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4.4, , 4.4, , The Definite Integral, , 387, , The Definite Integral, Definition of the Definite Integral, In Section 4.3 we saw that the area of the region under the graph of a continuous, nonnegative function f on an interval [a, b] is defined by a limit of the form, , Historical Biography, , n, , lim a f(ck)⌬x  lim [f(c1)⌬x  f(c2)⌬x  p  f(cn)⌬x], n→⬁, n→⬁, , (1), , SPL/Photo Researchers, Inc., , k1, , BERNHARD RIEMANN, (1826–1866), Bernhard Riemann was one of the few, mathematicians to impress his contemporary Carl Friedrich Gauss, and his work continues to deeply influence modern mathematics. Born the son of a poor country, pastor in Northern Germany, Riemann was, raised without family money to support his, education. Nevertheless, he was able to, secure a solid education and showed, exceptional mathematical insight at an, early age. While still in secondary school,, he studied the works of Euler (page 19) and, Legendre, mastering Legendre’s treatise on, number theory in less than a week. He, obtained his doctorate in 1851 from the University of Göttingen after writing a thesis, involving the theory of functions of a complex variable. In 1854, upon his appointment as Privatdozent (unpaid lecturer), Riemann was required to give a lecture to the, current professors. He submitted three topics to then department chair Gauss, who, in, past situations, had chosen whichever, topic was listed first. But Riemann had submitted the foundations of geometry as his, third topic, one that so interested Gauss, that it was the topic chosen. After two, months of preparation, Riemann presented, his lecture, and that work is now considered one of the great classical masterpieces of mathematics. It was documented, that even Gauss was impressed. Riemann’s, famous conjecture, the Riemann Hypothesis, remains unresolved to this day, and the, search for a solution to that problem is, still very active. The problem has been designated one of seven Prize Problems by the, Clay Mathematics Institute, and $1,000,000, will be awarded to the person who finds a, solution., , where ⌬x  (b  a)>n and ck is in [x k1, x k]. We also saw that the distance covered, by an object moving along a straight line with a positive velocity is found by evaluating a similar limit., In this section we will look at limits defined by Equation (1) in which f may take, on both positive and negative values. We will give a geometric interpretation for this, general case later on. We will also interpret such limits in terms of the position of an, object that moves with both positive and negative velocities. Looking ahead, we will, see that limits of this type arise when we try to find the length and mass of a curved, wire, the center of mass of a body, the volume of a solid, the area of a surface, the, pressure exerted by a fluid against the wall of a container, the amount of oil consumed, over a certain period of time, the net sales of a department store over a certain period,, and the total number of AIDS cases diagnosed over a certain period of time, just to, name a few applications., In the following definition we will assume, as before, that f is continuous. This, allows for a relatively simple development of the material ahead of us., , DEFINITION Definite Integral, Let f be a continuous function defined on an interval [a, b]. Suppose that [a, b], is divided into n subintervals of equal length ⌬x  (b  a)>n by means of, (n  1) equally spaced points, a  x0  x1  x2  p  xn  b, Let c1, c2, p , cn be arbitrary points in the respective subintervals with ck lying, in the kth subinterval [x k1, x k]. Then the definite integral of f on [a, b], denoted, by 兰ab f(x) dx, is, , 冮, , a, , b, , n, , f(x) dx  lim a f(ck)⌬x, n→⬁, , (2), , k1, , We also say that f is integrable on [a, b] if the limit (2) exists. The process of evaluating a definite integral is called integration. The number a in the definition is called, the lower limit of integration, and the number b is called the upper limit of integration. Together, the numbers a and b are referred to as the limits of integration. As in, the case of the indefinite integral, the function f to be integrated is called the integrand., The sum 兺nk1 f(ck)⌬x in the definition is called a Riemann sum in honor of the, German mathematician Bernhard Riemann (1826–1866). Actually, this sum is a special case of a more general form of a Riemann sum in which no assumption is made, requiring that f be continuous on [a, b] or that the interval be partitioned in such a way, that the resulting subintervals have equal length. For completeness we will discuss this, general case at the end of this section.

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388, , Chapter 4 Integration, , Notes, 1. The assumption that f is continuous on [a, b] guarantees that the definite integral, always exists. In other words, the limit in Equation (2) exists and is unique for all, choices of the evaluation points ck. Furthermore, if f is nonnegative, then the definite integral gives the area of the region under the graph of f on [a, b] since the, limit in Equation (2) reduces to the limit in Equation (1), page 380, in Section 4.3., 2. The symbol 兰 in the definition of the definite integral is the same as that used to, denote the indefinite integral of a function. (Remember that the definite integral is, a number, in contrast to the indefinite integral, which is a family of functions (the, antiderivatives of f ).), , EXAMPLE 1 Compute the Riemann sum for f(x)  4  x 2 on [1, 3] using five, , subintervals (n  5) and choosing the evaluation points to be the midpoints of the, subintervals., Solution, , Here, a  1, b  3, and n  5. So the length of each subinterval is, ⌬x , , 3  (1), ba, 4, , , n, 5, 5, , The partition points are, x 0  1,, , x 1  1 , 7, x3  ,, 5, , 4, 1,  ,, 5, 5, , x4 , , 11, ,, 5, , 4, 3, x 2  1  2a b  ,, 5, 5, x5  3, , and, , The midpoints of the subintervals are given by ck  12 (x k  x k1), or, 3, c1   ,, 5, , y, , y  4  x2, , c2 , , 1, ,, 5, , c3  1,, , c4 , , 9, ,, 5, , and, , c5 , , 13, 5, , (See Figure 1.), Therefore, the required Riemann sum is, , 3, , 5, , a f(ck)⌬x  f(c1)⌬x  f(c2)⌬x  f(c3)⌬x  f(c4)⌬x  f(c5)⌬x, , 1, c5, 3 2 1 c1, , c2 1, c3, , 2, c4, , k1, , 3, , 4 x, , 4, , FIGURE 1, The positive terms of the Riemann sum, are associated with the rectangles that, lie above the x-axis; the negative term, is associated with the rectangle that lies, below the x-axis., , 3, 1, 9, 13,  cf a b  f a b  f(1)  f a b  f a b d⌬x, 5, 5, 5, 5, 4, 3 2, 1 2,  a b e c4  a b d  c4  a b d  [4  (1)2], 5, 5, 5, 9 2, 13 2,  c4  a b d  c4  a b d f, 5, 5, 4,  a b (3.64  3.96  3  0.76  2.76), 5,  6.88, The Riemann sum computed in Example 1 is the sum of five terms. As you can, see in Figure 1, these terms are associated with the areas of the five rectangles shown., The positive terms give the areas of the rectangles that lie above the x-axis, while the, negative term is the negative of the area of the rectangle that lies below the x-axis.

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4.4, , EXAMPLE 2 Evaluate, , 冮, , The Definite Integral, , 389, , 3, , 1, , (4  x 2) dx., , Solution Here, a  1 and b  3. Furthermore, f(x)  4  x 2 is continuous on, [1, 3], so f is integrable on [1, 3]. To evaluate the given definite integral, let’s subdivide the interval [1, 3] into n equal subintervals of length, ⌬x , , 3  (1), ba, 4, , , n, n, n, , The partition points are, 4, x 2  1  2a b ,, n, , 4, x 1  1  ,, n, , x 0  1,, , 4, x k1  1  (k  1) a b ,, n, , 4, x k  1  ka b ,, n, , p,, , p,, xn  3, , Next, we pick ck to be the right endpoint of the subinterval [x k1, x k] so that, 4k, 4, ck  x k  1  ka b  1 , n, n, Then, , 冮, , 3, , 1, , (4  x 2)dx , , 冮, , 3, , n, , f(x) dx  lim a f(ck)⌬x, n→⬁, , 1, , k1, , n, , 4k 4,  lim a f a1  b a b, n, n, n→⬁ k1, n, 4k 2 4,  lim a c4  a1  b d a b, n, n, n→⬁ k1, , f(x)  4  x 2, , 16k 2, 4 n, 8k,  2 b,  lim a b a a3 , n, n→⬁ n k1, n,  lim c, n→⬁, , 4 n, 32 n, 64 n 2, 3, , k, , a, a, ak d, n k1, n 2 k1, n 3 k1, , 32 n(n  1), 64 n(n  1)(2n  1), 4,  lim c (3n)  2 ⴢ,  3 ⴢ, d, n→⬁ n, 2, 6, n, n, 1, 32, 1, 1,  lim c12  16a1  b , a1  b a2  b d, n, n, n, n→⬁, 3,  12  16 , , 64, 20, 2, , 6, 3, 3, 3, , 3, (Compare this with the approximate value of 兰1, (4  x 2) dx obtained in Example 1.), , Solution, , 冮, , b, , 1 2, (b  a 2)., 2, a, Let’s subdivide the interval [a, b] into n subintervals of length, , EXAMPLE 3 Show that, , x dx , , ⌬x , , ba, n

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390, , Chapter 4 Integration, , The partition points are, x 0  a,, , x1  a , , ba, ,, n, , x k  a  ka, , x 2  a  2a, , ba, b,, n, , p,, , ba, b,, n, , p,, , xn  b, , Next we choose the evaluation point ck to be the right endpoint of the subinterval, [x k1, x k], where 1  k  n; that is, we pick ck  x k for each 1  k  n. Then, b, , n, , 冮 x dx  lim a f(c )⌬x, n→⬁ k1, , a, , k, , n, ,  lim a ca  a, n→⬁, k1, , ba, ba, bkd a, b, n, n, ,  (b  a) lim, , 1 n, ba, bkd, ca  a, a, n k1, n, ,  (b  a) lim, , 1 n, ba n, caa a, b a kd, n k1, n, k1, ,  (b  a) lim, , n(n  1), ba, 1, cna  a, bⴢ, d, n, n, 2, , n→⬁, , n→⬁, , n→⬁, ,  (b  a) lim ca  a, n→⬁, ,  (b  a) ca  a,  (b  a) aa , , , n(n  1), ba, bⴢ, d, 2, n2, , ba, n1, d, b lim, n, 2, n→⬁, , ba, 2a  b  a, b  (b  a) a, b, 2, 2, , 1, 1, (b  a)(b  a)  (b 2  a 2), 2, 2, , EXAMPLE 4 Divide the interval [2, 5] into n subintervals of equal length, and let ck, be any point in [x k1, x k]. Write, n, , lim a 21  (ck)2 ⌬x, n→⬁, k1, , as an integral., y, y  f(x), , Solution Comparing the given expression with Equation (2), we see that it is the limit, of a Riemann sum of the function f(x)  21  x 2 on the interval [2, 5]. Next, since, f is continuous on [2, 5], the limit exists, so by Equation (2),, n, , lim a 21  (ck)2 ⌬x , n→⬁, , b, , A  y f(x) dx, , k1, , a, , 0, , a, , b, , FIGURE 2, If f(x) 0 on [a, b], then 兰ab f(x) dx, gives the area of the region under the, graph of f on [a, b]., , 5, , 冮 21  x, , 2, , dx, , 2, , x, , Geometric Interpretation of the Definite Integral, As was pointed out earlier, if f is a continuous, nonnegative function on [a, b], then the, definite integral 兰ab f(x) dx gives the area of the region under the graph of f on [a, b]., (See Figure 2.)

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4.4, , 391, , EXAMPLE 5 Evaluate the definite integral by interpreting it geometrically:, , y, 4, , a., , 3, , y  4  2x, , 4, , (4  2x) dx, , b., , x, , 2, , 冮 216  x, , 2, , dx, , 0, , Solution, a. The graph of the integrand f(x)  4  2x on [0, 2] is the straight line segment, shown in Figure 3. Since f(x) 0 on [0, 2], we can interpret the integral as the, area of the triangle shown. Thus,, , 1, 1, , 冮, , 2, , 0, , 2, , 0, , The Definite Integral, , 2, , 冮 (4  2x) dx  2 (2)(4)  4, , FIGURE 3, 兰02 (4  2x) dx  area of the triangle., , area , , 0, , 1, base ⴢ height, 2, , b. The integrand f(x)  216  x 2 is the positive root obtained by solving the equation x 2  y 2  16 for y, which represents the circle of radius 4 centered at the, origin; therefore, it represents the upper semicircle shown in Figure 4. Since, f(x) 0 on [0, 4], we can interpret the integral as the area of that part of the circle lying in the first quadrant. Since this area is 14 p(42)  4p, we see that, , y, 4, , 1, , y  √16  x2, , 4, , 4, , 0, , 4, , 冮 216  x, , x, , FIGURE 4, f(x)  216  x 2 represents the upper, semicircle., , 2, , dx  4p, , 0, , Next we look at a geometric interpretation of the definite integral for the case in, which f assumes both positive and negative values on [a, b]. Consider a typical Riemann sum of the function f,, n, , a f(ck)⌬x, k1, , corresponding to a partition P with points of subdivision, a  x 0  x 1  x 2  p  x k1  x k  p  x n1  x n  b, and evaluation points ck in [x k1, x k]. The sum consists of n terms in which a positive, term corresponds to the area of a rectangle of height f(ck) lying above the x-axis, and, a negative term corresponds to the area of a rectangle of height f(ck) lying below the, x-axis. (See Figure 5, where n  6.), , y, , y  f(x), , FIGURE 5, The positive (negative) terms in, the Riemann sum are associated, with the areas of the rectangles, that lie above (below) the x-axis., , c3, , c4, x3, , 0, , a  x0, , x1, c1, , c2, , x2, , x4, , x6  b, , x5, c5, , x, , c6, , As n gets larger and larger, the sums of the areas of the rectangles lying above the, x-axis seem to give a better and better approximation of the area of the region lying, above the x-axis. Similarly, the sums of the area of the rectangles lying below the

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392, , Chapter 4 Integration, , x-axis seem to give a better and better approximation of the area of the region lying, below the x-axis. (See Figure 6, where n  12.), y, y  f(x), , FIGURE 6, Approximating 兰ab f(x) dx, with 12 rectangles, , 0, , a, , b, , x, , This observation suggests that we interpret the definite integral, , 冮, , a, , b, , y, y  f(x), S1, 0, , a, , k1, , as a difference of areas. Specifically,, b, , 冮 f(x) dx  area of S, , 1, , S3, S2, , n, , f(x) dx  lim a f(ck)⌬x, n→⬁, ,  area of S2  area of S3, , a, , b, , x, , FIGURE 7, 兰ab f(x) dx  area of S1 , area of S2  area of S3, , where S2 is the region lying above the graph of f and below the x-axis. (See Figure 7.), More generally,, b, , 冮 f(x) dx  areas of the regions above [a, b]  areas of the regions below [a, b], a, , The Definite Integral and Displacement, In Section 4.3 we showed that if √(t) is a nonnegative velocity function of a car traveling in a straight line, then the distance covered by the car between t  a and t  b, is given by the area of the region under the graph of the velocity function on the time, interval [a, b]. Since the area of the region under the graph of a nonnegative function, √(t) on [a, b] is just the definite integral of √ on [a, b], we can write, b, , 冮 √(t) dt  displacement of the car between t  a and t  b, a, , If we denote the position of the car at any time t by s(t), then its position at t  a is, s(a). So we can then write its final position at t  b as, b, , s(b)  s(a) , , 冮 √(t) dt, a, , (See Figure 8.), , b, , ya √ (t) dt, FIGURE 8, The position of the car at t  b, is s(b)  s(a)  兰ab √(t) dt., , s(a), , s(b), , s, , Now suppose that √(t) assumes both positive and negative values on [a, b]. (See, Figure 9.)

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4.4, √, , The Definite Integral, , 393, , Then, b, , √  √(t), , 冮 √(t) dt  area of the regions above [a, b]  area of the region below [a, b], a, , S1, S3, 0, , a, , S2, , b, ,  displacement of the car between t  a and t  b, , t, , FIGURE 9, The area of S1 and the area of S3 give, the distance the car moves in the, positive direction, whereas the area of, S2 gives the distance it moves in the, negative direction., ,  distance covered by the car in the positive direction  distance covered, by the car in the negative direction, , In other words, the final position of the car at t  b is, b, , 冮 √(t) dt, , s(b)  s(a) , , a, , as before., , EXAMPLE 6 The velocity function of a car moving along a straight road is given by, √(t)  t  20 for 0  t  40, where √(t) is measured in feet per second and t in seconds. Show that at t  40 the car will be in the same position as it was initially., Solution, , The graph of √ is shown in Figure 10. We have, , 冮, , 40, , √(t) dt  area of S2  area of S1, , 0, , , , 1, 1, (20)(20)  (20)(20), 2, 2, ,  200  200  0, √ (ft/sec), , √(t)  t  20, , 20, , S2, 0, , 10, , 20, , 30, , 40, , t (sec), , S1, , FIGURE 10, The area of S1 is equal to the area of S2., , 20, , Therefore,, s(40)  s(0) , , 冮, , 40, , √(t) dt  s(0), , 0, , so the net change in the position of the car is zero, as was to be shown., We interpret this result as follows: The car moves a total of 200 ft in the negative, direction in the first 20 sec and then moves a total of 200 ft in the positive direction, in the next 20 sec, resulting in no net change in its position., Alternative Solution, , Let s(t) denote the position of the car at any time t. Then, ds,  √(t), dt

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394, , Chapter 4 Integration, , But √(t)  t  20, so, ds,  t  20, dt, Integrating with respect to t, we have, s(t) , , , 冮 (t  20) dt, 1 2, t  20t  C, 2, , C an arbitrary constant, , The position of the car at t  0 is s(0), and this condition gives, s(0) , , 1, (0)  20(0)  C, 2, , or, , C  s(0), , Therefore, the position of the car at any time t is, s(t) , , 1 2, t  20t  s(0), 2, , In particular, the position of the car at t  40 is, s(40) , , 1, (402)  20(40)  s(0)  s(0), 2, , its position at t  0, as was to be shown., Note The method used in the alternative solution of the problem in Example 6 hints, at the relationship between the definite integral of a function and the indefinite integral of the function. We will exploit this relationship in the next section., , Properties of the Definite Integral, When we defined the definite integral 兰ab f(x) dx, we assumed that a  b. We now, extend the definition to cover the cases a  b and a b., , DEFINITIONS Two Special Definite Integrals, a, , 1., , 冮 f(x) dx  0, a, , 2., , 冮, , a, , b, , a, , f(x) dx  , , 冮 f(x) dx,, , if a, , b, , b, , The first definition is compatible with the definition of the definite integral if we, observe that here,, ⌬x , , aa, ba, , 0, n, n

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4.4, , The Definite Integral, , 395, , The second definition is also compatible with the definition by observing that if we, interchange a and b, then the sign of the resulting Riemann sum changes because, ⌬x , , ba, ab, , n, n, , EXAMPLE 7 Evaluate the definite integral:, a., , 冮, , 2, , (x 2  2x  4) dx, , 1, , 冮, , b., , 2, , (4  x 2) dx, , 3, , Solution, 2, , a., , 冮 (x, , 2, ,  2x  4) dx  0, , 2, , b., , 冮, , 1, , (4  x 2) dx  , , 3, , 冮, , 3, , 1, , (4  x 2) dx  6, , 2, using the result of Example 2., 3, , In the expression 兰ab f(x) dx the variable of integration, x, is a dummy variable in, the sense that it may be replaced by any other letter without changing the value of the, integral. As an illustration, the results of Example 2 may be written, , 冮, , y, yc, b, , ya c dx, , 3, , 1, , 冮, , (4  x 2) dx , , 3, , 1, , (4  u 2) du , , 冮, , 3, , 1, , (4  s 2) ds  6, , 2, 3, , Suppose that c 0. Interpreting 兰 c dx as the area of the region under the graph, of f(x)  c on [a, b] gives, b, a, , b, , 0, , a, , b, , 冮 c dx  c(b  a), , x, , a, , FIGURE 11, If c 0, then interpreting 兰ab c dx, as the area of the region under the, graph of f(x)  c on [a, b] gives, 兰ab f(x) dx  c(b  a)., , (See Figure 11.), We will now look at some properties of the definite integral that will prove helpful later on when we evaluate integrals. Here we assume, as we did earlier, that all of, the functions under consideration are continuous., The Definite Integral of a Constant Function, If c is a real number, then, b, , 冮 c dx  c(b  a), a, , The special case where c, , 0 was discussed earlier., , 7, , EXAMPLE 8 Evaluate, , 冮 3 dx., 2, , Solution, , We use Equation (3) with c  3, a  2, and b  7, obtaining, 7, , 冮 3 dx  3(7  2)  15, 2, , (3)

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396, , Chapter 4 Integration, , The next two properties of the definite integral are analogous to the rules of integration for indefinite integrals (see Section 4.1)., Properties of the Definite Integral, 1. Sum (Difference), b, , 冮, , [f(x)  t(x)] dx , , a, , 冮, , b, , b, , f(x) dx , , a, , 冮 t(x) dx, a, , 2. Constant Multiple, b, , b, , 冮 cf(x) dx  c冮 f(x) dx,, a, , where c is any constant, , a, , Property 1 states that the integral of the sum (difference) is the sum (difference) of, the integrals. Property 2 states that the integral of a constant times a function is equal, to the constant times the integral of the function. Thus, a constant (and only a constant!) can be moved in front of the integral sign. These properties are derived by using, the corresponding limit laws. For example, to prove Property 2, we use the definition, of the definite integral to write, b, , n, , 冮 cf(x) dx  lim a cf(c )⌬x, n→⬁ k1, , a, , k, , n, ,  c lim a f(ck)⌬x, n→⬁, k1, , Constant Multiple, Law for limits, , b, , c, , 冮 f(x) dx, a, , 1, , 冮x, , EXAMPLE 9 Use the result, , 2, , dx , , 0, , 1, , a., , 冮 (x, , 1, of Example 8 in Section 4.3 to evaluate, 3, , 1, , 2, ,  4) dx, , b., , 0, , 冮 5x, , 2, , dx, , 0, , Solution, a., , 冮, , 1, , (x 2  4) dx , , 0, , 冮, , 1, , 0, , , , 冮, , 0, , 1, , 冮 4 dx, 0, , 1,  4(1), 3, , , b., , 1, , x 2 dx , , 11, 3, , 1, , 5x 2 dx  5, , 冮x, , 2, , dx, , 0, , 1, 5,  5a b , 3, 3, , Property 2, , Property 1

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4.4, y, c, , ya f(x) dx, , y  f(x), , c, , c, , 冮, , f(x) dx , , a, , x, , b, , b, , 冮, , b, , a, , 397, , Suppose that f is continuous and nonnegative on [a, b]. Then 兰ab f(x) dx gives the, area of the region under the graph of f on [a, b]. Next, if a  c  b, then 兰ac f(x) dx, and 兰cb f(x) dx give the area of the region under the graph of f on [a, c] and [c, b],, respectively. Therefore, as you can see in Figure 12,, , yc f(x) dx, 0, , The Definite Integral, , b, , f(x) dx , , a, , 冮 f(x) dx, c, , This observation suggests the following property of definite integrals., , FIGURE 12, , 冮, , a, , b, , f(x) dx , , 冮, , a, , c, , f(x) dx , , 冮, , c, , b, , f(x) dx, , Property of the Definite Integral, 3. If c is any number in [a, b], then, , 冮, , b, , a, , Note, , c, , 冮, , f(x) dx , , b, , f(x) dx , , a, , 冮 f(x) dx, c, , The conclusion of Property 3 holds for any three numbers a, b, and c., , EXAMPLE 10 Suppose that 兰16 f(x) dx  8 and 兰46 f(x) dx  5. What is 兰14 f(x) dx?, Solution, , Using Property 3, we have, , 冮, , 6, , 冮, , f(x) dx , , 1, , 4, , 6, , f(x) dx , , 1, , 冮 f(x) dx, 4, , from which we see that, , 冮, , 1, , 4, , f(x) dx , , 冮, , 6, , 6, , f(x) dx , , 1, , 冮 f(x) dx  8  5  3, 4, , The next three properties of the definite integral involve inequalities., Properties of the Definite Integral, 4. If f(x) 0 on [a, b], then, b, , 冮 f(x) dx, , 0, , a, , 5. If f(x), , t(x) on [a, b], then, , 冮, , b, , b, , f(x) dx, , a, , 冮 t(x) dx, a, , 6. If m  f(x)  M on [a, b], then, b, , m(b  a) , , 冮 f(x) dx  M(b  a), a, , The plausibility of Property 4 stems from the observation that the area of the region, under the graph of a nonnegative function is nonnegative. Also, if we assume that t, and therefore f are both nonnegative on [a, b], then Property 5 is a statement that the

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398, , Chapter 4 Integration, , area of the region under the graph of f is larger than the area of the region under the, graph of t. (See Figure 13.) The plausibility of Property 6 is suggested by Figure 14,, where m and M are the absolute minimum and absolute maximum values of f, respectively, on [a, b]: The area of the region under the graph of f on [a, b], 兰ab f(x) dx, is, greater than the area of the rectangle with height m, m(b  a), and smaller than the, area of the rectangle with height M, M(b  a)., y, , y, Absolute maximum, , M, y  f(x), y  g(x), , y  f(x), m, , 0, , a, , 0, , x, , b, , Absolute minimum, , FIGURE 13, If f(x) t(x) on [a, b], then the area of the, region under the graph of f is greater than the, area of the region under the graph of t., , a, , b, , x, , FIGURE 14, The area of the region under the graph of, f is greater than or equal to m(b  a) and, less than or equal to M(b  a)., , It should be mentioned that all of the properties of the definite integral can be proved, with mathematical rigor and without any assumption regarding the sign of f(x) (see, Exercise 60)., 3, , EXAMPLE 11 Use Property 6 to estimate, , 冮 23  x, , 2, , dx., , 1, , Solution The integrand f(x)  23  x 2 is increasing on [1, 3]. Therefore, its, absolute minimum value occurs at x  1 (the left endpoint of the interval), and its, absolute maximum value occurs at x  3 (the right endpoint of the interval). If we take, m  f(1)  2, M  f(3)  213, a  1, and b  3, then Property 6 gives, 3, , 2(3  1) , , 冮 23  x, , 2, , dx  213(3  1), , 2, , dx  423, , 1, , 3, , 4, , 冮 23  x, 1, , More General Definition of the Definite Integral, As was pointed out earlier, the points that make up a partition of an interval [a, b] need, not be chosen to be equally spaced. In general, a partition of [a, b] is any set, P  {x 0, x 1, p , x n} satisfying, a  x 0  x 1  x 2  p  x n1  x n  b, The subintervals corresponding to this partition of [a, b] are, [x 0, x 1],, , [x 1, x 2],, , p,, , [x k1, x k],, , p,, , [x n1, x n]

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4.4, , The Definite Integral, , 399, , The length of the kth subinterval is, ⌬x k  x k  x k1, Figure 15 shows one possible partition of [a, b]., x2, , x1, , FIGURE 15, A possible partition of [a, b], , x0  a, , x3, , x1 x2, , xk, , x3, , xk1, , xn, , xn1 xn  b x, , xk, , The length of the largest subinterval, denoted by 7 P 7 , is called the norm of P. For, example, in the partition shown in Figure 16,, 1, ⌬x 1  ,, 4, 1, ⌬x 5  ,, 8, , 1, ⌬x 2  ,, 4, 1, ⌬x 6  ,, 8, , 1, ⌬x 3  ,, 2, 1, ⌬x 7  ,, 4, , 1, ⌬x 4  ,, 4, ⌬x 8 , , and, , 1, 4, , so its norm is 12., x5  11, 8, , FIGURE 16, A possible partition of [0, 2], , x0  0 x1 , , 1, 4, , x2  12, , x3  1 x4 , , 5, 4, , x6 , , 3, 2, , x7 , , 7, 4, , x8  2, , x, , If the (n  1) points of a partition of [a, b] are chosen to be equally spaced so that, the resulting n subintervals have equal length, then the partition is regular. In a regular partition, the norm satisfies, 7 P 7  ⌬x , , ba, n, , For a general partition P,, 7P7, 0, , 1, 2, , 3, 4, , 7, 8, , ...1 x, , FIGURE 17, As the number of subintervals approach, infinity, 7 P 7 does not approach 0., , ba, n, , or, , n, , ba, 7P7, , From this inequality we see that as the norm of a partition approaches 0, the number, of subintervals approach infinity. The converse, however, is false. For example, the partition P of the interval [0, 1] in Figure 17 is given by, 0, , 1, 3, 7, 1, 1,    p  1  n1  1  n  1, 2, 4, 8, 2, 2, , has norm 12 for any positive integer n. Therefore, n → ⬁ does not imply that 7 P 7 → 0., But for a regular partition,, 7 P 7 → 0 if and only if, , n→⬁, , a fact that we will use shortly., We are now in a position to give a more general definition of the definite integral,, but first we observe that a function f is bounded on an interval [a, b] if there exists, some positive real number M such that 冟 f(x) 冟  M for all x in [a, b].

