., = log, tan x + 2, , Z, , 3x + 1, dx., (x − 1)2 (x + 3), B, 3x + 1, A, C, +, Solution. Let, =, +, 2, 2, x+3, (x − 1) (x + 3) x − 1 (x − 1), A(x − 1)(x + 3) + B(x + 3) + C(x − 1)2, =, (x − 1)2 (x + 3), , ST, , U, , Example 3.44. Evaluate, , Put x = 1., , ∴ 3x + 1 = A(x − 1)(x + 3) + B(x + 3) + C(x − 1)2 ., , 4B = 4 ⇒ B = 1., , Put x = −3., , −1, ., 2, Equating the coefficients of x2 on both sides we get,, 1, A + C = 0 ⇒ A = −C = ., 2, 16C = −8 ⇒ C =, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, 1, 1, 1, 3x + 1, = 2 +, − 2, ∴, (x − 1)2 (x + 3) x − 1 (x − 1)2 x + 3, Z, Z, Z, Z, 3x + 1, 1, 1, 1, 1, 1, ∴, dx, +, dx, dx, =, dx, −, 2, x−1, 2, x+3, (x − 1)2 (x + 3), (x − 1)2, Z, 1, 1, = log(x − 1) + (x − 1)−2 dx − · log(x + 3) + c, 2, 2, (x − 1)−1, 1, =, +c, log(x − 1) − log(x + 3) +, 2, −1, !, 1, x−1, 1, −, + c., = log, 2, x+3, x−1, , AP, P, , ., , 185, , 2x2 − x + 4, dx., x3 + 4x, 2, 2, 2x − x + 4 2x − x + 4 A Bx + C A(x2 + 4) + (Bx + C)x, Solution. Let 3, =, = + 2, =, ., x, x + 4x, x(x2 + 4), x +4, x(x2 + 4), , Example 3.45. Evaluate, , Z, , CO, R, , ∴ 2x2 − x + 4 = A(x2 + 4) + (Bx + C)x., Put x = 0., 4A = 4 ⇒ A = 1., , Equating the coefficients of x2 on both sides we get,, A + B = 2 ⇒ 1 + B = 2 ⇒ B = 1., , Equating the coefficients of x on both sides we get,, , ST, , U, , C = −1., 1, 1, x−1, x, 2x2 − x + 4 1, = + 2, = + 2, − 2, ., ∴, 3, x x + 4Z x x + 4 Z x + 4, x + 4x, Z, Z, x, 2x2 − x + 4, 1, 1, dx +, ∴, =, dx −, dx, 3, 2, 2, x Z, x + 4x, x +Z4, x +4, 1, 1 dt, −, = log x +, dx, 2, t 2, x + 22, , , x, 1, 1, +c, = log x + log t − tan−1, 2, 2, 2, 1 −1 x , 1, 2, + c., = log x + log(x + 4) − tan, 2, 2, 2, , Example 3.46. Evaluate, , Z, , In the second integral, let x2 + 4 = t, 2xdx = dt, dt, xdx = ., 2, , 2, dx., (1 − x)(1 + x2 ), , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 186, , Engineering Mathematics - I, , Solution. Let, , A, 2, Bx + C, =, +, 2, (1 − x)(1 + x ) 1 − x, 1 + x2, A(1 + x2 ) + (Bx + C)(1 − x), =, ., (1 − x)(1 + x2 ), , Put x = 1., , ∴ 2 = A(1 + x2 ) + (Bx + C)(1 − x)., , 2A = 2 ⇒ A = 1., , ., , A − B = 0 ⇒ A = B = 1., , AP, P, , Equating the coefficients of x2 on both sides we get,, , Equating the coefficients of x on both sides we get,, B−C = 0, , CO, R, , B = C ⇒ C = 1., 1, 1, 1, x+1, x, 2, =, +, =, +, +, ∴, (1Z− x)(1 + x2 ) 1 − x Z1 + x2 1 −Zx 1 + x2 Z, 1 + x2, 1, 1, 2, x, dx =, dx +, dx, ∴, dx +, 2, 2, 2, 1−x, (1 − x)(1 + x ), x +1, Z x +1, log(1 − x), 1 dt, +, + tan−1 x + c, =, −1, t 2, 1, = − log(1 − x) + log t + tan−1 x + c, 2, 1, = − log(1 − x) + log(1 + x2 ) + tan−1 x + c., 2, , In the second integral, let 1 + x2 = t, , Z, , 2xdx = dt, dt, xdx = ., 2, , U, , 10, dx., (x − 1)(x2 + 9), [A.U. Dec. 2015], A, Bx + C A(x2 + 9) + (Bx + C)(x − 1), 10, =, =, ., + 2, Solution. Let, (x − 1)(x2 + 9) x − 1, x +9, (x − 1)(x2 + 9), ∴ 10 = A(x2 + 9) + (Bx + C)(x − 1)., , ST, , Example 3.47. Evaluate using partial fractions, , Put x = 1., , 10A = 10 ⇒ A = 1., , Equating the coefficients of x2 on both sides we get,, A + B = 0 ⇒ B = −A = −1., Equating the constants on both sides we get,, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 187, , 9A − C = 10, , ⇒ C = 9A − 10 = 9 − 10 = −1., , −x − 1, 1, 10, + 2, =, 2, (x − 1)(x + 9) x − 1 x + 9, x, 1, 1, − 2, =, − 2, ., Zx − 1 x + 9Z x + 9, Z, Z, 1, 1, 10, x, dx =, dx −, dx, ∴, dx −, 2, 2, 2, x−1, (x − 1)(x + 9), x +9, x +9, Z, 1, 1 −1 x , 2x, = log(x − 1) −, dx, −, tan, +c, 2, 3, 3, x2 + 9, x, 1, 1, = log(x − 1) − log(x2 + 9) − tan−1, +c, 2, 3, 3, , , p, 1, x, +c, = log(x − 1) − log( x2 + 9) − tan−1, 3, 3, !, x, 1, (x − 1), − tan−1, = log √, + c., 3, 3, x2 + 9, , CO, R, , ., , AP, P, , ∴, , 4x2 − 3x + 2, dx., 4x2 − 4x + 3, 4x2 − 3x + 2, Solution. The integrand 2, is an improper fraction with the degrees, 4x − 4x + 3, of the Nr and Dr same., Example 3.48. Evaluate, , Z, , ST, , U, , ∴By the method of partial fractions we have, 4x2 − 3x + 2, Bx + C, = A+ 2, [∵4x3 − 4x + 3 is a nonfactorizable, 2, 4x − 4x + 3, 4x − 4x + 3, A(4x2 − 4x + 3) + Bx + C, second degree factor], =, 4x2 − 4x + 3, ∴ 4x2 − 3x + 2 = A(4x2 − 4x + 3) + Bx + C., Equating the coefficients of x2 ., 4A = 4 ⇒ A = 1., , Equating the coefficients of x, −4A + B = −3 ⇒ −4 + B = −3 ⇒ B = 4 − 3 = 1., Equating the constants, 3A + C = 2 ⇒ 3 + C = 2 ⇒ C = 2 − 3 = −1., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 188, , Engineering Mathematics - I, x−1, 4x2 − 3x + 2, =1+ 2, 2, 4x − 4x + 3, 4x − 4x + 3, x−1, =1+, 4(x2 − x + 3/4), x−1, x−1, =1+ , , =1+ , 3, 1, 1 2, 4 (x − 2 ) + 4 − 4, 4 (x − 21 )2 + 21, ∴, , , Z , , , x−1, 1 + , dx, 4 (x − 21 )2 + 21, Z, Z, x−1, 1, =, dx +, dx, 4, (x − 12 )2 + 21, Z, x − 12 − 21, 1, dx, = x+, 4, (x − 12 )2 + 21, Z, Z, x − 12, 1, 1, 1, = x+, dx, −, dx., 1, 1, 1, 4, 8, (x − 2 )2 + 2, (x − 2 )2 + 12, Z, Z, 1, 1 dt 1, 1, = x+, −, dx, 1, 1, 4, t+ 2 2 8, (x − 2 )2 + ( √1 )2, , 4x2 − 3x + 2, dx =, 4x2 − 4x + 3, , CO, R, , ., , Z, , AP, P, , ∴, , 2, , In the second integral, , ST, , U, , !2, 1, =t, Put x −, 2, !, 1, 2 x − dx = dt, 2, !, dt, 1, x − dx = ., 2, 2, , , x − 1 , 1, 1 1, 1, 2, log(t + ) − 1 tan−1 1 + c, 8, 2, 8 √, √, 2, 2, !, ! √, 2 −1 2x − 1 √, 1 2 1, 1, log (x − ) +, tan, × 2 +c, −, 8, 2, 2, 8, 2, !, !, 1, 1, 1, 1, −1 2x − 1, 2, +c, log x + − x +, − √ tan, √, 8, 4, 2, 4 2, 2, !, !, 2x − 1, 1, 1, 4x2 − 4x + 3, + c., log, − √ tan−1 √, 8, 4, 4 2, 2, , = x+, = x+, , = x+, = x+, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 189, , x4 − 2x2 + 4x + 1, dx., [A.U Nov 2016]., x3 − x2 − x + 1, Solution. The integrand is an improper fraction in which the degree of the Nr is, , Example 3.49. Evaluate, , 1 more than that of the Dr. Hence, by the method of partial fractions we have, x4 − 2x2 + 4x + 1, x4 − 2x2 + 4x + 1, x4 − 2x2 + 4x + 1, =, =, x3 − x2 − x + 1, x2 (x − 1) − (x − 1), (x − 1)(x2 − 1), x4 − 2x2 + 4x + 1, x4 − 2x2 + 4x + 1, =, =, ., (x − 1)(x − 1)(x + 1), (x − 1)2 (x + 1), x4 − 2x2 + 4x + 1, C, E, D, Let, = Ax + B +, +, +, 2, 2, x − 1 (x − 1), x+1, (x − 1) (x + 1), (Ax + B)(x − 1)2 (x + 1) + C(x − 1)(x + 1) + D(x + 1) + E(x − 1)2, =, (x − 1)2 (x + 1), , AP, P, , ., , Z, , ∴ x4 − 2x2 + 4x + 1 = (Ax + B)(x − 1)2 (x + 1) + C(x − 1)(x + 1) + D(x + 1) + E(x − 1)2 ., Put x = 1 : 2D = 1 − 2 + 4 + 1 = 4 ⇒ D = 2., Put x = −1 : 4E = 1 − 2 + 4 + 1 = 4, , ⇒ 4E = −4 ⇒ E = −1., , A = 1., , CO, R, , Equating the coefficients of x4 on both sides we get,, , Put x = 0 : B − C + D + E = 1, B − C + 2 − 1 = 1., , (1), , B − C = 0., , Put x = 2 : (2A + B) · 3 + 3C + 3D + E = 16 − 8 + 8 + 1, 6A + 3B + 3C + 3D + E = 17., , U, , 6 + 3B + 3C + 6 − 1 = 17, 3B + 3C = 17 − 11 = 6, , (2), , ST, , B + C = 2., , (1) + (2) ⇒ 2B = 2, , ⇒ B = 1., , (1) ⇒ B = C = 1., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 190, , Engineering Mathematics - I, , 1, 1, x4 − 2x2 + 4x + 1, 2, = x+1+, −, +, 3, 2, 2, x − 1 (x − 1), x+1, x −x −x+1, Z, Z, Z, Z, Z 4, Z, 1, 1, x − 2x2 + 4x + 1, 1, dx, =, xdx, +, dx, +, dx, −, ∴, dx, +, 2, dx, 3, 2, 2, x−1, x+1, x −x −x+1, (x − 1), Z, x2, + x + log(x − 1) + 2 (x − 1)−2 dx − log(x + 1) + c, =, 2, x2, (x − 1)−1, =, + x + log(x − 1) + 2, − log(x + 1) + c, 2, −1, !, 2, x−1, x2, + x + log, + c., −, =, 2, x+1, x+1, Z, x3, dx., Example 3.50. Evaluate, (x − 1)(x − 2), Solution. The integrand is an improper fraction in which the degree of the Nr is, , ., , 1 more than that of the Dr., , AP, P, , ∴, , CO, R, , ∴ By the method of partial fractions we have, x3, C, D, = Ax + B +, +, (x − 1)(x − 2), x−1 x−2, (Ax + B)(x − 1)(x − 2) + C(x − 2) + D(x − 1), =, (x − 1)(x − 2), , ∴ x3 = (Ax + B)(x − 1)(x − 2) + C(x − 2) + D(x − 1)., Put x = 1 : −C = 1 ⇒ C = −1., , Put x = 2 : D = 8., , Equating the coefficients of x3 on both sides we get,, , U, , A = 1., , Put x = 0 : 2B − 2C − D = 0, , ST, , 2B + 2 − 8 = 0., 2B − 6 = 0, , 2B = 6 ⇒ B = 3., 1, 8, x3, = x+3−, +, ∴, (x − 1)(x − 2), x−1 x−2, Z, Z, Z, Z, Z, 3, 1, 1, x, dx =, xdx + 3 dx −, dx + 8, dx, ∴, (x − 1)(x − 2), x−1, x−2, x2, =, + 3x − log(x − 1) + 8 log(x − 2) + c., 2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 191, , x3 + x, dx., x−1, Solution. The integrand is an improper fraction in which the degree of the Nr is, Example 3.51. Evaluate, , Z, , 2 more than that of the Dr., , ., , ∴ x3 + x = (Ax2 + Bx + C)(x − 1) + D., Put x = 1 : D = 2., , AP, P, , ∴ By the method of partial fractions we have, D, (Ax2 + Bx + C)(x − 1) + D, x3 + x, = Ax2 + Bx + C +, =, ., x−1, x−1, x−1, , Equating the coefficients of x3 on both sides we get,, A = 1., , Equating the coefficients of x2 on both sides we get,, −A + B = 0., B = A = 1., , CO, R, , Put x = 0 : −C + D = 0, C = D = 2., , 2, x3 + x, = x2 + x + 2 +, ., x−1, x−1, Z 3, Z, Z, Z, Z, x +x, 1, ∴, dx =, x2 dx +, xdx + 2 dx + 2, dx, x−1, x−1, x3 x2, +, + 2x + 2 log(x − 1) + c., =, 3, 2, , ST, , U, , ∴, , 3.5 Integration of irrational functions, Standard Results, , 1. Evaluate, , Z, , 1, , dx., a2 − x2, Solution. Let x = a sin θ, √, , ⇒, , dx = a cos θdθ., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 197, Z, , Z, , 1, dx., √, 2, 9x + 24x, Z, Z, 1, 1, Solution., dx =, √, q , dx, 9x2 + 24x, 9 x2 + 24x, 9, Z, 1, 1, dx, =, q, 3, 8x, 2, x + 3, Z, 1, 1, =, dx, q, , 3, 16, 4 2, x+ 3 − 9, Z, 1, 1, =, q, 2 2 dx, 3, x + 43 − 34, s, , , !, !2, , , 1, 16, 4, 4, = log x +, +, − + c, x+, 3, 3, 3, 9, r, , , !, 3x + 4, 1, 8 , , + x2 + x + c., = log , 3, 3, 3, Example 3.54. Evaluate, , ST, , U, , CO, R, , ., , AP, P, , 1, dx, p, 9 − (x − 4)2 + 16, Z, 1, dx, =, p, 25 − (x − 4)2, Z, 1, =, dx, p, 52 − (x − 4)2, !, −1 x − 4, = sin, + c., 5, =, , 3.5.2 Type II. Evaluation of integrals