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Chapter 4, , Integration, defined as a continuous summation process, is linked to dyferentiation by the fundamental theorem of calculus., , In everyday language, the word integration refers to putting things together,, while dgerentiation refers to separating, or distinguishing, things., The simplest kind of "differentiation" in mathematics is subtraction,, which tells us the difference between two numbers. We may think of differentiation in calculus as telling the difference between the values of a function at, nearby points in its domain., By analogy, the simplest kind of "integration" in mathematics is addition., Given two or more numbers, we can put them together to obtain their sum., Integration in calculus is an operation on functions, giving a "continuous, sum" of all the values of a function on an interval. This process can be, applied whenever a physical quantity is built up from another quantity which, is spread out over space or time. For example, in this chapter, we shall see that, the distance travelled by an object moving on a line is the integral of its, velocity with respect to time, generalizing the formula "distance = velocity x, time," which is valid when the velocity is constant. Other examples are that, the volume of a wire of variable cross-sectional area is obtained by integrating, this area over the length of the wire, and the total electrical energy consumed, in a house during a day is obtained by integrating the time-varying power, consumption over the day., , Summation, The symbol, , Cy=,ai is shorthand for, , a,, , + a, + . . + a,,., , To illustrate the basic ideas and properties of integration, we shall reexamine, the relationship between distance and velocity. In Section 1.1 we saw that, velocity is the time derivative of distance travelled, i.e.,, ~d - change in distance, velocity zz - At, change in time, , a, , Copyright 1985 Springer-Verlag. All rights reserved.

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202, , Chapter 4 The Integral, , In this chapter, it will be more useful to look at this relationship in the form, , ad w velocity x ~ t ., , (2), To be more specific, suppose that a bus is travelling on a straight highway, and that its position is described by a function y = F ( t ) , where y is the, position of the bus measured in meters from a designated starting position,, and r is the time measured in seconds. (See Fig. 4.1.1.) We wish to obtain the, , 5ocJ'o, , 700", , 40, , 40, , L, , Figure 4.1.1. What IS the, pos~t~on, of the bus ~nterms, of its velocity?, , I 111L t c r i, , 10, , 70, , 1, , I, , position y in terms of velocity 0. In Section 2.5 we did this by using the, formula v = d y / d t and the notion of an antiderivative. This time, we shall go, back to basic principles, starting with equation (2)., If the velocity is constant over an interval of length A t , then the, approximation ( z )in equation (2) becomes equality, i.e., A d = c a t . This, suggests another easily understood case: suppose that our time interval is, divided into two parts with durations A t , and At, and that the velocity during, these time intervals equals the constants c , and c,, respectively. (This situation, is slightly unrealistic, but it is a convenient idealization.) The d~stancetravelled during the first interval is c , A t , and that during the second is c , A t 2 ;, thus, the total distance travelled is, A d = c, At,, , + c,At2, , Continuing in the same way, we arrive at the following result:, , If a particle moves with a constant velocity r , for a time interval S t , ,, c, for a time interval At,, c, for a time interval At,, . . . and velocity, c,, for time interval At,,, then the total distance travelled is, , ., , A d = c,At, +c,At,+, , u,At,+, , ..., , + c,,At,,., , In (3), the symbol "+ . . . + " is interpreted as "continue summing until the, last term o,, At,, is reached.", The bus in Fig. 4.1.1 moves with the following velocities:, , Example I, , 4 meters per second for the first 2.5 seconds., 5 meters per second for the second 3 seconds,, 3.2 meters per second for the third 2 seconds, and, 1.4 meters per second for the fourth 1 second., How far does the bus travel?, Solution, , We use formula (3) with n, , =4, , and, , Copyright 1985 Springer-Verlag. All rights reserved.

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4.1 Summation, , 4,, u, = 5 ,, u, = 3.2,, u, = 1.4,, , U, =, , At,, , = 2.5,, , At,, , = 3,, , At,, , = 2,, , At,, , =, , 203, , 1, , to get, , Ad=4x2.5+5x3+3.2x2+, = 10+ 15 + 6 . 4 + 1.4, = 32.8 meters. B,, , 1.4x 1, , Integration involves a summation process similar to (3). T o prepare for the, development of these ideas, we need to develop a systematic notation for, summation. This notation is not only useful in the discussion of the integral, but will appear again in Chapter 12 on infinite series., Given n numbers, a , through a,, we denote their sum a ,, , + a, + . . . + a,, , by, , Here 2 is the capital Greek letter sigma, the equivalent of the Roman S (for, sum). We read the expression above as "the sum of a,, as i runs from 1 to n.", Example 2, , Solution, , (a) Find, , C:=,a,, if, , a l = 2, a2 = 3, a, = 4 , a4 = 6 . (b) Find, , (a) C : = , a , = a , + a , + a,+ a 4 = 2 + 3 + 4 + 6, , =, , 2:, , ,i2., , 15, , (b) Here a, = i2, so, , Notice that formula ( 3 ) can be written in summation notation as, , c c, a t , ., , =, , The letter i in (4) is called a dummy inde.~;we can replace it everywhere, by any other letter without changing the value of the expression. For instance,, n, , n, , a,, , and, , X=l, , a,, I =, , I, , have the same value, since both are equal to a , + . . ., A summation need not start a t I ; for instance, 6, , b, means, , i=2, , h,, , + a,,., , + h, + 6 , + b, + h,, , and, 3, , c, means, J=, , Example 3, , Solution, , Find, , c-,+c,+c~+c~+c~+c,., , -2, , ~i,,(k~ - k ) ., , 2 S , = , ( k 2 - k ) = ( 2 2 - 2 ) + (3', , -, , 3) + (4' - 4 ) + ( 5 2 - 5 ) = 2, , + 6 + 12 + 20, , = 40., , A, , Copyright 1985 Springer-Verlag. All rights reserved.

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204, , Chapter 4 The Integral, , where m < n are integers, and a, are real numbers, let i take each integer, value such that m S i S n. For each such i, evaluate a, and add the, resulting numbers. (There are n - m + 1 of them.), We list below some general properties of the summation operation:, , Properties of Summation, , 2., , n, , n, , i=m, , i=m, , C cai= c 2 a, ,, , 3. If m & n and n, P, , where c is a constant., , + 1 < p, then, P, , n, , 4. If a, = C for all i with m S i 6 n, where C is some constant, then, n, , 2 a,=, , C(n - m, , + 1)., , i= m, , 5. If a,, , < b, for all i, , with m ,< i 6 n, then, , These are just basic properties of addition extended to sums of many numbers at a time. For instance, property 3 says that a,, a m + , . . . + ap =, (a,, . . . a,) + (an+ . . . + up),which is a generalization of the associative law. Property 2 is a distributive Iaw; property 1 is a commutative law., Property 4 says that repeated addition of the same number is the same as, multiplication; property 5 is a generalization of the basic law of inequalities: if, a < b and c < d, then a c \< b + d., A useful formula gives the sum of the first n integers:, , +, , +, , +, , ,+, , +, , +, , + +, , +, , To prove this formula, let S = C:= ,i = 1 2 . . . n. Then write S again, with the order of the terms reversed and add the two sums:, , Copyright 1985 Springer-Verlag. All rights reserved.