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400, , Chapter 4 Integration, , DEFINITION Definite Integral (General Definition), Let f be a bounded function defined on an interval [a, b]. Then the definite inteb, , gral of f on [a, b], denoted by, , 冮 f(x) dx, is, a, , 冮, , b, , n, , f(x) dx  lim a f(ck)⌬x, 7P7→0, , (4), , k1, , a, , if the limit exists for all partitions P of [a, b] and all choices of ck in [x k1, x k]., , It can be shown that if f is continuous on [a, b], then the definite integral of f on, [a, b] always exists. Therefore, the limit (4) exists for all choices of P and ck. In particular, the limit exists if we choose a regular partition, as was done in our earlier presentation. In fact, for regular partitions, 7 P 7 → 0 if and only if n → ⬁ . So the limit (4), is equivalent to, b, , n, , 冮 f(x) dx  lim a f(c )⌬x, n→⬁ k1, , a, , k, , which is the definition of the definite integral given earlier., Finally, we note the following precise definition of the definite integral., , DEFINITION Precise Definition of the Definite Integral, The definite integral of f on [a, b] is, b, , 冮 f(x) dx, a, , if for every number e 0 there exists a number d 0 such that for every partition P of [a, b] with 7 P 7  d and every choice of points ck in [x k1, x k], the, inequality, n, , ` a f(ck)⌬x k , k1, , b, , 冮 f(x) dx `  e, a, , holds., , 4.4, , CONCEPT QUESTIONS, , 1. What is a Riemann sum of a continuous function f on an, interval [a, b]? Illustrate graphically the case in which f, assumes both positive and negative values on [a, b]., 2. Define the definite integral of a continuous function on the, interval [a, b]. Give a geometric interpretation of 兰ab f(x) dx, for the case in which (a) f is nonnegative on [a, b] and (b) f, assumes both positive and negative values on [a, b]. Illustrate your answers graphically., , 3. The following figure depicts the graph of the velocity function √ of an object traveling along a coordinate line over, the time interval [a, b]. The numbers c, d, and e satisfy, a  c  d  e  b. The areas of the regions S1, S2, S3, and, S4 are A1, A2, A3, and A4 respectively. Assume that the object, is located at the origin at t  a.

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4.4, , 0, , 4.4, , S3, , a, , S2, , c, , d, , e, , S4, , t (sec), , b, , EXERCISES, , 1. The graph of a function f on the interval [0, 8] is shown in, the figure. Compute the Riemann sum for f on [0, 8] using, four subintervals of equal length and choosing the evaluation, points to be (a) the left endpoints, (b) the right endpoints,, and (c) the midpoints of the subintervals., , In Exercises 3–6 you are given a function f defined on an, interval [a, b], the number n of subintervals of equal length, ⌬x  (b  a)>n, and the evaluation points ck in [x k1, x k]., (a) Sketch the graph of f and the rectangles associated with the, Riemann sum for f on [a, b], and (b) find the Riemann sum., , y, , 3. f(x)  2x  3,, , 5, , 4. f(x)  2x  1,, , 4, , 5. f(x)  1x  1,, , 3, , 6. f(x)  2 sin x,, , 2, , [0, 2], n  4,, , ck is the midpoint, , n  6,, , [1, 2],, , 1, , 1, 2, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , [0, 3], n  6, ck is the right endpoint, , C0, 5p, 4 D,, , n  5, ck is the right endpoint, , 2, , x, , 7., , 冮 x dx, , 8., , 9., , 3, , 冮, , 1, 2, , 4, , 11., 2. The graph of a function t on the interval [2, 4] is shown, in the figure. Compute the Riemann sum for t on [2, 4], using six subintervals of equal length and choosing the evaluation points to be (a) the left endpoints, (b) the right endpoints, and (c) the midpoints of the subintervals., , 冮, , (x  2) dx, , 10., , 冮, , (2x  1) dx, , 2, , 12., , 冮, , (x 3  2x) dx, , 2, , 1, , In Exercises 13–16, the given expression is the limit of a, Riemann sum of a function f on [a, b]. Write this expression, as a definite integral on [a, b]., n, , 0, 1, , [3, 1], , k1, , 2, , n, , 14. lim a 2ck(1  ck)2⌬x,, n→⬁, , [0, 3], , k1, , 1, , 1, , 2, , 3, , 4, , x, , x 2 dx, , 1, 1, , 冮 (3  2x ) dx, , 13. lim a (4ck  3)⌬x,, n→⬁, , y, , 2, , 1, 1, , 0, , 3, , 1, , ck is the left endpoint, , In Exercises 7–12, use Equation (2) to evaluate the integral., , 1, , 2, , 401, , a. Write the displacement of the object at t  c, t  d,, t  e, and t  b (i) in terms of A1, A2, A3, and A4 and, (ii) in terms of definite integrals., b. Write the distances covered by the object over the time, intervals [a, d] and [a, b]. Express your answer using A1,, A2, A3, and A4 and also using definite integrals., , √ (ft/sec), , S1, , The Definite Integral, , n, 2ck, ⌬x,, 15. lim a 2, n→⬁ k1 ck  1, n, , 16. lim a ck(cos ck)⌬x,, n→⬁, , 2, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , k1, , [1, 2], C0, p2 D

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4.4 The Definite Integral, , 37., , 冮, , p>4, , 冮, , p>2, , sin2 x cos x dx ⱕ, , 0, , 38., , 冮, , 52. a. Plot the graphs of f(x) ⫽ sin x and t(x) ⫽ x using the, viewing window C0, p2 D ⫻ [0, 2]., b. Prove that 0 ⱕ f(x) ⱕ t(x) ., c. Use the result of part (b) and Property 5 to show that, , p>4, , sin2 x dx, , 0, , 冮, , cos x dx ⱕ, , 0, , p>2, , (x 2 ⫹ 1) dx, , 0, , 0ⱕ, , In Exercises 39–44, use Property 6 of the definite integral to, estimate the definite integral., 2, , 39., , 冮 21 ⫹ 2x, , 3, , dx, , 40., , 1, , 41., , 冮, 冮, , 冮, , 2, , 1, , 2, , (x 2 ⫺ 2x ⫹ 2) dx, , 42., , ⫺1, p>4, , 43., , 冮, , 3, , 0, , sin x dx, , 冮, , 44., , p>6, , x ⫹5, , Hint: Use the result of Example 3., Note: The upper bound obtained here is better than that obtained, in Exercise 43., , x2 ⫹ 2, , dx, , p>2, , x sin x dx, , 46. a. Plot the graph of f(x) ⫽ sin3 x on the interval [0, 2p]., b. Prove that the area of the region above the x-axis is equal, to the area of the region below the x-axis., Hint: Look at f(p ⫹ t) for 0 ⱕ t ⱕ p., , c. Use the result of part (b) to show that 兰02p sin3 x dx ⫽ 0., 47. Suppose that f is continuous on [a, b] and f(x) ⱕ 0 on [a, b]., Prove that 兰ab f(x) dx ⱕ 0., , In Exercises 53 and 54, use Property 5 to prove the inequality., 4, , 53., , 冮 2x, , 4, , ⫹ x dx ⱖ, , 2, , b, , 冮 f(x) dx ` ⱕ 冮 冟 f(x) 冟 dx, a, , Hint: ⫺冟 f(x) 冟 ⱕ f(x) ⱕ 冟 f(x) 冟., , 49. Use the result of Exercise 48 to show that, 冟 兰ab x sin 2x dx 冟 ⱕ 12 (b 2 ⫺ a 2), where 0 ⱕ a ⬍ b., Hint: Use the result of Example 3., , (b ⫺ a)f(a) ⱕ, , 54., , 冮, , p>4, , x sin x dx ⱕ, , 0, , 冮 f(x) dx ⱕ 2 (b ⫺ a)[ f(a) ⫹ f(b)], 1, , a, , x, , 51. a. Plot the graphs of f(x) ⫽, , and t(x) ⫽ x using, , 21 ⫹ x, the viewing window [0, 1] ⫻ [0, 1]., b. Prove that 0 ⱕ f(x) ⱕ t(x) ., c. Use the result of part (b) and Property 5 to show that, , 55. Estimate the integral 兰 21 ⫹ x dx using (a) Property 6 of, the definite integral and (b) the result of Exercise 50. Which, estimate is better? Explain., 2, , 56. Show that 兰ab x 2 dx ⫽ 13 (b 3 ⫺ a 3)., 57. Find the constant b such that 兰0b (2 1x ⫺ x) dx is as large as, possible. Explain your answer., x, (t 4 ⫺ 2t 3) dt for x in, 58. Define the function F by F(x) ⫽ 兰⫺1, [⫺1, 2]., a. Plot the graph of f(t) ⫽ t 4 ⫺ 2t 3 on [⫺1, 2]., b. Use the result of part (a) to find the interval where F is, increasing and where F is decreasing on (⫺1, 2)., , 59. Determine whether the Dirichlet function, f(x) ⫽ e, , 1 if x is rational, 0 if x is irrational, , is integrable on the interval [0, 1]. Explain., 60. Prove Properties 4, 5, and 6 of the definite integral., In Exercises 61–66, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, explain why, or give an example to show why it is false., , 5, , b, , b, , b, , 冮 [ f(x) ⫹ ct(x)] dx ⫽ 冮 f(x) dx ⫹ c冮 t(x) dx, a, , a, , a, , 62. If f and t are continuous on [a, b], then, b, , b, , b, , 冮 f(x) t(x) dx ⫽ c 冮 f(x) dxd c 冮 t(x) dxd, a, , a, , a, , 63. If f is continuous on [a, b], then 兰ab xf(x) dx ⫽ x 兰ab f(x) dx., 64. If f is continuous on [a, b] and 兰ab f(x) dx ⬎ 0, then f must, be positive on [a, b]., , x, , 65. If f is continuous and decreasing on [a, b], then, (b ⫺ a)f(b) ⱕ 兰ab f(x) dx ⱕ (b ⫺ a)f(a)., , Hint: Use the result of Example 3., , 66. If f is nonnegative and continuous on [a, b] and, a ⬍ c ⬍ d ⬍ b, then 兰cd f(x) dx ⱕ 兰ab f(x) dx., , 0ⱕ, , 冮, , 0, , 1, , p3, 192, , 61. If f and t are continuous on [a, b] and c is constant, then, , 50. Suppose that f is continuous and increasing and its graph, is concave upward on the interval [a, b]. Give a geometric, argument to show that, b, , 56, 3, 1, 0, , 48. Suppose that f is continuous on [a, b]. Prove that, , a, , 5p2, 288, , sin x dx ⱕ, , 2, , 45. a. Plot the graph of f(x) ⫽ x2x 4 ⫹ 1 on the interval, [⫺1, 1]., b. Prove that the area of the region above the x-axis is equal, to the area of the region below the x-axis., c. Use the result of part (b) to show that, 1, x2x 4 ⫹ 1 dx ⫽ 0., 兰⫺1, , b, , 冮, , p>4, , p>6, , 1, dx, x, , p>4, , `, , 403, , 1, dx ⱕ, 5, 2, 21 ⫹ x

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404, , Chapter 4 Integration, , 4.5, , The Fundamental Theorem of Calculus, How Are Differentiation and Integration Related?, In Section 4.4 we defined the definite integral of a function by taking the limit of its, Riemann sums. But as we saw, the actual process of finding the definite integral of a, function based on this definition turned out to be rather tedious even for simple functions. This is reminiscent of the process of finding the derivative of a function by finding the limit of the difference quotient of the function. Fortunately, there are better and, easier ways of evaluating definite integrals., In this section we will look at what is undoubtedly the most important theorem in, calculus. Because it establishes the relationship between differentiation and integration,, it is called the Fundamental Theorem of Calculus. It was discovered independently, by Sir Isaac Newton (1643–1727) in England and by Gottfried Wilhelm Leibniz (1646–, 1716) in Germany. Before looking at this theorem, we need the results of the following theorem., , The Mean Value Theorem for Definite Integrals, Suppose that the velocity of a maglev traveling along a straight track is √(t) ft/sec for t, between t  a and t  b, where t is measured in seconds. What is the average velocity of the maglev over the time interval [a, b]?, To answer this question, let’s assume that √ is continuous on [a, b]. We begin by, partitioning the interval [a, b] into n equal subintervals of length, ⌬t , , ba, n, , by means of equally spaced points, a  t0  t1  t2  p  tn  b, Next, we choose the evaluation points c1, c2, p , cn lying in the subintervals [t 0, t 1],, [t 1, t 2], p , [t n1, t n], respectively, and compute the velocities of the maglev at these, points:, √(c1),, , √(c2),, , p,, , √(cn), , The average of these n numbers, √(c1)  √(c2)  p  √(cn), 1 n,  a √(ck), n, n k1, gives an approximation of the average velocity of the maglev over [a, b]. Since, n, , ba, ⌬t, , we can rewrite the expression in the form, n, n, 1, 1, 1 n, √(c, ), , √(c, ), , a k, a k, a √(ck)⌬t, n k1, b  a k1, b  a k1, ⌬t, , By letting n get larger and larger, we are approximating the average velocity of, the maglev using measurements of its velocity at more and more points over smaller

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4.5, , The Fundamental Theorem of Calculus, , 405, , and smaller time intervals. Intuitively, the approximations should improve with increasing n. This suggests that we define the average velocity of the maglev over the time, interval [a, b] to be, lim, , n→⬁, , n, 1, a √(ck)⌬t, b  a k1, , But by the definition of the definite integral, we have, lim, , n→⬁, , n, n, 1, 1, √(ck)⌬t , lim a √(ck)⌬t, a, b  a k1, b  a n→⬁ k1, , , , 1, ba, , b, , 冮 √(t) dt, a, , Thus, we are led to define the average velocity of the maglev over the time interval, [a, b] to be, 1, ba, , b, , 冮 √(t) dt, a, , More generally, we have the following definition of the average value of a function f, over an interval [a, b]., , DEFINITION Average Value of a Function, If f is integrable on [a, b], then the average value of f over [a, b] is the number, fav , , 1, ba, , b, , 冮 f(x) dx, , (1), , a, , If we assume that f is nonnegative, then we have the following geometric interpretation for the average value of a function over [a, b]. Referring to Figure 1, we see that, fav is the height of the rectangle with base lying on the interval [a, b] and having the, same area as the area of the region under the graph of f on [a, b]., y, y  f(x), , FIGURE 1, The area of the rectangle is, (b  a)fav  兰ab f(x) dx  area, of the region under the graph of f., , f av, , 0, , a, , b, , x, , Returning to the example involving the motion of the maglev, we see that if, we assume that √(t) 0 on [a, b], then the distance covered by the maglev over the, time period [a, b] is 兰ab √(t) dt, the area of the region under the graph of √ on [a, b]., But this area is equal to (b  a)√av, where √av is the average value of the velocity function √. Thus, we can cover the distance traveled by the maglev at a speed of √(t) ft/sec, from t  a to t  b by traveling at a constant speed, namely, at the average speed, √av ft/sec over the same time interval.

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406, , Chapter 4 Integration, , EXAMPLE 1 Find the average value of f(x)  4  x 2 over the interval [1, 3]., Solution, , Using Equation (1) with a  1, b  3, and f(x)  4  x 2, we find, fav , , y, y  f(x), f(c), f av  f(c), , 0, , a, , FIGURE 2, 1, fav , ba, , c, , b, , 冮 f(x) dx, a, , b, , b, , 冮 f(x) dx, , 1, ba, , a, , , , 1, 3  (1), , , , 1 20, a b, 4 3, , , , 5, 3, , 冮, , 3, , 1, , (4  x 2) dx, , Use the result of Example 2 in Section 4.4., , x, , If you look at Figure 1 again, you will see that there is a number c on [a, b] such, that f(c)  fav. (See Figure 2.), The following theorem guarantees that fav is always attained at (at least) one number in an interval [a, b] if f is continuous., , THEOREM 1 The Mean Value Theorem for Integrals, If f is continuous on [a, b], then there exists a number c in [a, b] such that, f(c) , , 1, ba, , b, , 冮 f(x) dx, a, , PROOF Since f is continuous on the interval [a, b], the Extreme Value Theorem tells, us that f attains an absolute minimum value m at some number in [a, b] and an absolute, maximum value M at some number in [a, b]. So m  f(x)  M for all x in [a, b]., By Property 6 of integrals we have, b, , m(b  a) , , 冮 f(x) dx  M(b  a), a, , If b, , a, then, upon dividing by (b  a), we obtain, m, , 1, ba, , b, , 冮 f(x) dx  M, a, , Because the number, 1, ba, , b, , 冮 f(x) dx, a, , lies between m and M, the Intermediate Value Theorem guarantees the existence of at, least one number c in [a, b] such that, f(c) , as was to be shown., , 1, ba, , b, , 冮 f(x) dx, a

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4.5, , The Fundamental Theorem of Calculus, , 407, , EXAMPLE 2 Find the value of c guaranteed by the Mean Value Theorem for Integrals for f(x)  4  2x on the interval [0, 2]., Solution The function f(x)  4  2x is continuous on the interval [0, 2]. Therefore,, the Mean Value Theorem for Integrals states that there is a number c in [0, 2] such that, 1, ba, , y, , b, , 冮 f(x) dx  f(c), a, , where a  0 and b  2. Thus,, , 4, y  4  2x, , 1, 20, , 3, 2, , 2, , 冮 (4  2x) dx  4  2c, 0, , but, f(1)  f av, , 1, , 2, , 冮 (4  2x) dx  4, , See Example 5a in Section 4.4., , 0, , 0, , 1, c1, , 2, , x, , 3, , So we have, , FIGURE 3, The number c  1 in [0, 2] gives, f(c)  fav as guaranteed by the Mean, Value Theorem for Integrals., , 1, (4)  4  2c, 2, or c  1. (See Figure 3.), , The Fundamental Theorem of Calculus, Part I, y, , Suppose that f is a continuous, nonnegative function defined on the interval [a, b]. If x, is any number in [a, b], let us put, , y  f(x), , x, , A(x) , , A(x), , 0, , a, , 冮 f(t) dt, a, , x, , x, , b, , FIGURE 4, A(x)  兰ax f(t) dt gives the area of the, region under the graph of f on [a, x]., , (We use the dummy variable t because we are using x to denote the upper limit of integration.) Since f is nonnegative, we can interpret A(x) to be the area of the region under, the graph of f on [a, x], as shown in Figure 4. Since the number A(x) is unique for each, x in [a, b], we see that A is a function of x with domain [a, b]., Let’s look at a specific example. Suppose that f(x)  x on the interval [0, 1]. If we, use the result of Example 3 in Section 4.4, with a  0 and b  x, we obtain, x, , A(x) , , y, , 冮 t dt  2 x, 1, , 0x1, , 2, , 0, , yt, , This result is also evident if you refer to Figure 5 and interpret the integral 兰0x t dt as, the area of the shaded triangle. Observe that, A¿(x) , , d, dx, , x, , 冮 t dt  dx a 2 x b  x  f(x), d, , 1, , 2, , 0, , so A(x) is an antiderivative of f(x)  x. Now if this result,, 0, , x, , FIGURE 5, The area of the triangle is, 1, 1 2, 2 (x)(x)  2 x ., , t, , d, dx, , x, , 冮 f(t) dt  f(x), a, , is true for all continuous functions f, then it is quite astounding because it provides a, link between the processes of differentiation and integration. Roughly speaking, this

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408, , Chapter 4 Integration, , equation says that differentiation undoes what integration does: The two operations are, inverses of one another. Thus, the two seemingly unrelated problems of differential calculus (that of finding the slope of a tangent line to a curve) and integral calculus (that, of finding the area of the region bounded by a curve) are indeed intimately related., As it turns out, the result is true. Because of its importance, it is called the Fundamental Theorem of Calculus., , THEOREM 2 The Fundamental Theorem of Calculus, Part 1, If f is continuous on [a, b], then the function F defined by, x, , F(x) , , 冮 f(t) dt, , axb, , a, , is differentiable on (a, b), and, x, , 冮 f(t) dt  f(x), , d, dx, , F¿(x) , , (2), , a, , PROOF Fix x in (a, b), and suppose that x  h is in (a, b), where h  0. Then, F(x  h)  F(x) , , 冮, , xh, , 冮, , x, , 冮, , xh, , x, , 冮 f(t) dt, , f(t) dt , , a, , , , a, , f(t) dt , , a, , , , 冮, , xh, , x, , f(t) dt , , x, , 冮 f(t) dt, , By Property 3, , a, , f(t) dt, , x, , By the Mean Value Theorem for Integrals there exists a number c between x and x  h, such that, , 冮, , xh, , f(t) dt  f(c) ⴢ h, , x, , Therefore,, F(x  h)  F(x), 1, , h, h, , 冮, , xh, , f(t) dt , , x, , f(c) ⴢ h,  f(c), h, , Next, observe that as h approaches 0, the number c, which is squeezed between x and, x  h, approaches x, and by continuity, f(c) approaches f(x). Therefore,, F(x  h)  F(x), 1,  lim, h→0, h, h→0 h, , F¿(x)  lim, , 冮, , xh, , f(t) dt  lim f(c)  f(x), h→0, , x, , which is the desired result., , EXAMPLE 3 Find the derivative of the function:, a. F(x) , , 冮, , x, , 1, , 1, 1  t2, , 3, , dt, , b. G(x) , , 冮 21  t, x, , 2, , dt

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4.5, , 409, , The Fundamental Theorem of Calculus, , Solution, a. The integrand, 1, , f(t) , , 1  t2, , is continuous everywhere. Using the Fundamental Theorem of Calculus, Part 1,, we find, F¿(x) , , d, dx, , 冮, , x, , 1, , dt  f(x) , , 2, 1 1  t, , 1, 1  x2, , b. The integrand 21  t 2 is continuous everywhere. Therefore,, G¿(x) , , d, dx, , , , 冮, , 3, , 21  t 2 dt , , x, , d, dx, , d, c, dx, , x, , 冮 21  t, , 2, , b, , dtd, , a, , 3, , a, , 冮 f(x) dx  冮 f(x) dx, b, , x, , 冮 21  t, , 2, , dt, , 3, ,  21  x 2, , EXAMPLE 4 If y , , 冮, , x3, , dy, ?, dx, , cos t 2 dt, what is, , 0, , Solution Notice that the upper limit of integration is not x, so the Fundamental Theorem of Calculus, Part 1, is not applicable as the problem now stands. Let’s put, u  x3, , du,  3x 2, dx, , so, , Using the Chain Rule and the Fundamental Theorem of Calculus, Part 1, we have, dy, dy du, d, , ⴢ, c, dx, du dx, du, , 冮, , u, , cos t 2 dtd ⴢ, , 0, , du, dx, ,  (cos u 2)(3x 2)  3x 2 cos x 6, , Fundamental Theorem of Calculus, Part 2, The following theorem, which is a consequence of Part 1 of the Fundamental Theorem of Calculus, shows how to evaluate a definite integral by finding an antiderivative, of the integrand, rather than relying on evaluating the limit of a Riemann sum, thus, simplifying the task greatly., , THEOREM 3 The Fundamental Theorem of Calculus, Part 2, If f is continuous on [a, b], then, b, , 冮 f(x) dx  F(b)  F(a), a, , where F is any antiderivative of f, that is, F¿  f., , (3)