of the form, , Z, , √, , ℓx + m, ax2 + bx + c, , dx, , , d 2, ax + bx + c + B. Find, dx, the values of A and B and then substitute for ℓx + m in the integral which will be, Evaluation procedure. Assume express ℓx + m = A ·, , evaluated easily., , ✎, , ☞, , ✍, , ✌, , Worked Examples, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 198, , Engineering Mathematics - I, Z, , 3x + 1, dx., √, 2x2 + x + 3, d, Solution. Let 3x + 1 = A (2x2 + x + 3) + B, dx, , Example 3.55. Evaluate, , AP, P, , ., , = A(4x + 1) + B., Equating the coefficients of x on both sides we get,, 3, 4A = 3 ⇒ A = ., 4, Equating the constants, 3, A+B=1⇒ +B=1, 4, 3 1, B=1− = ., 4 4, , ST, , U, , CO, R, , 1, 3, ∴ 3x + 1 = (4x + 1) + ., 4, 4, 3, Z, Z (4x + 1) + 1, 3x + 1, 4, 4 dx., ∴, dx =, √, 2, 2, 2x + x + 3, Z 2x + x + 3, Z, 4x + 1, 1, 1, 3, dx +, dx., =, √, √, 2+x+3, 2+x+3, 4, 4, 2x, 2x, Z, Z, −1, 1, 1, 3, (2x2 + x + 3) 2 d(2x2 + x + 3) +, =, s, ! dx, 4, 4, x 3, 2 x2 + +, 2 2, Z, 1, 1, 1, 3 (2x2 + x + 3) 2, +, =, dx, √, s, 1, !2, 4, 4 2, 2, 1, 1, 3, x+, + −, 4, 2 16, Z, p, 1, 1, 3, 2, dx, = × 2 2x + x + 3 + √, s, !2, 4, 4 2, 1, 24 − 1, x+, +, 4, 16, Z, p, 1, 3, 1, 2x2 + x + 3 + √, =, dx, s, !2 √ 2, 2, 4 2, , , 1, 23 , + , x+, , 4, 4, s, , , !2, , , 1, 23, 1, 1, 3p 2, + + c, x+, 2x + x + 3 + √ log x + +, =, 2, 4, 4, 16, 4 2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , AP, P, , r, , , 3p 2, 1, x 3 , 4x + 1, 2, =, 2x + x + 3 + √ log , + x + + + c., 2, 4, 2 2, 4 2, Z, 1, Example 3.56. Evaluate, dx by using trignometric substitution. Hence, √, a2 − x2, 6x + 5, dx., [A.U Nov 2016], use it in evaluating, 6 + x − 2x2, Solution. For the first part, refer Result(1). Let us evaluate the second part., d, Let 6x + 5 = A (6 + x − 2x2 ) + B, dx, = A(1 − 4x) + B., Equating the coefficients of x on both sides we get, 3, 6, −4A = 6 ⇒ A = − = − ., 4, 2, Equating the constants,we get, 3, A+B=5⇒− +B=1, 2, 3 13, B=5+ = ., 2, 2, , CO, R, , ., , 199, , ST, , U, , 13, 3, ∴ 6x + 5 = − (1 − 4x) + ., 2, 2, Z −3, Z, 13, (1, −, 4x), +, 6x + 5, 2, 2, dx =, dx., ∴, √, √, 2, 2, 6 + x − 2x, 6, +, x, −, 2x, Z, Z, 3, 1 − 4x, 1, 13, =−, dx +, dx., √, √, 2, 2, 2, 6 + x − 2x, 6 + x −Z2x2, Z, −1, 13, 1, 3, dx, (6 + x − 2x2 ) 2 d(6 + x − 2x2 ) +, =−, q, 2, 2, 2(3 + 2x − x2 ), Z, 1, 3 (6 + x − 2x2 ) 2, 1, 13, dx, =−, + √, q, 1, 2, 2 2, 2, 3 − (x2 − 2x ), Z, p, 13, 1, = −3 6 + x − 2x2 + √, q, o dx, n, 1, 1 2, 2 2, 3 − (x − 4 ) − 16, Z, p, 13, 1, dx, = −3 6 + x − 2x2 + √, q, 1, 2 2, 3 − (x − 14 )2 + 16, Z, p, 13, 1, 2, dx, = −3 6 + x − 2x + √, q, 2 2, ( 74 )2 − (x − 14 )2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 202, , Engineering Mathematics - I, , Solution., , Z p, , x2 − 4x + 6 dx =, , Z p, , (x − 2)2 + 6 − 4 dx, , Z p, , (x − 2)2 + 2 dx, Z q, √, =, (x − 2)2 + ( 2)2 dx, !q, √, x−2, (x − 2)2 + ( 2)2, =, 2, √, !, q, √, ( 2)2, 2, 2, +, log (x − 2) + (x − 2) + ( 2) + c, 2, !, , , p, x−2 p 2, 2, =, x − 4x + 6 + log (x − 2) + x − 4x + 6 + c., 2, , =, , AP, P, , ., , 3.5.4 Type IV.Integration by trigonometric substitution., , Certain integrals can be easily evaluated by trigonometric substitution., , CO, R, , There are three cases., , (i) Any integral involving quantities of the form, , √, a2 − x2 can be evaluated by, , the substitution x = a sin θ., , (ii) Integrals involving functions of the form, , √, a2 + x2 can be evaluated by the, , U, , substitution x = a tan θ or x = a sin hθ., , (iii) Integrals involving functions of the form, , √, , x2 − a2 can be evaluated by the, , ST, , substitution x = a sec θ or x = a cos hθ., ✎, , ☞, , ✍, , ✌, , Worked Examples, , Z √, 9 − x2, Example 3.60. Evaluate, dx., x2, √, √, Solution. Since 9 − x2 is of the form a2 − x2 , we can evaluate this integral by, , the substitution., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 217, , 1.3.5.7 π 35π, × =, ., 2.4.6.8 Z 2 256, Evaluation of integrals of the form, sinm x · cosn xdx, =, , Evaluation procedure, , AP, P, , ., , (i) If the power of cosine is odd (say n = 2k + 1)then write, Z, Z, m, n, sin x cos xdx =, sinm x · cos2k+1 xdx, Z, =, sinm x · cos2k x cos xdx, Z, =, sinm x · (cos2 x)k cos xdx, Z, =, sinm x · (1 − sin2 x)k cos xdx, , Now make the substitution t = sin x, and hence the integral is evaluated, , CO, R, , easily., , (ii) If, Z the power of sineZis odd (say n = 2k + 1)then write, sinm x cosn xdx =, sin2k+1 x · cosn x cos xdx, Z, =, sin2k x · sin x cosn xdx, Z, = (1 − cos2 x)k · cosn x sin xdx., , Now make the substitution t = cos x, and hence the integral is evaluated, , U, , easily., , ST, , (iii) If the powers of both sine and cosine are even, then use the relations, (1 − cos 2x), sin2 x =, 2, (1, +, cos, 2x), cos2 x =, 2, so that the resulting integrals can be evaluated easily., (iv) If the powers of both sine and cosine are odd, then apply either case(i) or, case(ii)., ✎, , ☞, , ✍, , ✌, , Worked Examples, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 222, , Engineering Mathematics - I, , Let us apply integration by parts, dv = sec2 xdx, Z, Z, du = sec x tan xdx, dv =, sec2 xdx ⇒ v = tan x, Z, Z, Z, 3, sec xdx =, udv = uv − vdu, Z, = sec x · tan x − tan x · sec x · tan xdx, Z, = sec x tan x − sec x tan2 xdx, Z, = sec x tan x − sec x(sec2 x − 1)dx, Z, Z, Z, 3, 3, sec xdx = sec x tan x − sec xdx + sec xdx, Z, Z, sec3 xdx + sec3 xdx = sec x tan x + log(sec x + tan x) + c, Z, 2 sec3 xdx = sec x tan x + log(sec x + tan x) + c, Z, , 1, sec3 xdx = sec x tan x + log(sec x + tan x) + c., 2, , CO, R, , ., , AP, P, , Choose, u = sec x, , 3.8 Improper integrals, , Definition of an improper integral Type-I, Rt, a, , f (x)dx exists for every number t ≥ a, then, , U, , (i) If, , R∞, , t→∞, , ST, , a, , f (x)dx = lim, , Rt, , f (x)dx, , a, , provided this limit exists (as a finite number)., , (ii) If, , Rb, t, , f (x)dx exists for every number t ≤ b, then, Rb, , −∞, , f (x)dx = lim, , t→−∞, , Rb, , f (x)dx, , t, , provided this limit exists (as a finite number)., R∞, Rb, The improper integrals f (x)dx and, f (x)dx are called convergent if the, a, , −∞, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 223, , corresponding limit exists and divergent if the limit does not exist., (iii) If both, , R∞, a, , Ra, , f (x)dx and, R∞, , f (x)dx are convergent then we define, , −∞, , f (x)dx =, , −∞, , Ra, , R∞, , f (x)dx +, , f (x)dx., , a, , −∞, , Here a is any real number., ✎, , ☞, , ✍, , ✌, , ., , Example 3.83. Discuss the convergence of, , R∞, 0, , Z∞, , Solution., , −x, , e dx = lim, , t→∞, , 0, , e−x, = lim, t→∞ −1, , !t, , 0, , , , −t, , e−x dx., , 0, , = − lim e − e, t→∞, , , , e−x dx is convergent., , Example 3.84. Determine whether the integral, 1, dx = lim, t→∞, x2, , Zt, , U, , Solution., , Z∞, , !, 1, = − lim t − 1 = −(−1) = 1, t→∞ e, , CO, R, , ∴, , e−x dx, , 0, , 0, , Z∞, , Zt, , AP, P, , Worked Examples, , 1, , 1, dx, x2, , R∞ 1, dx is convergent or divergent., 2, 1 x, , 1, , = lim, , ST, , t→∞, , Zt, , x−2 dx, , 1, , ∴, , Z∞, , x−1, = lim, t→∞ −1, , !t, , 1, , 1, = − lim, t→∞ x, , !t, , 1, , !, 1, = − lim, − 1 = −(−1) = 1., t→∞ t, , 1, dx is convergent., x2, , 1, , Example 3.85. Determine the convergence of, , R∞ 1, 1, , x, , dx., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 224, , Engineering Mathematics - I, , Solution., , Z∞, , 1, dx = lim, t→∞, x, , Zt, , 1, dx, x, , 1, , 1, , = lim (log x)t1 = lim (log t − log 1) = lim log t = ∞, t→∞, , t→∞, , t→∞, , ., , Example 3.86. Discuss the convergence of, , R∞, 0, , Solution., , Z∞, 0, , a2, , 1, dx = lim, t→∞, + x2, , Zt, , a2, , 0, , 1, dx, + x2, x t, , AP, P, , i.e the limit does not exist, as a finite number., R∞ 1, dx is divergent., ∴, 1 x, , a2, , 1, dx., + x2, , U, , CO, R, , 1 −1, = lim, tan, t→∞ a, a 0, , 1 −1 t, tan, − tan−1 0, = lim, t→∞ a, a, 1, −1, = (tan ∞ − 0), a, , π, 1 π, −0 =, which is a finite number., =, a 2, 2a, R∞ 1, dx is convergent., ∴, 2, 2, 0 a +x, , Example 3.87. Determine whether the integral, dx, √ = lim, x t→∞, , ST, , Solution., , Z∞, 1, , Zt, 1, , x, , − 21, , R∞ dx, √ convergent or divergent, x, 1, , 1 t, x 2 , √, dx = lim 1 = 2 lim ( t − 1) = 2 · ∞ = ∞., t→∞, , 2, , 0, , t→∞, , Here the limit does not exist as a finite number., R∞ 1, ∴ √ dx is divergent., x, 1, , Example 3.88. Discuss the convergence of the integral, , R∞, , log xdx., , 1, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, Z∞, , Solution., , 225, , log x dx = lim, , t→∞, , Zt, , log xdx, , 1, , 1, , , , Zt, , , , 1, , , x · dx, = lim log x · x t0 −, t→∞ , x , , [using integration by parts], , 1, , ., , i, , , = 2 lim t log t − (x)t1 = 2 lim t log t − t + 1 = ∞, t→∞, , t→∞, , log xdx is divergent., , 1, , AP, P, , ∴, , R∞, , h, , Z∞, , Example 3.89. Determine whether the integral, , log x, dx is convergent or, x, , 1, , divergent. Evaluate if the integral is convergent., Z∞, , Solution., , log x, dx = lim, t→∞, x, , 1, , Zt, , log x, dx = lim, t→∞, x, , 1, , Zt, 1, , [A.U. Dec. 2015], , log xd(log x), , CO, R, , !t, (log x)2, , = lim, , t→∞, , 2, , 1, , , 1, = lim (log t)2 − (log 1)2, 2 t→∞, 1, = lim (log t)2 = ∞,, [which is not a finite number.], 2 t→∞, , , ∴, , Z∞, , is divergent., , U, , 1, , log x, dx, x, , Example 3.90. For what values of p is the integral, , ST, , Z∞, , Solution., , 1, dx = lim, t→∞, xp, , 1, , Zt, , R∞ 1, dx convergent?, p, 1 x, , x−p dx, , 1, , = lim, , t→∞, , x−p+1, −p + 1, , !t, , 1, !, 1, 1, 1, 1−p, −1 ., lim (t, − 1) =, ·, =, 1 − p t→∞, 1 − p t p−1, , If p > 1, then p − 1 > 0 and hence t p−1 → ∞ as t → ∞ and hence, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , 1, t p−1, , → 0.