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205, , 4.1 Summation, , Since there are n terms in the sum, the right-hand side is n(n, 2 S = n(n + I), and S = n(n + I)., , 4, , Exampie 4, , Solution, , Example 5,, , Solution, , Example 6, , Solution, , + l),, , so, , Find the sum of the first 100 integers.', We substitute n, = 5050. A, Find the sum 4, , =, , +, , 100 into S = n(n, , + l),, , giving, , + . 100 . 101, , = 50, , . 101, , + 5 + 6 + . + 29., , This sum is ~ ! l , i . We may write it as a difference ~ ! l , i C j = , i using either, "common sense" or summation property 3. Using formula (5) twice gives, , Find ~ j % ( j 2)., We use the summation properties as follows:, , 102, , =, , 2, , C j - 21 j -, , j= 1, , ;=, , = +(102)(103) -, , 2(100), , (properties 3 and 4), , 3 - 200, , (formula (5)), , = 5050., , We can also do this problem by making the substitution i = j - 2. As j runs, from 3 to 102, i runs from 1 to 100, and we get, , The second method used in Example 6 is usually best carried out by thinking, about the meaning of the notation in a given problem. However, for reference,, we record the general formula for substitution of an index: With the substituq,, tion i = j, , +, , The following example illustrates a trick that utilizes cancellation., Example 7, , Solution, , Show that, , Cr=,[i3 - ( i - 113]= n3, , The easiest way to do this is by writing out the sum:, , + . . . + [(n - 1)' - (n - 2))] + [n3 - (n - I)'], and observing that we can cancel l 3 with - 13, Z3 with - 23, 33 with - 33, and, so on up to (n - 1)3 with -(n - l)3. This leaves only the terms, , ', , A famous story about the great mathematician C. F. Gauss (1777-1855) concerns a task his, class had received from a demanding teacher in elementary school. They were to add up the first, 100 numbers. Gauss wrote the answer 5050 on his slate immediately; had he derived, S = f n ( n + 1) in his head at age 10?, , Copyright 1985 Springer-Verlag. All rights reserved.

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206, , Chapter 4 The Integral, , The kind of sum encountered in Example 7 is called a telescoping, or, collapsing, sum. A similar argument proves the result in the following box., , The next example uses summation notation to retrieve a result which may, already be obvious, but the idea will reappear later in the fundamental, theorem of calculus., Suppose that the bus in Fig. 4.1.1 is at position yi at time ti, i = 0, . . . , n, and, that during time interval ( t i - , ,ti), the velocity is a constant, , Example 8, , Using a telescoping sum, confirm that the distance travelled equals the, difference between the final and initial position., Solution, , By formula (37, the distance travelled is, , Since vi = Ayi/Ati, we get, , This is a telescoping sum which, by (7), equals y, - y o ; i.e., the final position, minus the initial position (see Fig. 4.1.2 where n = 3). A, , 'y 1 1, --------A>,,, , Figure 4.1.2. Motion of the, bus in Example 8 (n = 3)., , Total, distance, travelled, , V, l'otal tlnie, , Copyright 1985 Springer-Verlag. All rights reserved.

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4.2 Sums and Areas, , 207, , Exercises for Section 4.1, In Exercises 1-4, a particle moves along a line with the, given velocities for the given time intervals. Compute, the total distance travelled., 1. 2 meters per second for the first 3 seconds,, 1.8 meters per second for the second 2 seconds,, 2.1 meters per second for the third 3 seconds,, 3 meters per second for the fourth 1.5 seconds., 2. 3 meters per second for the first 1.5 seconds,, 1.2 meters per second for the second 3 seconds,, 2.1 meters per second for the third 2.4 seconds,, 4 meters per second for the fourth 3 seconds., 3. 8 meters per second for the first 1.2 seconds,, 10 meters per second for the second 3.1 seconds,, 12 meters per second for the third 4.2 seconds., 4. 2 meters per second for the first 8.1 seconds,, 3.2 meters per second for the second 2 seconds,, 4.6 meters Der second for the third 1.1 seconds., Find the sums in Exercises 5-8., 6. Z ; = , i 3, 5. 24- l(i2+ 1 ), 7 . ~ 5 i- ( i - 1), 8. C!=l i ( i - 2), Find the sums in Exercises 9-12., 9. 1 + 2 + . . . + 2 5, 10. 3 + 4 + . . . + 3 9, 11. ~ 4 5 _ , i, 12. ~ 3 9 , , i, Find the sums in Exercises 13- 16., 14. CjO=8,(j - 7 ), 13. C F 4 ( j - 3 ), - i2] 16. Cl?l[(i + 1)' - i s ], 15. C39-,[(i +, 17. Find Cj, - 2 j3., 18. Find C ) z 10M) j5., 19. Find C F l ( j + 6)., 20. Find CF=-20 k., 2 1. Find a formula for I=$=, i, where rn and n are, positive integers., 22. Find 2::,ai, where ai is the number of days in, the ith month of 1987., 23. Show that ELFlI / ( I + k 2 ) < 1000., 24. Show that ~ : M ) , 3 / ( 1+ i ) < 300., Find the telescoping sums in Exercises 25-28., 25. c!M),[i4- ( i - 114], 26. 2;: {(3i12- [3(i - l)I2), 27. c;'2pl[(i 2)2 - (i + l)'], 28. 2:: ,[(i + 3)' - ( i + 21'1, , 29. Draw a graph like Fig. 4.1.2 for the data in, Exercise I., 30. Draw a graph like Fig. 4.1.2 for the data in, Exercise 2., Find the sums in Exercises 3 1-40., 31. 2 i E O ( 3 k- 2), 32. 27=,(2i + 1), 33. C", =,[(k, + 1)4 - k4], 34. 2;: ,[(k + 1)' - k 8 ], 35. ~ ! ? , [ ( i+ 2)2 - ( i - I)'], 36., ,[(2i+ 213 - (2q3], 37. 2;: _30[i5+ i + 21, 38. 2:: -,,[I9 + 51' - 1315 + I ], 39. Cq=,2', 40. -, , x;!, , a41. By the method of telescoping sums, we have, , C [ ( i + 1)3 - i 3 ] = ( n + I ) ~ 1.i= 1, , - i3 = 3i2 + 3i + 1 and use, (a) Write ( i +, properties of summation to prove that, , (b) Find a formula for, , in terms of m and n., (c) Using the method and result of (a), find a, formula for 27=,i3. (YOUmay wish to try, guessing an answer by experiment.), +42. (a) Prove that, , +, , n, , 2 i ( i + 1 ) = -31n ( n + 1)(n + 2 ), , i= l, , by writing, , and using a telescoping sum., (b) Find 2:-,i(i + l ) ( i 2)., (c) Find C:=,[l/i(i + I)]., , +, , Sums and Areas, Areas under graphs can be approximated by sums., , In the last section, we saw that the formula for distance in terms of velocity is, Ad = C',',,viAtiwhen the velocity is a constant vi during the time interval, (ti-, ,ti). In this section we shall discuss a geometric interpretation of this fact, which will be important in the study of integration., Let us plot the velocity of a bus as a function of time. Suppose that the, , Copyright 1985 Springer-Verlag. All rights reserved.