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410, , Chapter 4 Integration, , PROOF Let G(x)  兰ax f(t) dt. By Theorem 2 we know that G is an antiderivative, of f. If F is any other antiderivative of f, then Theorem 1 in Section 4.1 tells us that F, and G differ by a constant. In other words, F(x)  G(x)  C. To determine C, we put, x  a to obtain, a, , F(a)  G(a)  C , , 冮 f(t) dt  C  C, a, , a, , 冮 f(x) dx  0, a, , Therefore, evaluating F at b, we have, b, , F(b)  G(b)  C , , 冮 f(t) dt  F(a), a, , from which we conclude that, b, , F(b)  F(a) , , 冮 f(x) dx, a, , When applying the Fundamental Theorem of Calculus, it is convenient to use the, notation, CF(x) D a  F(b)  F(a), , “F(x) evaluated at b minus F(x), evaluated at a.”, , b, , For example, by using this notation, Equation (3) is written, b, , 冮 f(x) dx  CF(x) D, a, , b, a, ,  F(b)  F(a), , Also, by the Fundamental Theorem of Calculus, if F(x)  C is any antiderivative, of f, then, b, , 冮 f(x) dx  CF(x)  CD, a, , b, a, ,  [F(b)  C]  [F(a)  C],  F(b)  F(a)  CF(x) D a, b, , This result shows that we can drop the constant of integration when we use the Fundamental Theorem of Calculus., From now on, thanks to the Fundamental Theorem of Calculus, Part 2, we can use, our knowledge for finding antiderivatives to help us evaluate definite integrals., , EXAMPLE 5 Evaluate, a., , 冮, , 2, , (x 3  2x 2  1) dx, , b., , 1, , 冮, , 4, , 21x dx, , 0, , c., , 冮, , p>2, , cos x dx, , 0, , Solution, 2, , a., , 冮 (x, 1, , 3, , 2, 1, 2,  2x 2  1) dx  c x 4  x 3  xd, 4, 3, 1, ,  a4 , b., , 冮, , 0, , 4, , 21x dx , , 冮, , 0, , 4, , 16, 1, 2, 1,  2b  a   1b , 3, 4, 3, 12, , 4, 4, 4, 4, 32, 2x 1>2 dx  c x 3>2 d  (4)3>2  (0) , 3, 3, 3, 3, 0

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4.5, , c., , 冮, , p>2, , cos x dx  Csin xD 0, , p>2, , 0, , The Fundamental Theorem of Calculus, , 411, , 101, , The next example shows how to evaluate the definite integral of a function that is, defined piecewise., , EXAMPLE 6 Evaluate, , 2, , 冮, , f(x) dx, where, , 2, , f(x)  e, , Solution The graph of f is shown in Figure 6. Observe that f is continuous on [2, 2]., Since f is defined by different rules for x in the two subintervals [2, 0) and [0, 2], we, use Property 3 of definite integrals to write, , y, 9, y  f(x), , 冮, , 5, , 2, , 2, , f(x) dx , , , 1, 2, , x, , 2, , FIGURE 6, , 冮, , 2, , f(x) dx , , 冮, , 0, , 2, , f(x) dx , , 冮, , 0, , 冮, , 0, , 2, , 2, , f(x) dx , , 冮, , 2, , f(x) dx, , 0, 2, , (x 2  1) dx , , 冮 (x, , 3, ,  1) dx, , 0, , 0, 2, 1, 1,  c x 3  xd  c x 4  xd, 3, 4, 2, 0, , 3, 2, , x 2  1 if x  0, x 3  1 if x 0, , 冮, , 8, 16,  0  a  2b  (4  2)  0 , 3, 3, , 2, , f(x) dx, , 0, , Evaluating Definite Integrals Using Substitution, The next two examples show how the method of substitution can be used to help us, evaluate definite integrals., , EXAMPLE 7 Evaluate, , 冮, , 2, , x2x 2  4 dx., , 0, , Solution, , Method I: Consider the corresponding indefinite integral, I, , 冮 x2x, , 2, ,  4 dx , , 冮 x(x, , 2, ,  4)1>2 dx, , Let u  x 2  4, so that du  2x dx or x dx  12 du. Substituting these quantities into, the integral gives, I, , 冮2 u, 1, , 1>2, , du , , 1 3>2, 1, u  C  (x 2  4)3>2  C, 3, 3, , Armed with the knowledge of the antiderivative of the function f(x)  x2x 2  4, we, can evaluate the given integral as follows:, , 冮, , 0, , 2, , 2, 1, 1, 1, 8, x2x 2  4 dx  c (x 2  4)3>2 d  (8)3>2  (4)3>2  (2 12  1), 3, 3, 3, 3, 0, , Solution Method II: Changing the Limits of Integration As before, we make the, substitution u  x 2  4, so that du  2x dx or x dx  12 du. Next, we make the following intuitive observation: The given integral has lower and upper limits of integration, 0 and 2, respectively, and hence a range of integration given by the interval [0, 2]. In

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412, , Chapter 4 Integration, , making the substitution u  x 2  4, the original integral is transformed into another, integral in which the integration is carried out with respect to the new variable u., To obtain the new limits of integration, we note that if x  0, then u  0  4  4., This gives the lower limit of integration when integrating with respect to u. Similarly,, if x  2, then u  4  4  8, and this gives the upper limit of integration. Thus, the, range of integration when the integration is performed with respect to u is [4, 8]. In, view of this, we can write, , 冮, , 2, , x(x 2  4)1>2 dx , , 0, , 冮, , 8, , 4, , , , 8, 1 1>2, 1, u du  c u 3>2 d, 2, 3, 4, , 1, 1, 8, (8)3>2  (4)3>2  (2 12  1), 3, 3, 3, , as was obtained earlier., , 冮, , EXAMPLE 8 Evaluate, , p>4, , cos3 2x sin 2x dx., , 0, , Solution Let u  cos 2x, so that du  2 sin 2x dx or sin 2x dx  12 du. Also, if, x  0, then u  1, and if x  p>4, then u  0, giving 1 and 0 as the lower and upper, limits of integration with respect to u. Making these substitutions, we obtain, , 冮, , 0, , p>4, , cos3 2x sin 2x dx , , 0, , 冮 u a2 dub, 1, , 3, , 1, , 1 0,   u4 `, 8 1, 1, 1,  0  a b , 8, 8, Note Do not let the fact that the limits of integration with respect to u run from 1 to, 0 alarm you. This is not uncommon when we integrate using the method of substitution. Of course,, 0, , 冮, , 1, , 1, u 3 a dub  , 2, , 1, , 冮 u a2 dub, 3, , 1, , 0, , b, , a, , 冮 f(x) dx  冮 f(x) dx, a, , b, , as you can verify., , Definite Integrals of Odd and Even Functions, The following theorem makes use of the symmetry properties of the integrand to help, us evaluate a definite integral., , THEOREM 4 Integrals of Odd and Even Functions, Suppose that f is continuous on [a, a]., a. If f is even, then, , 冮, , a, , a, , a, , b. If f is odd, then, , 冮, , f(x) dx  2, , a, , a, , f(x) dx  0., , 冮 f(x) dx., 0

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4.5, , 413, , The Fundamental Theorem of Calculus, , PROOF We write, , 冮, , a, , a, , f(x) dx , , 冮, , 0, , f(x) dx , , a, , 冮, , a, , 冮, , a, , f(x) dx  , , 0, , 冮, , a, , a, , f(x) dx , , 0, , 冮 f(x) dx, , (4), , 0, , For the integral, f(x) dx, , 0, , let’s make the substitution u  x, so that du  dx. Also, if x  0, then u  0, and, if x  a, then u  a. So, a, , 冮, , f(x) dx , , 0, , 冮, , a, , a, , f(u)(du)  , , 0, , 冮 f(x) dx, 0, , Therefore, Equation (4) can be written as, , 冮, , a, , a, , f(x) dx , , a, , 冮, , f(x) dx , , 0, , 冮, , a, , a, , 冮 [ f(x)  f(x)] dx, , f(x) dx , , 0, , (5), , 0, , If f is even, then f(x)  f(x), so, using Equation (5), we have, , 冮, , a, , a, , f(x) dx , , 冮, , a, , a, , [f(x)  f(x)] dx  2, , 0, , 冮 f(x) dx, 0, , If f is odd, then f(x)  f(x), so Equation (5) gives, , 冮, , a, , a, , a, , f(x) dx , , 冮 [f(x)  f(x)] dx  0, 0, , Figure 7 gives a geometric interpretation of Theorem 4. In Figure 7a the area of the, region under the graph of the nonnegative function f from a to 0 is the same as that, under the graph of f from 0 to a, so the area of the region under the graph of f from a, to a is equal to twice that from 0 to a. But each of these areas is given by an appropriate, integral, leading to the first result in the theorem. In Figure 7b the area of the region above, the graph of f and under the x-axis from a to 0 is equal to the area of the region under, the graph of f from 0 to a; the former is given by the negative of the integral from 0 to a., y, , y, , a, a, , FIGURE 7, The integral of (a) an even function, and (b) an odd function, , a, , (a), , ya, , 0, , a, , 0, , a, , x, , x, , a, , f(x) dx  2 y f(x) dx, , (b) y, , a, , a, , 0, , f(x) dx  0, , EXAMPLE 9 Evaluate, a., , 冮, , 1, , 1, , (x 2  2) dx, , 2, , sin x, , 2, , 21  x 2, , 冮, , b., , dx, , Solution, a. Here, f(x)  (x)2  2  x 2  2  f(x), so f is even. Therefore, by Theorem 4,, , 冮, , 1, , 1, , (x 2  2) dx  2, , 冮, , 0, , 1, , 1, 14, 1, 1, (x 2  2) dx  2a x 3  2xb `  2a  2b , 3, 3, 3, 0

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414, , Chapter 4 Integration, , b. Here,, f(x) , , sin(x), 21  (x), , 2, , , , sin x, 21  x 2, ,  f(x), , so f is odd. Therefore, by Theorem 4,, 2, , sin x, , 2, , 21  x 2, , 冮, , dx  0, , The Definite Integral as a Measure of Net Change, In real-world applications we are often interested in the net change of a quantity over, a period of time. For example, suppose that P is a function giving the population,, P(t), of a city at time t. Then the net change in the population over the period from, t  a to t  b is given by, P(b)  P(a), , Population at t  b minus population at t  a, , If P has a continuous derivative P¿ on [a, b], then we can invoke the Fundamental Theorem of Calculus, Part 2, to write, b, , 冮 P¿(t) dt, , P(b)  P(a) , , P is an antiderivative of P¿., , a, , Thus, if we know the rate of change of the population at any time t, then we can calculate the net change in the population from t  a to t  b by evaluating an appropriate definite integral., , EXAMPLE 10 Population Growth in Clark County Clark County in Nevada, dominated, by Las Vegas, is one of the fastest-growing metropolitan areas in the United States., From 1970 through 2000 the population was growing at the rate of, R(t)  133,680t 2  178,788t  234,633, , 0t4, , people per decade, where t  0 corresponds to the beginning of 1970. What was the, net change in the population over the decade from the beginning of 1980 to the beginning of 1990?, Source: U.S. Census Bureau., , Solution The net change in the population over the decade from the beginning of, 1980 (t  1) to the beginning of 1990 (t  2) is given by P(2)  P(1), where P denotes, the population in the county at time t. But P¿  R, so, 2, , P(2)  P(1) , , 2, , 冮 P¿(t) dt  冮 R(t) dt, 1, , 1, , 2, , , , 冮 (133,680t, , 2, ,  178,788t  234,633) dt, , 1, ,  C44,560t 3  89,394t 2  234,633tD 1, 2, ,  [44,560(23)  89,394(22)  234,633(2)],  [44,560  89,394  234,633],  278,371, so the net change is 278,371 people.

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4.5, , The Fundamental Theorem of Calculus, , 415, , More generally, we have the following result. We assume that f has a continuous, derivative, even though the integrability of f ¿ is sufficient., Net Change Formula, The net change in a function f over an interval [a, b] is given by, b, , f(b)  f(a) , , 冮 f ¿(x) dx, , (6), , a, , provided that f ¿ is continuous on [a, b]., As another example of the net change of a function, let’s consider the motion of, an object along a straight line. Suppose that the position function and the velocity function of the object are s and √, respectively. Since s¿(t)  √(t), Equation (6) gives, , 冮, , s(b)  s(a) , , b, , b, , s¿(t) dt , , a, , a, , the net change in the position of the object over the time interval [a, b]. This net change, of position is the displacement of the object between t  a and t  b. (Recall that this, result was also discussed in Section 4.4.), To calculate the distance covered by the object between t  a and t  b, we, observe that if √(t) 0 on an interval [c, d], then the distance covered by the object, between t  c and t  d is given by its displacement 兰cd √(t) dt. On the other hand, if, √(t)  0 on an interval [c, d], then the distance covered by the object between t  c, and t  d is given by the negative of its displacement, that is, by  兰cd √(t) dt. But,  兰cd √(t) dt  兰cd √(t) dt. Since, , √, , 冟 √(t) 冟  e, S1, 0, , 冮 √(t) dt, , a, , S3, S2, , b, , FIGURE 8, Displacement is 兰ab √(t) dt  area of, S1  area of S2  area of S3, and, distance covered is 兰ab 冟 √(t) 冟 dt  area, of S1  area of S2  area of S3., , t, , √(t), √(t), , if √(t) 0, if √(t)  0, , we see that in either case the distance covered by the object is obtained by integrating, the speed 冟 √(t) 冟 of the object. Therefore, the distance covered by an object between, t  a and t  b is, b, , 冮 冟 √(t) 冟 dt, , (7), , a, , Figure 8 gives a geometric interpretation of the displacement of an object and the, distance covered by an object., , EXAMPLE 11 A car moves along a straight road with velocity function, √(t)  t 2  t  6, , 0  t  10, , where √(t) is measured in feet per second., a. Find the displacement of the car between t  1 and t  4., b. Find the distance covered by the car during this period of time., Solution, a. Using Equation (6), we see that the displacement is, 4, , s(4)  s(1) , , 4, , 冮 √(t) dt  冮 (t, 1, , 2, ,  t  6) dt, , 1, , 4, 1, 1, 1,  c t 3  t 2  6td  10, 3, 2, 2, 1, , That is, at t  4 the car is 10 12 ft to the right of its position at t  1.

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416, , Chapter 4 Integration, , b. Writing √(t)  t 2  t  6  (t  2)(t  3), we see that √(t)  0 on [1, 2] and, √(t) 0 on [2, 4]. (See Figure 9.) Using the integral in (7), we see that the distance covered by the car between t  1 and t  4 is given by, , √, 14, 10, , 冮, , √  t2  t  6, , 6, , 4, , 冟 √(t) 冟 dt , , 1, , 2, 0, , 1, , 2, , 3, , 冮, , 2, , 4, , (√(t)) dt , , 1, , , , t, , 4, , 冮, , 2, , 冮 √(t) dt, 2, , 4, , (t 2  t  6) dt , , 1, , 冮 (t, , 2, ,  t  6) dt, , 2, , 2, 4, 1, 1, 1, 1,  c t 3  t 2  6td  c t 3  t 2  6td, 3, 2, 3, 2, 1, 2, , 6, , FIGURE 9, √(t)  0 if, √(t) 0 if, , t 僆 [1, 2], and, t 僆 [2, 4]., ,  14, , 5, 6, , or 14 56 ft., , 4.5, , CONCEPT QUESTIONS, , 1. Define the average value of a function f over an interval, [a, b]. Give a geometric interpretation., 2. State the Mean Value Theorem for Integrals. Give a geometric interpretation., 3. State both parts of the Fundamental Theorem of Calculus., 4. State the Net Change Formula, and use it to answer the following:, a. If water is flowing through a pipe at the rate of R ft3/min,, what does 兰tt12 R(t) dt measure, where t 1 and t 2 are measured in minutes with t 1  t 2?, , 4.5, , b. If an object is moving along a straight line with an acceleration of a(t) ft/sec2, what does 兰tt12 a(t) dt measure if, t 1  t 2?, 5. Suppose that a particle moves along a coordinate line with, a velocity of √(t) ft/sec. Explain the difference between, 兰ab √(t) dt and 兰ab 冟 √(t) 冟 dt., , EXERCISES, , 1. Let F(x)  兰2x t 2 dt., a. Use Part 1 of the Fundamental Theorem of Calculus to, find F¿(x)., b. Use Part 2 of the Fundamental Theorem of Calculus to, integrate 兰2x t 2 dt to obtain an alternative expression for, F(x)., c. Differentiate the expression for F(x) found in part (b),, and compare the result with that obtained in part (a)., Comment on your result., , 9. t(x) , , 2, , 11. F(x) , , 13., 15., , 冮, , 0, x, , 5. t(x) , , 冮, , 2, , 7. F(x) , , 冮, , x, , 1, t2  1, , dt, , 4. G(x) , , 冮, , 6. h(x) , , 冮, , sin 2t dt, , 8. G(x) , , 17., , t, dt, 1t  1, , 19., , 冮, , x2, , t sin t dt, , 12. G(x) , , 冮, , 2, , 4 dx, , 14., , sin t 2 dt, 5, , sin t 2, dt, t, 1x, , 冮, , (t 2  4) dt, , 冮, , (3t  2)2 dt, , 冮, , 1, dx, 1x, , 20., , x1, dx, 1x, , 22., , 2, 4, 1, , 21., , 0, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 冮, , 4, , 9, , 冮, , 0, , 2, 2, , 冮, , 1, 1, , t 2t 2  1 dt, , 1, 3, , x, , p, , x, , t2, dt, t1, , 冮, , 0, , 3, 1, , In Exercises 3–12, find the derivative of the function., 13t  5 dt, , cos x, , 10. h(x) , , x2, , In Exercises 13–32, evaluate the integral., , 2. Repeat Exercise 1 with G(x)  兰 13t  1 dt., , x, , 冮, , sin t, dt, t, , 1, , x, 0, , 3. F(x) , , 冮, , 1x, , 16., , (2x  3) dx, , 冮 (2  4u  u ) du, 2, , 0, , 18., , 冮, , 2, , 冮, , 2, , 3, x3, , 1, , dx, , 3x 4  2x 2  1, 2x 2, , 1, 0, , 冮 (t, 1, , 1>2, ,  t 5>2) dt, , dx

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4.5, 0, , 23., , 冮 1x(x  1)(x  2) dx, , 24., , 2, , 25., , sec2 t dt, , 26., , p>6, p, , 27., , 冮, , sin 2x cos x dx, , 冮, , 28., , dx, , 冮, , 30., , sin2 x cos2 x, , csc u cot u du, , 冮, , 冮, , p, , 冮, , p, , 2sin x  sin3 x dx, , x  1 if x  0, 2x 2  1 if x 0, , 1, , 2, , 4, , dx, , 34., , 0, , 35., , 冮, , 8t(t 2  1)7 dt, , 36., , 1, , 冮 15  u du, 3, , 38., , 1, , 39., , 冮, , 4, , 冮, , p, , 冮, , 1>4, , 1, , dx, , 40., , 冮, , 1>p, 3, , 47., , 冮, 冮, , dt, , 1, x, , 冮, , 2, , 冮, , p>2, , sin, x, , 2, , x, 2x 2  1, , p>2, , csc2 u cot u du, , p>6, p, , dx, , 46., , 冮, , p, , x 2 sin x, 21  x, , 2, , dx, , p>4, , p>4, , 1x, , 2, , 60. lim, , 1 n k 1>3, a a n b ; [0, 1], n k1, , 61. lim, , 2 n, 2k 2, a2, , b ; [2, 4], a, n k1, n, , 62. lim, , p n, kp, a cosa 2n b ;, 2n k1, , n→⬁, , n→⬁, , 5, , C0, p2 D, , In Exercises 63–68, find the average value fav of the function, over the indicated interval., 63. f(x)  2x 2  3x; [1, 2], 64. f(x)  1  1x; [0, 4], 65. f(x)  x2x 2  4; [0, 2], , 3, , tan x, , n, , 4, a k ; [0, 1], n k1, , n→⬁, , 1cos u sin u du, , 1, , 59. lim, , n→⬁, , dx, , sin 2x dx, , 冮, 冮, , Cp2 , pD, , 57. Let f(x)  2x  x 2  2x., a. Plot the graph of f., b. Find the x-intercepts of f accurate to three decimal places., c. Use the results of parts (a) and (b) to find the area of the, region under the graph of f and above the x-axis., , 0, , 44., , C0, p4 D, , 55. f(x)  sec2 x;, , In Exercises 59–62, evaluate the limit by interpreting it as the, limit of a Riemann sum of a function on the interval [a, b]., , 12x  1 dx, , x, dx, 1x  1  15x  1, , 0, , 48., , 42., , sec pt tan pt dt, , 1>4, 2>p, , 45., , 0.2, , p>4, p>2, , 1, cosa xb dx, 41., 2, p>2, 43., , 冮, , 5, , 1, , 1x( 1x  1)2, , 1, , ; [1, 2], , 54. f(x)  2  1x  1; [0, 3], , 1, , 4, , 37., , 冮 (t  1), 0, , 2, , x2, , 58. Let f(x)  x 2  x  cos x., a. Plot the graph of f., b. Find the x-intercepts of f accurate to three decimal, places., c. Use the results of parts (a) and (b) to find the area of the, region under the graph of f and above the x-axis., , x 2  1 if x  0, f(x) dx where f(x)  e, cos x, if x 0, , 冮 (3  2x), , 1, , 417, , 4, , In Exercises 33–48, evaluate the integral., 33., , 53. f(x) , , 56. f(x)  冟 sin x 冟;, , 冟 cos x 冟 dx, , 0, , f(x) dx where f(x)  e, , 1, p>2, , 32., , p>4, , 0, , p>3, , p>4, 1, , 31., , 冮, , (sin x  1) dx, , p>6, p, , 0, , 29., , p>2, , 0, , p>4, , 冮, , 冮, , The Fundamental Theorem of Calculus, , dx, , 49. a. Prove that 0 , , 冮, , 1, , x, , 5, , 3, 2, 1  x4, , 0, , dx , , 1, ., 6, , cas b. Use a calculator or a computer to find the value of the, , integral accurate to five decimal places., , 66. f(x) , , x, , ; [0, 3], 2x  1, 67. f(x)  sin x; [0, p], 2, , C p3 , p2 D, , 68. f(x)  csc2 x;, , 2, ., 2, 3, 0 24  3x  x, cas b. Use a calculator or a computer to find the value of the, integral accurate to five decimal places., , In Exercises 69–72, (a) find the number c whose existence is, guaranteed by the Mean Value Theorem for Integrals for the, function f on [a, b], and (b) sketch the graph of f on [a, b] and, the rectangle with base on [a, b] that has the same area as that, of the region under the graph of f., , In Exercises 51–56, find the area of the region under the graph, of f on [a, b]., , 69. f(x)  x 2  2x; [0, 1], , 51. f(x)  x  2x  2;, , 71. f(x)  1x  3;, , 50. a. Prove that 0 , , 冮, , 1, , 2, , 52. f(x)  x  x;, 3, , [0, 1], , dx, , [1, 2], , , , 70. f(x)  x 3; [0, 2], 72. f(x)  cos x;, , [1, 6], , Cp3 , p3 D

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418, , Chapter 4 Integration, , 73. Distance Covered by a Car A car moves along a straight road, with velocity function, √(t)  2t 2  t  6, , 74. Average Acceleration of a Car A car moves along a straight road, with velocity function √(t) and acceleration function a(t)., The average acceleration of the car over the time interval, [t 1, t 2] is, , Source: The Los Angeles Times., , 80. Water Level in Boston Harbor The water level (in feet) in Boston, Harbor during a certain 24-hr period is approximated by the, formula, p, H  4.8 sinc (t  10)d  7.6, 6, , 75. Velocity of a Falling Hammer During the construction of a highrise apartment building, a construction worker accidentally, drops a hammer that falls vertically a distance of h ft. The, velocity of the hammer after falling a distance of x ft is, √  12tx ft/sec, where 0  x  h. Show that the average, velocity of the hammer over this path is √  23 12th., 76. Flow of Water in a Canal Water at a depth of x ft in a wide rectangular canal flows at a velocity of, x, √  √0  20 1hs a b, h, , 2, , 0  t  24, , where t  0 corresponds to 12 A.M. What is the average, water level in Boston Harbor over the 24-hr period on that, day?, 81. Predator-Prey Populations The wolf and caribou populations in, a certain northern region are given by, P1(t)  8000  1000 sin, , pt, 24, , and, P2(t)  40,000  12,000 cos, , feet per second, where √0 is the velocity of the water on the, surface, h is the depth of the canal, and s is its gradient., Find the average velocity of flow in a cross section of the, canal., 77. Flow of Blood in an Artery The velocity (in centimeters per second) of blood r cm from the central axis of an artery is, given by √(r)  k(R2  r 2), where k is a constant and R is, the radius of the artery. Suppose that k  1000 and R  0.2., Find the average velocity of the blood across a cross section, of the artery., R, , 78. Hotel Occupancy The occupancy rate of the all-suite Wonderland Hotel, located near a theme park, is approximated by, the function, 10 3 10 2 200, t , t , t  56, 81, 3, 9, , 0t7, , where A(t) is measured in pollutant standard index and t is, measured in hours with t  0 corresponding to 7 A.M. What, is the average level of nitrogen dioxide present in the atmosphere from 7 A.M. to 2 P.M. on that day?, , √(t 2)  √(t 1), t2  t1, , Show that a is equal to the average value of a(t) on [t 1, t 2]., , r(t) , , A(t)  0.03t 3 (t  7)4  62.7, , 0t8, , where √(t) is measured in feet per second., a. Find the displacement of the car between t  0 and t  3., b. Find the distance covered by the car during this period of, time., , a, , that impairs breathing, present in the atmosphere on a certain June day in downtown Los Angeles is approximated by, , 0  t  12, , where t is measured in months with t  0 corresponding to, January 1. What is the average occupancy rate of the hotel, over the year?, 79. Air Pollution According to the South Coast Air Quality Management District, the level of nitrogen dioxide, a brown gas, , pt, 24, , respectively, at time t, where t is measured in months. What, are the average wolf and caribou populations over the time, interval [0, 6]?, 82. Daylight Hours in Chicago The number of hours of daylight at, any time t in Chicago is approximated by, L(t)  2.8 sinc, , 2p, (t  79)d  12, 365, , where t is measured in days and t  0 corresponds to January 1. What is the daily average number of hours of daylight, in Chicago over the year? Over the summer months from, June 21 (t  171) through September 20 (t  262)?, 83. Global Warming The increase in carbon dioxide in the atmosphere is a major cause of global warming. Using data, obtained by Dr. Charles David Keeling, professor at Scripps, Institution of Oceanography, the average amount of carbon, dioxide in the atmosphere from 1958 through 2007 is, approximated by, A(t)  0.010716t 2  0.8212t  313.4, , 1  t  50, , where A(t) is measured in parts per million volume (ppmv), and t in years with t  1 corresponding to the beginning of, 1958. Find the average amount of carbon dioxide in the, atmosphere from 1958 through 2007., Source: Scripps Institution of Oceanography.