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 226, , Engineering Mathematics - I, R∞ 1, 1, 1, dx =, (0 − 1) =, , which is a finite number., p, 1− p, p−1, 1 x, R∞ 1, dx is convergent if p > 1., ∴, p, 1 x, If p < 1, then p − 1 < 0 or 1 − p > 0., 1, Hence p−1 = t1−p → ∞ as t → ∞., t, R∞ 1, 1, dx =, (∞ − 1) = ∞., ∴, p, x, 1, −, p, 1, R∞ 1, ∴, dx diverges, if p < 1., p, 1 x, R∞ 1, dx = ∞, [Example 3.96], Also, when p = 1,, 1 x, R∞ 1, dx is divergent if p = 1., ∴, p, 1 x, R∞ 1, Hence,, dx is convergent if p > 1 and divergent if p ≤ 1., p, 1 x, , CO, R, , ., , AP, P, , ∴, , Z0, , Example 3.91. Discuss the convergence of, R0, , Solution., , −∞, , =, , e x dx = lim, , t→−∞, , lim [e x ]0t, t→−∞, , R0, , e x dx, , −∞, , e x dx, , t, , = lim [e0 − et ] = 1 − 0 = 1, which is a finite number, t→−∞, , U, , ∴ The given integral is convergent., , ST, Solution., , Z0, , −∞, , 1, (1 − 3x), , R0, , 1, dx., (1, −, 3x)2, −∞, Z0, dx = lim, (1 − 3x)−2 dx, 2, , Example 3.92. Evaluate, , t→−∞, , t, , = lim, , t→−∞, , Example 3.93. Evaluate, , R0, , ", , (1 − 3x)−1, (−1)(−3), , #0, t, , ", #0, 1, 1, 1, 1, lim [1 − 0] = ., =, t→−∞ 3 1 − 3x, t→−∞, 3, 3, t, , = lim, , xe x dx., , −∞, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 228, , Engineering Mathematics - I, , Example 3.96. Discuss the convergence of, , Solution., , Z0, , x sin xdx = lim, , t→−∞, , Z0, , Z0, , x sin xdx., , −∞, , xd(− cos x), , t, , −∞, , , , Z0, , , , , 0, = lim [−x cos x]t − − cos xdx, t→−∞ , , t, , ., , AP, P, , = lim [0 − t cos t + (sin x)0t ], t→−∞, , = lim (−t cos t + 0 − sin t), t→−∞, , Since cos x and sin x oscillates between −1 and 1,, , R0, , x sin xdx oscillates finitely., , −∞, , CO, R, , Definition of an improper integral type II, , (i) If f is continuous on [a, b) and is discontinuous at b, then, Zb, Zt, f (x)dx = lim−, f (x)dx, t→b, , a, , = lim, , ∈→0, , a, b−∈, Z, , f (x)dx, , a, , U, , if the limit exists (as a finite number)., , ST, , (ii) If f is continuous on (a, b] and is discontinuous at a, then, Zb, Zb, f (x)dx = lim+, f (x)dx, t→a, , a, , t, , = lim, , Zb, , ∈→0, a+∈, , f (x)dx, , if the limit exists (as a finite number)., Rb, The improper integral f (x)dx is called convergent if the corresponding limit, a, , exists and divergent if the limit does not exist., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 229, , (iii) If f hss a discontinuity at c where a < c < b, and both, , Rc, , f (x)dx and, , a, , a, , c, , ✎, , ☞, , ✍, , ✌, , Worked Examples, , Z1, , 1, √ dx if possible., x, , AP, P, , ., , Example 3.97. Evaluate, , f (x)dx, , c, , a, , are convergent, then we define, Rb, Rc, Rb, f (x)dx = f (x)dx + f (x)dx., , Rb, , 0, , Solution. The integrand has an infinite discontinuity at the lower limit 0., 1 1, Z1, Z1, √, 1, x 2 , − 12, x dx = lim 1 = 2 lim (1 − ∈) = 2., ∴, √ dx = lim, ∈→0, ∈→0, ∈→0, x, 2 ∈, 0+∈, , 0, , Example 3.98. Determine whether the integral, , Z3, , dx, √ is convergent or, x, , 0, , [A.U. Dec. 2015], , CO, R, , divergent. Evaluate if it is convergent., , Solution. The integrand has an infinite discontinuity at the lower limit 0., 1 3, Z3, Z3, x 2 , √, √, √, 1, 1, ∴, x− 2 dx = lim 1 = 2 lim ( 3 − ∈) = 2 3,, √ dx = lim, ∈→0, ∈→0, ∈→0, x, 2 ∈, 0, , 0+∈, , U, , [which is a finite number.], √, Hence, the given integral is convergent and its values is 2 3., Example 3.99. Find the value of the improper integral, , Z5, 2, , √, , 1, x−2, , dx if possible., , ST, , Solution. The integrand has an infinite discontinuity at the lower limit 2., , ∴, , Z5, 2, , √, , 1, , x−2, , dx = lim, , Z5, , ∈→0, 2+∈, , (x − 2), Z1, , − 12, , , 5, (x − 2) 21 , , dx = lim , 1, ∈→0, , 2, , 2+∈, , √, √, √, = 2 lim ( 3 − ∈) = 2 3., ∈→0, , 1, dx if possible., √, 1 − x2, 0, Solution. The integrand has an infinite discontinuity at the upper limit 1., , Example 3.100. Evaluate, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, −1 and 1., ∴, , Z1, , −1, , 1, x, , 2, 3, , 231, , dx =, , Z0, , −1, , 1, x, , 2, 3, , Z1, , dx +, , 1, 2, , dx, , x3, , 0, , Z0−∈, Z1, 2, 2, = lim, x− 3 dx + lim, x− 3 dx, ∈→0, , ∈→0, 0+∈, , −1, , 1 1, 1 −∈, x 3 , x 3 , = lim 1 + lim 1 , ∈→0, , ∈→0, , ., , ∈→0, , AP, P, , 3 −1 1, 3 ∈ , , 1, = 3 lim (− ∈) 3 + 1 + 3 lim 1 − (∈) 3 = 3 × 1 + 3 × 1 = 6., ∈→0, , Example 3.104. Evaluate the improper integral, , Z3, , 1, dx if possible., x−1, , 0, , Solution. The integrand has an infinite discontinuity at x = 1 which lies between, 0 and 3., , 0, , Z1, , Z3, , CO, R, , ∴, , Z3, , 1, dx =, x−1, , 0, , 1, dx +, x−1, , = lim, , ∈→0, , Z1−∈, 0, , , , 1, , 1, dx + lim, ∈→0, x−1, , = lim log |x − 1|, ∈→0, , 1, dx, x−1, Z3, , 1+∈, , 1−∈, 0, , 1, dx, x−1, , , , + lim log |x − 1| 31+∈, ∈→0, , ST, , U, , , = lim log | ∈ | − log | − 1| + log 2 − log | ∈ |, , ∴, , R3, 0, , ∈→0, , = log 0 + log 2 − log 0, = −∞ + log 2 + ∞, , which is an indeterminate, , 1, dx diverges., x−1, , Example 3.105. Evaluate, , Z1, , 1, dx if possible., x, , −1, , Solution. The integrand has an infinite discontinuity at x = 0 which lies between, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 232, , Engineering Mathematics - I, , −1 and 1., , Z1, , ∴, , 1, dx =, x, , Z0, , −1, , −1, , Z1, , 1, dx +, x, , = lim, , ∈→0, , 0, , Z0−∈, , −1, , 1, dx + lim, ∈→0, x, , ., , Z1, , 1, dx, x, , 0+∈, , h, , = lim log |x|, ∈→0, , 1, dx, x, , i−∈, −1, , h, i1, + lim log |x|, , −∈, , ∈→0, , AP, P, , , , = lim log |− ∈ | − log | − 1| + lim log 1 − log | ∈ |, ∈→0, , ∈→0, , , , = lim log | ∈ | − log(∈), ∈→0, !, |∈|, = lim log 1 = 0,, = lim log, ∈→0, ∈→0, |∈|, , ∴, , R1 1, dx converges., −1 x, , which is a finite number, , CO, R, , A comparison test for improper integrals., , Comparison theorem. Suppose that f and g are continuous functions with, f (x) ≥ g(x) ≥ 0 for x ≥ a., R∞, a, , (ii) If, , R∞, , a, , R∞, g(x)dx is divergent, then f (x)dx is divergent., a, , ST, , a, , R∞, f (x)dx is convergent, then g(x)dx is convergent., , U, , (i) If, , Example 3.106. Show that, , Solution., , R∞, , −x2, , e, , 0, , R1, , dx =, , R1, 0, , −x2, , e, , ✎, , ☞, , ✍, , ✌, , Worked Examples, , Z∞, , 2, , e−x dx is convergent., , 0, , −x2, , e, , dx +, , R∞, , 2, , e−x dx, , 1, , dx =a definite value and hence it is convergent., , 0, , Let us consider the integral, , R∞, , 2, , e−x dx., , 1, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Integral Calculus, , 233, , For all x ≥ 1,we have x2 ≥ x., ⇒ −x2 ≤ −x, 2, , 2, , ⇒ e−x ≤ e−x ⇒ e−x ≥ e−x ., Z∞, Zt, t, −x, e dx = lim, e−x dx = lim −e−x 1, Now, t→∞, , 1, , t→∞, , 1, , −t, , −1, , = − lim e − e, t→∞, , ., , ∴, , R∞, , e−x dx converges., , 1, , By comparison theorem, ⇒, , R∞, , R∞, , , , !, 1, 1, = ,, =− 0−, e, e, , 2, , e−x dx converges., , 0, , −x2, , e, , which is finite., , AP, P, , , , dx is convergent., , 0, , Example 3.107. Prove that the integral, , Z∞, , 1 + e−x, dx is divergent., x, , CO, R, , 1, , 1 + e−x 1 e−x 1, Solution. We have for all x ≥ 1,, = +, >, x, x, x, x, R∞ 1, dx diverges, [∴ p = 1], We know from Example 3.101,, 1 x, R∞ 1 + e−x, ∴ By comparison theorem,, dx is divergent., x, 1, , 1, , U, , Example 3.108. Determine the convergency or divergency of the integral, Z∞, 1, dx, using comparison test., 3, x +1, 1, 1, 1, 1, < 3 ⇒ 3 > 3, ., +1 x, x, x +1, ∞, R 1, dx converges, We have from Example 3.101,, 3, 1 x, R∞ 1, ∴ By comparison theorem, dx is convergent., 3, 1 x +1, , ST, , Solution. For all x ≥ 1, we have,, , x3, , [p > 1]., , Example 3.109. Determine the convergency or divergency of the integral, Z∞, cos2 x, dx., x2, 2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 234, , Solution. We know that for all x ≥ 2 cos2 x ≤ 1., 1, cos2 x, ≤ 2., [∴ x2 > 0], ⇒, 2, x, x, 1, cos2 x, or 2 ≥, ., x, x2, R∞ 1, dx converges, [p > 1]., By Example 3.101,, 2, 2 x, R∞ cos2 x, dx also converges., ∴ By comparison theorem, x2, 2, , Z∞, , AP, P, , ., , Engineering Mathematics - I, , Example 3.110. Discuss the convergency or divergency of the integral, , 2, , 1, dx., x + ex, , 1, 1, <, = e−x ., Solution. For all x > 2, we have, x + ex ex, 1, ⇒ e−x >, ., x + ex, Zt, Z∞, t, −x, Now e dx = lim, e−x dx = lim −e−x 2, t→∞, , t→∞, , 2, , CO, R, , 2, , , , −t, , , , R∞, , 1, dx is convergent., x + ex, , = − lim e − e, t→∞, , ∴, , R∞, 2, , e−x dx is convergent., , ∴ By comparison theorem,, , !, 1, 1, =− 0− 2 = 2, e, e, , −2, , ST, , U, , 2, , which is finite., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , ., , 4.1 Evaluation of double integrals, Definition., , Let f (x, y) be a single, , valued and bounded function of two, independent variables x and y which, is defined in a closed region R of, , Y, , R, , Divide the region, , CO, R, , the xy plane., , AP, P, , 4 Multiple Integrals, , R into rectangles by drawing lines, parallel to the coordinate axes. Let, Ai be the ith rectangle., , Let (xi , yi ), , be any point inside the ith rectangle, and ∆Ai be its area., , X, , O, , Let there be, , U, , n rectangles A1 , A2 , . . . , An which lie, , ST, , completely inside R., , Let us consider the sum, f (x1 , y1 )∆A1 + f (x2 , y2 )∆A2 + · · · + f (xn , yn )∆An =, , n, X, , f (xi , yi )∆Ai ., , (1), , i=1, , As n → ∞, the number of rectangles increase indefinitely such that the largest, linear dimension of the rectangle which is the diagonal of ∆Ai tends to zero. The, limit of the sum (1) if exists is defined to be the double integral of f (x, y) over the, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 236, , Engineering Mathematics - I, , region R and it is denoted by, , ", , f (x, y)dA., , R, , i.e., lim, , n, X, , n→∞, ∆Ai →0 i=1, , f (xi , yi )∆Ai =, , ", , f (x, y)dA., , R, , If A is a typical rectangle whose dimensions parallel to the coordinate axes are dx, , R, , AP, P, , ., , and dy then dA = dxdy. Then the above double integral can be written as, ", ", ", f (x, y)dA =, f (x, y)dxdy =, f (x, y)dydx., R, , R, , In cartesian coordinates the following cases are to be taken into account while, evaluating the double integrals., , Y, , Case (i) Consider the case where, Let the, , CO, R, , the limits are constants., , R, , region R be a rectangle given by, R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d},, , where, , a, b, c, d are constants, then, the double, integral is given by, , R, , , , Zb Zd, Zd, , , , f (x, y)dy dx or =, f (x, y)dxdy =, , , , U, , Z Z, , a, , c, , c, , X, , O, , , b, , Z, , , f, (x,, y)dx, dy., , a, , i.e., if the limits are constants, the order of integration, is immaterial, provided, , ST, , proper limits are taken., , Case (ii) Consider the case when, , Y, , the limits of x are constants and, , the limits of y are functions of x., , h(x), , In this case the region R is given, by, , R = {(x, y) : a ≤ x ≤ b, g(x) ≤ y ≤ h(x)},, , where a, b are constants., , Now the, , g(x), x=a, , x=b, , double integral becomes, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , X