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208, , Chapter 4 The Integral, , total time interval in question is [ a ,b]; i.e., t runs from a to b, and this interval, is divided into n smaller intervals so that a = to < t , < . . . < t, = b. The ith, . ~ form of u is, interval is ( t i - , ,ti), and v is a constant ui on this i n t e r ~ a l The, shown in Fig. 4.2.1 for n = 5., , Figure 4.2.1. The velocity, of the bus., , to, , r ? [ 3, , f4, , ts, , We notice that viAti is exactly the area of the rectangle over the ith, interval with base At, and height vi (the rectangle for i = 3 is shaded in the, figure). Thus,, n, , Ad, , vi At, is the total area of the rectangles under the graph of v., , =, i= l, , Figure 4.2.2. The region, under the graph off on, [a,61., , This suggests that the problem of finding distances in terms of velocities, should have something to do with areas, even when the velocity changes, smoothly rather than abruptly. Turning our attention to areas then, we go, back to the usual symbol x (rather than t ) for the independent variable., The area under the graph of a functionf on an interval [ a ,b] is defined to, be the area of the region in the plane enclosed by the graph y = f(x), the x, axis, and the vertical lines x = a and x = b. (See Fig. 4.2.2.) Here we assume, that f(x) > 0 for x in [a,b]. (In the next section, we shall deal with the, possibility that f might take negative values.), Let us examine certain similarities between properties of sums and areas., ,ai of sums, there corresponds the, To the property Cf=,ai = C:'=mai+ Cf=n+, additive property of areas: if a plane region is split into two parts which, overlap only along their edges, the area of the region is the sum of the areas of, the parts. (See Fig. 4.2.3.) Another property of sums is that if a, S bi for, , Figure 4.2.3. Area (A) =, Area (A ,) t Area (A2)., , Figure 4.2.4., Area ( A ) > Area (B)., , We are deliberately vague about the value of u at the end points, when the bus must suddenly, switch velocities. The value of Ad does not depend on what v is at each t,, so we can safely ignore, these points., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.2 Sums and Areas, , Figure 4.2.5. The graph of, a step fynction on [a, b], with n = 3., , Example 1, , Solution, , 209, , i = m,m + 1, . . . , n, then C:=,ai < 'jJ,,b,; the counterpart for areas is the, inclusion property: if one plane region is contained in another, the containing, region has more area. (See Fig. 4.2.4.), The connection between areas and sums becomes more explicit if we, consider step functions. A function g on the interval [a, b ] is called a step, function if [a, b ] can be broken into smaller intervals (called subintervals) with, g constant on each part. More precisely, there should be numbers x,,, x , , . . . , x,, with a = x 0 < x , < x 2 <, < x , - , < x , = b , such that g is, constant on each of the intervals ( x , , ~ , )( x, ,,x,), . . . , ( x , - , ,x,), as in Fig., 4.2.5. The values of g at the endpoints of these intervals will not affect any of, our calculations. The list (x,, x , , . . . , x,) is called apartition of [a, b ] ., , Draw a graph of the step function g defined on [2,4] by, , The graph of g on [2,2.5]is a horizontal line with height 1 on this interval. The, endpoints on the graph are drawn as solid dots to indicate that g takes the, value 1 at the endpoints x = 2 and x = 2.5. Continuing through the remaining, subintervals and using open dots to indicate endpoints which do not belong to, the graph, we obtain Fig. 4.2.6. A, , Figure 4.2.6. The graph of, the step function g in, Example 1., , Figure 4.2.7. The shaded, area is the sum of k,Ax,,, k, Ax2 and k, Ax,., , If a step function is non-negative, then the region under its graph can be, broken into rectangles, and the area of the region can be expressed as a sum., It is common to write Ax, for length xi - x i - , of the ith partition interval; if, the value of g on this interval is k , 2 0, then the area of the rectangle from, x i - , to xi with height k , is k i Ax,. Thus the total area under the graph is, k, Ax, + k , A x , + . . + k , Ax, = Cy-',,kiAx,, as in Fig. 4.2.7., , -, , Example 2, , What are the xi's, Ax,'s, and k,'s for the step function in Example l? Compute, the area of the region under its graph., , Solullon, , Looking at Figs. 4.2.6 and 4.2.7, we begin by labelling the left endpoint as x,;, i.e., x, = 2. The remaining partition points are x , = 2.5, x , = 3.5, and x , = 4., The Axj's are the widths of the intervals: Ax, = x , - x, = 0.5, Ax, = 1 , and, Ax, = 0.5. Finally, the ki9s are the heights of the rectangles: k, = 1, k , = 3,, and k , = 2. The area under the graph is, , Copyright 1985 Springer-Verlag. All rights reserved.

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210, , Chapter 4 The Integral, , If each k, is positive (or zero), the area under the graph of g is, , In deriving our formula for the area under the graph of a step function, we, used the fact that the area of a rectangle is its length times width, and the, additive property of areas. By using the inclusion property, we can find the, areas under graphs of general functions by comparison with step functionsthis idea, which goes back to the ancient Greeks, is the key to defining the, integral., Given a non-negative functionf, we wish to compute the area A under its, graph on [ a ,b]. A lower sum for f on [ a ,b] is defined to be the area under the, graph of a non-negative step function g for which g(x) < f(x) on [ a ,b]. If, g(x) = k, on the ith subinterval, then the inclusion property of areas tells us, that C:=, k, Ax, < A . (Fig. 4.2.8)., , Figure 4.2.8. The shaded, area Cy=,k, Ax, is a lower, sum for f on [ a ,b]., , Figure 4.2.9. The shaded, area C,"= ,I, Axj is an upper, sum for f on [ a ,b]., , Similarly, an upper sum for f on [ a ,b] is defined to be, ,l.AxJ, where h, is a step function with f(x) < h(x) on [ a ,b], and h(x) = I, on the jth subinterval of a partition of [ a ,b] (Fig. 4.2.9). By the inclusion property for areas,, A < C,"= ,l, Ax,, so the area lies between the upper and lower sums., , XI"!, , Example3, , Let f ( x ) = x 2 + 1 for O, , < x < 2., , Let, , (2, , O < x <+,, , Draw a graph showing f(x), g(x), and h ( x ) . What upper and lower sums for f, can be obtained from g and h?, Solution, , The graphs are shown in Fig. 4.2.10., For g we have Ax, = 1, k, = O and Ax, = 1, k, = 2. Since g(x) < f (x) for, all x in the open interval (0,2) (the graph of g lies below that off), we have as, a lower sum,, , Copyright 1985 Springer-Verlag. All rights reserved.

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4.2 Sums and Areas, , 211, , Figure 4.2.10. The area, under the graph of h is a n, upper sum for f ; the area, under the graph of g is a, lower sum., , For h we have A x , = 4 , I, = 2, Ax, = f , 1, = 4, and Ax, = $, l3 = 5 . Since, the graph of h lies above that o f f , h(x) > f(x) for all x in the interval (0,2),, we get the upper sum, , Using partitions with sufficiently small subintervals, we hope to find step, functions below and above f such that the corresponding lower and upper, sums are as close together as we wish. Notice that the difference between, lower and upper sums is the area between the graphs of the step functions (Fig., 4.2.1 1). We expect this area to be very small if the subintervals are small, enough and the values of the step functions are close to the values off., , Figure 4.2.111. The dark, shaded area is the, difference between upper, and lower sums for f on, [a,b ] .The area under the, graph is between the upper, and lower sums., , Suppose that there are lower sums and upper sums which are arbitrarily, close to one another. Then there can only be one number A such that, L < A < U for every lower sum L and every upper sum U, and this number, must be the area under the graph., , To calculate the area under the graph of a non-negative function f, we, try to find upper and lower sums (areas under graphs of step functions, lying below and above f) which are closer and closer together. (See, Example 6 below for a specific instance of what is meant by "closer and, closer.") The area A is the number which is above all the lower sums and, , Copyright 1985 Springer-Verlag. All rights reserved.