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4.5, 84. Projected U.S. Gasoline Use The White House wants to cut, gasoline use from 140 billion gallons per year in 2007 to, 128 billion gallons per year in 2017. But estimates by the, Department of Energy’s Energy Information Agency suggest, that this will not happen. In fact, the agency’s projection of, gasoline use from the beginning of 2007 to the beginning of, 2017 is given by, A(t)  0.014t 2  1.93t  140, , 0  t  10, , where A(t) is measured in billions of gallons per year and t, is in years with t  0 corresponding to the beginning of, 2007., a. According to the agency’s projection, what will be the, gasoline consumption at the beginning of 2017?, b. What will be the average consumption per year from the, beginning of 2007 to the beginning of 2017?, Source: U.S. Department of Energy, Energy Information Agency., , 85. Air Purification To test air purifiers, engineers ran a purifier in, a smoke-filled 10-ft 20-ft room. While conducting a test, for a certain brand of air purifier, it was determined that the, amount of smoke in the room was decreasing at the rate of, R(t) percent of the (original) amount of smoke per minute,, t min after the start of the test, where R is given by, R(t)  0.00032t 4  0.01872t 3  0.3948t 2  3.83t  17.63, 0  t  20, How much smoke was left in the room 5 min after the start, of the test? How much smoke was left in the room 10 min, after the start of the test?, Source: Consumer Reports, , 86. Voltage in AC Circuits The voltage in an AC circuit is given by, V  V0 sin vt, a. Show that the average (mean) voltage from t  0 to, t  p>v (a half-cycle) is Vav  (2>p)V0, which is 2>p, 1 about 23 2 times the maximum voltage V0., b. Show that the average voltage over a complete cycle is 0., Explain., 87. If a feet of fencing are used to enclose a rectangular garden,, show that the average area of such a garden is a 2>24 ft2., 88. Find dx>dy if, , 冮, , x, , 13  2 cos t dt , , 0, , 冮, , y, , sin t dt  0, , 0, , 89. Find the x-coordinates of the relative extrema of the function, F(x) , , 冮, , 0, , x, , sin t, dt, t, , x, , 0, , 冮, , 0, , f(t) dt , , 1, , 冮 f(t) dt, x, , for every x 僆 (0, 1), , 419, , 91. Let, f(x)  e, , 1  x if 0  x  1, x  1 if 1  x  3, , a. Find F(x)  兰0x f(t) dt., b. Plot the graph of F, and show that it is continuous on, [0, 3]., c. Where is f differentiable? Where is F differentiable?, 1, h→0 h, , 92. Evaluate lim, 93. Evaluate, , 冮, , 1>2, , 冮, , 2h, , 25  t 2 dt., , 2, , x 7  2x 5  3x 3  2x 2  x  2, x2  1, , 1>2, 1, , 94. Evaluate, , 冮, , 2x 5  x 4  3x 3  2x 2  8x  1, x2  1, , 1, p>4, , 95. Evaluate, , 冮, , dx., dx., , (cos x  1) tan3 x dx., , p>4, , 96. Show that, , 冮, , 1, , 2x 2  1 sec x dx  2, , 1, , 冮, , 1, , 2x 2  1 sec x dx, , 0, , 97. a. Show that 兰0p x f(sin x) dx  (p>2) 兰0p f(sin x) dx., Hint: Use the substitution x  p  u., , b. Use the result of part (a) to evaluate 兰0p x sin x dx., p, f(x) cos nx dx, 98. a. If f is even, what can you say about 兰p, p, and 兰p f(x) sin nx dx if n is an integer? Explain., p, b. If f is odd, what can you say about 兰p, f(x) cos nx dx, p, and 兰p f(x) sin nx dx? Explain., , 99. Use the identity, sin 1 n  12 2 x, 2 sin 2x, , , , to show that, , 冮, , p, , 1,  cos x  cos 2x  p  cos nx, 2, sin 1 n  12 2 x, sin 2x, , 0, , dx  p, , 100. a. Show that if f is a continuous function, then, , 冮, , a, , a, , f(x) dx , , 0, , 冮 f(a  x) dx, 0, , and give a geometric interpretation of this result., b. Use the result of part (a) to prove that, , 冮, , p, , 0, , 90. Find all functions f on [0, 1] such that f is continuous on, [0, 1] and, x, , The Fundamental Theorem of Calculus, , sin 2kx, dx  0, sin x, , where k is an integer., c. Plot the graph of, f(x) , , sin 2kx, sin x, , for k  1, 2, 3, and 4. Do these graphs support the, result of part (b)?

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420, , Chapter 4 Integration, 103. Let f be continuous on (⬁, ⬁), and let c be a constant., Show that, , d. Prove that the graph of, f(x) , , sin 2kx, sin x, , 冮, , 101. A car travels along a straight road in such a way that the, average velocity over any time interval [a, b] is equal to, the average of its velocities at a and at b., a. Show that its velocity √(t) satisfies, , 冮, , b, , a, , (1), , b. Show that √(t)  ct  d for some constants c and d., Hint: Differentiate Equation (1) with respect to a and with, , a, , 4.6, , a, , In Exercises 104–107, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, explain why, or give an example to show why it is false., 104. Assuming that the integral exists, then, a, f(x 2) dx  2 兰0a f(x 2) dx., 兰a, 105. Assuming that the integral exists and that f is even, then, a, f(x 3) dx  2 兰0a f(x 3) dx., 兰a, 106. Assuming that the integral exists and that f is odd, then, a, f(x 3) dx  0., 兰a, , 102. Let f be continuous on (⬁, ⬁). Show that, , 冮 f(x  h) dx  冮, , 冮 f(cx) dx, , 107. Assuming that the integral exists and that f is even and t is, odd, then, , respect to b., , b, , b, , f(x) dx  c, , ca, , on [0, p] is antisymmetric with respect to the line, x  p>2 by showing that f 1 x  p2 2  f 1 x  p2 2 for, 0  x  p2 , and use this result to explain part (b)., , 1, √(t) dt  [√(a)  √(b)](b  a), 2, , cb, , 冮, , bh, , a, , a, , f(x) dx, , f(x)[t(x)]2 dx  2, , 冮, , a, , f(x)[t(x)]2 dx, , 0, , ah, , Numerical Integration, Approximating Definite Integrals, Table 1 gives the daily consumption of oil in the United States in millions of barrels,, in two-year intervals from 1987 through the year 2007. Suppose that we want to determine the average daily consumption of oil over the period in question. From our earlier work, we know that the solution is obtained by computing, 1, 20, , 冮, , 20, , f(t) dt, , 0, , where f(t) is the oil consumption in year t and t  0 corresponds to 1987. But the problem here is that we do not know the algebraic rule defining the integrand f for all values of t in [0, 20]. We are given its values only at a discrete set of points in that interval! Here, the Fundamental Theorem of Calculus cannot be used to help us evaluate, the integral, since we cannot find an antiderivative of f. Other situations also arise (for, example, f(t)  sin t 2) in which, although the integrand of a definite integral is defined, algebraically, we are not able to find its antiderivative in terms of elementary functions., In each of these situations the best we can do is to obtain an approximation to the definite integral. (We will return to the problem of finding the average daily consumption, of petroleum in Example 5.), TABLE 1, Year, , 1987 1989 1991 1993 1995 1997 1999 2001 2003 2005 2007, , Consumption, , 16.7, , 17.3, , 16.7, , 17.2, , Source: U.S. Energy Information Administration., , 17.7, , 18.6, , 19.5, , 19.6, , 20.0, , 20.8, , 20.7

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438, , Chapter 5 Applications of the Definite Integral, , 5.1, , Areas Between Curves, A Real-Life Interpretation, Two cars are traveling in adjacent lanes along a straight stretch of a highway. The velocity functions for Car A and Car B are √  f(t) and √  t(t), respectively. The graphs, of these functions are shown in Figure 1., √  f(t), , √, , √  t(t), , S, , FIGURE 1, The shaded area S gives the, distance that Car A is ahead, of Car B at time t  b., , A, B, , 0, , t, , b, , The area of the region under the graph of f from t  0 to t  b gives the total distance covered by Car A in b seconds over the time interval [0, b]. The distance covered by Car B over the same period of time is given by the area under the graph of t, on the interval [0, b]. Intuitively, we see that the area of the (shaded) region S between, the graphs of f and t on the interval [0, b] gives the distance that Car A will be ahead, of Car B at time t  b., Since the area of the region under the graph of f on [0, b] is, b, , 冮 f(t) dt, 0, , and the area of the region under the graph of t on [0, b] is, b, , 冮 t(t) dt, 0, , we see that the area of the region S is given by, b, , b, , b, , 冮 f(t) dt  冮 t(t) dt  冮 [ f(t)  t(t)] dt, 0, , 0, , 0, , Therefore, the distance that Car A will be ahead of Car B at t  b is, b, , 冮 [ f(t)  t(t)] dt, 0, , This example suggests that some applied problems can be solved by finding the, area of a region between two curves, which in turn can be found by evaluating an, appropriate definite integral. Let’s make this notion more precise., , The Area Between Two Curves, Suppose f and t are continuous functions with f(x)  t(x) for all x in [a, b], so that, the graph of f lies on or above that of t on [a, b]. Let’s consider the region S bounded, by the graphs of f and t between the vertical lines x  a and x  b as shown in Figure 2. To define the area of S, we take a regular partition of [a, b],, a  x0  x1  x2  x3  p  xn  b

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5.1, y, , xa, , y  f(x), , n, , a [f(ck)  t(ck)]⌬x, , y  g(x), b, , FIGURE 2, The region S between the graphs of f, and t on [a, b], , 439, , and form the Riemann sum of the function f  t over [a, b] with respect to this partition:, , xb, , S, , 0 a, , Areas Between Curves, , x, , k1, , where ck is an evaluation point in the subinterval [x k1, x k] and ⌬x  (b  a)>n. The, kth term of this sum gives the area of a rectangle with height [ f(ck)  t(ck)] and width, ⌬x. As you can see in Figure 3, this area is an approximation of the area of the subregion of S that lies between the graphs of f and t on [x k1, x k]., y, , (ck , f (ck )), , y, y  f (x), y  f(x), , f (ck)  g(ck), ck, 0, , a xk1, , xk, , y  g(x), , y  g(x), x, , b, , 0 a, , (ck , g(ck )), x, , FIGURE 3, The kth term of the Riemann sum of f  t gives, the area of the kth rectangle of width ⌬x., , x, , b, , FIGURE 4, The Riemann sum of f  t approximates the area of S., , Therefore, the Riemann sum provides us with an approximation of what we might, intuitively think of as the area of S (see Figure 4). As n gets larger and larger, we might, expect the approximation to get better and better. This suggests that we define the area, A of S by, n, , A  lim a [ f(ck)  t(ck)]⌬x, n→⬁, , (1), , k1, , Since f  t is continuous on [a, b], the limit in Equation (1) exists and is equal to the, definite integral of f  t from a to b. This leads us to the following definition of the, area A of S., , DEFINITION Area of a Region Between Two Curves, Let f and t be continuous on [a, b], and suppose that f(x)  t(x) for all x in, [a, b]. Then the area of the region between the graphs of f and t and the vertical lines x  a and x  b is, b, , A, , 冮 [ f(x)  t(x)] dx, , (2), , a, , Notes, 1. If t(x)  0 for all x in [a, b], then the region S is just the region under the graph, of f on [a, b], and its area is, b, , b, , 冮 [ f(x)  0] dx  冮 f(x) dx, a, , as expected (see Figure 5a)., , a

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440, , Chapter 5 Applications of the Definite Integral, , 2. If f(x)  0 for all x in [a, b], then the region S lies on or below the x-axis, and its, area is, , 冮, , b, , b, , [0  t(x)] dx  , , a, , 冮 t(x) dx, a, , This shows that we can interpret the definite integral of a negative function as the, negative of the area of the region above the graph of t on [a, b] (see Figure 5b)., y, , y, , xb, , xa, , xa, , xb, , y  f (x), S, 0, , y  g(x)  0, , a, , a, , 0, , x, , b, , y  f (x)  0, , b, , x, y  g(x), , S, , b, , FIGURE 5, , Historical Biography, , b, , (a) If g(x)  0 on [a, b], then ya f (x) dx, gives the area of S., , (b) If f(x)  0 on [a, b], then ya g(x) dx gives the, area of S., , The following guidelines are useful in setting up the integral in Equation (2)., , GILLES PERSONE DE ROBERVAL, (1602–1675), Born near Beauvais, France, Gilles Persone, (sometimes spelled Personier) took the, name de Roberval from the village where he, was born. He began studying mathematics, at the age of 14, later earned a living as a, traveling teacher of mathematics, and was, eventually appointed Chair of Mathematics, at the Royal College of France. This competitive position required the current chair to, periodically propose mathematical questions, and request solutions. If a submitted solution was found to be better than the solution presented by the current chair, he was, required to resign. Roberval was able to, keep the chair for the remainder of his life—, 41 years—but he published very few of his, discoveries, preferring to keep them private, so he could pose problems that others could, not solve. One of his many significant contributions was the discovery of a method of, deriving one curve from another that can be, applied to find finite areas between certain, curves and their asymptotes., , Finding the Area Between Two Curves, 1. Sketch the region between the graphs of f and t on [a, b]., 2. Draw a representative rectangle with height [ f(x)  t(x)] and width ⌬x, and note that its area is, ⌬A  [f(x)  t(x)]⌬x, 3. Observe that the height of the rectangle, [f(x)  t(x)], is the integrand in, Equation (2). The width ⌬x reminds us to integrate with respect to x. Thus,, b, , A, , a, , (See Figure 6.), , y, , xa, , xb, , (x, f(x)), , f(x)  g(x), , y  f(x), y  g(x), , 0, , FIGURE 6, The area of the vertical rectangle, is ⌬A  [ f(x)  t(x)]⌬x., , 冮 [ f(x)  t(x)] dx, , a, , b, (x, g(x)), x, , x

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5.1, , Areas Between Curves, , 441, , EXAMPLE 1 Find the area of the region between the graphs of y  x 2  2 and, , y  x  1 and the vertical lines x  1 and x  2., y, x  1, , x2, , Solution First, we make a sketch of the region and draw a representative rectangle., (See Figure 7.) Observe that the graph of y  x 2  2 lies above that of y  x  1., Therefore, if we let f(x)  x 2  2 and t(x)  x  1, then f(x)  t(x) on [1, 2]. Also,, from the figure we see that the area of the vertical rectangle is, , y  x2  2, , 5, yx1, , ⌬A  [f(x)  t(x)]⌬x  [(x 2  2)  (x  1)]⌬x  (x 2  x  3)⌬x, (upper function  lower function)⌬x, , 3, , So the area of the required region is, 2 1 0, , 1, , 2, , 3, , A, , x, , 4, , 冮, , b, , [f(x)  t(x)] dx , , a, , 2, , 冮, , 2, , 1, , (x 2  x  3) dx, , 2, 1, 1,  c x 3  x 2  3xd, 3, 2, 1, , FIGURE 7, The graph of f(x)  x 2  2 lies above, that of t(x)  x  1 on [1, 2]., , 8, 1, 1, 21,  a  2  6b  a   3b , 3, 3, 2, 2, , or 10, , 1, 2, , EXAMPLE 2 Find the area of the region bounded by the graphs of y  2  x 2 and, , y  x., x  1, , y, , Solution We first make a sketch of the desired region and draw a representative rectangle. (See Figure 8.) The points of intersection of the two graphs are found by solving the equations y  2  x 2 and y  x simultaneously. Substituting the second, equation into the first yields, , x2, , 2, y  2  x2, (1, 1), , 1, , x  2  x 2, 1, , 0, , 1, , 2, , x, , x2  x  2  0, , 1, 2, , (2, 2), y  x, , FIGURE 8, The graph of f(x)  2  x 2 lies above, that of t(x)  x on [1, 2]., , (x  1)(x  2)  0, giving x  1 and x  2 as the x-coordinates of the points of intersection. We can, think of the region in question as being bounded by the vertical lines x  1 and, x  2. This gives the limits of integration as a  1 and b  2 in Equation (2). Next,, if we let f(x)  2  x 2 and t(x)  x, then f(x)  t(x) on [1, 2], and the representative rectangle has area, ⌬A  [ f(x)  t(x)]⌬x  [(2  x 2)  (x)]⌬x  (x 2  x  2)⌬x, Therefore, the area of the required region is, A, , 冮, , a, , b, , [f(x)  t(x)] dx , , 冮, , 2, , 1, , (x 2  x  2) dx, , 2, 1, 1,  c x 3  x 2  2xd, 3, 2, 1, , 8, 1, 1, 27,  a  2  4b  a   2b , 3, 3, 2, 6, , or 4, , 1, 2

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442, , Chapter 5 Applications of the Definite Integral, , EXAMPLE 3 Find the area of the region bounded by the graphs of x  y 2 and, , y  x  2., y, , yx2, x  y2, (4, 2), , 2, 1, S1, , y  y2  2, , S2, , 0, , y2  y  2  0, 2, , 3, , 4, , 1, 2, , Solution The region S is shown in Figure 9. The points of intersection of the two, curves are found by solving the equations x  y 2 and y  x  2 simultaneously. Substituting the first equation into the second yields, , (1, 1), , FIGURE 9, The region S is the union of the two, nonoverlapping regions S1 and S2., , x, , (y  1)(y  2)  0, giving y  1 and y  2 as the y-coordinates of the points of intersection. The corresponding x-coordinates are x  1 and x  4., Observe that for 0  x  1 a representative rectangle lies between the half, parabola described by the function t(x)   1x (solve x  y 2 for y) and the half, parabola described by the function f(x)  1x; whereas for 1  x  4 a representative rectangle lies between the straight line y  h(x)  x  2 and the half parabola,, y  f(x)  1x. This observation suggests that we consider the area of S to be the, sum of the areas of S1 and S2., The area of a representative rectangle in S1 is, ⌬A  [ f(x)  t(x)]⌬x  [1x  ( 1x)]⌬x  21x ⌬x, Therefore, the area of the region S1 is, 1, , A1 , , 冮 21x dx, 0, , Similarly, the area of a representative rectangle in S2 is, ⌬A  [ f(x)  h(x)]⌬x  [1x  (x  2)]⌬x  ( 1x  x  2)⌬x, This tells us that the area of S2 is, 4, , A2 , , 冮 ( 1x  x  2) dx, 1, , Therefore, the area of the region S is, A1  A2  2, , 冮, , 1, , 4, , 1x dx , , 0, , 冮 ( 1x  x  2) dx, 1, , 1, , 4, 4, 2, 1,  c x 3>2 d  c x 3>2  x 2  2xd, 3, 3, 2, 0, 1, , , , 4, 16, 2, 1, 9,  a  8  8b  a   2b , 3, 3, 3, 2, 2, , or 4, , 1, 2, , Integrating with Respect to y, Sometimes it is easier to find the area of a region by integrating with respect to y rather, than with respect to x. Consider, for example, the region S bounded by the graphs of, x  f(y) and x  t(y), where f(y)  t(y), and the horizontal lines y  c and y  d,, where c  d, as shown in Figure 10.

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5.1, y, , x  f( y), , x  g(y), , yd, , d, , y, yc, , 0, , 443, , Observe that the condition f(y)  t(y) implies that the graph of f lies to the right, of the graph of t. Considering the horizontal rectangle of length [f(y)  t(y)] and, width ⌬y, we see that its area is, ⌬A  [ f(y)  t(y)]⌬y, , S, c, , Areas Between Curves, , This suggests that the area of S is, x, , d, , f(y)  g(y), , A, , FIGURE 10, The region S is bounded on the left by, the graph of x  t(y) and on the right, by that of x  f(y) on [c, d]., , 冮 [ f(y)  t(y)] dy, , (3), , c, , Since a rigorous derivation of Equation (3) proceeds along lines that are virtually, identical to that of Equation (2), it will be omitted here., , EXAMPLE 4 Find the area of the region of Example 3 by integrating with respect, to y., y  x  2 ( f (y)  y  2), , y, 2, , y, , 1, S, 0, 1, , 1, , 2, , 3, , x  y2 (g(y)  y2), 4, , x, , 2, , FIGURE 11, The horizontal rectangle has area, [ f(y)  t(y)]⌬y., , Solution We view the region S as being bounded by the graphs of the functions, f(y)  y  2 (solve y  x  2 for x), t(y)  y 2, and the horizontal lines y  1 and, y  2. See Figure 11. Observe that f(y)  t(y) for y in [1, 2]. The area of the representative horizontal rectangle is, ⌬A  [f(y)  t(y)]⌬y  [(y  2)  y 2]⌬y  (y  2  y 2)⌬y, (right function  left function)⌬y, , This implies that, A, , 2, , 2, 1, 1, (y  2  y 2) dy  c y 2  2y  y 3 d, 2, 3, 1, 1, , 冮, , 8, 1, 1, 9,  a2  4  b  a  2  b , 3, 2, 3, 2, , or 4, , 1, 2, , Note Sometimes we prefer to use Equation (3) instead of Equation (2) or vice versa., In general, the choice of the formula depends on the shape of the region. Often one, would integrate with respect to the variable that results in the minimal splitting of the, region. But sometimes the use of one formula leads to an integral(s) that is difficult to, evaluate, in which case the other formula should be used., , What Happens When the Curves Intertwine?, Sometimes we are required to find the area of a region S between two curves in which, the graph of one function f lies above that of another function t for some values of x, ( f(x)  t(x)) and lies below it for other values of x ( f(x)  t(x)). You will be asked, to give a physical interpretation of a problem involving precisely such a situation in, Exercise 46., To find the area of the region S, we divide it into subregions S1, S2, p , Sn, each of, which is described by the sole condition f(x)  t(x) or f(x)  t(x). Figure 12 illustrates the case in which n  3. We then use the guidelines developed earlier to calculate the area of each subregion. Adding up these areas gives the area of S. Thus, the

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444, , Chapter 5 Applications of the Definite Integral, , area of the region S shown in Figure 12 between the graphs of f and t and between the, vertical lines x  a and x  b is, c, , A, , d, , b, , 冮 [ f(x)  t(x)] dx  冮 [t(x)  f(x)] dx  冮 [ f(x)  t(x)] dx, a, , c, , d, , y, xa, , y  f (x), , xb, y  g(x), , S1, , FIGURE 12, The region S is the union of S1, where, f(x)  t(x); S2, where f(x)  t(x),, and S3, where f(x)  t(x)., , 0, , S2, , S3, , c, , x, , d, , Since, 冟 f(x)  t(x) 冟  e, , f(x)  t(x), t(x)  f(x), , if f(x)  t(x), if f(x)  t(x), , we can also write A in the abbreviated form, b, , A, , 冮 冟 f(x)  t(x) 冟 dx, , (4), , a, , When using Equation (4), however, we still need to determine the subintervals of [a, b], where f(x)  t(x) and/or where t(x)  f(x) and write A as the sum of integrals giving, the areas of the subregions on these subintervals., , EXAMPLE 5 Find the area of the region S bounded by the graphs of y  cos x and, y  (2>p)x  1 and the vertical lines x  0 and x  p., 2, y  πx  1, , y, 1, , y  cos x, xπ, , S1, 0, , Solution The region S is shown in Figure 13. To find the points of intersection of the, graphs of y  cos x and y  (2>p)x  1, we solve the two equations simultaneously., Substituting the first equation into the second, we obtain, 2, x1, p, , cos x , π, 2, , S2, , π, , x, , 1, , FIGURE 13, The area of S is the sum of the areas, of S1 and S2., , By inspecting the graphs, we see that x  p>2 is the only solution of the equation., Therefore, the point of intersection is 1 p2 , 0 2 . Let f(x)  cos x and t(x)  (2>p)x  1., Referring to Figure 13, we see that the areas A1 and A2 of the subregions S1 and S2 are, A1 , , 冮, , p>2, , 冮, , p>2, , [f(x)  t(x)] dx, , f(x)  t(x), , 0, , , , 0, , ccos x  a, ,  csin x , , 2, x  1b d dx , p, p>2, , 1 2, x  xd, p, 0, , 1, , 冮, , 0, , p>2, , acos x , , 2, x  1b dx, p, , 4p  p2  2p2, 4p, 1 p 2 p, a b  , , p 2, 2, 4p, 4

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5.1, , Areas Between Curves, , 445, , and, A2 , , 冮, , p, , 冮, , p, , [t(x)  f(x)] dx, , t(x)  f(x), , p>2, , , , p>2, , c, , a, , 2, x  1  cos xb dx, p, , p, 1 2, 1, 1 p 2 p, x  x  sin xd  c (p2)  p  0d  c a b   1d, p, p, p 2, 2, p>2, , pp, , p, p, 4p,  1, 4, 2, 4, , Therefore, the required area is, A  A1  A2 , , 4p, 4p, 4p, p, , , 2, 4, 4, 2, 2, , The following example, drawn from the field of study known as the theory of elasticity, gives yet another physical interpretation of the area between two curves., , EXAMPLE 6 Elastic Hysteresis Figure 14 shows a stress–strain curve for a sample, of vulcanized rubber that has been stretched to seven times its original length. The, function f whose graph is the upper curve gives the relationship between the stress and, the strain as the load (the stress) is applied to the material. Because the material is elastic, the rubber returns to its original length when the load is removed. However, when, the load is decreased, the graph of f is not retraced. Instead, the stress–strain curve, given by the graph of the function t is obtained., , FIGURE 14, A stress–strain curve for a, sample of vulcanized rubber: The, upper curve shows what happens, when the load is applied, and the, lower curve shows what happens, when the load is decreased., , Stress, , y (kg), , y  f (x), y  g(x), , 0, , Strain, , 700 x (%), , The lack of coincidence of the curves for increasing and decreasing stress is known, as elastic hysteresis. The graphs of f and t on the interval [0, 700] form the hysteresis, loop for the material. It can be shown that the area of the region enclosed by the hysteresis loop is proportional to the energy dissipated within the rubber. Thus, the elastic hysteresis of the rubber is given by, , 冮, , 700, , [ f(x)  t(x)] dx, , Since f(x)  t(x) on [0, 700], , 0, , Certain types of rubber have large hysteresis, and these materials are often used as, vibration absorbers. Most of the internal energy is dissipated in the form of heat, thereby, minimizing the transmission of the energy of vibration to the mediums to which the, machinery is mounted.