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , 237, , ", , , , Zb Zh(x), , , , f (x, y)dxdy =, f (x, y)dy dx., , , , a, , R, , g(x), , i.e., if the limits of the innermost integral are functions of x, then the order of, evaluation of the integral is, first with respect to y and finally with respect to x., , AP, P, , ., , Case (iii) Consider the case when, , Y, , the limits of y are constants and, the, y., by, , limits, , of, , are, , x, , functions, , y=d, , of, , Now the region R is given, , g(y), , R = {(x, y) : g(y) ≤ x ≤ h(y), c ≤ y ≤ d},, , where c, d are constants., , The double, , h(y), , y=c, , integral now takes the form, , CO, R, , X, , ", , , , Zd Zh(y), , , , f (x, y)dxdy =, f (x, y)dx dy., , , c, , R, , g(y), , i.e., if the limits of the innermost integral are functions of y, the order of, , U, , integration is first with respect to x and finally with respect to y., , ST, , Example 4.1. Evaluate, y = 2., , Solution., , ", , dxdy =, , R, , ", , ✎, , ☞, , ✍, , ✌, , Worked Examples, , dxdy over the region R bounded by x = 0, x = 2, y = 0,, , R, , [Jan 1996], Z2 Z2, 0, , dxdy., , 0, , Since the limits are constants, the order of integration is immaterial., ∴, , ", R, , dxdy =, , Z2, , (x)20 dy, , 0, , =, , Z2, 0, , 2dy = 2, , Z2, , dy = 2[y]20 = 4., , 0, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 242, , ∴, , Engineering Mathematics - I, , ", , xydxdy =, , R, , Za, , y=0, , 1, =, 2, , √2 2, Za −y, x=0, , Za, , y=0, , y=0, , ", #a, ", #, ", #, 1 a4 a4, 1 2 y2 y4, a4, 1, a4, =, y(a − y )dy =, a, −, −, =, 1−, = ., 2, 2, 4 0 2 2, 4, 4, 2, 8, 2, , 2, , !, Example 4.10. Evaluate, xydxdy over the region in the positive quadrant, x y, bounded by + = 1., a b, Solution. The region of integration is bounded by the coordinate axes y = 0, x = 0, x y, and the line + = 1., a b, Y, Since the innermost integration, (0, b), , AP, P, , ., , √, Za " 2 # a2 −y2, x, xydxdy =, y, dy, 2 0, , B, , CO, R, , is w.r.t. x, divide the region into strips, , x, a, , x=0, , parallel to the x-axis. Along a typical, , , strip PQ, x varies from 0 to a 1 − by . y, , varies from 0 to b. a(1− y ), Zb Z b, ", xydxdy, ∴, xydxdy =, R, , y, b, , =1, , Q, , P, (0, 0), , +, , y=0, , A, (a, 0), , y=0 x=0, , x2, y·, 2, , !a(1− by ), , U, , Zb, , =, , y=0, , #, Zb " , y 2, y a2 1 −, − 0 dy, b, , ST, , 1, =, 2, , dy, , 0, , y=0, , a2, =, 2, , a2, =, 2, , Zb, , y=0, , !, y2 2y, dy, y 1+ 2 −, b, b, , Zb, , !, y3 2y2, dy, y+ 2 −, b, b, , y=0, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , X
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , ", #b, 2y3, a2 y2, y4, −, =, +, 2 2 4b2, 3b 0, " 2, #, 2, 4, 2b3, a b, b, =, + 2−, 2 2, 3b, 4b, " 2, #, 2, 2, b, 2b2, a b, +, −, =, 2 2, 4, 3, " 2, #, 2, 2, a 6b + 3b − 8b2, =, 2, 12, 2, 2, 2, a b2, a b, =, =, 2 12, 24, ", Example 4.11. Evaluate, xydxdy where R is the region bounded by the, , AP, P, , ., , 243, , R, , parabola y2 = x, the x−axis and the line x + y = 2 lying on the first quadrant., Solution. Solving y2 = x and x + y = 2 we get the points of intersection., , CO, R, , y2 + y − 2 = 0 ⇒ (y + 2)(y − 1) = 0 ⇒ y = 1, −2. When y = 1, x = 1., Y, , Divide the region into strips parallel to, the x−axis ., varies from, , Along a typical strip, x, , y2, , to 2−y and finally y varies, , from 0 to 1., , xydxdy =, , Z1 Z2−y, , y=0 x=y2, , xydxdy =, , ST, , R, , P, , U, , ", , =, , (0, 2), , Z1, , y=0, , Z1, , x2, y, 2, , y=0, , !2−y, , , y, 1, (2 − y)2 − y4 dy =, 2, 2, , y2 = x, , (1, 1), Q, (2, 1), , X, x=2−y, , dy, , y2, , Z1 , , y 4 + y2 − 4y − y4 dy, , y=0, , Z1, , !1, , , 1 y2 y4, y3 y6, 3, 2, 5, 4 +, −4 −, 4y + y − 4y − y dy =, 2 2, 4, 3, 6 0, 0, !, ", #, 9, 3, 1 24 + 3 − 16 − 2, 1, 1 4 1 4 1, + − −, = ., =, = ×, =, 2 2 4 3 6, 2, 12, 2 12 8, 1, =, 2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 244, , Engineering Mathematics - I, , Example 4.12. Evaluate, , ", , xydxdy where A is the region bounded by x = 2a and, , A, , the curve x2 = 4ay., , [Jan 2006], , Solution., Y, Let us evaluate this double integral first, , region into strips parallel to the y−axis., PQ is one such strip., , x2 = 4ay, , x2, Along this strip, y varies form 0 to, 4a, and finally x varies from 0 to 2a., , ", A, , xydxdy =, , x = 2a, , X, , P, , x2, , Z2a Z4a, , Q, , xydydx, , CO, R, , ., , AP, P, , w.r.t. y and then w.r.t. x. Divide the, , x=0 y=0, , x2, Z2a " 2 # 4a, y, x, =, dx, 2 0, , =, , x=0, Z2a, , U, , x=0, , !, Z2a 5, x, x4, dx =, dx, x, 2, 32a, 32a2, x=0, , #2a, x6, , a4, 1, 1, 6, (64a, −, 0), =, =, ., 3, 32a2 6 0, 32a2 × 6, ", Example 4.13. Evaluate, xy(x + y)dxdy over the area between y = x2 and y = x., , ST, , =, , ", , R, , Solution. Solving the two equations we get the points of intersection., y = x2 , y = x., =⇒ x2 = x =⇒ x2 − x = 0 =⇒ x(x − 1) = 0 =⇒ x = 0 & x = 1., When x = 0, y = 0, when x = 1, y = 1., , ∴ The points of intersection are (0, 0) and (1, 1)., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , 245, , Y, Let us evaluate this double, integral, first w.r.t. y and then w.r.t. x., y=x, , Divide this region into strips parallel to, Q, , y = x2, , the y−axis. PQ is one such strip. Along, P, , this strip, y varies from x2 to x and x, varies from 0 to 1., ", Z1 Zx, xy(x + y)dxdy =, (x2 y + xy2 )dydx, R, , x=0 y=x2, , =, , y3, x, 3, , !x, , x2, , +x, , 2, , y2, 2, , !x !, x2, , dx, , x=0, Z1, 0, , !, x 3, x2 2, 6, 4, (x − x ) + (x − x ) dx, 3, 2, , Z1, , !, x4 x6 x4 x7, dx., −, +, −, 2, 2, 3, 3, , CO, R, , =, , Z1, , =, , 0, , x5 x7 x5 x8, =, −, +, −, 10 14 15 24, , !1, 0, , =, , 1, 1, 1, 1, 3, −, +, −, = ., 10 14 15 24 56, , ydxdy over the region R bounded by y = x and, , U, , Example 4.14. Evaluate, , ", , X, , (0, 0), , AP, P, , ., , (1, 1), , R, , y = 4x − x2 ., , ST, , Solution. The equation of the parabola is, y = −(x2 − 4x) = −((x − 2)2 − 4) = −(x − 2)2 + 4, , ⇒ y − 4 = −(x − 2)2, vertex is (2, 4) and it is open downwards., , Solving the two equations we get, x = 4x − x2 ⇒ x2 + x − 4x = 0 ⇒ x2 − 3x = 0 ⇒ x(x − 3) = 0 ⇒ x = 0, x = 3., When x = 0, y = 0., When x = 3, y = 12 − 9 = 3., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 246, , Engineering Mathematics - I, , The points of intersection are (0, 0), (3, 3)., Y, , Let us evaluate the double integral first, , (2, 4), , w.r.t. y and then w.r.t. x. Divide the, , Q, , (3, 3), , region into strips parallel to the y−axis., y=x, , PQ is one such strip. Along this strip, y, P, , varies from x to 4x − x2 . Finally x varies, , X, (0, 0), , from 0 to 3., ", R, , Zx=3 4x−x, Z 2, Z3 2 !4x−x2, y, ydxdy =, ydydx =, dx, 2 x, x=0 y=x, , 1, =, 2, , Z3, , x=0, , , , x=0, , 1, (4x − x ) − x dx =, 2, 2 2, , x5, , , , !3, x4, , Z3 , 0, , , 16x2 + x4 − 8x3 − x2 dx, , #, 35, 1, 1, 3, 4, 15 +, −8, 5×3 +, −2×3, =, =, 2, 3, 5, 4 0 2, 5, !, !, 1, 27 4 54, 9, 27 25 + 9 − 30, = 27 5 + − 6 =, =, = ., 2, 5, 2, 5, 2 5, 5, , CO, R, , x3, , 2, , AP, P, , ., , y = 4x − x2, , ", , U, , 4.2 Change of order of integration, Consider the integral, , Za2, , y=g, Z 2 (x), , f (x, y)dydx., , x=a1 y=g1 (x), , ST, , To evaluate this integral, since the innermost integral has functions of x as, , limits, first we integrate w.r.t. y and then we integrate w.r.t x. Sometimes, it is, very difficult to evaluate in this order. By changing the order of integration we, , can easily evaluate. In this example the original order of integration is first w.r.t., , y and then w.r.t. x. By the change of order of integration we need to integrate, first w.r.t. x and then w.r.t. y. For this, divide the region into strips parallel to the, x-axis. Consider one such strip. The limits for x will be functions of y which are the, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , 247, , values of x corresponding to the curves in which the ends of the strip lies. Finally, the values of y will be constants according to the movement of this strip along the, given region. If the first integration is with respect to y then divide the region into, strips parallel to the y−axis and then find the corresponding limits for y and x and, then evaluate., , ✎, , ☞, , ✍, , ✌, , ., , AP, P, , Worked Examples, , √, , Example 4.15. Change the order of integration in, , 2, Z1 Z, , x, , f (x, y)dydx., , [Jan 2014], , 0 0, √, Solution. The region of integration is given by y = 0, y = 2 x, x = 0, x = 1., , i.e., y = 0, y2 = 4x, x = 0, x = 1., , Y, changing, , the, , order, , of, , CO, R, , By, , (1, 2), , integration, the first integration is w.r.t., , x. Hence, divide the region into strips, , 4, y =, 2, , x=0, , ∴, , 0, , Z2 Z1, , f (x, y)dydx =, , y=0, , X, , y=0, , f (x, y)dxdy., , 2, , x= y4, , ST, , 0, , x, , U, , √, , 2, Z1 Z, , Q, x=1, , P, , parallel to the x−axis. PQ is one such, y2, strip. Along this strip, x varies from, 4, to 1. Finally y varies from 0 to 2., , x, , Example 4.16. Change the order of integration in, evaluate., , Za Za, 0, , y, , x2, , x, dxdy and hence, + y2, , [Jan 2014, Jun 2013, Jun 2011], , Solution. The region of integration is given by x = y, x = a, y = 0, y = a., i.e., the region of integration is bounded by the lines x = 0, x = a, y = 0, y = x, By changing the order of integration, the first integration is w.r.t y. The double, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 248, , Engineering Mathematics - I, , integral takes the form, Za Za, 0, , y, , x, dxdy =, 2, x + y2, , ", R, , x2, , x, dydx., + y2, , Divide the shaded region into strips parallel to the y-axis. PQ is one such strip., , Y, , y, , x, , x=0, Za, , x, dydx, 2, x + y2, 1 −1 y, tan, x, x, , !x, , (a, a), , Q, , x=0, , dx, , 0, , =, , x=a, , X, , (0, 0), , P, , CO, R, , =, , x=0 y=0, Za, , y=a, , y, , 0, , x, dxdy =, 2, x + y2, , Za Zx, , =, , ∴, , Za Za, , x, , ., , AP, P, , Along this strip, y varies from 0 to x and finally x varies from 0 to a., , (a, 0), , π, πa, π, dx = (x)a0 =, ., 4, 4, 4, , x=0, , Example 4.17. Evaluate, , Z∞ Z∞, 0, , e−y, dydx by changing the order of integration., y, , x, , [May 2011, Jan 1999], , U, , Solution. The region of integration is y = x, y = ∞, x = 0, x = ∞., , ST, , Here, the first integration is w.r.t y and then we integrate w.r.t x., Y, , By changing the order of integration,, the innermost integration must be, w.r.t.x., , So, we divide the region of integration, , x=0, y=x, , into strips parallel to the x−axis., Along a typical strip, x varies from 0 to, , X, (0, 0), , y=0, , y and finally y varies from 0 to ∞., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , ∴, , Z∞ Z∞, , e−y, dydx =, y, , x, , 0, , =, , 249, , , , Z∞ Zy −y , Z∞ −y, e, , e, dx dy =, , , y, y, , y=0 x=0, Z∞ −y, e, , y, , y=0, , y, (x)0 dy, , =, , 0, , x=0, , 0, , e−y, e dy =, −1, −y, , !