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212, , Chapter 4 The Integral, , What we have done here for areas has a counterpart in our distance-velocity, problem. Suppose that v = f(t) defined for a < t < b gives the velocity of a, moving bus as a function of time, and that there is a partition ( t o ,t,, . . . , t,), of [a,b] and numbers k,, . . . , k, such that k, < f(t) for t in the ith interval, (ti-,,ti). Taking for granted that a faster moving object travels further in a, given time interval, we may conclude that the bus travels a distance at least, ki(ti - ti- ,) in the ith time interval. Thus the total distance travelled must be, at least k, At, + - . + k, At, = Cy=,ki Ati (where, as usual, we write Ati for, ti - t i - ,), so we have a lower estimate for the distance travelled between t = a, and t = b. Similarly, if we know that f(t) < li on (ti-,,ti) for some numbers, E l , . . . , I,, we get an upper estimate C?=,IiAti for the distance travelled. By, making the time intervals short enough we hope to be able to find ki and li, close together, so that we can estimate the distance travelled as accurately as, we wish., Example 4, , Solution, , The velocity of a moving bus (in meters per second) is observed over periods, of 10 seconds, and it is found that, 4<v<5, when 0 < t < 10,, when 10 < t < 20,, 5.5 < v < 6.5, when 20 < t < 30., 5 < v < 5.7, Estimate the distance travelled during the interval 0 < t < 30., , A lower estimate is 4 - 10 + 5.5 . 10 + 5 - 10 = 145, and an upper estimate is, 5 . 10 + 6.5 - 10 5.7 - 10 = 172, so the distance travelled is between 145 and, 172 meters. A, , +, , Example 5, , The velocity of a snail at time t seconds is (0.001)(t2+ I ) meters per second at, time t. Use the calculations in Example 3 to estimate how far the snail crawled, between t = 0 and t = 2., , Solution, , We may use the comparison functions g and h in Example 3 if we multiply, their values by 0.001 (and change x to t). The lower sum and upper sum are, also multiplied by 0.001, and so the distance crawled is between 0.002 and, 0.00733 . . . meters, i.e., between 2 and 74 millimeters. ~(r,, When we calculate derivatives, we seldom use the definition in terms of limits., Rather, we use the rules for derivatives, which are much more efficient., Likewise, we will not usually calculate areas in terms of upper and lower sums, but will use the fundamental theorem of calculus once we have learned it., Now, however, to reinforce the idea of upper and lower sums, we shall do one, area problem "the hard way.", , Example 6, , Solution, , Use upper and lower sums to find the area under the graph of f(x), [0, 11., , =, , x on, , The area is shaded in Fig. 4.2.12., We will look for upper and lower sums which are close together. The, simplest way to do this is to divide the interval 10, I] into equal parts with a, partition of the form (0,l/n,2/n, . . . , (n - l)/n, 1). A step function g(x), below f(x) is given by setting g(x) = (i - I)/n on the interval [(i - l)/n, i/n),, while the step function with h(x) = i/n on ((i - l)/n,i/nj is above f(x) (Fig., 4.2.13)., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.2 Sums and Areas, , 2113, , The difference between the upper and lower sums is equal to the total, area of the chain of boxes in Fig. 4.2.14, on which both g(x) and h ( x ) are, graphed. Each of the n boxes has area (1 / n ) . ( 1 / n ) = 1/ n 2 , so their total area, is n . ( 1 / n 2 ) = 1/ n , which becomes arbitrarily small as n + CQ, so we know, that the area under our graph will be precisely determined. To find the area,, we compute the upper and lower sums. For the lower sum, g(x) = (i - l ) / n, = ki on the ith subinterval, and Ax, = I / n for all i, so, , Y, , n, , Figure 4.2.12. The region, under the graph of f ( x ) = x, on [O, I]., , xk,Axi=, i= 1, , x -' -. I, , i=l, , n, , I =I, , n, , n2, , i(i-l)=, , The upper sum is, , I, Figure, , 4.2.14. Difference, between the upper and, lower sums., , Figure 4.2.13. Lower and upper sums for f(x) = x on [0, 11., , The area under the graph is the unique number A which satisfies the, inequalities 1 / 2 - 1/2n < A < 1/2 1/2n for all n (see Fig. 4.2.15). Since, the number 1 satisfies the condition, we must have A = A, , +, , 0, , L-L L 1, 7, , 211, , 7, , -c, , I, , 1, , lt?;, , Figure 42-15. The area lies, in the interval [ 1 / 2 - 1/ 2 n ,, +, for all n. The, length of this interval -0, as n-oo., , 4., , The result of Example 6 agrees with the rule from elementary geometry that, the area of a triangle is half the base times the height. The advantage of the, method used here is t h t itd~hbeapplied to more general graphs. (Another case, is given in Exercise 20.) This method was used extensively during the century, before the invention of calculus, and is the basis for the definition of the, integral., , Exercises for Section 4.2, Draw the graphs of the step functions in Exercises 1-4., , Copyright 1985 Springer-Verlag. All rights reserved.