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446, , Chapter 5 Applications of the Definite Integral, , 5.1, , CONCEPT QUESTIONS, , 1. Write an expression that gives the area of the region completely enclosed by the graphs of f and t in Figures 1 and 2, in terms of (a) a single integral and (b) two integrals., y, y  g(x), , 2. Two cars start out side by side moving down a straight road., The velocity functions for Car A and Car B are f and t,, respectively. Their graphs are shown in Figure 3. Suppose, that f(t) and t(t) are measured in feet per second and t in, seconds, where t lies in the interval [0, 10]. Answer the following questions using definite integral(s) if appropriate., , y  f (x), , a, , 0, , b, , c, , y (ft/sec), y  f(t), , x, , y  g(t), , FIGURE 1, y, , 0, e, , x  f( y), , x, , FIGURE 2, , 5.1, , EXERCISES, , In Exercises 1–6, find the area of the shaded region., 1., , 32 1 0, , 4, yx1, , y  x  2 √x, , 1, 1, , 0, , x, , 2 3, , 1, , 4., , 2, , 3, , x, , 4, , y, y  x2, , y, , y, , x2, , 1, , 1, , 0, 1, , y  x3  4, 1, , 2, , y  x1/3, , 1, , yx2, , 4, 3, 2, 1, 2 1 0, , y  x2  4x, , 3, 2, , 2, 3, 4, 5, , 2., , 3. y, , y, 3, 2, 1, , 10 t (sec), , FIGURE 3, , x  g( y), 0, , 7, , a. By what distance is Car A ahead of Car B after 3 sec?, After 7 sec? After 10 sec?, b. Is one car always ahead of the other after the start of, motion?, c. What is the greatest distance between the two cars over, the 10-sec interval?, , d, , c, , 3, , x, , 2, 3, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 1, , x

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5.1, 5., , y, , y ($/week), x  2y2  1, , 0, , 1, , 447, , y  g(t), , y  x3, 1, , Areas Between Curves, , y  f(t), , x, , 1, , 6. x  y2/3 y, , 0, , t (week), , T, , T1, , x  y4  2y2  2, , In Exercises 9–38, sketch the region bounded by the graphs of, the given equations and find the area of that region., , 1, , 1, , 0 1, , 2, 9. y  x  3,, , x, , 2, , 10. y  x 3  1,, , 1, , 7. Oil Production Shortfall Energy experts disagree about when, global oil production will begin to decline. In the following, figure, the function f gives the annual world oil production, in billions of barrels from 1980 to 2050 according to the, U.S. Department of Energy projection. The function t gives, the world oil production in billions of barrels per year over, the same period according to longtime petroleum geologist, Colin Campbell. Find an expression in terms of definite integrals involving f and t giving the shortfall in the total oil, production over the period in question heeding Campbell’s, dire warnings., , 11. y  x 2  4,, , y  3x  4, , 12. y  x  4x,, , y  x  4, , 2, , 13. y  x  4x  3, y  x 2  2x  3, , Billions of barrels/year, , y  4  x2, , 14. y  (x  2)2,, 15. y  x,, , y  x3, , 16. y  x 2,, , y  x4, , 17. y  1x, y  x 2, 18. y  x 3  6x 2  9x, y  x 2  3x, 1, y   x  1,, 2, , 19. y  1x,, , 21. y , , 50, y  f(t), , y  f(t), , 1, , y  x 2,, , ,, , x2, , x  1,, , x4, , x3, , y  x1x  1, , 22. y  2x,, , 23. y  x  6x  5,, 2, , 30, , y  x2  5, , 24. y  x24  x 2, y  0, y  g(t), , 25. y , , 10, 0, , x1, , x1, , 2, , 60, , 20, , x  1,, , x  1,, , 20. y  21x  x, y   1x, , y, , 40, , y  x  1,, y  x  1,, , 1990 2000 2010 2020 2030 2040 2050, Year, Source: U.S. Department of Energy and Colin Campbell., , t, , 8. Rate of Change of Revenue The rate of change of the revenue, of Company A over the (time) interval [0, T] is f(t) dollars, per week, whereas the rate of change of the revenue of, Company B over the same period is t(t) dollars per week., Suppose the graphs of f and t are as depicted in the following figure. Find an expression in terms of definite integrals, involving f and t giving the additional revenue that Company B will have over Company A in the period [0, T]., , x, 216  x 2, , 26. x  y 2  1,, , ,, , y  0,, , x  0,, , 27. x  y ,, , x  y  3,, , 28. x  y 2,, , x  2y  3, , 2, , x3, , y  1, y  2, y  1,, , y2, , 29. y  x 3  x, y  x 4  1, 30. 1x  1y  1, x  y  1, 31. y  冟 x 冟,, , y  x2  2, , 32. y  sin x, y , , 2, p, x  1, x   ,, p, 2, , 33. y  sin 2x, y  cos x, x , , p, ,, 6, , x, , x, p, 2, , p, 2

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448, , Chapter 5 Applications of the Definite Integral, , 34. y  cos 2x,, , y  sin x,, , x  0, x , , 3p, 2, , picted in the figure. Let A1 and A2 denote the areas of the, regions shown shaded., , p, p, 35. y  sec2 x, y  2, x   , x , 4, 4, 36. y  sec2 x,, , p, y  cos x, x   ,, 3, , 37. y  2 sin x  sin 2x,, 38. x  sin y  cos 2y,, , y  0,, x  0,, , y (ft/sec), , x, , x  0,, , p, 3, , 40. Find the area of the region enclosed by the curve, y 2  x 2(1  x 2) ., In Exercises 41 and 42, use integration to find the area of the, triangle with the given vertices., 41. (0, 0) , (1, 6), (4, 2), 42. (2, 4), (0, 2), (6, 2), In Exercises 43 and 44, find the area of the region bounded, by the given curves (a) using integration with respect to x and, (b) using integration with respect to y., y  2x  4,, , 44. y  1x,, , y, , 1, x,, 2, , x0, y  1,, , T, , T1, , t (sec), , a. Write the number, T, , 冮 [t(t)  f(t)] dt  冮, T1, , T1, , [ f(t)  t(t)] dt, , 0, , in terms of A1 and A2., b. What does the number obtained in part (a) represent?, In Exercises 47–52, use a graphing utility to (a) plot the graphs, of the given functions and (b) find the x-coordinates of the points, of intersection of the curves. Then find an approximation of the, area of the region bounded by the curves using the integration, capabilities of the graphing utility., y  4  x4, , 48. y  x 3  3x 2  1, y  x 2  4, , y2, , R, Dollars per dollar, , 0, , 47. y  x 2,, , 45. Effect of Advertising on Revenue In the accompanying figure,, the function f gives the rate of change of Odyssey Travel’s, revenue with respect to the amount x it spends on advertising with its current advertising agency. By engaging the, services of a different advertising agency, Odyssey expects, its revenue to grow at the rate given by the function t. Give, an interpretation of the area A of the region S, and find an, expression for A in terms of a definite integral involving f, and t., , 0, , A1, , p, 2, , 39. Find the area of the region in the first quadrant bounded, by the parabolas y  x 2 and y  14 x 2 and the line y  2., , 43. y  x 3,, , A2, , xp, , y  0, y , , y  g(t), , y  f (t), , R  g(x), , 49. y  x 3  4x 2, y  x 3  9x, 50. y  x 4  2x 2  2, y  4  x 2, 51. y  x 2,, , y  sin x, , 52. y  cos x,, , y  冟x冟, , 53. Turbocharged Engine Versus Standard Engine In tests conducted, by Auto Test Magazine on two identical models of the, Phoenix Elite, one equipped with a standard engine and the, other with a turbocharger, it was found that the acceleration, of the former (in ft/sec2) is given by, a  f(t)  4  0.8t, , 0  t  12, , t sec after starting from rest at full throttle, whereas the, acceleration of the latter (in ft/sec2) is given by, , S, R  f (x), , b, , x ($), , 46. Two cars start out side by side and travel along a straight, road. The velocity of Car A is f(t) ft/sec, and the velocity of, Car B is t(t) ft/sec over the interval [0, T], where 0  T1  T., Furthermore, suppose that the graphs of f and t are as de-, , a  t(t)  4  1.2t  0.03t 2, , 0  t  12, , How much faster is the turbocharged model moving than, the model with the standard engine at the end of a 10-sec, test run at full throttle?, 54. Velocity of Dragsters Two dragsters start out side by side. The, velocity of Dragster A, VA, and the velocity of Dragster B,, VB, for the first 8 sec of the race are shown in the following, table, where VA and VB are measured in feet per second. Use, Simpson’s Rule with n  8 to estimate how far Dragster A, is ahead of Dragster B 8 sec after the start of the race.

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5.1, , t (sec), , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , VA (ft/sec), , 0, , 22, , 46, , 70, , 94 118 142 166 190, , VB (ft/sec), , 0, , 20, , 44, , 66, , 88 112 138 160 182, , 55. Surface Area of the Jacqueline Kennedy Onassis Reservoir The, reservoir located in Central Park in New York City has the, shape depicted in the figure below. The measurements, shown were taken at 206-ft intervals. Use Simpson’s Rule, with n  10 to estimate the surface area of the reservoir., , Areas Between Curves, , 449, , 57. Profit Functions The weekly total marginal cost incurred by, the Advance Visuals Systems Corporation in manufacturing, x 19-inch LCD HDTVs is, C¿(x)  0.000006x 2  0.04x  120, dollars per set. The weekly marginal revenue realized by the, company from the sale of x sets is, R¿(x)  0.008x  200, dollars per set., a. Plot the graphs of C¿ and R¿ using the viewing window, [0, 10,000] [0, 300]., b. Find the area of the region bounded by the graphs of C¿, and R¿ and the vertical lines x  2000 and x  5000., Interpret your result., 58. Find the area of the region bounded by the curve, y 2  x 3  x 2 and the line x  2., 59. Find the area of the region bounded by the graph of, f(x)  1x, the y-axis, and the tangent line to the graph of f, at (1, 1) ., 60. Find the number a such that the area of the region bounded, by the graph of x  (y  1)2 and the line x  a is 92., , 1030 ft 1498 ft 1910 ft 2304 ft 2323 ft, 1349 ft 1817 ft 1985 ft 2585 ft 1592 ft, , 61. Find the area of the region bounded by the x-axis and the, graph of f(x)  x 4  2x 3 and to the right of the vertical line, that passes through the point at which f attains its absolute, minimum., 62. The area of the region in the right half plane bounded by, the y-axis, the parabola y  x 2  2x  3, and a line tangent to the parabola is 83. Find the coordinates of the point, of tangency., , Source: The Boston Globe, , 56. Estimating the Rate of Flow of a River A stream is 120 ft wide., The following table gives the depths of the river measured, across a section of the river in intervals of 6 ft. Here, x, denotes the distance from one bank of the river, and y, denotes the corresponding depth (in feet). The average, rate of flow of the river across this section of the river, is 4.2 ft/sec. Use Simpson’s Rule to estimate the rate of, flow of the river., x (ft), , 0, , 6, , 12 18 24 30 36 42 48 54 60, , y (ft) 0.8 1.2 3.0 4.1 5.8 6.6 6.8 7.0 7.2 7.4 7.8, x (ft) 66, , 72, , 78, , 84, , 90, , 96 102 108 114 120, , y (ft) 7.6 7.4 7.0 6.6 6.0 5.1 4.3, , 3.2, , 2.2, , 1.1, , 63. The region S is bounded by the graphs of y  1x, the, x-axis, and the line x  4., a. Find a such that the line x  a divides S into two subregions of equal area., b. Find b such that the line y  b divides S into two subregions of equal area., 64. Find the value of c such that the parabola y  cx 2 divides, the region bounded by the parabola y  19 x 2, and the lines, y  2, and x  0 into two subregions of equal area., 65. Let A(x) denote the area of the region in the first quadrant, completely enclosed by the graphs of f(x)  x m and, t(x)  x 1>m, where m is a positive integer., a. Find an expression for A(m)., b. Evaluate lim m→1 A(m) and lim m→⬁ A(m). Give a geometric interpretation., c. Verify your observations in part (b) by plotting the, graphs of f and t.

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450, , Chapter 5 Applications of the Definite Integral, 1, and t(x)  冟 x 冟., x2  1, a. Plot the graphs of f and t using the viewing window, [1, 1] [0, 1.5]. Find the points of intersection of the, graphs of f and t accurate to three decimal places., b. Use a calculator or computer and the result of part (a) to, find the area of the region bounded by the graphs of f, and t., , 66. Let f(x) , , 67. The curve with equation y 2  4x 3  4x 4  0 is called a, piriform., a. Plot the curve using the viewing window, [1, 1] [1, 1]., b. Find the area of the region enclosed by the curve accurate to five decimal places., 68. The curve with equation 4y 2  4xy 2  x 2  x 3  0 is, called a right strophoid., a. Plot the curve using the viewing window, [1.5, 1.5] [0.5, 0.5]., b. Find the area of the region enclosed by the loop of the, curve., , 5.2, , In Exercises 69–72, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, explain why, or give an example to show why it is false., 69. If A denotes the area bounded by the graphs of f and t on, [a, b], then, b, , A2 , , 冮 [ f(x)  t(x)] dx, 2, , a, , 70. If f and t are continuous on [a, b] and 兰ab[ f(t)  t(t)] dt, then f(t)  t(t) for all t in [a, b]., , 0,, , 71. Two cars start out traveling side by side along a straight, road at t  0. Twenty seconds later, Car A is 30 ft behind, Car B. If √1 and √2 are continuous velocity functions for, Car A and Car B, respectively, where √1(t) and √2 (t) are, measured in feet per second, then, , 冮, , 0, , 20, , √2 (t) dt , , 冮, , 20, , √1(t) dt  30, , 0, , 72. Suppose that the acceleration of Car A and Car B along a, straight road are a1 (t) ft/sec2 and a2(t) ft/sec2, respectively,, over the time interval [t 1, t 2], where a1 and a2 are continuous, functions with a1 (t)  a2(t) on [t 1, t 2]. Then at time t  t 2,, Car A will be traveling 兰tt12[a1(t)  a2 (t)] dt ft/sec faster than, Car B. (Assume that t is measured in seconds.), , Volumes: Disks, Washers, and Cross Sections, In Section 5.1 we saw the role played by the definite integral in finding the area of, plane regions. In the next two sections we will see how the definite integral can be, used to help us find the volumes of solids such as those shown in Figure 1., , FIGURE 1, , (a) Wine barrel, , (b) Pyramid, , (c) Pontoon, , Figure 1c depicts a pontoon for a seaplane. In designing a pontoon, the engineer, needs to know the volume of water displaced by the part of the pontoon that lies, below the waterline in order to determine the buoyancy of the pontoon (Archimedes’, Principle).

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5.2, , Volumes: Disks, Washers, and Cross Sections, , 451, , Solids of Revolution, A solid of revolution is a solid obtained by revolving a region in the plane about a, line in the plane. The line is called the axis of revolution. For example, if the region, R under the graph of f on the interval [a, b] shown in Figure 2a is revolved about the, x-axis, we obtain the solid of revolution S shown in Figure 2b. Here, the axis of revolution of the solid is the x-axis., y, , y, , y  f(x), S, , R, 0, , FIGURE 2, , a, , b, , x, , x, , 0, , (a) Region R under the graph of f, , (b) Solid obtained by revolving R about the x-axis, , The Disk Method, To define the volume of a solid of revolution and to devise a method for computing, it, let’s consider the solid S generated by the region R shown in Figure 3a. Let, P  {x 0, x 1, p , x n} be a regular partition of [a, b]. This partition divides the region, R into n nonoverlapping subregions R1, R2, p , Rn. When these regions are revolved, about the x-axis, they give rise to the n nonoverlapping solids S1, S2, p , Sn, whose, union is S. (See Figure 3b.), y, , y, , y  f(x), , FIGURE 3, A partition of [a, b] produces, n subregions R1, R2, p , Rn, that are revolved about the x-axis, to obtain the n solids S1, S2, p , Sn, that together form S. (Here n  8.), , R 1 R2, 0, , x0  a x1 x2 · · ·, xk1, , (a) The region R, , Rk, , Rn, · · · xn  b, xk, , S1 S2 · · · Sk · · · Sn, x, , x, , 0, , (b) The solid S, , Let’s concentrate on the part of the solid of revolution that is generated by the region, Rk under the graph of f on the interval [x k1, x k]. This region is shown enlarged for the, sake of clarity in Figure 4. If ck is an evaluation point in [x k1, x k], then the area of Rk, is approximated by the rectangle of height f(ck) and width ⌬x  (b  a)>n. If this rectangle is revolved about the x-axis, it generates the disk Dk having radius f(ck) and width, ⌬x; therefore, its volume is, ⌬Vk  p[ f(ck)]2 ⌬x, , p(radius)2 ⴢ width

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452, , Chapter 5 Applications of the Definite Integral, Δx, , Δx, , Rk, , Δx, , Dk, , Sk, , f (ck), , xk1 ck xk, , FIGURE 4, The region Rk , shown shaded, is, approximated by the rectangle., The volume of Sk is approximated, by the volume of the disk Dk., , The kth region and the, approximating rectangle, , The kth solid of, revolution, , The kth disk, , The volume of Dk provides us with an approximation of the volume of Sk. Therefore, by approximating the volume of each solid S1, S2, p , Sn with the volume of a, corresponding disk D1, D2, p , Dn, we see that the volume V of S is approximated by, the sum of the volumes of these disks. (See Figure 5.) Thus,, n, , n, , V ⬇ a ⌬Vk  a p[f(ck)]2 ⌬x, k1, , k1, , y, , D1 D2, , FIGURE 5, The volume V of the solid of revolution, S is approximated by the sum of the, volume of the n disks D1, p , Dn., , · · · Dk · · · Dn, x, , 0, , Recognizing this sum to be the Riemann sum of the function pf 2 on the interval [a, b],, we see that, b, , n, , lim a p[f(ck)]2 ⌬x , n→⬁, k1, , 冮 p[f(x)] dx, 2, , a, , DEFINITION Volume of a Solid of Revolution, (Region revolved about the x-axis), Let f be a continuous nonnegative function on [a, b], and let R be the region, under the graph of f on the interval [a, b]. The volume of the solid of revolution, generated by revolving R about the x-axis is, n, , V  lim a p[ f(ck)]2 ⌬x , n→⬁, k1, , b, , 冮 p[f(x)] dx, 2, , (1), , a, , Just as we were able to recall the formulas for finding the area under a curve by, looking at the area of a representative rectangle, so can we recall Formula (1) by looking at the volume of the disk obtained by revolving a representative rectangle about, the x-axis., We proceed as follows: Having made a sketch of the region R under the graph of, y  f(x) on [a, b], draw a representative vertical rectangle of height f(x), or y, corre-

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5.2, , Volumes: Disks, Washers, and Cross Sections, , 453, , sponding to a value of x in [a, b], and width ⌬x. (See Figure 6.) We can regard this, disk with volume, ⌬V  p[ f(x)]2 ⌬x  py 2 ⌬x, , p(radius)2 ⴢ width, , as representing an element of volume of a solid. Now observe that the expression next, to ⌬x, py 2, is the integrand in Formula (1)., y, Δx, , y  f (x), , f (x) or y, R, , FIGURE 6, If a representative vertical rectangle is, revolved about the x-axis, it generates a, disk of radius f(x), or y, and width ⌬x., , 0, , a, , x, , b, , Δx, , Volume by Disk Method (Region revolved about the x-axis), Vp, , 冮, , b, , b, , [ f(x)]2 dx  p, , a, , 冮y, , 2, , dx, , f0, , a, , From now on, when we introduce a notion and/or derive a formula through the use, of Riemann sums, we will often use the heuristic approach of looking at a representative element associated with the general term of the Riemann sum (without the subscripts) to help us recall the appropriate formula., , EXAMPLE 1 Find the volume of the solid obtained by revolving the region under the, graph of y  1x on [0, 2] about the x-axis., Solution From the graph of y  1x sketched in Figure 7a, we see that the radius of, the representative disk corresponding to a particular value of x in [0, 2] (the height of, the representative rectangle) is y, or 1x. Therefore, the volume of the disk is, ⌬V  py 2 ⌬x, , Here y  f(x)  1x., ,  p( 1x) ⌬x  px(⌬x), 2, , y, , y, y  √x, S, y, R, , 0, Δx, , FIGURE 7, If R is revolved about the x-axis,, we obtain the solid of revolution S., , (a) The region R, , 2, , x, , 0, , (b) The solid S, , x

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454, , Chapter 5 Applications of the Definite Integral, , Summing the volumes of the disks and taking the limit, we find that the volume of the, solid is, V, , 冮, , 2, , 2, , px dx  p, , 0, , , , 冮 x dx, 0, , 2, 1, 1, px 2 `  p(4  0), 2, 2, 0, , or 2p, , EXAMPLE 2 By revolving the region under the graph of y  2r 2  x 2 on [r, r],, , show that the volume of a sphere of radius r is V  43 pr 3., , Solution The graph of y  2r 2  x 2 is a semicircle, as shown in Figure 8a. We can, see that the radius of a representative disk is y, the height of the vertical rectangle., Therefore, the volume of the disk is, ⌬V  py 2 ⌬x,  p(r 2  x 2)⌬x, , Since y  2r 2  x 2, , y, , y, , r, y  √r  x, 2, , 2, , y, R, r, , FIGURE 8, By revolving the region R about the, x-axis, we obtain the sphere of radius r., , Δx, , x, , r, , x, , (a) The region R, , (b) The solid S, , Summing the volumes of the disks and taking the limit, we obtain the required volume, as, V, , 冮, , r, , p(r 2  x 2) dx, , r, , p, , 冮, , r, , r, , (r 2  x 2) dx, , r, ,  2p, , 冮 (r, , 2, ,  x 2) dx, , 0, ,  2pcr 2x , , 1 3 r, x d, 3, 0, ,  2par 3 , , 1 3, r b, 3, , , , 4, pr 3, 3, , Use the symmetry of the region.