∞, 0, , = −[e−∞ − e0 ] = −[0 − 1] = 1, , √, Z3 Z4−y, Example 4.18. Evaluate, (x + y)dxdy by changing the order of integration., 0, , AP, P, , ., , Z∞, , y , Z, , , , dx dy, , 1, , [Jan 2003], p, Solution. The region of integration is bounded between x = 1, x = 4 − y and, y = 0, y = 3. i.e.,x = 1, x2 = 4 − y and y = 0, y = 3., , CO, R, , x = 1, x2 = −(y − 4) and y = 0, y = 3., , The region is bounded between x = 1, , Y, , and the parabola x2 = −(y−4) with vertex, , (0, 4), , (0, 4) and y = 0 and y = 3. By changing, the order of integration the innermost, , y=3, , (1, 3), , integration is w.r.t.y. Hence, divide this, , x2 = 4−y, , x=1, , region into strips parallel to the y−axis., , U, , Along a typical strip y varies from 0 to, , X, (0, 0), , ST, , 4 − x2 and finally x varies from 1 to 2., √, 4−x2, !4−x2, Z3 Z4−y, Z2, Z2 Z, y2, dx, xy +, (x + y)dydx =, (x + y)dxdy =, 2 0, 0, , =, , =, , Z2, , !, (4 − x2 )2, x(4 − x ) +, dx, 2, , 1, , !, 16 + x4 − 8x2, dx, 4x − x +, 2, , x=1, Z2, , (2, 0), y=0, , x=1, , x=1 y=0, , 1, , (1, 0), , 2, , 3, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , 253, , y2, , =, , 1, a, , Za Za, , y2, r , 2 2, y, a, , y=0 x=0, , dxdy, −, , x2, , ya2, Za , 2 −1 x , 1, y sin 2 dy, =, y, a, 1, a, , a, , x=0, , y2 (sin−1 1)dy, , 0, , π y3, =, 2a 3, =, , a3, , !a, 0, , π, πa2, =, ., 2a 3, 6, , AP, P, , ., , =, , y=0, Za, , Note. While changing the order of integration, we may come across situations, , CO, R, , where the region has to be divided into several parts and evaluation has to be, done accordingly. The following examples illustrate this aspect., Z1 Z2−y, Example 4.22. Change the order of integration and hence evaluate, xydxdy., , U, , Solution. The region of integration is given by x = 0,, , 0, , 0, , [Dec 2011], x = 2 − y,, , y = 0, y = 1., , x+y=2, , Y, , ST, , The region of integration is, , OACE., , After changing the order of, , B(0, 2), x+y=2, , integration, the first integration is w.r.t., y and final integration is w.r.t. x., Divide the region into two regions, ODCE and DAC., Z1 Z2−y, ", ", ∴, xydxdy =, xydydx+, xydydx., 0, , 0, , ODCE, , E, , Q, , x=0, , C(1, 1), x=1, , y=1, , Q′, A(2, 0), , O(0, 0), , P, , D(1, 0), y=0, , P′, , DAC, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , X
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 254, , Engineering Mathematics - I, , Consider the region ODCE, Divide the region into strips parallel to the y−axis. PQ is one such strip., Along this strip, y varies from 0 to 1 and finally x varies from 0 to 1., ", Z1 Z1, ∴, xydydx =, xydydx., ODCE, , x=0 y=0, , ., , y2, x·, 2, , x=0, , 1, =, 2, , !1, , dx, , 0, , Z1, , x(1 − 0)dx, , Z1, , xdx, , AP, P, , =, , Z1, , 0, , 1, =, 2, , 0, , CO, R, , !1, 1 x2, = ·, 2 2 0, 1, = (1 − 0), 4, 1, = ., 4, Consider the region DAC, , Divide this region into strips parallel to y−axis. P′ Q′ is one such strip., , U, , Along this strip, y varies from 0 to 2 − x and finally x varies from 1 to 2., Z2 Z2−x, ", xydydx., ∴, xydydx =, , ST, , DAC, , x=1 y=0, , =, , Z2, , y2, x·, 2, , x=1, , !2−x, , dx., , 0, , 1, =, 2, , Z2, , x(2 − x)2 dx., , 1, 2, , Z2, , x(4 + x2 − 4x)dx., , 1, , =, , 0, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , 255, , 1, =, 2, , Z2, , (4x + x3 − 4x2 )dx., , 0, , , !2, !2, !2 , x2, 1 , x4, x3 , = 4 ·, +, −4·, , 2, 2 1, 4 1, 3 1, , =, , ., , =, =, , !, , AP, P, , =, , 4, 1, 1, 2[4 − 1] + [16 − 1] − (8 − 1), 2, 4, 3, !, 1, 15 28, −, 6+, 2, 4, 3, !, 1 72 + 45 − 112, 2, 12, 5, 1, (5) =, 24, 24, , Z1 Z2−y, 5, 6 + 5 11, 1, =, = ., ∴, xydxdy = +, 4 24, 24, 24, 0, , CO, R, , 0, , Z1 Z2−y, Example 4.23. Change the order of integration in, xydxdy and hence, evaluate., , y, , 0, , [Jan 2001], , Solution. The region of integration is x = y, x = 2 − y, y = 0, y = 1., i.e.,x = y, x + y = 2, y = 0, y = 1. Solving we get 2x = 2 ⇒ x = 1., , U, , B is (1, 1), A is (2, 0)., , Y, 2, =, , (0, 2), , y, x, , the innermost integration is w.r.t.y and, , the next integration is w.r.t.x., , 0, , y, , OCB, , B(1, 1), , x=0, Q, , Divide this region into two regions OCB, and CBA., Z1 Z2−y, ", ", xydxdy =, xydydx +, xydydx., , y, =, , After the change of order of integration, , x+, , ST, , The region of integration is OAB., , x=1, , Q′, A(2, 0), , O(0, 0), , P, , C(1, 0), y=0, , P′, , CBA, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , X
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 256, , Engineering Mathematics - I, , Consider the region OCB, Divide this region into strips parallel to the y−axis., One such strip is PQ., Along this strip y varies from 0 to x., Finally x varies from 0 to 1., , ., , xydydx =, , OCB, , Z1 Zx, , xydydx =, , x=0 y=0, , 1, =, 2, , Z1, , 1, x(x − 0)dx =, 2, 2, , !x, , dx, , 0, , Z1, , x3 dx, , 0, , !1, x4, , 1, 2 4, 1, = ., 8, , y2, x, 2, , x=0, , 0, , 0, , CO, R, , =, , Z1, , AP, P, , ", , Consider the region CBA, , Divide this region into strips parallel to the y−axis., , One such strip is P′ Q′ . Along this strip y varies from 0 to 2 − x and finally x varies, from 1 to 2., , !2−x, Z2 Z2−x, Z2, y2, xydydx =, xydydx =, x, dx, 2 0, , U, , ", , ST, , CBA, , x=1 y=0, , 1, =, 2, , Z2, , x=1, , 1, x(2 − x)2 dx =, 2, , x=1, Z2, , Z2, , x(4 + x2 − 4x)dx, , x=1, , , !2, !2, !2 , 1 x2, x4, x3 , (4x + x − 4x )dx = 4, +, −4, , 2, 2 1, 4 1, 3 1, x=1, ", ", #, #, 1, 15 28, 4, 1, 1, 2(4 − 1) + (16 − 1) − (8 − 1) =, 6+, −, =, 2, 4, 3, 2, 4, 3, !, !, 1 72 + 45 − 112, 1 5, 5, =, =, = ., 2, 12, 2 12, 24, 1, =, 2, , 3, , 2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , 257, , Z1 Z2−y, 1, 5, 3+5, 8, 1, ∴, xydxdy = +, =, =, = ., 8 24, 24, 24 3, 0, , y, , √a, , √, , Z 2 Za2 −x2, Example 4.24. Evaluate, y2 dydx by changing the order of integration., 0, , x, , ., , a, i.e.,y = x, x2 + y2 = a2 , x = 0, x = √ ., 2, The required region is OAB. Draw AC, perpendicular to OB., Divide this region into two parts OAC, and CAB. After changing the order of, , must be w.r.t.x., √a, , ∴, , √, , a, a2 − x2 , x = 0, x = √ ., 2, , Y, , B(0, a), P′, C, , Q′, , A, , P, , Q, , O, , x=, , √a, 2, , , , √a , √a, 2, 2, , X, , x=0, , CO, R, , integration, the innermost integration, , p, , AP, P, , Solution. The region of integration is bounded by y = x, y =, , ", ", Z 2 Za2 −x2, 2, 2, y dydx =, y dxdy+, y2 dxdy., 0, , x, , OAC, , CAB, , Consider the region OAC, , Divide this region into strips parallel to the x−axis., , U, , Let PQ be one such strip., , ST, , a, Along this strip x varies from 0 to y and finally y varies from 0 to √ ., 2, √a, √a, ", Z 2 Zy, Z2, y, y2 dxdy =, y2 (x)0 dy, y2 dxdy =, OAC, , y=0 0, √a, 2, , =, , Z, 0, , y4, y dy =, 4, 3, , y=0, , ! √a, , 2, , =, , 0, , a4, 1 a4, = ., 4 4, 16, , Consider the region CAB, Divide this region into strips parallel to the x−axis., , STUCOR APP, , DOWNLOADED FROM STUCOR APP, ,
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , 259, , √, , x, , 0, , integration., , x, p, , x2 + y2, , dydx by changing the order of, , Solution. The region of integration is y = x, y =, i.e., y = x, x2 + y2 = 2, x = 0, x = 1., After changing the order of, , Y, , √, B(0, 2), P′, C, , integration the inner most integration, is w.r.t.x and the next integration is, , P, , y=0, , w.r.t. y. The given region is OAB. Draw, , Divide this region into two parts OAC, and CAB., √, , 0, , x, , p, , x2, , +, , y2, , dydx =, , ", , x, , dxdy+, , ", , x, , CO, R, , ∴, , x, , A(1, 1), Q, , O, , x=1, , X, , x=0, , AC perpendicular to OB., , Z1 Z2−x2, , Q′, , AP, P, , ., , p, 2 − x2 , x = 0, x = 1., x, , Example 4.25. Evaluate, , y=, , Z1 Z2−x2, , OAC, , p, , x2, , y2, , +, , CAB, , p, , x2, , + y2, , dxdy, , Consider the region OAC, , Divide this region into strips parallel to the x−axis., , Let PQ be one such strip. Along this strip x varies from 0 to y and finally y varies, ", , x, , p, , x2, , +, , y2, , dxdy =, , ST, , OAC, , Z1 Zy, , U, , from 0 to 1., , y=0 x=0, , 1, =, 2, , Z1, , x, , 1, dxdy =, p, 2, 2, 2, x +y, , Zy, , 2, , 2 − 21, , (x + y ), , Z1 Zy, , y=0 x=0, , 2x, p, , x2 + y2, , 1, d(x + y )dy =, 2, 2, , 2, , y=0 0, , Z1 2, 1 y, (x + y2 ) 2 , dy, , 1, , y=0, , Z1 , Z1 √, , 2 21, =, (2y ) − y dy = ( 2y − y)dy, y=0, , dxdy, , 2, , y=0, , √, !, !1, √ y2 1, 2 1 1 √, y2, = 2, −, =, − = ( 2 − 1)., 2 0, 2 0, 2, 2 2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , 0
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 260, , Engineering Mathematics - I, , Consider the region CAB, Divide this region into strips parallel to the x−axis., , CAB, , =, , y=1 x=0, √, Z2, , y=1, , AP, P, , ., , p, One such strip is P′ Q′ . Along this strip x varies from 0 to 2 − y2 and finally y, √, varies from 1 to 2., √ √, √, √ 2, 2, 2−y, ", Z 2 Z2−y, Z 2 2, 1 , 2, , 2, x, 1, x, (x + y ) , dxdy =, dxdy =, dy, , , p, p, 1, 2, x2 + y2, x2 + y2, 2, 0, √, y2, 2y −, (2 − y)dy =, 2, 1, 2, , y=1, , !, , √, , 2, , 1, , √ √, √ √, 1, 1, 2( 2 − 1) − (2 − 1) = 2( 2 − 1) −, 2, 2, √, √, √, 1, 2−1, Value of the integral =, + 2( 2 − 1) −, 2, 2, √ √, √, 2 − 1 + 2 2( 2 − 1) − 1, =, 2 √, √, √, 2−1+4−2 2−1 2− 2, =, =, ., 2, 2, Za 2a−x, Z, xydydx and, Example 4.26. Change the order of integration in the integral, 0, , x2, a, , [Jan 2013], x2, Solution. The region of integration is given by y = , y = 2a − x, x = 0, x = a., a, x2 = ay, x + y = 2a, x = 0, x = a., , ST, , U, , evaluate it., , CO, R, , =, , Y, , After changing the order of, , x2 = ay, , B(0, 2a), , integration, the first integration is w.r.t., P′, C, , y and the final integration is w.r.t. x., OAB is the required region. Divide this, region into OAC and CAB., , y=0, , ", ", Za 2a−x, Z, xydydx =, xydxdy+, xydxdy., ∴, 0, , x2, a, , OAC, , Q′, A(a, a), , P, , Q, , O, , x=a, , x=0, , CAB, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , x+, , y=, , 2a, , X
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , 277, , !, a2 π, a2 −1, sin (1) = 4, = πa2 ., =4, 2, 2 2, Y, √, y = a2 − x 2, , (0, a), , ., , AP, P, , x=0, , X, , O, , y= 0, , (a, 0), , x2 +y2 = a2, , x2 y2, +, = 1 using double integration., a2 b2, [Jun 2013], , CO, R, , Example 4.47. Find the area of the ellipse, , Solution., , Y, , Consider the area in the first quadrant., Divide this area into strips parallel to, , (0, b), , the y-axis. Along a typical strip, y varies, bp 2, from 0 to, a − x2 and finally x varies, a, from 0 to a., , x=0, , y = ab, , √, a2 − x 2, , U, , y=0, , ST, , By symmetry, Area of the ellipse = 4× Area in the first quadrant., Area = 4, , Za, , x=0, , b, a, , √, , Za, , 2 −x2, , dydx = 4, , y=0, , Za, , b, a, , (y)0, , √, , x=0, , a2 −x2, , dx = 4, , Za, 0, , bp 2, a − x2 dx, a, , !a, Za p, 4b x p 2, a2 −1 x , 4b, 2, 2, 2, a − x dx =, a −x +, =, sin, a, a 2, 2, a 0, , =, , 0, 4b a2, , a 2, , sin−1 (1) = 2ab, , π, = πab., 2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , (a, 0), , X
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 280, , Engineering Mathematics - I, , when x = 0, 9y = 0 ⇒ y = 0., One point of intersection is O(0, 0)., where x = 3, 9y = 5 × 9, y = 5., ∴ The other point of intersection is (3, 5)., , 3 y, √, , ∴ Area =, , Z5 Z 5, 0, , Z5, , dxdy =, , 3y2, 25, , Z5, , (x), , 0, , √, 3 y, √, 5, 3y2, 25, , dy, , AP, P, , ., , The shaded portion is the region of integration. Divide the region, into strips, √, 3 y, 3y2, to √ and finally y, parallel to the x−axis. Along one such strip, x varies from, 25, 5, varies from 0 to 5. √, , CO, R, , !, 3y2, 3 √, =, dy, √ y−, 25, 5, 0, , 5, !5, √, 2, y3, 3, 3 y3/2 , 1, ·, = √ [5 5] −, · 53 = 10 − 5 = 5 Sq.units., = √ 3 −, 25, 5 2 0 25 3 0, 5, , Example 4.51. Find the smaller of the areas bounded by the ellipse 4x2 + 9y2 = 36, and the straight line 2x + 3y = 6., , [Jan 2012], , U, , Solution. Solving the given two equations, we get the points of intersection., , 2, , 2, , ST, , 4x + 9y = 36, 2x + 3y = 6, , From (2), 3y = 6 − 2x, 6 − 2x, ., i.e., y =, 3, Substituting in (1) we get, 4x2 +, , Y, ,, B(0, , (1), , 2), 4x2 +9y2 = 36, , 2 x+, 3y=, 6, , (2), , X, , O, , A(3, ,, , 9(6 − 2x)2, = 36, 9, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , 0)