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214, , Chapter 4 The Integral, , In Exercises 5-8 compute the xi's, Axi9s,and k,'s for the, indicated step function and compute the area of the, region under its graph., 5. For g in Exercise 1., 6. For g in Exercise 2., 7. For g in Exercise 3., 8. For g in Exercise 4., In Exercises 9 and 10, draw a graph showing f, g, and h, and compute the upper and lower sums for f obtained, from g and h ., 9. f(x) = x2, 1 < X < 3;, , In Exercises 15-18, use upper and lower sums to find, the area under the graph of the given function., 15. f(x) = x for 1 < x < 2., 16. f(x)==2x for 0 < x < 1., 17. f(x) = 5x for a < x < b, a > 0., 18. f(x) = x + 3 for a < x < 6 , a > 0., , 19. Using upper and lower sums, find the area under, the graph of f(x) = 1 - x between x = 0 and, x = 1., *20. Using upper and lower sums, show that the area, under the graph of f(x) = x 2 between x = a and, x = b is ?j-(b3- a3). (YOUwill need to use the, result of Exercise 41(a) from Section 4.1.), x,, 1<x<2., Find the area under the graph off on [O, 21, using, the results of the Exercises 15 and 20., 922. Let, , 11. The velocity of a moving bus (in meters per, second) is observed over periods of 5 seconds, and it is found that, , 5.0 < v < 6.0, when 0 < t < 5,, 4.0 < v < 5 5, when 5 < t < 10,, 6 . 1 ~ 0 ~ 7 . 2 when 1 0 < t < 1 5 ,, when 15 < t < 20., 3.2 < v < 4.7, Estimate the distance travelled during the interval t = 0 to t = 20., 12. The velocity of a moving bus (in meters per, second) is observed over periods of 7.5 seconds, and it is found that, when 0 < t < 7.5,, 4.0 < u < 5.1, when 7.5 < t < 15,, 3.0 g v g 5.0, when 15 < t < 22.5,, 4.4 < v < 5.5, when 22.5 < t < 30., 3.0 < v < 4.1, Estimate the distance travelled during the interval t = 0 to t = 30., 13. The velocity of a snail at time t is (0.002)t2, meters per second at time t . Use the functions g, and h in Exercise 9 to estimate how far the snail, crawled between t = 1 and t = 3., 14. The velocity of a snail at time t is given by, (0.0005)(t3+ 1) meters per second at time t. Use, the functions g and h in Exercise 10 to estimate, how far the snail crawled between t = 1 and, t = 3., , Using the results of Exercises 17 and 19, find the, area under the graph off on [0,4]., +23. By combining the results of Example 6 and Exercise 20, find the area of the shaded region in Fig., 4.2.16. (Hint:Write the area as a difference of, known areas.), , Figure 4.2.16. Find the, shaded and striped areas., , *24. Using the results of previous exercises, find the, area of the striped region in Fig. 4.2.16., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.3 The Definition of the Integral, , 215, , 4.3 The Definition, , of the integral, The integral of a function is a "signed" area., , In the previous section, we saw how areas under graphs could be approximated by the areas under graphs of step functions. Now we shall extend this, idea to functions that need not be positive and shall give the formal definition, of the integral., Recall that if g is a step function with constant value k, > 0 on the, interval ( x i _, , x i ) of width Ax, = xi - xi- then the area under the graph of g, is, , ,,, , n, , Area, , =, , 2 k, Ax, ., i= l, , Figm4.3.1. The product, is the negative of the, shaded area., , This formula is analogous to the formula for distance travelled when the, velocity is a step function; see formula (31, Section 4.1. In that situation, it is, reasonable to allow negative velocity (reverse motion). Likewise, in the area, formula we wish to allow negative k,. To do so, we shall have to interpret, "area" correctly. Suppose that g ( x ) is a negative constant k, on an interval of, width Ax,. Then k, Ax; is the negative of the area between the graph of g and, the x axis on that interval. (See Fig. 4.3.1.), To formalize this idea, we introduce the notion of signed area. If R is any, region in the xy plane, its signed area is defined to be the area of the part of R, lying above the x axis, minus the area of the part !ying below the axis., If f is a function defined on the interval [ a ,b], the region between the, graph off and the x axis consists of those points ( x , y) for which x is in [ a ,b], a n d y lies betwen 0 and f ( x ) . It is natural to consider the signed area of such a, region, as illustrated in Fig. 4.3.2. For a step function g with values k, on, intervals of length Ax,, the sum C:=,k, Ax, gives the signed area of the region, between the graph of g and the x axis., , Figure 4.3.2. The signed, area between the graph off, and the x axis on [ a ,b] is, regions, the area of the, minus the area of the regions., , +, , minus the area of the portion below the x axis., For the region between the x axis and the graph of a step function, g , this signed area is C:= k, Ax,., , ,, , Copyright 1985 Springer-Verlag. All rights reserved.

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216, , Ghariplter 4 The Integral, , Example I Draw a graph of the step function g on [O,1] defined by, , Compute the signed area of the region between its graph and the x axis., , Solution, , 't, , The graph is shown in Fig. 4.3.3. There are three intervals, with A x , = f ,, Ax,= - I3 = 1-,, 12 and Ax 3 -- 1 - 14 = I, 4 ; k , = -2, k2 = 3, and k 3 = 1. Thus, the signed area is, 3, , k,Axl= ( - 2 ) ( f ), , + (3)(&) + ( I ) ( + ), , = -f, , + $ + 4= 2 . A, , r=l, , --C, , The counterpart of signed area for our distance-velocity problem is directed, distance, explained as follows: If the bus is moving to the right, then u > O, and distances are increasing. If the bus is moving to the left, then v < O and, the distances are decreasing. In the formula Ad = C ~ = , v , A t ,Ad, , is the, displacement, or the net distance the bus has moved, not the total distance, travelled, which would be C:= ,lull At. Just as with signed areas, movement to, the left is considered negative and is subtracted from movement to the right., (See Fig. 4.3.4.), , Figwe 4.3.3. The graph of, the step function in, Example 1., , rnovemerit, to left, , F i w e 4.3.4. Ad is the, displacement; i.e., net, distance travelled., , To find the signed area between the graph and the x axis for a function, which is not a step function, we can use upper and lower sums. Just as with, positive functions, if g is a step function lying below f, i.e., g ( x ) < f ( x ) for x, in [ a ,b], we call, , L=, , 2 k, Ax,, , a lower sum for f. Likewise, if h is a step function lying above f, with values, on intervals of width Ax, for j = 1, . . . , m , then, , I,, , is an upper sum for f. If we can find L's and U's arbitrarily close together,, lying on either side of a number A , then A must be the signed area between, the graph off and the x axis on [ a ,61. (See Fig. 4.3.5.), , Copyright 1985 Springer-Verlag. All rights reserved.

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4.3 The Definition of the Integral, , 217, , Figure 4.3.5. The signed, area of the region R lies, between the upper and, lower sums., , We are now ready to define the integral of a function f., Definition, , Let f be a function defined on [ a ,b]. We say that f has an integral or that f is, integrable if upper and lower sums for f can be found which are arbitrarily, close together. The number I such that L < I < U for all lower sums L and, upper sums U is called the integral off and is denoted, , We call jthe integral sign, a, b , the endpoints or limits of integration, and f the, integrand., The precise meaning of "afbitrariiy close together'' is the same as in Example, 6, Section 4.2, namely, that there should exist sequences Ln and Un of lower, and upper sums such that limn,,(Un - Ln) = 0. (Limits of sequences will be, treated in detail in Chapter 11.), , Given a function f on [ a ,b], the integral off, if it exists, is the number, , which separates the upper and lower sums. This number is the signed, , The notation for the integral is derived from the notation for sums. The Greek, letter C has turned into an elongated S ; ki and $ have turned into function, values f ( x ) ; Axi has become d x ; and the limits of summation (e.g., i goes from, 1 to n ) have become limits of integration:, , Just as with antiderivatives, the "x" in "dx" indicates that x is the variable of, integration., , Copyright 1985 Springer-Verlag. All rights reserved.