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5.2, , Volumes: Disks, Washers, and Cross Sections, y, , y, , d, , x, or, t(y), , x  t(y), , FIGURE 9, If a representative horizontal, rectangle is revolved about the, y-axis, it generates a disk of, radius t(y), or x, and width ⌬y, and hence volume ⌬V  px 2 ⌬y., , 455, , R, Δy, , Δy, c, x, , t(y), , 0, , x, , Formula (1) is used to find the volume of a solid of revolution when the axis of, revolution is the x-axis. To derive a formula for the volume V of a solid of revolution, obtained by revolving a region about the y-axis, consider the region R bounded by the, graphs of x  t(y), x  0, y  c, and y  d as shown in Figure 9., If R is revolved about the y-axis, then a representative horizontal rectangle (perpendicular to the axis of revolution) with length x, or t(y), and width ⌬y generates a, disk with volume, ⌬V  p[t(y)]2 ⌬y  px 2 ⌬y, Summing the volumes of the disks and taking the limit, we obtain the following formula., Volume by Disk Method (Region revolved about the y-axis), Vp, , 冮, , d, , d, , [t(y)]2 dy  p, , c, , 冮x, , 2, , t0, , dy, , c, , EXAMPLE 3 Find the volume of the solid obtained by revolving the region bounded, by the graphs of y  x 3, y  8, and x  0 about the y-axis., Solution The region R in question together with the solid generated by revolving that, region about the y-axis is shown in Figure 10. A representative horizontal rectangle, sweeps out a disk of radius x and width ⌬y. Therefore, its volume is, ⌬V  px 2 ⌬y,  p(y 1>3)2 ⌬y, , Solve y  x 3 for x., ,  py 2>3 ⌬y, y, , y, y8, , 8, R, , Δy, , FIGURE 10, If a horizontal rectangle is revolved, about the y-axis, it generates a disk of, radius t(y)  y 1>3, or x, and width ⌬y., , y  x3 or x  y1/3, x, , x, , 0, , x, , Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

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456, , Chapter 5 Applications of the Definite Integral, , If we sum the volume of these disks and take the limit, we find that the required volume is, 8, , Vp, , 冮y, , 2>3, , dy, , 0, , 3 5>3 8 3, py `  p(85>3), 5, 5, 0, , , , 96p, 5, , or, , The Washer Method, Let R be the region between the graphs of the functions f and t and between the vertical lines x  a and x  b, where f(x)  t(x)  0 on [a, b]. If R is revolved about, the x-axis, we obtain a solid of revolution with a hole in it. (See Figure 11.) Observe, that when a representative vertical rectangle between the curves is revolved about the, x-axis, the resultant element of volume of the solid has the shape of a washer with, outer radius f(x) and inner radius t(x). Therefore, the volume of this element is, ⌬V  p[ f(x)]2 ⌬x  p[t(x)]2 ⌬x, p(outer radius)2 ⴢ width  p(inner radius) 2 ⴢ width, ,  p{[ f(x)]2  [t(x)]2}⌬x, y, , y, Δx, , y  f(x), R, , y  t(x), t(x), , Δx, , FIGURE 11, When a vertical rectangle is revolved, about the x-axis, it generates a, washer of outer radius f(x),, inner radius t(x), and width ⌬x., , 0, , a, , x, , b, , x, , f(x), x, , Summing the volumes of the washers and taking the limit, we see that the volume V, of the solid S is given by the following., Volume by Washer Method (Region revolved about the x-axis), b, , Vp, , 冮 {[f(x)], , 2, ,  [t(x)]2} dx, , ft0, , a, , EXAMPLE 4 Find the volume of the solid obtained by revolving the region bounded, by y  1x and y  x about the x-axis., Solution The region bounded by y  1x and y  x is shown in Figure 12. The curves, y  1x and y  x intersect at (0, 0) and (1, 1) , as may be verified by solving the equations simultaneously. The outer and inner radius of the washer generated by the representative vertical rectangle shown are f(x)  1x and t(x)  x, respectively. Therefore,, its volume is, ⌬V  p{[ f(x)]2  [t(x)]2}⌬x,  p(x  x 2)⌬x, Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

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5.2, , Volumes: Disks, Washers, and Cross Sections, y, , y, 1, , (1, 1), , y  √x, , yx, , x, , FIGURE 12, If a vertical rectangle is revolved, about the x-axis, it generates a, washer with outer radius 1x,, inner radius x, and width ⌬x., , 0, , 457, , Δx, , √x, , x, , 1, , x, , Summing the volumes of the washers and taking the limit, we find that the required, volume is, 1, , V, , 冮 p(x  x ) dx, 2, , 0, , 1, , p, , 2, , 0, , Historical Biography, Science Source/Photo Researchers, Inc., , 冮 (x  x ) dx, , 1, 1, 1, 1, 1,  pc x 2  x 3 d  pa  b, 2, 3, 2, 3, 0, , or, , p, 6, , EXAMPLE 5 Find the volume of the solid generated by revolving the region of Example 4 about the line y  2., , EVANGELISTA TORRICELLI, (1608–1647), Evangelista Torricelli was born to a family, that had no money to educate him. Fortunately, his uncle was a Camaldolese monk,, who, through his connections with the, church, was able to send Torricelli to Rome, to study under Benedictine Benedetto, Castelli, a professor of mathematics at the, Collegio della Sapienza. Castelli introduced, Torricelli to the works of Galileo, which led, to Torricelli’s corresponding with Galileo, and eventually traveling to France to work, with him until Galileo’s death three months, later. Torricelli established many connections supporting the inverse relationship, between the tangent and quadrature problems that were later developed as differentiation and integration. He also invented, the barometer and a figure known as, Gabriel’s Horn (also called Torricelli’s Trumpet), which has infinite surface area but, finite volume. The name Gabriel’s Horn, refers to the horn blown by the biblical, archangel Gabriel to announce Judgment, Day, associating the infinite with the, divine., , Solution The region and the resulting solid of revolution are shown in Figure 13. If, a representative vertical rectangle is revolved about the line y  2, the resultant solid, is a washer with outer radius 2  x, inner radius 2  1x, and width ⌬x. Therefore,, its volume is, ⌬V  p[(2  x)2  (2  1x)2]⌬x, p[(outer radius)2  (inner radius) 2]⌬x, ,  p(x 2  5x  41x)⌬x, y, 4, , 3, y2, 1, y  √x, 0, yx, , Δx, , 1, , 2x, , 2  √x, , x, , √x, , x, , x, , FIGURE 13, If a vertical rectangle is revolved about the line y  2, it generates a washer with outer radius, 2  x, inner radius 2  1x, and width ⌬x., , Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

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458, , Chapter 5 Applications of the Definite Integral, , Summing the volumes of the washers and taking the limit, we find that the required, volume is, 1, , V, , 冮 p(x, , 2, ,  5x  4 1x) dx, , 0, , 1, 1, 5, 8,  pc x 3  x 2  x 3>2 d, 3, 2, 3, 0, , 1, 5, 8,  pa   b, 3, 2, 3, , or, , p, 2, , EXAMPLE 6 Find the volume of the solid generated by revolving the region of Example 4 about the y-axis., Solution The region together with the solid of revolution is shown in Figure 14. When, a horizontal rectangle is revolved about the y-axis, the resultant solid is a washer with, outer radius y, inner radius y 2, and width ⌬y. Therefore, the volume of the solid is, ⌬V  p(y 2  y 4)⌬y, y, , y, y  √x (x  y2), 1, , y, , yx, , FIGURE 14, If a horizontal rectangle is revolved, about the y-axis, it generates a, washer with outer radius y,, inner radius y 2, and width ⌬y., , y2, 0, , 1, , x, , 1, , x, , 0, , Summing the volumes of the washers and taking the limit, we find that the volume of, the solid is, V, , 冮, , 1, , 1, , p(y 2  y 4) dy  p, , 0, , 冮 (y, , 2, ,  y 4) dy, , 0, , 1, 1, 1, 1, 1,  pc y 3  y 5 d  pa  b, 3, 5, 3, 5, 0, , or, , 2p, 15, , The Method of Cross Sections, We now turn to the more general problem of defining the volume of an irregularly, shaped object. Consider, for example, the solid that is the part of a pontoon that lies, below the waterline. The side view of one such pontoon is shown in Figure 15. A cross, section of the pontoon (by a plane perpendicular to the x-axis) at the point x is shown, on the right., y, , b, , FIGURE 15, A(x) is the area of a cross, section of a pontoon at x., , a, , x, , x, , (waterline), A (x)

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5.2, , 459, , Volumes: Disks, Washers, and Cross Sections, , To find the volume of the pontoon, let’s take a regular partition P  {x 0, x 1, p , x n}, of the interval [a, b]. The planes that are perpendicular to the x-axis at the partition, points will slice the pontoon into “slabs” much like the way one slices a loaf of bread., The volume ⌬V of the kth slab between x  x k1 and x  x k is approximated by the, volume of the cylinder with constant cross-sectional area A(ck) and height ⌬x, where, ck lies in [x k1, x k]. (See Figure 16.) Thus,, ⌬V ⬇ A(ck)⌬x, , A(ck), , FIGURE 16, The volume of the kth “slab”, is approximately A(ck)⌬x., , Δx, , If we add up these n terms, we obtain an approximation to the volume V of the pontoon. We can expect the approximations to get better and better as n → ⬁ . Recognizing this sum to be the Riemann sum of the function A(x) on the interval [a, b], we are, led to the following definition., , DEFINITION Volume of a Solid with Known Cross Section, Let S be a solid bounded by planes that are perpendicular to the x-axis at x  a, and x  b. If the cross-sectional area of S at any point x in [a, b] is A(x), where, A is continuous on [a, b], then the volume of S is, b, , n, , V  lim a A(ck)⌬x , n→⬁, k1, , 冮 A(x) dx, , (2), , a, , EXAMPLE 7 A solid has a circular base of radius 2. Parallel cross sections of the, solid perpendicular to its base are equilateral triangles. What is the volume of the solid?, Solution Suppose that the base of the solid is bounded by the circle with equation, x 2  y 2  4. The solid is shown in Figure 17a, where we have highlighted a typical, cross section. To find the area of the cross section, observe that the base of the triangular cross section is 2y, as shown in Figure 17b. Using the Pythagorean Theorem, we, see that the height of the cross section is 13y (see Figure 17c). Therefore, the area, A(x) of a typical cross section is, A(x) , , 1, (2y)( 13y)  13y 2  13(4  x 2), 2, , y2  4  x 2, , y, y, , x2  y2  4, , 2, , (x, y), , x, , 2, , 2, , 2y, , √ 3y, , y, x, , 2y, , 2, , FIGURE 17, , (a) The solid, , (b) The base of a cross section, , (c) A cross section

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460, , Chapter 5 Applications of the Definite Integral, , Using Formula (2), we see that the volume of the solid is, V, , 冮, , 2, , 2, , A(x) dx , , 冮, , 2, , 2, , 2, , 13(4  x 2) dx  2, , 冮 13(4  x ) dx, 2, , 0, , The integrand is even., 2, 1, 32 13,  2 13 c4x  x 3 d , 3, 3, 0, , EXAMPLE 8 Find the volume of a right pyramid with a square base of side b and, height h., Solution Let’s place the center of the base of the pyramid at the origin as shown in, Figure 18a. A typical cross section of the pyramid perpendicular to the y-axis is a square, of dimension 2x by 2x. From Figure 18b we see by similar triangles that, hy, x, , b, h, 2, or, x, , b, (h  y), 2h, y, , y, , h–y, , (x,y), , (x,y), , h, y, , x, , FIGURE 18, , (a) A right pyramid, , x, , b, __, 2, , (b) A side view of the pyramid, , Therefore, the area of the cross section is, A(y)  (2x)(2x)  4x 2 , , b2, , (h  y)2, , h2, , The pyramid lies between y  0 and y  h. Therefore, its volume is, V, , 冮, , h, , A(y) dy , , 0, ,  c, , 冮, , 0, , h, , b2, 3h, , 2, , b2, h2, h, , (h  y)2 dy, , (h  y)3 d  0  a, 0, , b2, 3h, , 2, , b (h3) , , 1 2, b h, 3

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5.2, , 5.2, , 5.2, , 2. Write the integral that gives the volume of a solid using the, method of cross sections., , EXERCISES, 9., , In Exercises 1–12, find the volume of the solid that is obtained, by revolving the region about the indicated axis or line., 1. y, , 2., , 2, , 5., , 3 2 1, , 1, , 0, , x, , 1, , 4. y, , 0, , y  (x , , 2, , 1, , 3, 2, 1, , x, , 2, , 1, , 0, , y4x, , 2, , 1, , 0, , 1, , x, , 2, , 12. y, , 3, 2, , y, , y  32, , 3, 2, , 1, , yx, , yx, , y  x3, y  x3, x, , 1 2, , 0, , x, , 1, , 0, , 1, , 6. y, In Exercises 13–30, find the volume of the solid generated by, revolving the region bounded by the graphs of the equations, and/or inequalities about the indicated axis., , 1, y  x2, , x  tan y, , y  x3, , 13. y  x 2,, 14. y  x ,, 3, , 0, , x, , 2, , 11. y, , 4, 1)2, , y, π, 4, , 2, , x, , 2, , 2, 0, , y  x2, , 3, , 1, , 6, , 1, , y2  8x, , 4, , y  √1  x, , y, , 4, , y, , x  y  4y  5, , 3, , 1, 1, , 10., , y, 2, , y, , y  12 x, , 1, , 3., , x, , 1, , 0, , x, , 1, , y  0,, , x  2; the x-axis, , y  0,, , x  1; the x-axis, , 15. y  x  2x, y  0; the x-axis, 2, , 16. y  1x  1,, 7. y, , 8., , y, , 1, y  x2, , 4, y  x3, 2, , 1, 17. x  ,, y, , y  12 x2  2, y, , x  0,, , 18. x  y 3>2,, , x2, , y  0,, , x  2,, , y  1, y  2;, , x  5;, , the x-axis, , the y-axis, , x  0, y  1; the y-axis, , 19. x  24  y 2, x  0, y  0; the y-axis, 20. x  y 2  2y, x  0; the y-axis, , 0, , 461, , CONCEPT QUESTIONS, , 1. Write the integral that gives the volume of a solid of revolution using (a) the disk method and (b) the washer method., Illustrate each case graphically by drawing the region R,, indicating the axis of revolution, and drawing a representative rectangle that helps you to derive the formula., , 0, , Volumes: Disks, Washers, and Cross Sections, , 1, , x, , 2, , 0, , 2, , x, , 21. x 2  y 2  4,, 22. y , , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 1, (x  1)3>2, , x  0, y  2, y  2; the y-axis, ,, , y  0,, , x  0, x  3;, , the x-axis, , x

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462, , Chapter 5 Applications of the Definite Integral, , 23. x  y24  y 2, x  0; the y-axis, 24. y  1sin x,, 25. y  cos x,, , y  0,, x  0,, , p, ; the x-axis, 2, , x, , y  x; the x-axis, , 27. y  x ,, , y  1x; the x-axis, , 28. y  x ,, , y  2  x 2; the x-axis, , 2, , p, ; the x-axis, 2, , y  0, x , , 26. y  x 2,, 2, , 44. a. Find the volume of the solid (a prolate spheroid), generated by revolving the upper half of the ellipse, 9x 2  25y 2  225 about the x-axis., b. Find the volume of the solid (an oblate spheroid), generated by revolving the right half of the ellipse, 9x 2  25y 2  225 about the y-axis., 45. Find the volume of the solid obtained by revolving the, region enclosed by the curve y 2  14 (2x 3  x 4), where, y  0, about the x-axis., 46. Find the volume of the solid generated by revolving the, region enclosed by the astroid x 2>3  y 2>3  a 2>3 about the, x-axis., , 3, 29. x  y  1, y  x, y  0; the x-axis;, 2, (the smaller region), 2, , 2, , 2, , 2, , 2, , In Exercises 31 and 32 use a graphing utility to (a) plot the, graphs of the given functions and (b) find the approximate, x-coordinates of the points of intersection of the graphs. Then, find an approximation of the volume of the solid obtained by, revolving the region bounded by the graphs of the functions, about the x-axis., 1, 31. y  x 5,, 2, , y  2x  x, 2, , 3, , 32. y  x ,, 5, , 0, , a, , x, , a, , y  sin(x ), , 35. y  4  x 2,, y, , if 0  x  1, 1x, x 2  2x  2 if 1  x  2, , 48. Verify the formula for the volume of a right circular cone by, finding the volume of the solid obtained by revolving the triangular region with vertices (0, 0) , (0, r) , and (h, 0) about, the x-axis., , y  0; the line y  5, 1 2, x  2; the line y  5, 2, , 37. x  y 2  4y  5,, , fe, , Find the volume of the solid generated by revolving the, region under the graph of f on [0, 2] about the x-axis., , y  0; the line y  2, , y  x 2; the line y  2, , 38. y  x 2,, , a, , 47. The function f is defined by, , 33. y  x 2  2x,, , 36. y  x 2,, , a, , 2, , In Exercises 33–38, find the volume of the solid generated by, revolving the region bounded by the graphs of the equations, about the indicated line., 34. y  x,, , y, , 3, y  x; the y-axis; (the smaller region), 2, , 30. x  y  1,, 2, , x  2; the line x  1, , y 2  8x; the line x  2, , 49. Find the volume of a frustum of a right circular cone with, height h, lower base radius R, and upper radius r., r, , In Exercises 39–42, sketch a plane region, and indicate the axis, about which it is revolved so that the resulting solid of revolution has the volume given by the integral. (The answer is not, unique.), 39. p, , 1, , p>2, , 冮, , sin2 x dx, , 40. p, , 0, , 冮y, , 2>3, , dy, , R, , 0, , 1, , 41. p, , h, , 冮 (x, 0, , 1, , 2, ,  x 4) dx, , 42. p, , 冮 [(1), , 2, ,  (x 2  1)2] dx, , 0, , 43. Find the volume of the solid generated by revolving the, region enclosed by the graph of x 1>2  y 1>2  a 1>2 and, the coordinate axes about the x-axis., , 50. Verify the formula for the volume of a sphere of radius r by, finding the volume of the solid obtained by revolving the, region bounded by the graph of x 2  y 2  r 2, x  0, and, the y-axis about the y-axis.

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5.2, 51. Find the volume of a cap of height h formed from a sphere of, radius r., , Volumes: Disks, Washers, and Cross Sections, , 55. Cross section: an equilateral triangle, y, 4, , h, , y x  2, , R, , (1, 1), , r, , 463, , 1, , (2, 4), , y  x2, x, , 1 2, , 56. Cross section: a quarter circle, 52. Newton’s Wine Barrel Find the capacity of a wine barrel with, the shape of a solid that is obtained by revolving the region, bounded by the graphs of x  R  ky 2, x  0, y  h>2,, and y  h>2 about the y-axis., y, , y, 2, , y, R, 2, , _h, 2, , x  R  ky2, 0, , R, , x, , x, ,  _h2, , In Exercises 53–56, find the volume of the solid with the given, base R and the indicated shape of every cross section taken perpendicular to the x-axis., 53. Cross section: a square, , x  y2, 2, R, x, , 4, 2, , x  4  y2, , 57. The curve defined by y 4  1  0 x>2 0 4 is called a hyperellipse., a. Plot the curve using the viewing window, [3, 3] [2, 2]., b. Estimate the volume V of the solid obtained by revolving, the region enclosed by the hyperellipse for y  0 about, the x-axis., c. Use a calculator or computer to find V accurate to four, decimal places., Hint: The hyperellipse is almost rectangular in shape., , y, , 61. The base of a solid is the region bounded by the graphs of, y  4  x 2 and y  0. The cross sections perpendicular to, the y-axis are equilateral triangles. Find the volume of the, solid., , 2, , 2, , R, x2  y2  4, , 59. The curve defined by 2y 2  x 3  x 2  0 is called a, Tschirnhausen’s cubic., a. Plot the curve using the viewing window, [1.5, 1.5] [1.5, 1.5]., b. Find the volume of the solid obtained by revolving the, region enclosed by the loop of the curve about the x-axis., 60. A solid has a circular base of radius 2, and its parallel cross, sections perpendicular to its base are isosceles right triangles, oriented so that the endpoints of the hypotenuse of a triangle, lie on the circle. Find the volume of the solid., , 54. Cross section: a semicircle, , 2, , x, , 58. A solid has a circular base of radius 2, and its parallel cross, sections perpendicular to its base are rectangles of height 2., Find the volume of the solid., , y, , 2, , 4, , x, , 62. The base of a solid is an isosceles triangle with a base of, 6 in. and a height of 8 in. The cross sections perpendicular, to the altitude of the triangle are semicircles. Find the volume of the solid.

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464, , Chapter 5 Applications of the Definite Integral, , 63. The base of a wooden wedge is in the form of a semicircle, with radius a, and its top is a plane that passes through the, diameter of the base and makes a 45° angle with the plane, of the base. Find the volume of the wedge., , square meters measured from the front to the back of the, tank at 1-m intervals are summarized in the following table., x (distance, from front), , 0, , A(x), , 0, , x (distance, from front), , 5, , A(x), , 1, , 2, , 3, , 4, , 0.3041 0.6206 0.8937 0.8937, , 6, , 7, , 8, , 0.8937 0.6206 0.3041, , 0, , 64. The axes of two right cylinders, each of radius r, intersect at, right angles. Find the volume of the resulting solid that is, common to both cylinders. (The figure shows one eighth of, the solid.), Use Simpson’s Rule to estimate the capacity (in liters) of, the fuel tank., 67. The Volume of a Pontoon A pontoon is 12 ft long. The areas of, the cross sections in square feet measured from the blueprint, at intervals of 2 ft from the front to the back of the part of, the pontoon that is under the waterline are summarized in, the following table., , 65. Cavalieri’s Theorem Cavalieri’s Theorem states that if two, solids have equal altitudes and all cross sections parallel to, their bases and at equal distance from their bases have the, same area, then the solids have the same volume., , R1, , R2, , x, , 0, , 2, , 4, , 6, , 8, , 10, , 12, , A(x), , 0, , 3.82, , 4.78, , 3.24, , 2.64, , 1.80, , 0, , Use Simpson’s Rule to estimate the volume of the pontoon., , r, h, , (a) area of R1 = area of R2, , (b) An oblique circular cylinder, , a. Prove Cavalieri’s Theorem., b. Use Cavalieri’s Theorem to find the volume of the, oblique circular cylinder shown in part (b) of the figure., 66. Capacity of a Fuel Tank The external fuel tank for a fighter, aircraft is 8 m long. The areas of the cross sections in, , 68. a. Let S be a solid bounded by planes that are perpendicular, to the x-axis at x  0 and x  h. If the cross-sectional, area of S at any point x in [0, h] is A(x), where A is a, polynomial of degree less than or equal to three, show, that the volume of the solid is, V, , h, h, cA(0)  4Aa b  A(h)d, 6, 2, , b. Use the result of part (a) to verify the result of Exercise 50.

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474, , Chapter 5 Applications of the Definite Integral, center of the lake. Assume that readings taken along other, radial directions produce similar data, so that the capacity, of the lake can be approximated by the volume of the, solid obtained by revolving the region bounded above by, the x-axis and below by the graph of y  f(x) about the, y-axis. Use Simpson’s Rule to approximate this capacity., , 48. Land Reclamation A hill will be leveled, and the earth will be, used in a land reclamation project that includes the construction of additional landing strips for an existing airport. The, hill resembles the solid of revolution obtained by revolving, the region under the graph of the function f on [0, 240], about the y-axis. Use Simpson’s Rule to approximate the, amount of earth that can be recovered., , y (ft), , y (ft), 2, , 0, , 4, , 6, , 8, , 10, 20, , 46 45 41 40, , 30, , 10, , 12, , 14, , 36 29 25, , 16 18, 19, , 8, , 20, 160, , x (hundred ft), , 120, y  f(x), 80, , 40, 50, , 40, , 146, , 104, , 88, 20, , 44, 0, , 5.4, , 40, , 80, , 120, , 160, , 200, , 240, , x (ft), , Arc Length and Areas of Surfaces of Revolution, Upon leaving port, an oil tanker sails along a course given by the curve C shown in, Figure 1, where the port is taken to be located at the origin of a coordinate system., What is the distance traveled by the tanker when it reaches a point on the course that, is located 4 mi to the east and 2 mi to the north of the port?, N, W, y (mi), P(4, 2), , 2, 1, , FIGURE 1, The curve C gives the course, taken by an oil tanker., , O, , E, S, , C, , 1, , 2, , 3, , 4, , x (mi), , Intuitively, we see that this distance is given by the length of the curve C between, the points O and P. So to answer this question, we must (a) define what we mean by, the length of a curve and (b) devise a way of computing it. (We will solve this problem in Example 1.), , Definition of Arc Length, Suppose that C is the graph of a continuous function f on a closed interval [a, b]. Let, P  {x 0, x 1, p , x n} be a regular partition of [a, b]. If yk  f(x k), then the points, Pk(x k, yk) divide C into n arcs that we denote by +, P0P1 , +, P1P2 , p , +, Pn1Pn . (See Figure 2.)