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , 281, , 4x2 + 36 + 4x2 − 24x = 36, 8x2 − 24x = 0, x2 − 3x = 0, x(x − 3) = 0, , ⇒ x = 0 or x, , = 3., , 6, = 2., 3, One point of intersection is (0, 2), 6−6, When x = 3, y =, = 0., 3, The other point of intersection is (3, 0)., , The shaded portion is the smaller area. Divide the regionrinto, 6 − 2x, to 4 −, the y−axis. Along one such strip, y varies from, 3, varies from 3 to 0., q, , strips parallel to, 4 2, x and finally x, 9, , CO, R, , ., , AP, P, , When x = 0, (2) ⇒ y =, , ∴ Area =, , Z0, , 4− 94 x2, , Z, , dxdy =, , x=3 y= 6−2x, 3, , Z0, 3, , q, , 4− 94 x2, dx, y 6−2x, 3, , !, Z0 r, 4 2, 6 − 2x , , =, 4 − x −, dx, 9, 3, 3, , 4, (9 − x2 )dx −, 9, , U, , =, , Z0 r, 3, , Z0, 3, , 2, 2dx +, 3, , Z0, , xdx, , 3, , ST, , #0, ", " #0, 9 −1 x , 2 x2, 2 xp, 0, 2, 9 − x + sin, − 2[x]3 +, =, 3 2, 2, 3 3, 3 2 3, ", #, 9π, 1, −3π, 3π, 2, 0−, − 2(0 − 3) + [0 − 9] =, +6−3=3−, =, 3, 22, 3, 2, 2, , , π, 3(2 − π), =3 1−, =, Sq.units., 2, 2, , Example 4.52. Find the smaller of the areas bounded by y = 2 − x and x2 + y2 = 4, using double integral., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 282, , Engineering Mathematics - I, , Solution. Solving the given two equations, we get the points of intersection., We have x2 + y2 = 4 ⇒ x2 + (2 − x)2 = 4 ⇒ x2 + 4 + x2 − 4x = 4 ⇒ 2x2 − 4x = 0, ⇒ x2 − 2x = 0 ⇒ x(x − 2) = 0 ⇒ x = 0, x = 2., Y, When x = 0, y = 2, when x = 2, y = 0., , AP, P, , ., , (0, 2), , The points of intersection are (0, 2) and, , y = 2− x, , x=0, , (2, 0). The shaded portion is the required, region., , Divide this region into strips, , O, , parallel to the y-axis. Along one such, √, typical strip, y varies from 2−x to 4 − x2, and finally x varies from 0 to 2., , R, , (2, 0), , √, , Z2 √, Z2 Z4−x2, Z2 p, 4−x2, dydx = (y)2−x dx = ( 4 − x2 − (2 − x))dx, dxdy. =, , CO, R, , Now area =, , ", , X, , y= 0, , x=0 y=2−x, , x=0, , x !2, , x=0, , !, 2 2, , 4, xp, x, 4 − x2 + sin−1, − 2(x)20 +, 2, 2, 2 0, 2 0, π, 4, = 2 sin−1 (1) − 2(2 − 0) + = 2 − 4 + 2 = π − 2., 2, 2, =, , U, , Example 4.53. Find the area bounded by the parabola y2 = 4 − x and y2 = 4 − 4x as, a double integral and evaluate it., , [Jan 2001], , ST, , Solution. The parabola y2 = 4 − x = −(x − 4). Its vertex is at (4, 0)., , Consider the parabola y2 = 4 − 4x = −4(x − 1) . Its vertex is at (1, 0)., , Both the parabolas are open to the left. Let us solve the two equations to find the, , points of intersection, y2 = 4 − x and y2 = 4 − 4x., , ∴ 4 − x = 4 − 4x ⇒ 4x − x = 0 ⇒ 3x = 0 ⇒ x = 0 ⇒ y2 = 4 ⇒ y = ±2., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , 283, , Y, The, , points, , of, , intersection, , are, , (2, 0), (−2, 0). The shaded portion is the, required region. Divide this region into, , Now area =, , ", , dxdy. =, , R, , =, , Z2, , y=−2, , Z2, , y=−2, , 2, x=4−y, Z, , dydx =, , 2, x=1− y4, , (4, 0), , X, , AP, P, , ., , strips parallel to the x-axis. Along one, y2, typical strip, x varies from 1 −, to 4−y2 ., 4, Finally y varies from −2 to 2., , (1, 0), O, , Z2, , (x), , y=−2, , 4−y2, , 2, , 1− y4, , dy =, , Z2, , y2, 4−y − 1−, 4, 2, , y=−2, , !!, , dy, , !, !2, !2, y2, 1 y3, y3, 2, 2, 4−y −1+, dy = 4(y)−2 −, − (y)−2 +, 4, 3 −2, 4 3 −2, 2, , CO, R, , 1, 16, 16, 1, −4+, = 4(2 + 2) − (8 + 8) − (2 + 2) + (8 + 8) = 16 −, 3, 12, 3, 12, 12, 16 4, + = 12 −, = 12 − 4 = 8., = 16 − 4 −, 3, 3, 3, Example 4.54. Find the area of the region bounded by y = x − 2 and y2 = 2x + 4., Solution. The parabola is y2 = 2x + 4 = 2(x + 2)., The vertex is (−2, 0)., , U, , To find the points of intersection, let us solve the two equations., The curves are y = x − 2, y2 = 2x + 4., , ST, , (x − 2)2 = 2x + 4., , x2 − 4x + 4 − 2x − 4 = 0, x2 − 6x = 0, , x(x − 6) = 0, When x = 0, y = −2., , ⇒ x = 0, x = 6., , When x = 6, y = 4., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 284, , Engineering Mathematics - I, , Y, The points of intersection are (0, −2) and, (6, 4). The shaded portion is the required, (6, 4), , ., , Divide the region into strips, , parallel to the x-axis. Along one such, y2 − 4, strip, x varies from, to y + 2 and, 2, finally y varies from −2 to 4., , Now, Area =, , ", , dxdy =, , R, , −2, , y=−2, , Zy+2, , dxdy =, , (0, −2), , Z4, , y=−2, , 2, , x= y 2−4, , (x), , y+2, , 2, , x= y 2−4, , dy, , !, !!, !, Z4, Z4, y2, y2, y2, dy, − 2 dy =, + 2 dy =, y+2−, y+4−, y+2−, 2, 2, 2, , CO, R, , =, , Z4, , Z4, , X, , O, , (−2, 0), , AP, P, , region., , −2, , −2, , !4, y2, y3, 1, 1, =, + 4y −, = (16 − 4) + 4(4 + 2) − (64 + 8) = 6 + 24 − 12 = 18., 2, 6 −2 2, 6, Area as double integral in polar coordinates, !, Result. Area in polar coordinates is given by rdrdθ., R, , ✎, , ☞, , ✍, , ✌, , U, , Worked Examples, , Example 4.55. Find the area of the cardioid r = a(1 − cos θ)., , ST, , Solution., , θ, , Since the curve is symmetric about the, P, , initial line, Area = 2× area above the, initial line., , θ=π, O, , θ=0, , Along this region, r varies from 0 to, a(1 − cos θ) and θ varies from 0 to π., , r = a(1−cos θ), , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , r
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 296, , Engineering Mathematics - I, Z1 Z1−x", , =, , 0, , 0, , #1−x−y, , dydx, , 0, , 1, =−, 2, , Z1 Z1−x, , 2−2 − (1 + x + y)−2 dydx, , 1, =−, 2, , Z1 Z1−x, 0, , 0, , 0, , !, 1, −2, − (1 + x + y) dydx, 4, , AP, P, , 0, , ., , (1 + x + y + z)−2, −2, , ", #1−x , Z1 , , (1 + x + y)−1, 1, 1 1−x, dx, =−, [y]0 −, 2, 4, −1, 0, 0, , 1, =−, 2, , Z1, 0, , !, 1, (1 − x) + 2−1 − (1 + x)−1 dx, 4, , U, , CO, R, , , , !1, , 1 1 1 1 x2, 1 1, 1, = − (x)0 − ·, + [x]0 − (log(1 + x))0 , 2 4, 4 2 0 2, !, 1 1 1 1, − + − log 2, =−, 2 4 8 2, !, 1 2−1+4, =−, − log 2, 2, 8, !, 1 5, 1, 5, =−, − log 2 = log 2 − ., 2 8, 2, 16, $, Example 4.70. Find the value of, xyzdxdydz through the positive spherical, octant for which x2 + y2 + z2 ≤ a2 ., , [Jan 2014, Jun 2010], , ST, , Solution. The limits for x, y, z are as follows., q, z : 0 to a2 − x2 − y2, p, y : 0 to a2 − x2, x : 0 to a., , ∴, , $, , xyzdxdydz =, , Za, , x=0, , √, Za2 −x2, y=0, , √, , 2 −x2 −y2, aZ, , xyzdzdydx., , z=0, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 298, , Engineering Mathematics - I, , 4.7 Volume as triple integral, Result. If V is the volume bounded by a definite, $ region D in space, then the, volume is calculated by the triple integral V =, dxdydz with proper limits., D, , integral., , x2 y2 z2, +, +, = 1 using triple, a2 b2 c2, , AP, P, , ., , Example 4.71. Find the volume of the ellipsoid, , Solution. We know that the ellipsoid is symmetric about the, coordinate planes., $, ∴ Volume of the ellipsoid = 8×Volume in the first octant = 8, , dxdydz., , V, , x2 y2 z2, The ellipsoid is given by 2 + 2 + 2 = 1, a, b, c r, , ∴ Volume = 8, , q, 2, b 1− x2 c, a, Z Z a, , r, , x=0, , U, , y=0, q, 2, b 1− x, Za Z a2, , ST, , =8, , 1−, , x2 y2, −, a2 b2, , CO, R, , In the first octant, z varies from 0 to c, r, x2, y varies from 0 to b 1 − 2, a, and x varies from 0 to a., 2, , 2, , 1− x2 − y2, a, , b, , Z, , dzdydx, , z=0, , c, , r, , (z)z=0, , 2, , 2, , 1− x2 − y2, a, , b, , dydx, , x=0, , = 8c, , y=0, q, 2, b, 1− x, Za Z a2, , 1−, , 1, b, , s, , y=0, q, 2, (1− x ), b, Za Z a2, , x=0, , = 8c, , r, , x=0, , y=0, , x2 y2, − dydx, a2 b2, , b2, , !, x2, 1 − 2 − y2 dydx, a, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 300, , Engineering Mathematics - I, , x=0, , =8, , √, Za2 −x2, , √2 2 2, a −x −y, dydx, (z)0, , y=0, √, Za Za2 −x2 q, , x=0, Za, , y=0, , a2, , =8, , x=0, Za, , ., , =8, , a2 − x2 − y2 dydx, , −, 2, , x2, , √, , sin−1, , a2 − x2 −1, sin (1)dx, 2, , x=0, , π, =4, 2, , ! a2 −x2, q, y, y, 2, 2, 2, dx, a −x −y, +, √, a2 − x2 2, 0, , AP, P, , =8, , Za, , Za, , (a2 − x2 )dx, , x=0, , !a, x3, = 2π a x −, 3 0, !, 3, 4πa3, a, 3, =, ., = 2π a −, 3, 3, , Note, , CO, R, , 2, , ST, , U, , The volume of the portion of the sphere that lies in the first octant is, 1 4πa3 πa3, 1, × Volume of the sphere = ×, =, ., 8, 8, 3, 6, , Example 4.73. Find the volume of the tetrahedron bounded by the plane, x y z, + + = 1 and the coordinate planes., [Jun 2006], a b c, Solution. Let D be the region in space bounded by the tetrahedron., x y z, The region of integration is given by + + = 1, x = 0, y = 0, z = 0., a b c, , x y, Along the region of integration z varies from 0 to c 1 − − ., a b, , x, y varies from 0 to b 1 − ., a, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , 301, , and x varies from 0 to a., x y, x, Za−b), Za b(Z1− a ) c(1−, dzdydx., ∴ Volume =, , y=0, , x=0, , z=0, , 1− ax, , Za b(Z ), c(1− x − y ), (z)0 a b dydx., =, y=0, , x=0, , ., , y=0, , x=0, , =c, , Za, , x=0, Za, , =c, , !b(1− ax ), 0, , dx., , n , x 2 o, x 1 2 , x x , − b 1−, − b 1−, dx, b 1−, a, a, a, 2b, a, , CO, R, , x=0, , xy y2, −, y−, a, 2b, , AP, P, , x, Za b(Z1− a ) , x y, c 1 − − dydx., =, a b, , bc, =, 2, , Za , x 2, 1−, dx, a, , x=0, , , a, 1 − x 3 , bc , a , =, 2 3 −1 , a, 0, , U, , −abc, [0 − 1], =, 6, abc, =, ., 6, , ST, , Example 4.74. Find the volume of the portion of the cylinder x2 +y2 = 1 intercepted, , between the plane x = 0 and the paraboloid x2 + y2 = 4 − z., , [Jun 2012], , Solution. The surfaces are x2 + y2 = 1, x = 0 and x2 + y2 = 4 − z., For the required volume integral, , x varies from 0 to 1, √, y varies from 0 to 1 − x2, , z varies from 0 to 4 − x2 − y2 ., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
Page 314 : DOWNLOADED FROM STUCOR APP, , AP, P, , STUCOR APP, , ST, , U, , CO, R, , ., , **Note: Other Websites/Blogs Owners Please do not Copy (or) Republish, this Materials, Students & Graduates if You Find the Same Materials with, EasyEngineering.net Watermarks or Logo, Kindly report us to,