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218, , Chapter 4 The Integral, , Example 2, , Compute JP3f(x)dx for the function f sketched in Fig. 4.3.6. How is the, integral related to the area of the shaded region in the figure?, , Figure 4.3.6. The signed, area of the shaded region is, an integral., , Solullon, , The integral, , 2:J, , f(x)dx is the signed area of the shaded region., , f(x)dx= (1)(1), Example 3, , + (-1)(2) + (2)(2) = 1 - 2 + 4 =, , 3., , Write the signed area of the region in Fig. 4.3.7 as an integral., , Figure 4.3.7. The signed, area of this region equals, what integral?, , Solution, , The region is that between the graph of y, x = 1, so the signed area is, , =, , x3 and the x axis from x, , = -, , 4 to, , The next example shows how upper and lower sums can be used to approximate an integral. (In Chapter 6, we will learn how to compute this integral, exactly.), Example 4, , Using a division of the interval [ l ,2] into three equal parts, find, , (1 / x ) dx to, , &, &, we must find lower and upper sums which, , within an error of no more than, Solution, , X2, , To estimate the integral within, are within, of one another. We divide the interval into three equal parts and, use the step functions which give us the lowest possible upper sum and highest, possible lower sum, as shown in Fig. 4.3.8. For a lower sum we have, , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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4.3 The Definition of the Integral, , Figure, 4.3.8. Illustrating, -, , upper and lower sums for, I / x on [1,2]., , 1, , I, , ,, , ,, , ,, , 1, , 413, , 513, , ?, , 219, , *, Y, , For an upper sum we have, , It follows that, , Since the integral lies in the interval [%,%I, whose length is +,we may take, the midpoint % = as our estimate; it will differ from the true integral by no, = &, which is less than &. A, more than, , 4, , We have been calculating approximations to integrals without knowing, whether some of those integrals actually exist or not. Thus it may be, reassuring to know the following fact whose proof is given in more advanced, cour~es.~, Existence, , Iff is continuous on [ a ,b],then it has an integral., In particular, all differentiable functions have integrals, but so do step, functions and functions whose graphs have corners (such as y = 1x1); thus,, integrability is a more easily satisfied requirement than differentiability or, even continuity. The reader should note, however, that there do exist some, "pathological" functions that are not integrable. (See Exercise 36)., It is possible to calculate integrals of functions which are not necessarily, positive by the method used in Example 6 of the previous section, but this is a, tedious process. Rather than doing any such examples here, we shall wait until, we have developed the machinery of the fundamental theorem of calculus to, assist us., Let us now interpret the integral in terms of the distance-velocity, problem. We saw in our previous work that the upper and lower sums, represent the displacement of vehicles whose velocities are step functions and, which are faster or slower than the one we are studying. Thus, the displaceSee, for instance, Calculus Unlimited by J . Marsden and A. Weinstein, Benjamin/Cummings, (1981), p. 159, or one of the other references given in the Preface., , Copyright 1985 Springer-Verlag. All rights reserved.

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220, , Chapter 4 The Integral, , ment, like the integral, is sandwiched between upper and lower sums for the, velocity function, so we must have, displacement =, , Example 5, , ib/(r), dt., , A bus moves on the line with velocity v, Write formulas in terms of integrals for:, , = ( t 2 - 4t, , + 3) meters per, , (a) the displacement of the bus between t = 0 and t = 3;, (b) the actual distance the bus travels between t = 0 and t, Solution, , second., , = 3., , +, , (a) The displacement is ,$(t2 - 4t 3)dt., (b) We note that v can be factored as ( t - l ) ( t - 3), so it is positive on (0,1), and negative on (1,3). The total distance travelled is thus, , We close this section with a discussion of a different approach to the integral,, called the method of Riemann sums. Later we shall usually rely on the step, function approach, but Riemann sums are also widely used, and so you, should have at least a brief exposure to them., The idea behind Riemann sums is to use step functions to approximate, the function to be integrated, rather than bounding it above and below. Given, a function f defined on [ a ,b] and a partition (x,,x,, . . . , x,) of that interval,, we choose points c,, . . . , c, such that ci lies in the interval [xi- ,,xi]. The step, function which takes the constant value f(ci) on (xi- ,,xi)is then an approximation to /; the signed area under its graph, namely,, , is called a Riemann sum.4 It lies above all the lower sums and below all the, upper sums constructed using the same partition, so it is a good approximation to the integral off on [a, b] (see Fig. 4.3.9). Notice that the Riemann sum, , Figure 4.3.9. The area of, the shaded region is a, Riemann sum for f on, [ a , bl., , is formed by "sampling" the values off at points c,, . . . , c,, "weighting" the, samples according to the lengths of the intervals from which the ci's are, chosen, and then adding., If we choose a sequence of partitions, one for each n, such that the, lengths Ax, approach zero as n becomes larger, then the Riemann sums, approach the integral JY(x)dx in the limit as n + co.From this and Fig. 4.3.9,, we again see the connection between integrals and areas., Just as the derivatives may be defined as a limit of difference quotiehts,, so the integral may be defined as a limit of Riemann sums; the integral as, defined this way is called the Riemann integral., After the German mathematician Bernhard Riemann (1826-1866)., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.3 The Definition of the Integral, , 221, , Choose, for each n, a partition of [a,b] into n subintervals such that the, maximum of Ax, in the nth partition approaches zero as n + w . If c, is a, point chosen in the interval [ x ,- ,xi], then, , ,, , Example 6, Solution, , Write, , 6', , x 3 d x as a limit of sums., , As in Example 6, Section 4.2, divide [0,1] into n equal parts by the partition, (0,l/n, 2/n,. . . ,(n - l)/n, 1). Choose ci= i/n, the right endpoint of the, interval [(i - l)/n, i/n]. (We nay choose any point we wish; the left endpoint, or midpoint would have been just as good.) Then with f ( x ) = x 3 , we get, , Therefore,, 1, lim -, , 9 So x 3 d x ., i3=, , 1, , "-)an4 ;= 1, , Thus, we can find fAx3dx if we can evaluate this limit, or vice versa. A, , Supplement to Section 4.3, Solat Energy, Besides the distance-velocity and area problems, which we used to introduce, the integral, there are other physical problems that could be used in the same, way. Here, we consider the problem of computing solar energy and shall see, how it, too, leads naturally to the integral in terms of upper and lower sums., Consider a solar cell attached to an energy storage unit (such as a, battery) as in Fig. 4.3.10. When light shines on the solar cell, it is converted, into electrical energy which is stored in the battery (as electrical-chemical, energy) for later use., , Figure 4.3.10. The storage, unit accumulates the power, received by the solar cell., , Solar cell, , Ftirrgy .;torage unit, , We will be interested in the relation between the amount E of energy, stored and the intensity I of the sunlight. The number E can be read off a dial, on the energy storage device; I can be measured with a photographer's light, meter. (The units in which E and I are measured are unimportant for this, discussion.), Experiments show that when the solar cell is exposed to a steady source, of sunlight, the change AE in the amount of energy stored is proportional to, , Copyright 1985 Springer-Verlag. All rights reserved.