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5.4, y, , Pk1, , y  f (x), P0, , 475, , Arc Length and Areas of Surfaces of Revolution, , Pk1Pk, , P2, , Pk, , Pn1Pn, , Pn, , Pn1, , P1, P1P2, , P0P1, , FIGURE 2, The graph of f on [a, b] is the union, of the arcs +, Pn1Pn ., P0P1 , +, P1P2 , p , +, , 0, , a  x0, , x1, , x2, , ..., , xk . . . xn1 xn  b, , xk1, , x, , Since these arcs are disjoint (except for their endpoints), we see that the length L, of C from P0 to Pn is just the sum of the lengths of these arcs. Now the length of the, arc +, Pk1Pk can be approximated by the length d(Pk1Pk) of the line segment joining, Pk1 and Pk (shown in red in Figure 2). Therefore, approximating the length of each, arc with the length of the corresponding line segment, we see that, n, , L ⬇ a d(Pk1Pk), k1, , This approximation improves as n gets larger and larger. This observation suggests that, we define the length of C as follows., , DEFINITION Arc Length of a Curve, Let f be a continuous function defined on [a, b], and let P  {x 0, x 1, p , x n} be, a regular partition of [a, b]. The arc length of the graph of f from P(a, f(a)) to, Q(b, f(b)) is, n, , L  lim a d(Pk1Pk), n→⬁, , (1), , k1, , if the limit exists., , Note, , L is also called the arc length of the graph of f on the interval [a, b]., , Length of a Smooth Curve, A function f is smooth on an interval if its derivative f ¿ is continuous on that interval., The continuity of f ¿ implies that a small change in x produces a small change in the, slope f ¿(x) of the tangent line to the graph of f at any point (x, f(x)). Consequently, the, graph of f cannot have an abrupt change in direction. In other words, the graph of f has, no cusps or corners and is a smooth curve. (See Figure 3.), y, , y, y  f(x), y  f(x), , 0, , FIGURE 3, , [, a, , (a) The function f is smooth., , ], b, , x, , 0, , [, a, , (b) The function f is not smooth., , ], b, , x

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476, , Chapter 5 Applications of the Definite Integral, , The length of the graph of a smooth function can be found by integration. To derive, a formula for finding the length L of such a graph, suppose f is a smooth function, defined on the closed interval [a, b] and P  {x 0, x 1, p , x n} is a regular partition of, [a, b]. Then by Equation (1),, n, , L  lim a d(Pk1Pk), n→⬁, k1, , Using the distance formula, we have, d(Pk1Pk)  2(x k  x k1)2  (yk  yk1)2,  2(x k  x k1)2  [ f(x k)  f(x k1)]2, , yk  f(x k), , Applying the Mean Value Theorem to f on the interval [x k1, x k], we see that, f(x k)  f(x k1)  f ¿(ck)(x k  x k1), where ck is a number in the interval (x k1, x k). Therefore,, d(Pk1Pk)  2(x k  x k1)2  [ f ¿(ck)(x k  x k1)]2,  2{1  [f ¿(ck)]2}(x k  x k1)2,  21  [ f ¿(ck)]2 ⌬x, , ⌬x  (b  a)>n, , So, n, , L  lim a d(Pk1Pk), n→⬁, k1, n, ,  lim a 21  [ f ¿(ck)]2 ⌬x, n→⬁, k1, , Recognizing this expression as the Riemann sum of the continuous function, t(x)  21  [f ¿(x)]2 leads to the following result., The Arc Length Formula, Let f be smooth on [a, b]. Then the arc length of the graph of f from P(a, f(a)), to Q(b, f(b)) is, b, , L, , 冮 21  [f ¿(x)] dx, 2, , (2), , a, , N, W, , y (mi), P(4, 2), , 2, , b, , L, , 冮 B 1  a dx b, dy, , 2, , dx, , (3), , a, , C, , 1, O, , E, S, , Note If the equation defining the function f is expressed in the form y  f(x), then, Equation (2) is sometimes written, , EXAMPLE 1 Distance Traveled by a Tanker The graph C of the equation y  14 x 3>2, 1, , 2, , 3, , 4, , FIGURE 4, The course taken by the oil tanker, , x (mi), , gives the course taken by an oil tanker after leaving port, which is taken to be located, at the origin of a coordinate system. (See Figure 4.) Find the distance traveled by the, tanker when it reaches a point on the course that is located 4 mi to the east and 2 mi, to the north of the port.

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5.4, , Arc Length and Areas of Surfaces of Revolution, , Solution The required distance is given by the length L of the curve C from x  0 to, x  4. To use Equation (3), we first find, , Historical Biography, WILLIAM NEILE, , dy, d 1 3>2, 3, , a x b  x 1>2, dx, dx 4, 8, , (1637–1670), Englishman William Neile, at the age of 19,, was the first to give what was called the, rectification of a curve. Rectification of a, curve is the concept of constructing a, straight line segment that is equal in length, to a given curve, and prior to Neile’s discovery in 1657 it was thought to be impossible, for algebraic curves. Neile’s work on this, appeared in mathematician John Wallis’s De, Cycloide in 1659 and was a major advance in, the development of what would become, infinitesimal calculus. Neile was a very talented and promising mathematician, but,, unfortunately, his personal life derailed his, professional life. His father would not, approve of Neile’s proposed marriage, and, Neile never recovered from the emotional, devastation this caused. He did not publish, any additional works and died at the age of, 32. The curve he worked with was y2  x3,, now known as the Neilian parabola., , and, 1a, , 2, dy 2, 3, 9, b  1  a x 1>2 b  1 , x, dx, 8, 64, , Then, 4, , L, , 冮B, 0, , a, , , 1a, , dy 2, b dx , dx, , 64 2, 9, b a b a1 , xb, 9, 3, 64, , 128, 9, c a1  b, 27, 16, , 4, , 冮 B 1  64 x dx, 9, , 0, , 3>2 4, , `, , 0, , 3>2, ,  1d , , 128 125, 122, a,  1b , ⬇ 4.52, 27 64, 27, , So the oil tanker will have traveled approximately 4.52 mi when it reaches the point, in question., , EXAMPLE 2 Find the length of the graph f(x) , , y, 12, f(x)  1 x3  1, 3, 4x, , 8, , 477, , 1 3, 1, x , on the interval [1, 3]., 3, 4x, , Solution The graph of f is sketched in Figure 5., We first find, f ¿(x) , , 4, , d 1 3 1 1, 1, c x  x d  x2  2, dx 3, 4, 4x, , Using Equation (2) with, 0, , 1, , 2, , 3, , x, , 4, , 1  [ f ¿(x)]2  1  ax 2 , , FIGURE 5, 1, 1, The graph of f(x)  x 3 , 3, 4x, , 4x, , 2, , 1, , , d, x  g(y), , b 1a, , 16x  8x  1, 8, , 4, , 16x, , y, , 4, , 4x 4  1, 4x, , 2, , b, , 2, , 16x  8x 4  1, 8, , , , 16x 4, , (4x 4  1)2, 16x 4, , we see that the required length is, L, , C, , 冮, , 3, , 冮, , 3, , 3, , 21  [f ¿(x)]2 dx , , 1, , c, 0, , 2, , 1, , , x, , FIGURE 6, The curve C is the graph of x  t(y), for c  y  d., , 冮B, , 1, , 1, , ax 2 , , (4x 4  1)2, 16x 4, , dx , , 冮, , 1, , 3, , 4x 4  1, 4x 2, , dx, , 1 2, 1, 1 3, 1, 1, 1, 53, x b dx  c x 3  d  a9  b  a  b , 4, 3, 4x 1, 12, 3, 4, 6, , By interchanging the roles of x and y in Equation (2), we obtain the following formula for finding the arc length of the graph of a smooth function defined by x  t(y), on the interval [c, d]. (See Figure 6.)

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478, , Chapter 5 Applications of the Definite Integral, , Arc Length: Integrating with Respect to y, , 冮, , L, , d, , d, , 21  [t¿(y)]2 dy , , c, , 冮B, c, , EXAMPLE 3 Find the length of the graph of x , , Q, , 1, , P, , ( , 2), 67, 24, , 1a, , ( , 1), 7, 12, , 1, , 2, , 3, , (4), , 1 3, 1, y , from P 1 127 , 1 2 to Q 1 67, 24 , 2 2 ., 3, 4y, , dx 2, 1 2, b  1  ay 2  2 b, dy, 4y,  1  y4 , , 0, , dx 2, b dy, dy, , Solution The graph of x  t(y) is shown in Figure 7. Here x is a function of y, so, we use Equation (4). First, we compute, , y, 2, , 1a, , x, ,  ay 2 , , FIGURE 7, The graph of the function, 1, 1, x  t(y)  y 3 , 3, 4y, , 1, 4y, , 1, 1, 1, 1, ,  y4  , 4, 2, 2, 16y, 16y 4, b, 2, , 2, , Then observing that y runs from y  1 to y  2 and using Equation (4), we find that, the required length is, 2, , L, , 冮B, 1, , ay 2 , , 1, 4y, , 2, , b dy , 2, , 2, , 冮 ay, , 2, , , , 1, , 1, 4y 2, , b dy, , 1, 1 2, 8, 1, 1, 1, 59,  c y3  d  c a  b  a  b d , 3, 4y 1, 3, 8, 3, 4, 24, , The Arc Length Function, y, s(x), , y  f (x), Q(x, f (x)), , Suppose that C is the graph of a smooth function f defined by y  f(x) on the closed, interval [a, b]. If x is a point in [a, b], we can use Equation (2) to express the length, of the arc of the graph of f from P(a, f(a)) to Q(x, f(x)). (See Figure 8.) Denoting this, length by s(x) (since it depends on x), we have, , C, , P(a, f(a)), , x, , s(x) , 0, , a, , x, , b, , x, , FIGURE 8, s(x) is the length of the arc of the graph, of f from P(a, f(a)) to Q(x, f(x))., , 冮 21  [ f ¿(t)] dt, 2, , a, , This equation enables us to define the following function., , DEFINITION Arc Length Function, Let f be smooth on [a, b]. The arc length function s for the graph of f is defined, by, x, , s(x) , , 冮 21  [ f ¿(t)] dt, 2, , a, , with domain [a, b]., , (5)

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5.4, , Arc Length and Areas of Surfaces of Revolution, , 479, , If we use the Fundamental Theorem of Calculus, Part 1, to differentiate Equation, (5), we obtain, s¿(x) , , d, dx, , 冮, , x, , 21  [f ¿(t)]2 dt  21  [ f ¿(x)]2 , , a, , B, , 1a, , dy 2, b, dx, , (6), , The quantity ds  s¿(x) dx is the differential of arc length. In view of Equation (6) we, can express ds in the following forms., , Arc Length Differentials, y, , ds, , dy, , (7), , (ds) 2  (dy)2  (dx)2, , (8), , B, , 1a, , or, equivalently,, , y  f(x), , dx, , 0, , dy 2, b dx, dx, , ds , , x, , FIGURE 9, The relationship between ds, dy, and dx, follows from the Pythagorean Theorem., , Figure 9 gives a geometric interpretation of the differential of arc length in terms, of the differentials dx and dy. Observe that if dx  ⌬x is small, then ds affords a, good approximation of the arc length of the graph of f corresponding to the change,, ⌬x, in x., , EXAMPLE 4 Use differentials to obtain an approximation of the arc length of the, graph of y  2x 2  x from P(1, 3) to Q(1.1, 3.52) ., Solution, , Using Equation (7), we find, ds , , B, , 1a, , dy 2, b dx  21  (4x  1)2 dx, dx, , Letting x  1 and dx  0.1, we obtain the approximation, ds ⬇ 21  52(0.1)  0.1126 ⬇ 0.51, Note The expression in Equation (8) provides us with an easy way of recalling the, formula for the arc length L of the graph of a function f on [a, b]. From, (ds)2  (dy)2  (dx)2, we see that, ds , , B, , 1a, , dy 2, b dx, dx, , and, x, , s(x) , , 冮 ds, a, , Therefore,, L  s(b) , , 冮, , a, , b, , b, , ds , , 冮B, a, , 1a, , dy 2, b dx , dx, , b, , 冮 21  [ f ¿(x)] dx, 2, , a

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480, , Chapter 5 Applications of the Definite Integral, , Surfaces of Revolution, A surface of revolution is a surface that is obtained by revolving the graph of a continuous function about a line. For example, if the graph C of the function f on the interval [a, b] shown in Figure 10a is revolved about the x-axis, we obtain the surface of, revolution S shown in Figure 10b., y, , y, , y = f(x), , C, S, 0, , FIGURE 10, If we revolve the graph of f, about the x-axis in (a), we, obtain the surface S in (b)., , a, , b, , x, , (a), , 0, , a, , b, , x, , (b), , Our immediate objective is to devise a formula for finding the surface area of S., To do this, we need the formula for the lateral surface area of a frustum of a right circular cone (Figure 11)., If the upper and lower radii of a frustum are r1 and r2, respectively, and its slant, height is l, then the surface area S of the frustum is, , r1, l, r2, , FIGURE 11, The frustum of a cone obtained, by cutting off its top using a, plane parallel to its base, , S  2prl, , (9), , where r  12 (r1  r2) is the average radius of the frustum. (You will be asked to establish this formula in Exercise 53.), Next, consider the surface S generated by revolving the graph C of a smooth nonnegative function f about the x-axis from x  a to x  b. Let P  {x 0, x 1, p , x n}, be a regular partition of [a, b]. If yk  f(x k) , then the points Pk(x k, yk) divide C into n, disjoint (except at their endpoints) arcs +, P0P1 , +, P1P2 , p , +, Pn1Pn whose union is C (Figure 12a). The surface S is the union of the surfaces S1, S2, p , Sn obtained by revolving these arcs about the x-axis (Figure 12b)., Pn1 Pn, y  f(x), Pk, P២P, , y, , y, , n1 n, , Pk1, , P0, , P1 P2, ២, P២, 0 P1 P1P2, 0, , FIGURE 12, (a) A partition of [a, b] produces n, arcs +, P0P1 , +, P1P2 , p , +, Pn1Pn ,, which, when revolved about the, x-axis, give n surfaces S1, S2, p , Sn,, which together form S. (b) Here, n  5., , x0  a, , x1, , ២P, Pk1, k, S1, x, , x2· · · xk1 xk· · · xn1, , 0, , xn  b, , (a), , (b), , a, , S2, , S3, , S4, , S5, b, , x

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5.4, , Arc Length and Areas of Surfaces of Revolution, , 481, , Let’s concentrate on the part of the surface generated by the arc of the graph of f, on the interval [x k1, x k]. This arc is shown in Figure 13. If ⌬x is small, then the arc, +, Pk1Pk may be approximated by the line segment joining Pk1 and Pk. This suggests, that the surface area of the frustum that is generated by revolving this line segment, about the x-axis will provide us with a good approximation of the surface area of Sk., (See Figure 13.), Pk, , Pk, , Pk1, , Pk1, , Pk1, , FIGURE 13, In part (a) the arc +, Pk1Pk is, approximated by the line segment, joining Pk1 to Pk. So the area of, Pk1Pk in, the surface generated by +, (b) is approximated by the lateral, surface area of the frustum generated, by the line segment in (c)., , Pk, , xk1 xk, , f (xk), , f (xk1), , Sk, x, , Δx, , (a), , (c), , (b), , Since the frustum has an average radius of r  12[f(x k1)  f(x k)] and a slant height, of l  d(Pk1Pk), Formula (9) tells us that its surface area is, ⌬S  2pc, , f(x k1)  f(x k), d d(Pk1Pk), 2, , But as in the computations leading to Equation (2), we have, d(Pk1Pk)  21  [f ¿(ck)]2 ⌬x, where ck is a number in the interval (x k1, x k). Also, if ⌬x is small, the continuity of f, implies that f(x k1) ⬇ f(ck) and f(x k) ⬇ f(ck) . Therefore,, ⌬S  2pc, , f(ck)  f(ck), d21  [f ¿(ck)]2 ⌬x, 2, , Approximating the area of each surface Sk by the area of the corresponding frustum,, we see that*, n, , S ⬇ a 2pf(ck)21  [ f ¿(ck)]2 ⌬x, k1, , This approximation can be expected to improve as n gets larger and larger. Finally, recognizing this sum to be the Riemann sum of the function t(x)  2pf(x)21  [f ¿(x)]2, on the interval [a, b], we see that, n, , lim a 2pf(ck)21  [ f ¿(ck)]2 ⌬x , n→⬁, k1, , b, , 冮 2pf(x)21  [f ¿(x)] dx, 2, , a, , This discussion leads to the following definition., , *It is conventional to denote the area of a surface by S, and we will do so here even though we have used, this very letter to denote the surface of revolution itself.

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482, , Chapter 5 Applications of the Definite Integral, , DEFINITION Surface Area of a Surface of Revolution, Let f be a nonnegative smooth function on [a, b]. The surface area of the surface obtained by revolving the graph of f about the x-axis is, b, , S  2p, , 冮 f(x)21  [f ¿(x)] dx, 2, , (10), , a, , Note, , If we use Equation (7), then we can write Equation (10) in the form, b, , S  2p, , 冮 y ds, , (11), , a, , which is the arc length differential form of Equation (10)., This formula can be remembered as follows: If ⌬x is small, the differential of arc, length, ds, gives an approximation of the slant height of the frustum of a cone of average radius approximated by y (or f(x)). So 2py ds represents an element of area of the, surface (Figure 14). By summing and taking the limit, we then obtain Equation (11)., y, ds, , f (x), 0, , x  Δx, , x, , FIGURE 14, If ⌬x is small, ds approximates, the slant height of the frustum,, and f(x) approximates the, average height of the frustum., , x, , Δx, , EXAMPLE 5 Find the area of the surface obtained by revolving the graph of, f(x)  1x on the interval [0, 2] about the x-axis., y, , Solution The graph of f and the resulting surface of revolution are shown in Figure 15., We have, , y  √x, , f ¿(x) , 0, , 1, , 2, , x, , 1, 21x, , Using Equation (10), we find that the required area is given by, 2, , S  2p, , 冮 f(x)21  [f ¿(x)] dx, 2, , 0, , FIGURE 15, The graph of y  1x on [0, 2] and, the resulting surface of revolution, obtained by revolving the graph, about the x-axis, ,  2p, , 冮, , 0, , 2, , 2, , p, , 1x, , B, , 1a, , 冮 14x  1 dx, 0, , 1 2, b dx  2p, 21x, , 2, , 冮 1x B 1  4x dx, 0, , 1

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5.4, , Arc Length and Areas of Surfaces of Revolution, , 483, , We evaluate this integral using the method of substitution with u  4x  1, so that, du  4 dx or dx  14 du. The lower and upper limits of integration with respect to u, are 1 and 9, respectively. We obtain, S, , p, 4, , 冮, , 9, , 1u du , , 1, , p 2 3>2 9 p, 2, 13p, c u d  a18  b , 4 3, 4, 3, 3, 1, , By interchanging the roles of x and y in Equation (10), we obtain the following formula for finding the area of the surface obtained by revolving the graph of a smooth, function defined by x  t(y) on the interval [c, d] about the y-axis., Surface Area: Integrating with Respect to y, S  2p, , 冮, , d, , d, , t(y)21  [t¿(y)]2 dy  2p, , c, , 冮 x ds, , (12), , c, , EXAMPLE 6 Find the area of the surface obtained by revolving the graph of x  y 3, on the interval [0, 1] about the y-axis., Solution Here, x  t(y)  y 3 and so t¿(y)  3y 2. Therefore, Equation(12) gives the, required surface area as, , y, , 1, , S  2p, , 2, , 0, , 1, ,  2p, xy, , 冮, , 1, , 1, , y 3 21  (3y 2)2 dy  2p, , 0, , 3, , 1, , FIGURE 16, The graph of x  y 3 on [0, 1] and, the surface obtained by revolving, it about the y-axis, , 5.4, , 冮 t(y)21  [t¿(y)] dy, , x, , 冮 y 21  9y, 3, , 4, , dy, , 0, , To evaluate the integral, we use the method of substitution with u  1  9y 4 so that, du  36y 3 dy. The lower and upper limits of integration are 1 and 10, respectively. We, obtain, S, , 2p, 36, , 冮, , 10, , 1u du , , 1, , p 2 3>2 10, p 2 3>2 2, p, c u d , a 10  b , (10 110  1), 18 3, 18, 3, 3, 27, 1, , The surface is shown in Figure 16., , CONCEPT QUESTIONS, , 1. a. Write an integral that gives the arc length of (1) a, smooth function y  f(x) on the interval [a, b] and (2) a, smooth function x  t(y) on the interval [c, d]., b. Write two different integrals that give the arc length L of, the curve defined by the equation y  x 2>3  1 from the, point P(0, 1) to the point Q(5 15, 4). Which integral, would you choose to compute L? Explain. Then use your, choice of integral to compute L., , 2. Write the formulas for finding the surface area of a surface, of revolution obtained by (a) revolving the graph of a nonnegative smooth function y  f(x) on the interval [a, b], about the x-axis and (b) revolving the graph of a smooth, function x  t(y) on the interval [c, d] about the y-axis.

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484, , Chapter 5 Applications of the Definite Integral, , 5.4, , EXERCISES, , In Exercises 1–4, find the arc length of the graph from A to B., 1., , 2., , y, , In Exercises 15–20, write an integral giving the arc length of the, graph of the equation from P to Q or over the indicated interval., (Do not evaluate the integral.), , y, , B (4, 9), 1.5, , 8, 6, , 3., , 4, 3, , ), , 1, , 2, , 3, , A (0, 0) 0.5, , x, , 4, , 4., , 6, , B (8, 5), , 4, , B, , 4, , 6, , ( , 2), 19, 12, , A, , 0, , 1, , 2, 3, , 2, , 3, , In Exercises 5 and 6, find the length of the line segment joining, the two given points by finding the equation of the line and, using Equation (2). Then check your answer by using the distance formula., 5. (0, 0) and (3, 8), , x, , 22. f(x)  2x 2  x 4; [0, 1], 23. f(x)  x  21x; [0, 4], 24. f(x) , , 9. y  2(x  1)3>2;, , P(1, 0), Q(5, 16), , 1, 1, 10. x  y 4  2 ;, 4, 8y, , P1 1 2, Q1, 3, 8,, , 2 2, (x  1)3>2;, 3, , [1, 4], , 12. y  (2  x 2>3)3>2;, , [1, 2], , 13. (y  3)  4(x  2) ;, 1, x3,  ;, 14. y , 3, 4x, , 1  x4, , ; [0, 1], , 25. The graph of the equation x 2>3  y 2>3  a 2>3, where a 0,, shown in the following figure, is called an astroid. Find the, arc length of the astroid., the curve joining P to Q(a, 0). To find the coordinates of P, find the, point of intersection of the astroid with the line y  x., y, , P(1, 5), Q(2, 1), P 1 4, 133 2 , Q(9, 17), , 2, , x2, , Hint: By symmetry the arc length is equal to 8 times the length of, , 2 3>2, x  1;, 3, , 11. y , , ; P 1 1, 12 2 , Q 1 2, 15 2, , 19. y  tan x; P(0, 0), Q 1 p4 , 1 2, , In Exercises 7–14, find the arc length of the graph of the given, equation from P to Q or on the specified interval., , 8. y , , x2  1, , [0, 1], , 18. y  cos x; [0, p], , 6. (1, 2) and (3, 6), , 7. y  2x  3;, , 1, , 21. f(x)  2x 3  x 4; [0, 2], , x, , 8, , 16. y  x  1;, , In Exercises 21–24, (a) plot the graph of the function f, (b) write, an integral giving the arc length of the graph of the function, over the indicated interval, and (c) find the arc length of the, curve accurate to four decimal places., , ( , 1), , 1, , P(1, 1), Q(2, 4), , 3, , 20. x  sec y; P 1 12, p4 2 , Q(1, 0), , 1, x 6 , 2y, , 2, A (0, 1), , 2.0 x, , 1.5, , y3, , 2, , y  x2/3  1, , 2, , 1.0, , y, 3, , 15. y  x 2;, , 17. y , , A (1, 2), , y, , 0, , (, , B 8 √9 3,, , 0.5, , 2, 0, , 1.0, , y  x3/2  1, , 4, , y  x2/3, , 3, , 129, 32 ,, , x2/3  y2/3  a2/3, , P, Q(a, 0), , 22, , P(2, 3), Q(2, 13), , yx, , a, , 0, , a, , a, , x, , a, , 26. Use the fact that the circumference of a circle of radius 1 is, 2p to evaluate the integral, , [1, 3], , 冮, , 0, , 12>2, , dx, 21  x 2, , Hint: Interpret 兰012>221  (y¿)2 dx, where y  21  x 2., , V Videos for selected exercises are available online at www.academic.cengage.com/login.

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5.4, In Exercises 27 and 28, use differentials to approximate the arc, length of the graph of the equation from P to Q., 27. y  x 3  1;, , P(1, 2), Q(1.2, 2.728), , 28. y  1x  1;, , Arc Length and Areas of Surfaces of Revolution, , 485, , 46. Find the area of the spherical zone formed by revolving the, graph of y  2r 2  x 2 on [a, b], where 0  a  b  r,, about the x-axis., , P(4, 3), Q(4.3, 3.074), , In Exercises 29–38, find the area of the surface obtained by, revolving the given curve about the indicated axis., 29. y , , r, , 1, x  2 for 0  x  2;, 2, , 30. y  1x on [4, 9];, 31. y  x on [0, 1];, 3, , 32. y  x, , 1>3, , on [1, 8];, , x-axis, , x-axis, x-axis, a, , y-axis, , 33. y  4  x 2 on [0, 2];, , b, , y-axis, , 47. A Pursuit Curve The graph C of the function, , 1 3, 1, y , for 1  y  2;, 6, 2y, , y-axis, , 35. 2x  3y  6 for 2  y  1;, , y-axis, , 34. x , , 36. y , , 1 4, 1, x  2 on [1, 2];, 4, 8x, , 37. y , , 1, 2x 2  x 4 on [0, 1];, 2 12, , x-axis, x-axis, , 1, 38. x  2y(3  y)2 on 0  y  3;, 3, , y, , x 1>2 4, x 3>2, 2, a1  b  2a1  b , 3, 2, 2, 3, , gives the path taken by Boat A as it pursues and eventually, intercepts Boat B (x  2). Initially, Boat A was at the origin,, and Boat B was at the point (2, 0), heading due north. Find, the distance traveled by Boat A during the pursuit., y (mi), , y-axis, C, , In Exercises 39 and 40, write an integral giving the area of the, surface obtained by revolving the curve about the x-axis. (Do not, evaluate the integral.), 39. y , , 1, on [1, 2], x, , A, , 40. y  sin x on C0, p2 D, , 41. Refer to Exercise 21. Use a calculator or computer to find, the area of the surface formed by revolving the graph of, f(x)  2x 3  x 4, where 0  x  2, about the x-axis, accurate to four decimal places., 42. Refer to Exercise 22. Use a calculator or computer to find, the area of the surface formed by revolving the graph of, f(x)  2x 2  x 4, where 0  x  1, about the x-axis, accurate to four decimal places., 43. Verify that the lateral surface area of a right circular cone of, height h and base radius r is S  pr2r 2  h2 by evaluating a definite integral., , B, , 0, , 2, , x (mi), , 48. Motion of a Projectile Refer to Exercise 29 in Section 2.4. A, projectile is fired from a cannon located on a horizontal, plane. If we think of the cannon as being located at the, origin O of an xy-coordinate system, then the path of the, projectile is, y  13x , , x2, 400, , where x and y are measured in feet. Estimate the distance, traveled by the projectile in the air., y (ft), , Hint: The cone is generated by revolving the region bounded by, y  (h>r)x, y  h, and x  0 about the y-axis., , 44. Verify that the surface area of a sphere of radius r is, S  4pr 2 by evaluating a definite integral., Hint: Generate this sphere by revolving the semicircle x 2  y 2  r 2,, , ¨, , where y  0, about the x-axis., , 45. Find the area of the surface obtained by revolving the graph, of y  24  x 2 on [0, 1] about the x-axis. This surface is, called a spherical zone., , O, , A, , x (ft)

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486, , Chapter 5 Applications of the Definite Integral, , 49. Flight Path of an Airplane The path of an airplane on its final approach to landing is described by the equation y  f(x) with, f(x)  4.3403, , 1010x 3  1.5625, , 105x 2  3000, 0  x  24,000, , 52. Let f be a smooth nonnegative function on [a, b]. Show that, the area of the surface obtained by revolving the graph of f, about the line y  L is given by, b, , S  2p, , 冮 冟 f(x)  L 冟21  [ f ¿(x)] dx, 2, , a, , where x and y are both measured in feet. Estimate the distance traveled by the airplane during the landing approach., , y, y  f (x), , 50. Area of a Roof A hangar is 100 ft long and has a uniform cross, section that is described by the equation y  10  0.0001x 4,, where both x and y are measured in feet. Estimate the area of, the roof of the hangar., , yL, , y, 0, , 10, x, , 0, , 51. Manufacturing Corrugated Sheets A manufacturer of aluminum, roofing products makes corrugated sheets as shown in the, figure. The cross section of the corrugated sheets can be, described by the equation, y  sina, , px, b, 10, , 0  x  30, , a, , 53. Show that the lateral surface area of a frustrum of a right, circular cone of upper and lower radii r1 and r2, respectively,, and slant height l is S  2prl, where r  12 (r1  r2) ., 54. Let L denote the length of the graph of y  f(x) connecting, the points (0, 0) and (l, 0), and let D  L  l (see the figure). Show that, 1, 2, , l, , 0, , 2, , dx  D , , 1, 2, , l, , 冮 (y¿), , 2, , dx, , 0, , assuming that y¿ is continuous on (0, l)., y, y  f (x), , 0, , 5.5, , (y¿)2, , 冮 21  (y¿), , where x and y are measured in inches.If the corrugated sheets, are made from flat sheets of aluminum using a stamping, machine that does not stretch the metal, find the width w of a, flat aluminum sheet that is needed to make a 30-in. panel., , w, , x, , b, , l, , x, , 30 in., , Work, The term work, as used in physics and engineering, is the transference of energy that, results when the application of a force causes a body to move. Scientists and engineers, need to know precisely how much energy is required to perform certain tasks. For, example, a rocket scientist needs to know the amount of energy required to put an artificial satellite into an orbit around the earth, and a power engineer needs to know the, amount of energy derived from water flowing through a dam., , Work Done by a Constant Force, We begin by defining the work done by a constant force in moving an object along a, straight line.