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Multiple Integrals, , 307, , Solution. The given region is bounded by, y2 = x, , (1), , x+y+z=2, , (2), , z = 0., , (3), , Whenz = 0, (2) ⇒ x + y = 2, x = 2 − y., y2 = 2 − y, y2 + y − 2 = 0, (y − 1)(y + 2) = 0, ∴ y = −2, 1., For the given region we have,, , AP, P, , ., , Substituting in (1) we get, , CO, R, , y varies from −2 to 1,, , x varies from y2 to 2 − y and, , z varies from 0 to 2 − x − y., , Z1 Z2−y 2−x−y, Z, ∴ Volume =, dzdxdy, y=−2 x=y2 z=0, , U, , Z1 Z2−y, 2−x−y, =, [z]0, dxdy, y=−2 x=y2, , ST, , Z1 Z2−y, =, (2 − x − y)dxdy, y=−2 x=y2, , , " 2 #2−y, Z1 , , x, , 2−y , 2−y, − y[x]y2 dy, =, 2 · [x]y2 −, 2 y2, y=−2, , =, , Z1, , −2, , !, 1, 2, 4, 2, 2{2 − y − y } − {(2 − y) − y } − y{2 − y − y } dy., 2, 2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , ., , 5.1 Introduction, , AP, P, , 5 Differential Equations, , A differential equation is an equation involving one dependent variable and its, derivatives with respect to one or more independent variables., Ordinary Differential Equation, , An ordinary differential equation is an equation in which there is only one, , derivatives., , CO, R, , independent variable and so the derivatives involved in it are ordinary, , Partial differential equation, , A partial differential equation is an equation in which there are two or more, independent variables and partial differential coefficients with respect to any one, of them., , ST, , U, , Examples, ∂2 u, ∂2 u, (i) 2 = a2 2 ., ∂x, ∂t, 2, 2, ∂2 u, 2∂ u, 2∂ u, +, y, (ii) x, +, 2xy, = 0., ∂x∂y, ∂x2, ∂y2, ∂u, ∂u, = 5u., (iii) x + y, ∂x, ∂y, Order of a differential equation, , The order of a differential equation is the order of the highest derivative that, , occurs in it., Degree of a differential equation, The degree of a differential equation is the degree of the highest derivative, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 310, , Engineering Mathematics - I, , occurring in it after the equation has been reduced to a form free from radicals, and fractions as far as the derivatives are concerned., Examples, d2 x, + a2 x = 0, the order is 2 and the degree is 1, dt2, , !2 23, 2, , , dy, = c d y , the order is 2 and degree is 2,, (ii) For the differential equation 1 +, dx, dx2, , !, !2 3, 2y 2, , , dy, d, 2, = c, ., since it can be reduced to the form 1 +, dx , dx2, Formation of a differential equation, , ., , AP, P, , (i) For the differential equation, , An ordinary differential equation can be formed by eliminating certain arbitrary, , constants from a relation involving variables and constants., , In the study of, , applied mathematics, every geometrical and physical problems when translated, into a mathematical model will always give rise to a differential equation., ✎, , Worked Examples, , ✌, , CO, R, , ✍, , ☞, , Example 5.1. Form the differential equation of the family of straight lines passing, through the origin., , Solution. The equation of the family of straight lines passing through the origin, is given by, , (1), , y = mx., , dy, = m., dx, , ST, , U, , Differentiating w.r.t. x, we get, , Substituting the value of m in (1) we obtain, !, dy, x, y=, dx, which is the required differential equation., Example 5.2. Obtain the differential equation of the family of circles, x2 + y2 + 2ax + r2 = 0 by eliminating the arbitrary constant a., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential Equations, , 313, , Solution of a differential equation, A solution of a differential equation is a relation between the variables that, satisfies the given differential equation. A solution of a differential equation is, also called as the integral of the equation., Examples, 1. x = a cos nt is a solution of the differential equation, For the differential equation, , dy, d2 y, + 3 + 2y = 0, y = e−x and y = e−2x are, 2, dx, dx, , solutions., General Solution, , AP, P, , ., , 2., , d2 x, + n2 x = 0., dt2, , A general or complete solution of a differential equation is the one in which the, number of arbitrary constants is equal to the order of the differential equation., Example. Consider, , (1), , CO, R, , y = A cos αx + B sin αx, , where A and B are the arbitrary constants., Differentiating w.r.t. x we get, , U, , dy, = −Aα sin αx + Bα cos αx., dx, d2 y, = −Aα2 cos αx − Bα2 sin αx., dx2, , = −α2 (A cos αx + B sin αx) = −α2 y., , d2 y, + α2 y = 0., dx2, , ST, , (2), , Hence, (1) is a general solution of (2) as the number of arbitrary constants A, B is, the same as the order of (2)., Particular Solution, A particular solution is a solution that can be obtained by giving particular, values to the arbitrary constants in the general solution., Example. y = 2 cos αx + sin αx is a particular solution of (2), since it can be derived, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 318, , Engineering Mathematics - I, , Example 5.8. Find the particular integral of (D2 − 4)y = cosh 2x., PI =, =, =, =, , ., , =, =, , 1, cosh 2x, −4, " 2x, #, e + e−2x, 1, (D + 2)(D − 2), 2, ", #, 1, 1, 1, e2x +, e−2x, 2 (D + 2)(D − 2), (D + 2)(D − 2), , , x −2x, 1 x 2x, e +, e, 2 4, −4, x 2x, [e − e−2x ], 8, x, sinh 2x., 4, D2, , AP, P, , Solution., , [May 2011], , [May 2005], , Example 5.10. Find the particular integral of (D − 1)2 = sinh x., , [Nov 2003], , Solution., , CO, R, , Example 5.9. Find the particular integral of (D3 − 1)y = e2x ., 1, e2x, e2x, Solution. PI = 3, ., e2x = 3, =, 7, D −1, 2 −1, , #, " x, e − e−x, 1, 1, sinh, x, =, 2, (D − 1)2, (D − 1)2, ", #, 1, 1, 1, =, ex −, e−x, 2, 2 (D − 1), (D − 1)2, #, ", 1 x2 x 1 −x, e − e, =, 2 2, 4, 2, −x, x, e, = ex −, ., 4, 8, , ST, , U, , PI =, , Example 5.11. Solve, , d2 y, dy, + 4 + 5y = −2 cosh x., 2, dx, dx, , Solution. The A.E is, m2 + 4m + 5 = 0 ⇒ (m + 2)2 + 5 − 4 = 0, ⇒ (m + 2)2 = −1 ⇒ m + 2 = ±i ⇒ m = −2 ± i., yc = e−2x (c1 cos x + c2 sin x)., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential Equations, , 319, , 2, [−(e x + e−x )], + 4D + 5), −1, 1, = 2, ex − 2, e−x, D + 4D + 5, D + 4D + 5, −e x, e−x, =, −, 1+4+5 1−4+5, −e x e−x, −, ., =, 10, 2, , yp =, , The general solution is y = yc + y p, , AP, P, , ., , 2(D2, , y = e−2x (c1 cos x + c2 sin x) −, , e x e−x, −, ., 10, 2, , Example 5.12. Find the particular integral of (D + 2)(D − 1)2 y = e−2x + 2 sinh x., [May 2006], , PI =, =, =, =, , U, , =, , 1, [e−2x + 2 sinh x], (D + 2)(D − 1)2, 1, [e−2x + e x − e−x ], 2, (D + 2)(D − 1), 1, 1, 1, e−2x +, ex −, e−x, 2, 2, (D + 2)(D − 1), (D + 2)(D − 1), (D + 2)(D − 1)2, x −2x x2 x e−x, e + e −, 9, 6, 1(4), 2, x −2x x x e−x, e + e −, ., 9, 6, 4, , CO, R, , Solution., , Example 5.13. Solve (D2 + 1)2 y = 0., +, , 1)2, , [May 2008], , = 0., , ST, , Solution. A.E is, , (m2, , m2 = −1, m2 = −1 =⇒ m = ±i, m = ±i., The complex roots are repeated., ∴ yc = CF = (c1 + c2 x) cos x + (c3 + c4 x) sin x., y p = PI = 0., , Solution is y = yc + y p, y = (c1 + c2 x) cos x + (c3 + c4 x) sin x., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 322, , Engineering Mathematics - I, , 5.2.2 Type II, Q(x) = sin ax or cos ax where a is a constant., y p = P.I. =, , 1, 1, sin ax or, cos ax., f (D), f (D), , Working rule. Replace D2 by −a2 and evaluate till D is eliminated., 1, x, P.I. = 2, sin ax =, 2, 2, D +a, , and, 1, x, cos ax =, 2, 2, 2, D +a, Another method, , sin axdx =, , cos axdx =, , −x, cos ax., 2a, , x, sin ax., 2a, , 1, 1 iax, sin ax = Imaginary Part of, e, f (D), f (D), , CO, R, , P.I. =, and, , Z, , Z, , AP, P, , ., , If f (D) = D2 + a2 , then f (D) = 0 when D2 is replaced by −a2 . In this case, , P.I. =, , 1, 1 iax, cos ax = Real Part of, e, f (D), f (D), , which can be evaluated using Type I., Note, , ST, , U, , Suppose f (D) = 0 when D2 is replaced by −a2 , then f (D) is of the form f (D2 )., 1, cos ax, x cos ax, Then, P.I. =, cos ax =, . If f (−a2 ) = 0 then P.I. = ′, if f ′ (−a2 ) , 0., 2, 2, f (D ), f (−a ), f (−a2 ), x2 cos ax, if f ′′ (−a2 ) , 0, and so on. In a similar way the, If f ′ (−a2 ) = 0, then P.I. = ′′, f (−a2 ), particular integral for sin ax can be evaluated replacing cos ax., ✎, , ☞, , ✍, , ✌, , Worked Examples, , Example 5.20. Find the particular integral of (D2 + 4)y = sin 2x., [Dec 2009, Dec 2011], Solution., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential Equations, 1, x, sin 2x =, 2, 2, D +4, !, x − cos 2x, =, 2, 2, x, = − cos 2x., 4, , PI =, , 323, Z, , sin 2xdx, , [∵ f (−a2 ) = 0], , Example 5.21. Find the particular integral of (D2 + 1)y = sin x sin 2x., D2, , AP, P, , ., , 1, (sin x sin 2x), +1", #, −1, cos 3x − cos x, = 2, 2, D +1, ", #, −1, 1, 1, =, cos 3x − 2, cos x, 2 D2 + 1, D +1, ", #, Z, −1 cos 3x, x, cos, xdx, =, −, 2 −32 + 1 2, #, ", 1 cos 3x x, − sin x, =−, 2 −8, 2, cos 3x x sin x, +, ., =, 16, 4, , PI =, , CO, R, , Solution., , [May 1997], , Example 5.22. Find the particular, integral of (D2 + 1)y = sin x., Z, x, 1, −x, sin x =, Solution. P.I = 2, cos x., sin xdx =, 2, 2, D +1, , [Jun 2010], , Example 5.23. Find the particular integral of (D2 + 4D + 2)y = sin 3x. [May 1998], 1, 1, 1, sin 3x =, sin 3x =, sin 3x, −9 + 4D + 2, 4D − 7, + 4D + 2, (4D + 7), 4 cos 3x(3) + 7 sin 3x, =, sin 3x =, (4D + 7)(4D − 7), 16D2 − 49, 12 cos 3x + 7 sin 3x, =, 16(−9) − 49, 12 cos 3x + 7 sin 3x, =, −144 − 49, 1, [12 cos 3x + 7 sin 3x]., =−, 193, , PI =, , U, , Solution., , ST, , D2, , 2, Example 5.24. Find the particular, Z integral of (D + 4)y = cos 2x., 1, x sin 2x x sin 2x, x, Solution. PI = 2, =, ., cos 2xdx =, cos 2x =, 2, 2 2, 4, D +4, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , [May 2001]
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 324, , Engineering Mathematics - I, , Example 5.25. Find the particular integral of (D2 + 1)2 y = sin 2x., sin 2x, sin 2x, 1, sin 2x =, =, ., Solution. PI = 2, 2, 2, 9, (D + 1), (−4 + 1), , AP, P, , ., , Example 5.26. Find the particular integral of (D2 + 1)y = sin2 x., !, Solution., 1, 1 − cos 2x, 1, 2, sin x = 2, PI = 2, 2, D +1, D +1, #, ", 1, 1, 1, −, cos 2x, =, 2 D2 + 1 D2 + 1, ", #, 1, cos 2x, 1, 0x, =, e −, 2 D2 + 1, −4 + 1, ", #, 1, cos 2x, =, 1+, 2, 3, 1 cos 2x, ., = +, 2, 6, , CO, R, , Example 5.27. Find the particular integral of (D"2 + 1)y = sin(2x, + 5)., #, sin(2x + 5), 1, sin(2x + 5), sin(2x + 5) =, ., =−, Solution. PI = 2, −4 + 1, 3, D +1, Example 5.28. Find the particular integral of (D2 + 4D + 8)y = cos(2x + 3)., Solution., , 1, 1, cos(2x + 3) =, cos(2x + 3), −4 + 4D + 8, D2 + 4D + 8, 1 1, 1, cos(2x + 3) =, cos(2x + 3), =, 4D" + 4, 4 D + 1#, 1, (D − 1), =, cos(2x + 3), 4 (D − 1)(D + 1), ", #, 1 −2 sin(2x + 3) − cos(2x + 3), =, 4, D2 − 1, ", #, 1 2 sin(2x + 3) + cos(2x + 3), =−, 4, −4 − 1, 1, [2 sin(2x + 3) + cos(2x + 3)] ., =, 20, , ST, , U, , PI =, , Example 5.29. Solve (D2 − 4D + 3)y = sin 3x cos 2x., , [May 2007], , Solution. The A.E is, m2 − 4m + 3 = 0 ⇒ (m − 1)(m − 3) = 0 ⇒ m = 1, 3., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential Equations, , 331, , 5.2.5 Type V, Q(x) = xm cos ax or xm sin ax. In this case, 1 m, 1 m, x cos ax or, x sin ax., f (D), f (D), 1 m iax, 1 iax m, e x or I.P, x e ., = R.P, f (D), f (D), 1, 1, = R.P eiax, xm or I.P eiax, xm, f (D + ia), f (D + ia), , ., , which can be evaluated by earlier methods., ✎, , AP, P, , P.I =, , Worked Examples, , ✍, , ☞, , ✌, , Example 5.39. Find the particular integral of (D2 + 4D + 4)y = xe−2x . [May 2005], 1, 1, xe−2x =, xe−2x, + 4D + 4, (D + 2)2, 1, 1, = e−2x, x = e−2x 2 x, 2, (D − 2 + 2), D, Z, 1, 1, 1, = e−2x, x = e−2x, xdx, DD, D, !, Z, 2, e−2x, −2x 1 x, =, x2 dx, =e, D 2, 2, x3 e−2x, e−2x x3, =, ., =, 2 3, 6, , PI =, , D2, , CO, R, , Solution., , ST, , U, , Example 5.40. Find the particular integral of (D2 + 1)y = xe x ., 1, 1, 1, Solution. P.I =, xe x = e x, x = ex 2, x, 2, 2, D +1, (D + 1) + 1, D + 2D + 1 + 1, 1, 1, ix, = ex 2, x = ex h, 2, D + 2D + 2, 2 1 + D +2D, 2, ", #−1, ex, D2 + 2D, =, 1+, x, 2, 2, ", #, D2 + 2D, ex, ex, 1−, =, + · · · x = [1 − D · · · ]x, 2, 2, 2, ex, = [x − 1]., 2, , STUCOR APP, , DOWNLOADED FROM STUCOR APP, , [Jun 2003]