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222, , Chapter 4 The Integral, , the product of the intensity I and the length At of the period of exposure., Thus, , A E = K IAt,, where K is a constant depending on the apparatus and on the units used to, measure energy, time, and intensity. (We can imagine K being told to us as a, manufacturer's specification.), The intensity I can change-for example, the sun can move behind a, cloud. If during two periods, At, and At,, the intensity is, respectively, I , and, I,, then the total change in energy is the sum of the energies stored over each, individual period. That is,, AE = K I At,, ~, , + K I , A =~ ~K ( IAt,, ~ + 12At2)., , Likewise, if there are n periods, At,, . . . , At,, during which the intensity is, I , , . . . , I, (as in Fig. 4.3.1 1(a)), the energy stored will be the sum of n terms,, , Notice that this sum is exactly K times the integral of the step function g,, where g ( t ) = I, on the interval of length At,., In practice, as the sun moves gradually behind the clouds and its, elevation in the sky changes, the intensity I of sunlight does not change by, jumps but varies continuously with t (Fig. 4.3.1 1(b)). The change in stored, , Figure 4.3.11. The intensity, of sunlight varying with, time., , energy AE can still be measured on the energy storage meter, but it can no, longer be represented as a sum in the ordinary sense. In fact, the intensity now, takes on infinitely many values, but it does not stay at a given value for any, length of time., If I = f(t), the true change in stored energy is given by the integral, , which is K times the area under the graph I = f(t). If g(t) is a step function, with g(t) < f(t), then the integral of g is less than or equal to the integral of, f(t). This is in accordance with our intuition: the less the intensity, the less the, energy stored., The passage from step functions to general functions in the definition of, the integral and the interpretation of the integral can be carried out in many, contexts; this gives integral calculus a wide range of applications., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.3 The Definition of the Integral, , 223, , Exercises for Section 4.3, In Exercises 1-4, draw a graph of the given step function, and compute the signed area of the region between its graph and the x axis., , Figure 4.3.12. Graphs for, Exercises 9- 12., 13. Find L 4 ( l / x ) d x to within an error of no more, than &., 14. If you used the method in Example 4 to calculate, In Exercises 5-8, compute the indicated integrals., 5. i i g ( x ) d x , g as in Exercise 1., , i 2 ( l / x ) d x to within, , @ 15. Estimate, , 1 16. Estimate, 8. r 2 g ( x ) d x . g as in Exercise 4., In Exercises 9-12, write the signed areas of the shaded, regions in terms of integrals. (See Figure 4.3.12.), , how many subintervals, , would you need?, , 6, fO,g(x)dx, g as in Exercise 2., 7, f02X(x)dx, g as in Exercise 3., , i&j,, , i2, J2, , ( l / x 2 ) d x to within, , A., , ( I / x 2 ) d x to within, , 17. A bus moves on the line with velocity given by, c = 5(r2- 51 + 6). Write a formula in terms of, integrals for:, (a) the displacement of the bus between t = 0, and r = 3;, (b) the actual disiance the bus travels between, t = 0 and t = 3., 18. A bus moves on the line with velocity given by, c = 6 r 2 - 301 + 24. Write a formula in terms of, integrals for:, ( a ) the displacement of the bus between t = 0, and t = 5;, (b) the actual distance the bus travels beiween, t = 0 and t = 5., In Exercises 19-22, write the given integral as a limit of, sums., I 9 J'x5dx., , 20. i ' 9 x 3 d x ., , 21. J 4 L d x ., 2 I+x2, , 22., , 23. Show that, 24. Show that, , -3, , ,<, , 1', , J2, , tI0dt, , I;~, , I+x, , ( t 3 - 4)dl, , <, , dx., , < 4., , 1., , 25. Let f(t) be defined by, f(t)=, , i, , 2, 0, , -1, , if, if, if, , O<t<l,, l<t<3,, 3<t<4., , For any number x in (0,4], f(t1 is a step function, on [0, x]., (a) Find ff(r)dt, , as a function of x. (You will, , need to use different forinulas on different, intervals.), , Copyright 1985 Springer-Verlag. All rights reserved.

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224, , Chapter 4 The Integral, , d, , (b) Let F(x) = Xf(t)dt, for x in (0,4]. Draw a, graph of F., (c) At which points is F differentiable? Find a, formula for F'(x)., 26. Let f be the function defined by, , *30. Suppose that f(x) is a step function on [a, b ] , and, let g(x) = f(x) + k, where k is a constant., (a) Show that g(x) is a step function., (b) Find j b g ( x ) d x in terms of bbf(x)dx., a, , *31. Let h(x) = kf(x), where f(x) is a step function, on [a,b]., (a) Show that h(x) is a step function., (b) Find I b h ( x ) dx in terms of bbf(x) dx., a, , (a) Find Jlof(x)dx., , *32. For x E [0, 11 let f(x) be the first digit after the, decimal point in the decimal expansion of x., , (b) Find J9f(x) dx., , (a) Draw a graph off. (b) Find j l f ( x ) d x ., , 2, , (c) Suppose that g is a function on [ l , 101 such, that g(x) < f(x) for all x in [ l , 101. What, , 0, , *33. Define the functions f and g on [O, 31 as follows:, , inequality can you derive for 1log(x)dx?, (d) With g(x) as in part (c), what inequalities can you obtain for, , 1;", , -, , J;, , 'O, , 2g(x)dx and, , g(x) dx? [Hint: Find functions like f, , with which you can compare 2g and -g.], 27. Let f(t) be the "greatest integer function"; that, is, f(t) is the greatest integer which is less than or, equal to t-for example, f(n) = n for any integer,, f(54)= 5, f(-54) = -6, and so on., (a) Draw a graph of f(t) on the interval - 4,4]., (b) Find, , ', , f, , b6f(t)dt, I_:/(t)dt., , and, , (a) Draw the graph of f(x), , (b) Compute, , X2, , [f(x), , + g(x) and compute, , + g(x)] dx., , (c) Compare l32/(x) dx with 2 'f(x)dx., , i,, , (d) Show that, , L4'5, , f (t) dt., , (c) Find a general formula for, n is any positive integer., (d) Let F(x) =, , i", , I", , f(t) dt, where x, , f(t)dt, where, , > 0., , Draw a, , graph of F for x E [0, 41, and find a formula, for F1(x), where it is defined., 28. A rod 1 meter long is made of 100 segments of, equal length such that the linear density of the, kth segment is 30k grams per meter. What is the, total mass of the rod?, 29. The volume of a rod of uniform shape is A Ax,, where A is the cross-sectional area and Ax is the, length., (a) Suppose that the rod consists of n pieces,, with the ith piece having cross-sectional area, A, and length Ax,. Write a formula for the, volume., (b) Suppose that the cross-sectional area is, A = f(x), where f is a function on [0, L], L, being the total length of the rod. Write a, formula for the volume of the rod, using the, integral notation., , (e) Is the following true?, , *34. Suppose that f is a continuous function on [a, b], and that f(x) # 0 for all x in [a, b]. Assume that, a # b and that f((a b)/2) = 1. Prove that, , +, , Lbf(x)dx, , > 0. [Hint: Find a lower sum.], , *35. Compute the exact value of, , I', , x5dx by using, , Riemann sums and the formula, , *36. Let the function f be defined on [0, 31 by, =, , (, , if, if, , x is a rational number,, x is irrational., , (a) Using the fact that between every two real, numbers there lie both rationals and irrationals, show that every upper sum for f on, [0, 31 is at least 6., (b) Show that every lower sum for f on [0, 31 is, at most 0., (c) Is f integrable on [0, 3]? Explain., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.4 The Fundamental Theorem of Calculus, , 225, , 4.4 The Fundamental, Theorem of Calculus, The processes of integration and difSerentiation are inverses to one another., We now know two ways of expressing the solution of the distance-velocity, problem. Let us recall the problem and these two ways., Problem, , First Solution, , A bus moves on a straight line with given velocity v = f ( t ) for a, the displacement Ad of the bus during this time interval., , < t < b. Find, , The first solution uses antiderivatives and was presented in Section 2.5. Let, y = F ( t ) be the position of the bus at time t. Then since u = dy/dt, i.e.,, f = F ' , F is an antiderivative of f. The displacement is the final position, minus the initial position; i.e.,, Ad = F ( b ) - F ( a ) ,, ('1, the difference between the values of the antiderivative at f = a and t = b., , Second Solullon, , The second solution uses the integral as defined in the previous section. We, saw that, , Ad, , =, , I,"f ( I ) d t ., , We arrived at formulas (1) and (2) by rather different routes. I-Iowever,, the displacement must be the same in each case. Equating (1) and (2), we get, , This equality is called the fundamental theorem of calculus. It expresses the, integral in terms of an antiderivative and establishes the key link between, differentiation and integration., The argument by which we arrived at (3) was based on a physical model., Later, in this section, we shall also give a purely mathematical proof., With a slight change of notation, we restate (3) in the following box., , Suppose' that the function F is differentiable everywhere on [ a ,b ] and, that F' is integrable on [ a ,bj. Then, , In other words, i f f is integrable on [ a ,b ] and has an antiderivative F,, , We may use this theorem to find the integral which we previously computed, "by hand" (Example 6 , Section 4.2)., , Copyright 1985 Springer-Verlag. All rights reserved.