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5.5, , Work, , 487, , DEFINITION Work Done by a Constant Force, The work W done by a constant force F in moving a body a distance d in the, direction of the force is, W  Fd, , work  force ⴢ distance, , The unit of work in any system is the unit of force times the unit of distance. In the, English system the unit of force is the pound (lb), the unit of distance is the foot (ft),, and so the unit of work is the foot-pound (ft-lb). In the International System of Units,, abbreviated SI (for Système international d’unités), the unit of force is the newton (N),, the unit of distance is the meter (m), and so the unit of work is the newton-meter, (N-m). A newton-meter is also called a joule (J)., , EXAMPLE 1, a. Find the work done in lifting a 25-lb object 4 ft off the ground., b. Find the work done in lifting a 2.4-kg package 0.8 m off the ground. (Take, t  9.8 m/sec2.), Solution, a. The force F required to do the job is 25 lb (the weight of the object). Therefore,, the work done by the force is, W  Fd  25(4)  100, or 100 ft-lb., b. The magnitude of the force required is F  mt  (2.4)(9.8)  23.52, or 23.52 N., So the work done is, W  Fd  (23.52)(0.8) ⬇ 18.8, or 18.8 J., , Work Done by a Variable Force, Suppose that a body moves along the x-axis in the positive direction from x  a to, x  b under the action of a force F(x) that depends on x. Suppose also that the function F is continuous on the interval [a, b] with the graph depicted in Figure 1. Next,, let P  {x 0, x 1, p , x n} be a regular partition of [a, b]., y, y  F(x), , F(ck), , FIGURE 1, The graph of a variable force, defined by the function F, , ck, 0, , x0  a x1 x2 ... xk1 xk, , ..., , xn1 xn  b, , x, , Let’s concentrate on the subinterval [x k1, x k]. If ⌬x  (b  a)>n is small, then, the continuity of F guarantees that the values of F(x) at any two points in [x k1, x k]

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488, , Chapter 5 Applications of the Definite Integral, , do not differ by much. Therefore, if ck is any point in [x k1, x k], we can approximate, F(x) by F(ck) for all x in [x k1, x k]. Physically, we are saying that the force F(x) is, approximately constant when measured over a small distance. So if we assume that, F(x)  F(ck) in [x k1, x k], then the work done by F in moving the body along the, x-axis from x  x k1 to x  x k is, ⌬Wk ⬇ F(ck)⌬x, , constant force ⴢ distance, , (This is the area of a rectangle of height F(ck) and width ⌬x.) It follows that the work, W done by F in moving the body from x  a to x  b is, n, , W ⬇ a F(ck)⌬x, k1, , Intuitively, we see that the approximation improves as n gets larger and larger. This, suggests that we define the work done by F by taking the limit of the sum, n, , a F(ck)⌬x, k1, , as n → ⬁ . But this sum is just the Riemann sum of F on the interval [a, b]. Therefore,, our discussion leads to the following definition., , DEFINITION Work Done by a Variable Force, Suppose that a force F, where F is continuous on [a, b], acts on a body moving, it along the x-axis. Then the work done by the force in moving the body from, x  a to x  b is, b, , n, , W  lim a F(ck)⌬x , n→⬁, k1, , 冮 F(x) dx, , (1), , a, , Note When we derived Equation (1), we assumed that b, necessary and may be dropped., , a. This condition is not, , EXAMPLE 2 Find the work done by the force F(x)  3x 2  x (measured in pounds), , in moving a particle along the x-axis from x  2 to x  4 (measured in feet)., , Solution Here, F(x)  3x 2  x, so the work done by F in moving the body from, x  2 to x  4 is, W, , 冮, , 4, , 4, , F(x) dx , , 2, , 冮 (3x, 2, , 2, ,  x) dx  cx 3 , , 1 2 4, x d  72  10  62, 2, 2, , or 62 ft-lb., , Hooke’s Law, As another application of Equation (1), let’s find the work done in stretching or compressing a spring. Recall that Hooke’s Law states that the force F required to stretch, or compress a spring x units past its natural length is proportional to x. That is,, F(x)  kx, where k, the constant of proportionality, is called the spring constant, or the stiffness., Hooke’s Law is valid provided that 冟 x 冟 is not too large.

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5.5, , Work, , 489, , EXAMPLE 3 A force of 30 N is required to stretch a spring 4 cm beyond its natural, length of 18 cm. Find the work required to stretch the spring from a length of 20 cm, to a length of 24 cm., x, , 0, (a) The unstretched spring, , 30  k(0.04), , x, 0, , Solution Suppose that the spring is placed on the x-axis with the free end at the origin as shown in Figure 2. According to Hooke’s Law, the force F(x) required to stretch, the spring x meters beyond its natural length is F(x)  kx. Since a 30-N force is required, to stretch the spring 4 cm, or 0.04 m, beyond its natural length, we see that, , x, , x, , k  750, , or, , that is, 750 N/m. Therefore, F(x)  750x for this spring. Using Equation (1), we find, that the work required to stretch the spring from 20 cm to 24 cm is, , (b) The spring stretched x units beyond its, natural length, , W, , 冮, , 0.06, , 0.02, , FIGURE 2, , 0.06, 1, 750x dx  750c x 2 d, 2, 0.02, ,  375[(0.06) 2  (0.02)2]  1.2, or 1.2 J., , Moving Nonrigid Matter, The next two examples involve the computation of the work involved in moving nonrigid matter, such as the evacuation of fluid from a container and the hoisting of an, object., , EXAMPLE 4 A tank has the shape of an inverted right circular cone with a base of, radius 5 ft and a height of 12 ft. If the tank is filled with water to a height of 8 ft, find, the work required to empty the tank by pumping the water over the top of the tank., (Water weighs 62.4 lb/ft3)., Solution We think of the tank as being placed on a coordinate system with its vertex, at the origin and its axis along the y-axis as shown in Figure 3a. Think of the water as, being subdivided into slabs by planes perpendicular to the y-axis from y  0 to y  8., The volume ⌬V of a representative slab is approximated by a disk of radius x and width, ⌬y, that is,, ⌬V ⬇ px 2 ⌬y, You can see how to express x in terms of y by referring to Figure 3b. By similar triangles,, y, x, , 5, 12, , x, , or, , 5, y, 12, , y, 5, , 5, 12  y, 12, , 12 ft, , x, , 8 ft, , FIGURE 3, We wish to find the amount of work, required to pump all of the water, out of the top of the conical tank., , y, , Δy, x, (a), , y, x, (b)

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490, , Chapter 5 Applications of the Definite Integral, , so, ⌬V ⬇ pa, , 2, 5, 25p 2, yb ⌬y , y ⌬y, 12, 144, , Since water weighs 62.4 lb/ft3, or 62 25 lb/ft3, the weight of a representative slab (the, force required to lift this slab) is, ⌬F ⬇ 62, , 2 25p 2, 65p 2, ⴢ, y ⌬y , y ⌬y, 5 144, 6, , Since this slab is transported a distance of approximately (12  y) ft, the work done, by the force is, ⌬W ⬇ ⌬F(12  y), , , 65p 2, y (12  y) ⌬y, 6, , Finally, summing the work done in lifting each slab to the top of the tank and taking, the limit, we see that the work required to empty the tank is, W, , 冮, , 0, , 8, , , , , 65p 2, y (12  y) dy, 6, , 65p, 6, , 8, , 冮 (12y, , 2, ,  y 3) dy, , 0, , 8, 65p, 1, 65p, c4y 3  y 4 d , (1024), 6, 4, 6, 0, , or approximately 34,851 ft-lb., , EXAMPLE 5 A ship’s anchor, weighing 800 lb, is attached to a chain that weighs, 10 lb per running foot. Find the work done by the winch if the anchor is pulled in from, a height of 20 ft. (See Figure 4.), Solution The work done by the winch is W  WA  WC, where WA is the work, required to hoist the anchor to the top of the ship and WC is the work required to pull, the cable to the top of the ship. To find WA, observe that the force required to lift the, anchor is 800 lb and that it will be applied over a distance of 20 ft. Therefore,, WA  (800)(20)  16,000, or 16,000 ft-lb., y, , y, , y0, (20  y) ft, 20, , 20 ft, Δy, y, , FIGURE 4, The anchor is hoisted to the top of, the ship by means of a cable chain., , y0, (a), , 0, (b)

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5.5, , Work, , 491, , To find WC, think of the chain as being subdivided into pieces. The length of a representative piece is ⌬y ft, and its weight is 10 ⌬y lb (weight per running foot times, length). This element is to be lifted a distance of approximately (20  y) ft, so the, work required is, ⌬WC ⬇ 10 ⌬y(20  y), Summing the work done in lifting each piece of the chain to the top and taking the, limit, we see that the work required is, WC , , 冮, , 20, , 10(20  y) dy, , 0, , 1 2 20, y d  2000, 2, 0, ,  10c20y , , or 2000 ft-lb. So the work required to pull in the anchor from a height of 20 ft is, W  WA  WC  16,000  2,000  18,000, or 18,000 ft-lb., Note A 2-horsepower winch with a capacity of 1100 ft-lb/sec can pull in this anchor, in approximately 16 sec., , Work Done by an Expanding Gas, EXAMPLE 6 Figure 5 shows the cross section of a cylindrical casing of internal radius r. When the confined gas expands, the resulting increase in pressure exerts a force, against the piston, moving it and thus causing work to be done. If the confined gas has, a pressure of p lb/in.2 and the gas expands from a volume of V0 in.3 to V1 in.3, show, that the work done by the expanding gas is, W, , 冮, , V1, , p dV, , V0, , r, gas, , FIGURE 5, Work is done by the expanding gas against the piston., , Solution Draw the x-axis parallel to the side of the casing as shown in Figure 6, and, suppose that the piston has initial and final positions x  a and x  b, respectively., , FIGURE 6, The work done by the expanding, gas in moving the piston from, x to x  ⌬x is ⌬W, which is, approximately p(x)(pr 2)⌬x., , 0, , x, , x  Δx, , x

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492, , Chapter 5 Applications of the Definite Integral, , The force exerted by the expanding gas against the piston head at x (a  x  b) is, F(x)  p(x)(pr 2), , pressure ⴢ area, , so the work done by the force in moving the piston a distance of ⌬x from x to x  ⌬x is, ⌬W ⬇ p(x)(pr 2)⌬x, , constant force ⴢ distance, , Summing the work done by the force in moving the piston over each of the subintervals in the interval [a, b] and taking the limit, we see that the work done is, b, , W, , 冮 p(x)pr, , 2, , dx, , a, , To express this integral in terms of the volume of the gas, observe that the volume V, of the gas is related to x by V  pr 2x, so dV  pr 2 dx. Furthermore, observe that, when x  a, V  V0, and when x  b, V  V1. Therefore,, W, , 冮, , V1, , p dV, , V0, , 5.5, , CONCEPT QUESTIONS, , 1. a. A force of 3 lb moves an object along a coordinate line, from x  0 to x  10 (x is measured in feet). What is, the work done by the force on the object?, b. A force of magnitude 3 lb acts on an object in the negative direction with respect to a coordinate line as the, object moves from x  0 to x  10 (x is measured in, feet). What is the work done by the force on the object?, c. As an object moves in the coordinate plane from the, point A(0, 0) to the point B(10, 0) along the x-axis, a, , 5.5, , force of magnitude 5 lb acts on the body in the positive, y-direction. What is the work done by the force on the, object? Explain., 2. a. Can the work done on a body by a force be negative?, Explain with an example., b. A force acts on an object situated on a coordinate line. If, the work done by the force on the object is 0 ft-lb, does, this mean that the force has magnitude 0 and/or the distance moved by the object is 0 ft? Explain., , EXERCISES, , 1. Find the work done in lifting a 50-lb sack of potatoes to a, height of 4 ft above the ground., 2. How much work is done in lifting a 4-kg bag of rice to a, height of 1.5 m above the ground?, 3. A particle moves a distance of 100 ft along a straight line., As it moves, it is acted upon by a constant force of magnitude 5 lb in a direction opposite to that of the motion. What, is the work done by the force?, 4. An engine crane is used to raise a 400-lb engine a vertical, distance of 2 ft so that it can be placed in an engine dolly., Find the work done by the crane., 5. Find the work done by the force F(x)  2x  1 (measured, in pounds) in moving an object along the x-axis from, x  2 to x  4 (x is measured in feet)., , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 6. Find the work done by the force f(x)  4>x 2 (measured in, pounds) in moving a particle along the x-axis from x  1 to, x  6 (x is measured in feet)., 7. When a particle is at the point x on the x-axis, it is acted, upon by a force of x 2  2x newtons. Find the work done by, the force in moving the particle from the origin to the point, x  3 (x is measured in meters)., 8. A particle moves along the x-axis from x  1 to x  3. As, it moves, it is acted upon by a force F(x)  3x 2  x. If x, is measured in meters and F(x) is measured in newtons, find, the work done by the force., 9. When a particle is at the point x on the x-axis, it is acted, upon by a force of sin px newtons. Find the work done by, the force in moving the particle from x  1 to x  2 (x is, measured in meters).

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5.5, 10. A force of 8 lb is required to stretch a spring 2 in. beyond, its natural length. Find the work required to stretch the, spring 3 in. beyond its natural length., 11. A force of 20 N is required to stretch a spring 3 cm beyond, its natural length of 24 cm. Find the work required to stretch, the spring from 30 to 35 cm., 12. Suppose that it takes 3 J of work to stretch a spring 5 cm, beyond its natural length. How much work is required to, stretch the spring from 2 cm beyond its natural length to, 4 cm beyond its natural length?, 13. A spring has a natural length of 8 in. If it takes a force of, 14 lb to compress the spring to a length of 6 in., how much, work is required to compress the spring from its natural, length to 7 in.?, 14. A chain with length 5 m and mass 30 kg is lying on the, ground. Find the work done in pulling one end of the chain, vertically upward to a height of 2 m., 15. A chain weighing 5 lb/ft hangs vertically from a winch, located 12 ft above the ground, and the free end of the chain, is just touching the ground. Find the work done by the, winch in pulling in the whole chain., 16. A chain weighing 5 lb/ft hangs vertically from a winch, located 16 ft above the ground, and the free end of the chain, is 3 ft from the ground. Find the work done by the winch in, pulling in 4 ft of the chain., 17. A steel girder weighing 200 lb is hoisted from ground, level to the roof of a 60-ft building using a chain that, weighs 2 lb/running foot. Find the work done., , 20. Leaking Bucket A bucket weighing 4 lb when empty and, attached to a rope of negligible weight is used to draw, water from a well that is 30 ft deep. Initially, the bucket, contains 40 lb of water, but as it is pulled up at a constant, rate of 2 ft/sec, the water leaks out of the bucket at the rate, of 0.2 lb/sec. Find the work done in pulling the bucket to, the top of the well., 21. Leaking Bucket A bucket weighing 4 lb when empty and, attached to a rope of negligible weight is used to draw water, from a well that is 40 ft deep. Initially, the bucket contains, 40 lb of water and is pulled up at a constant rate of 2 ft/sec., , 493, , Halfway up, the bucket springs a leak and begins to lose, water at the rate of 0.2 lb/sec. Find the work done in pulling, the bucket to the top of the well., 22. A tank having the shape of a right circular cylinder with a, radius of 5 ft and a height of 6 ft is filled with water weighing 62.4 lb/ft3. Find the work required to empty the tank by, pumping the water out of the tank through a pipe that, extends to a height of 2 ft beyond the top of the tank., 23. A tank has the shape of an inverted right circular cone with, a base radius of 2 m and a height of 5 m. If the tank is filled, with water to a height of 3 m, find the work required to, empty the tank by pumping the water over the top of the, tank. (The mass of water is 1000 kg/m3.), 24. Consider the tank described in Exercise 23. If water is, pumped in through the bottom of the tank, find the work, required to fill the empty tank to a depth of 2 m., 25. Emptying a Storage Tank A gasoline storage tank in the shape, of a right cylinder of radius 3 ft and length 12 ft is buried in, the ground in a horizontal position. If the top of the tank is, 4 ft below the surface, find the work required to empty a, full tank of gasoline weighing 42 lb/ft3 by pumping it, through a pipe that extends to a height of 2 ft above the, ground., , 2 ft, 4 ft, 3 ft, , 18. An aquarium has the shape of a rectangular tank of length, 4 ft, width 2 ft, and height 3 ft. If the tank is filled with, water weighing 62.4 lb/ft3, find the work required to empty, the tank by pumping the water over the top of the tank., 19. A tank having the shape of a right-circular cylinder with a, radius of 4 ft and a height of 6 ft is filled with water weighing 62.4 lb/ft3. Find the work required to empty the tank by, pumping the water over the top of the tank., , Work, , 12 ft, , 26. Emptying a Trough An 8-ft-long trough has ends that are equilateral triangles with sides that are 2 ft long. If the trough is, full of water weighing 62.4 lb/ft3, find the work required to, empty it by pumping the water through a pipe that extends, 1 ft above the top of the trough., 1 ft, , 8 ft, 2 ft

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494, , Chapter 5 Applications of the Definite Integral, , 27. Emptying a Trough An 8-ft-long trough has ends that are semicircles of radius 2 ft. If the trough is full of water weighing, 62.4 lb/ft3, find the work required to empty it by pumping, the water through a pipe that extends 1 ft above the top of, the trough., , rocket and the earth, respectively, so that F  GmM>r 2, where, R  r  R  h. At r  R the force will be the weight of the rocket,, that is, mt  GmM>R2. Therefore, G  tR2>M, so F  mtR2>r 2., (ii) W  兰RRhF dr., , h, R, , 1 ft, , 2 ft, , 8 ft, , 28. A boiler has the shape of a (lower) hemisphere of radius, 5 ft. If it is filled with water weighing 62.4 lb/ft3, find the, work required to empty the boiler by pumping the water, over the top of the boiler., 29. Refer to Example 6. Suppose that the pressure P and volume V of the steam in a steam engine are related by the law, PV 1.4  100,000, where P is measured in pounds per square, inch and V is measured in cubic inches. Find the work done, by the steam as it expands from a volume of 100 in.3 to a, volume of 400 in.3., 30. Refer to Example 6. The pressure P and volume V of, the steam in a steam engine are related by the equation, PV 1.2  k, where k is a constant. If the initial pressure of, the steam is P0 lb/in.2 and its initial volume is V0 in.3, find, an expression for the work done by the steam as it expands, to a volume of four times its initial volume., 31. Launching a Rocket Newton’s Law of Gravitation states that, two bodies having masses m 1 and m 2 attract each other with, a force, FG, , m 1m 2, r2, , where G is the gravitational constant and r is the distance, between the two bodies. Assume that the mass of the earth, is 5.97 1024 kg and is concentrated at the center of the, earth, the radius of the earth is 6.37 106 m, and, G  6.67 1011 N-m2/kg2. Find the work required to, launch a rocket of mass 500,000 kg vertically upwards, to a height of 10,000 km., 32. Launching a Rocket Show that the work W required to launch, a rocket of mass m from the ground vertically upward to a, height h is given by the formula, W, , 33. Launching a Lunar Landing Module A lunar landing module with, a weight of 20,000 lb, as measured on the earth, is to be, launched vertically upward from the surface of the moon, to a height of 20 mi. Taking the radius of the moon to be, 1100 mi and its gravitational force to be one sixth that of, the earth, find the work required to accomplish the task., Hint: See Exercise 32., , 34. Work Done by a Repulsive Charge Coulomb’s Law states that the, force exerted on two point charges q1 and q2 separated by a, distance r is given by, F, , 1 q1q2, 4pe0 r 2, , where e0 is a constant known as the permittivity of free, space. Suppose that an electrical charge q1 is concentrated, at the origin of the coordinate line and that it repulses a like, charge q2 from the point x  a to the point x  b. Show, that the work W done by the repulsive force is given by, W, , q1q2 1, 1, a  b, 4pe0 a, b, , 35. Work Done by a Repulsive Charge An electric charge Q distributed uniformly along a ring-shaped conductor of radius a, repulses a like charge q along the line perpendicular to the, plane of the ring, through its center. The magnitude of the, force acting on the charge q when it is at the point x is, given by, F, , qQx, 1, ⴢ, 4pe0 (x 2  R2)3>2, , and the force acts in the direction of the positive x-axis., Find the work done by the force of repulsion in moving the, charge q from x  a to x  b., , R, Q, , a, , q, , b, , x, , mtRh, Rh, , where R is the radius of the earth., Hint: Use Newton’s Law of Gravitation given in Exercise 31, and follow these steps: (i) Let m and M denote the mass of the, , 36. The following table shows the force F(x) (in pounds), exerted on an object as it is moved along a coordinate axis, from x  0 to x  10 (x is measured in feet). Use Simpson’s Rule to estimate the work done by the force.

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5.6, , x (ft), , 0, , 1, , 2, , 3, , 4, , 5, , F(x) (lb), , 0, , 0.69, , 1.61, , 2.28, , 2.88, , 3.20, , Fluid Pressure and Force, , where √1 and √2 are the velocities of the body when it is at, x  x 1 and x  x 2, respectively., Hint: Use Newton’s Second Law of Motion aF  m, Chain Rule to write, , x (ft), F(x) (lb), , 6, , 7, , 8, , 9, , 10, , 3.58, , 3.95, , 4.20, , 4.38, , 4.64, , 冮, , x2, , x1, , 5.6, , F(x) dx , , d√, b and the, dt, , d√, d√ dx, d√, , ⴢ, √, dt, dx dt, dx, The quantity 12 m√2 is the kinetic energy of a body of mass m, moving with a velocity √. Thus, the work done by the force, is equal to the net change in the kinetic energy of the body., , 37. Work and Kinetic Energy A force F(x) acts on a body of, mass m moving it along a coordinate axis. Show that the, work done by the force in moving the body from x  x 1, to x  x 2 is, W, , 495, , 38. Refer to Exercise 37. A 4-kg block is attached to a horizontal spring with a spring constant of 400 N/m. The spring is, compressed 5 cm from equilibrium and released from rest., Find the speed of the block when the spring is at its equilibrium position., , 1 2 1 2, m√2  m√1, 2, 2, , Fluid Pressure and Force, , DAJ/Getty Images, , Corbis/SuperStock, , FIGURE 1, , Brand X/PhotoLibrary, , Whether designing a hydroelectric dam, an aquarium, or a submarine, an engineer must, consider the pressure exerted by the water on the walls or surfaces of the object. (See, Figure 1.), , Fluid Pressure, Consider a thin horizontal plate of area A ft2 submerged to a depth of h ft in a liquid, of weight density d lb/ft3 (Figure 2a). The force acting on the surface of the plate is, just the weight of the column of liquid above it (Figure 2b). Since the volume of this, column of liquid is Ah ft3 and its weight density is d lb/ft3, we see that the force exerted, on the plate by the liquid is given by, F  dAh, , weight density ⴢ volume, , h, , FIGURE 2, The force exerted by the fluid on, the horizontal plate (a) is the weight, of the column of liquid above it., , h, , A, (a), , A, (b)