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 332, , Engineering Mathematics - I, , Example 5.41. Find the particular integral of y′′ + 2y′ + 5y = e−x cos 2x.[May 2005], 1, 1, Solution. P.I =, e−x cos 2x = e−x, cos 2x, 2, 2, D + 2D + 5, (D − 1) + 2(D − 1) + 5, 1, 1, = e−x 2, cos 2x = e−x 2, cos 2x, DZ− 2D + 1 + 2D − 2 + 5, D +4, x sin 2x, x, cos 2xdx = e−x, = e−x, 2, 2 2, e−x x sin 2x, =, ., 4, , AP, P, , CO, R, , ., , Example 5.42. Find the particular integral of (D2 − 2D + 2)y = e x cos x. [Dec 2010], 1, Solution. P.I =, e x cos x., D2 − 2D + 2, 1, cos x., = ex ·, 2, (D + 1) − 2(D + 1) + 2, 1, cos x., = ex · 2, ✟+ 1 −✟, ✟− 2, D +✟, 2D, 2D, ✁+2, ✁, cos x, 1, cos x. = e x · x, = ex · 2, 2D, D, Z +1, xe x sin x, xe x, ., cos xdx =, =, 2, 2, , Example 5.43. Find the particular integral of (D2 − 2D + 4)y = e x cos x. [Dec 2001], 1, 1, e x cos x = e x, cos x, 2, − 2D + 4, (D + 1) − 2(D + 1) + 4, 1, 1, = ex 2, cos x = e x 2, cos x, D + 2D + 1 − 2D − 2 + 4, D +3, e x cos x, cos x, =, ., = ex, −1 + 3, 2, , PI =, , D2, , U, , Solution., , ST, , Example 5.44. Find the particular integral of (D − 1)2 y = e x sin x., Solution., , [May 2003, Jun 2012], , 1, 1, e x sin x = e x, sin x, PI =, 2, (D − 1), (D + 1 − 1)2, 1, 1 1, sin x, = e x 2 sin x = e x, DD, DZ, 1, 1, = ex, sin xdx = e x (− cos x), D, D, Z, = −e x cos xdx = −e x sin x., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential Equations, , 335, , 1, 1, e−x, ex, x, −, x, 2 (D + 1)2 − 4, 2 (D − 1)2 − 4, e−x, ex, 1, 1, x, −, x, =, 2, 2, 2 D + 2D + 1 − 4, 2 D − 2D + 1 − 4, ex, e−x, 1, 1, =, x, −, x, 2 D2 + 2D − 3, 2 D2 − 2D − 3, ex, e−x, 1, 1, =, x−, x, 2, 2 (−3)(1 − D +2D ), 2 (−3)(1 − D2 −2D ), =, , 3, , The solution is y = yc + y p, , ex h, 2 i e−x h, 2i, x+, +, x−, 6, 3, 6, 3, , CO, R, , y = c1 e2x + c2 e−2x −, , AP, P, , ., , 3, , ex , D2 + 2D −1, e−x , D2 − 2D −1, =, 1−, x+, 1−, x, −6, 3, 6, 3, i, i, −e x h, D2 + 2D, e−x h, D2 − 2D, =, 1+, + ··· x +, 1+, + ··· x, 6, 3, 6, 3, −e x h, 2 i e−x h, 2i, P.I. =, x+, +, x− ., 6, 3, 6, 3, , e−x, d2 y, dy, +, y, =, +, 2, ., [Jun 2013, May 1989 ], dx, dx2, x2, Solution. A.E is m2 + 2m + 1 = 0 ⇒ (m + 1)2 = 0 ⇒ m = −1, −1., , Example 5.48. Solve, , yc = e−x (c1 + c2 x)., , e−x, 1, 1, 1, = e−x, 2, 2, 2, (D + 1) x, (D − 1 + 1) x2, !, Z, 1 1, 1, = e−x 2 2 = e−x, x−2 dx, D, D x, Z, h −1 i, 1, −x, −x 1 x, = −e, dx, =e, D −1, x, , ST, , U, , yp =, , y p = −e−x log x., , The general solution is y = yc + y p, y = e−x (c1 + c2 x) − e−x log x., , Example 5.49. Solve (D4 − 1)y = e x cos x., , [Apr 2009], , Solution. A.E is m4 − 1 = 0 ⇒ (m2 − 1)(m2 + 1) = 0, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential Equations, , 341, , dy, d2 y, − 2 + y = xe x sin x., [Dec 2013, May 2010], 2, dx, dx, 2, 2, Solution. A.E. is m − 2m + 1 = 0 ⇒ (m − 1) = 0 ⇒ m = 1, 1., Example 5.55. Solve, , yc = e x (C1 + C2 x)., 1, xe x sin x, (D − 1)2, 1, = ex, x sin x, (D + 1 − 1)2, 1, = e x 2 x sin x, DZ, x1, =e, x sin xdx, D, Z, 1, xd(− cos x), = ex, D, Z, i, 1h, − x cos x − − cos xdx, = ex, D, i, 1h, − x cos x + sin x, = ex, D, Z, h Z, i, x, =e −, x cos xdx + sin xdx, h Z, i, x, =e −, xd(sin x) − cos x, ), Z, h (, i, x, = e − x sin x − sin xdx − cos x, , CO, R, , ., , AP, P, , yp =, , U, , h, i, = e x − x sin x − cos x − cos x, h, i, = −e x x sin x + 2 cos x ., , ST, , The solution is y = yc + y p, h, i, y = e x (C1 + C2 x) − e x x sin x + 2 cos x ., , Example 5.56. Solve (D2 − 4D + 4)y = x2 e2x cos 2x., , [Jan 2007], , Solution. A.E. is (m − 2)2 = 0 ⇒ m = 2, 2., yc = e2x (C1 + C2 x)., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 346, , Engineering Mathematics - I, 1 h 1, 1 i, −, tan ax, 2ai D − ai D + ai, i, 1, 1 h 1, tan ax −, tan ax ., =, 2ai D − ai, D + ai, 1, PI1 =, tan ax, D −Zai, Z, i, h, 1, ax, aix, −iax, f (x) = e, f (x)e−ax dx, =e, tan axe dx, ∵, D−a, Z, = eaix tan ax(cos ax − i sin ax)dx, Z, sin2 ax, iax, =e, (sin ax − i, )dx, cos ax, Z, − cos ax, 1 − cos2 ax , −i, dx, = eiax, a, cos ax, − cos ax i, sin ax , − log(sec ax + tan ax) + i, = eiax, a, a, a, , −eiax , cos ax − i sin ax + i log(sec ax + tan ax), =, a, , −1 iax −iax, =, e e, + ieiax log(sec ax + tan ax), a, , −1 , 1 + ieiax log(sec ax + tan ax) ., =, a, =, , CO, R, , AP, P, , ., , changing i into −i we get, , 1, −1 , tan ax =, 1 − ie−iax log(sec ax + tan ax), D + ai, a, , ST, , U, , 1 1, −1 1, (1 + ieiax log(sec ax + tan ax)) +, (1 − ie−iax log(sec ax + tan ax)), 2ai a, 2ai a, −1, 1, = 2 (−i + eiax log(sec ax + tan ax)) + 2 (−i − e−iax log(sec ax + tan ax)), 2a, 2a, i, 1 h, iax, = 2 (i − e log(sec ax + tan ax) − i − e−iax log(sec ax + tan ax)), 2a, 1, = − 2 [(cos ax + i sin ax) log(sec ax + tan ax), 2a, , yp =, , + (cos ax − i sin ax) log(sec ax + tan ax)], , −1, log(sec ax + tan ax)2 cos ax, 2a2, −1, y p = 2 cos ax log(sec ax + tan ax)., a, =, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential Equations, , 367, , = c1 cos log(1 + x) + c2 sin log(1 + x)., 4, cos z, +1, z, = 4 cos z, 2θZ, , yp =, , θ2, , = 2z, , cos zdz, , ., , = 2 log(1 + x) sin log(1 + x)., Solution is y = yc + y p, , AP, P, , = 2z sin z, , y = c1 cos log(1 + x) + c2 sin log(1 + x) + 2 log(1 + x) sin log(1 + x)., , 5.4 Method of variation of parameters, , CO, R, , Consider the equation, , dy, d2 y, + P(x) + Q(x)y = R(x)., 2, dx, dx, , (1), , Let C.F = c1 y1 (x) + c2 y2 (x) where c1 and c2 are arbitrary constants, then y1 (x) and, y2 (x) are two independent solutions of, , dy, d2 y, + P(x) + Q(x)y = 0., dx, dx2, , (2), , U, , By the method of variation of parameters, y p is evaluated by y p = u(x)y1 + v(x)y2, where u(x) and v(x) are evaluated by the following way., , ST, , Define the Wronskian of y1 and y2 by, W=, , y1 y2, y′1 y′2, , , 0,, , Z, , Z, y2 R(x), y1 R(x), Then, u(x) = −, dx and v(x) =, dx., W, W, Then, the general solution is given by y = yc + y p ., ✎, , ☞, , ✍, , ✌, , Worked Examples, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential Equations, , 371, , Example 5.89. Solve by the method of variation of parameters (D2 + a2 )y = tan ax., [Dec 2014, Jun 2009], Solution. A.E is m2 + a2 = 0 ⇒ m = ±ai., yc = c1 cos ax + c2 sin ax., y1 = cos ax, y2 = sin ax., , y1 y2, y′1 y′2, , =, , cos ax, , sin ax, , −a sin ax a cos ax, , = a., , y p = u(x)y1 + v(x)y2 ., Z, Z, y2 R(x), sin ax tan ax, u(x) = −, dx = −, dx, W, a, Z, Z, sin2 ax, 1, 1 − cos2 ax, 1, dx = −, dx, =−, a, cos ax, a, cos ax, Z, Z, 1, 1, sec axdx +, cos axdx, =−, a, a, 1 log(sec ax + tan ax), 1, =−, + 2 sin ax., a Z, a, Za, y1 R(x), cos ax tan ax, v(x) =, dx =, dx, W, a, Z, 1, 1, sin axdx = − 2 cos ax., =, a, a, , U, , CO, R, , ., , W=, , AP, P, , R(x) = tan ax., , y p = u(x)y1 + v(x)y2, cos ax sin ax, 1, (log(sec ax + tan ax) − sin ax) cos ax −, ., 2, a, a2, , ST, , =−, , Solution is y = yc + y p ., , y = c1 cos ax + c2 sin ax −, , cos ax sin ax, 1, (log(sec ax + tan ax) − sin ax) cos ax −, ., 2, a, a2, , Example 5.90. Solve by the method of variation of parameters 2, , d2 y, + 8y = tan 2x., dx2, [Dec 2013], , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 374, , Engineering Mathematics - I, , =, , e−x, , xe−x, , −e−x e−x (1 − x), , = e−x × e−x, , 1, , x, , −1 1 − x, , = e−2x (1 − x + x) = e−2x ., , AP, P, , CO, R, , ., , y p = u(x)y1 + v(x)y2 ., Z, y2 R(x), dx, u(x) = −, W, Z, xe−x · e−x, dx., =−, 2 −2x, Z x e, 1, =−, dx = − log x., x, Z −x −x, Z, e e, y1 R(x), dx =, dx, v(x) =, W, x2 e−2x, Z, Z, 1, x−1, 1, −2, =− ., dx, =, x, dx, =, =, 2, −1, x, x, Now y p = u(x)y1 + v(x)y2, , 1, = − log x(e−x ) − xe−x, x, = −e−x (log x + 1), , Solution is y = yc + y p, , U, , y = e−x (c1 + c2 x) − e−x (log x + 1)., , ST, , Example 5.92. Solve by the method of variation of parameters the differential, d2 y, equation 2 + y = cosecx., [Dec 2012, May 2011], dx, Solution. The auxiliary equation is, m2 + 1 = 0, m2 = −1, m = ±i., yc = c1 cos x + c2 sin x., , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , Differential Equations, , 375, , y1 = cos x, y2 = sin x., R(x) = cosecx., W=, , y1 y2, y′1 y′2, , =, , cos x, , sin x, , − sin x cos x, , CO, R, , ., , y p = u(x)y1 + v(x)y2 ., Z, y2 R(x), dx, Now, u(x) = −, W, Z, = − sin x · cosecxdx., Z, = − dx = −x., Z, y1 R(x), dx, v(x) =, W, Z, =, cos x · cosecxdx, Z, =, cot xdx = log sin x., , AP, P, , = cos2 x + sin2 x = 1., , ∴ y p = u(x)y1 + v(x)y2, , = −x cos x + log(sin x) · sin x., , Solution is y = yc + y p, , U, , y = c1 cos x + c2 sin x − x cos x + sin x log sin x., , ST, , 5.5 Simultaneous linear differential equations with constant, coefficients, , We have seen so far, the method of solving a single differential equation involving, , one independent variable x and one dependent variable y. Quite often we come, across linear differential equations in which there will be two or more dependent, variables and a single independent variable., , Such equations are known as, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
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DOWNLOADED FROM STUCOR APP, STUCOR APP, , 376, , Engineering Mathematics - I, , simultaneous linear equations. In this section, we consider linear differential, equations with one independent variable t and two dependent variables x and y., We need two differential equations to solve for x and y. Hence, we will be given a, system of two linear differential equations which need not be of the same order., We shall consider here only first order linear differential equations with constant, , ., , 5.5.1 Type I, , AP, P, , coefficients and we consider three types of equations., , We consider simultaneous equations of the form, a1, , dx, + b1 y = f (t),, dt, , a2, , dy, + b2 x = g(t)., dt, , First we eliminate one of the dependent variables from the two equations which, , CO, R, , results in a second order linear differential equation with constant coefficients in, the other dependent variable and the independent variable t., ✎, , ☞, , ✍, , ✌, , Worked Examples, , Example 5.93. Solve for x and y if, , dx, dy, = x,, = y., dt, dt, , [May 2004], , Solution. The given equations are, dy, =x, dt, dx, = y., dt, Differentiating (2) w. r. t. x we get, , ST, , U, , (1), (2), , d2 x dy, =, = x[from (1)], dt, dt2, , d2 x, −x=0, dt2, , (D2 − 1)x = 0where D =, , d, ., dt, , A.E is m2 − 1 = 0, m2 = 1, , STUCOR APP, , DOWNLOADED FROM STUCOR APP
Page 405 : DOWNLOADED FROM STUCOR APP, , AP, P, , STUCOR APP, , ST, , U, , CO, R, , ., , **Note: Other Websites/Blogs Owners Please do not Copy (or) Republish, this Materials, Students & Graduates if You Find the Same Materials with, EasyEngineering.net Watermarks or Logo, Kindly report us to,
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