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226, , Chapter 4 The Integral, , Example 1, , Using the fundamental theorem of calculus, compute, , i, , 1, , x dx., , Solution By the power rule, an antiderivative for f ( x ) = x is F ( x ) = f x2. (You could, also have found F ( x ) by guessing, and you can always check the answer by, differentiating f x2.) The fundamental theorem gives, , which agrees with our earlier result. A, Next, we use the fundamental theorem to obtain a new result., Example 2, Solution, , Using the fundamental theorem of calculus, compute i\2dx., Let f ( x ) = x 2 ; again by the power rule, we may take F ( x ) = + x 3 . By the, fundamental theorem, we have, ibx2dx=, , ib, , f(X)d~=, F ( b ) - F ( a ) = +b3- +a3., , We conclude that Jb,x2dx= l3 ( b 3 - a3). This gives the area under a segment, of the parabola y = x 2 (Fig. 4.4.1). A, We can summarize the integration method provided by the fundamental, theorem as follows:, Figure 4.4.1. The shaded, area is Jtx2dx = f ( b 3- a 31., , To integrate the function f ( x ) over the interval [ a ,b ] : find an antiderivative F ( x ) for f ( x ) , then evaluate F at a and b and subtract the results:, , Notice that the fundamental theorem does not specify which antiderivative to, use. However, if F , and F2 are two antiderivatives off on [a,b], they differ by, a constant (see Section 3.6); F,(t) = F2(t)+ C , and so, F,(b) - F , ( a ) = [ F 2 ( b )+ C] - [ F2(a)+ C] = F,(b) - F2(a)., (The C's cancel.) Thus all choices of F give the same result., Expressions of the form F(b) - F(a) occur so often that it is useful to, have a special notation for them., , Example 3, , Solution, , Find ( x 3+ 5)Ii., Here F ( x ) = x 3 + 5 and, ( x 3+ 5)Ii = F(3) - F(2), = (33, , + 5 ) - (23 + 5 ), , = 32 - 13 = 19. A, , Copyright 1985 Springer-Verlag. All rights reserved.

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4.4 The Fundamental Theorem of Calculus, , 227, , In terms of this new notation, we can write the formula of the fundamental, theorem of calculus in the form, , where F is an antiderivative off on [ a ,b ] ., Example 4, Solution, , Example 5, Solution, , Find f ( x 2, , + 1)dx., , By the sum and power rules for antiderivatives, and antiderivative for x 2, is f x3 X . By the fundamental theorem,, , +, , Evaluate, , 1, , 2, , +1, , 1 dx., , An antiderivative of l / x 4 = x - is~ - 1 / 3 x 3 , since, , Hence, , We will now give a complete proof of the fundamental theorem of calculus., The basic idea is as follows: letting F be an antiderivative for f on [ a ,b ] , we, will show that the number F ( b ) - F ( a ) lies between any lower and upper, sums for f on [ a ,b ] . Since f is assumed integrable, it has upper and lower sums, arbitrarily close together, and the only number with this property is the, integral off (see page 217). Thus, we will have F ( b ) - F ( a ) = f ( x ) d x ., , Jt, , Proof of the For the lower sums, we must show that any step function g below f on ( a ,b ), Fundamental has integral at most F ( b ) - F(a). So let k , , k,, . . . , k, be the values of g on, Theorem the partition intervals ( x o , x , ) , ( x , , x 2 )., . . , ( x,-,, x,) (See Fig. 4.4.2). On, , Figure 4.4.2. The integral, of g is a lower sum for J on, , ( x i - , , x i ) , ki < f ( x ) = F f ( x ) , so the difference quotient for F satisfies the, inequality k, < [ F ( x i )- F ( x i _, ) ] / [ x i- x i _ ,I, by the first consequence of the, mean value theorem (Section 3.6). Thus, k, Ax, < F(x,) - F(x,- ,). Summing, from i = 1 to n, we get, , Copyright 1985 Springer-Verlag. All rights reserved.

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228, , Chapter 4 The Integral, , The lekhand side is just the integral of g on [ a ,bj, while the right-hand side is, a telescoping sum which collapses to F(x,) - F(x,); so we have proven that, J i g ( x ) d x ,< F ( b ) - F(a)., An identical argument works for upper sums: Pf h is a step function, above f on ( a ,b), then F ( b ) - F ( a ) < J i h ( x ) d x (see Exercise 49). Thus the, proof of the fundamental theorem is complete. W, Here are two more examples illustrating the use of the fundamental, theorem. Notice that any letter can be used as the variable of integration, just, like the "dummy variable" in summation., Example 6, Solution, , Find, , L4(z2+, , 3t7/')dr., , By the sum, constant multiple, and power rules for antiderivatives, an antiderivative for t 2 + 3t7/2is ( t 3 / 3 ) 3 . ( 2 / 9 ) t 9 l 2 .Thus,, , +, , In the next example, some algebraic manipulations are needed before the, integral is computed., , 9, 2, , Example 7 Compute, Solution, , I;', , ds ., , The integrand may be broken apart:, , We can find an antiderivative term by term, by the power rule:, , Next we use the fundamental theorem to solve area and distance-velocity, problems. Let us first recall, from Sections 4.2 and 4.3, the situation for areas, under graphs., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.4 The Fundamental Theorem of Calculus, , If f(x), , > 0 for x, , 229, , in [ a ,b], the area under the graph o f f between x = a, , 1;, , Iff is negative at some points of [ a ,b], then f(x)dx is the signed, area of the region between the graph off, the x axis, and the lines x = a, , Example 8, , (a) Find the area of the region bounded by the x axis, the y axis, the line, x = 2, and the parabola y = x2. (b) Compute the area of the region shown in, Fig. 4.4.3., , Figure 4.4.3. Compute this, area., Figure 4.4.4. The shaded, area equals j i x 2 dx., , Solution, , (a) The region described is that under the graph of f(x) = x 2 on [0,2] (Fig., 4.4.4). The area of the region is Jix2dx = 'x3I2, 3, o = 83 ., (b) The region is that under the graph of y = x3 from x = 0 to x = 1, so its, area is j;x3dx. By the fundamental theorem,, , Thus, the area is, Example 9, , (a) Interpret, , 4. A, , 8', , (x2 - 1)dx in terms of areas and evaluate. (b) Find the shaded, , area in ~ i ~ u r e - 4 . 4 . 5 ., , F i m e 4.4.6. j i ( x 2 - 1) dx, is the difference between, the areas of R2 and R,., , Figwe 4.4.5. Find the area, of this region., , Solution, , (a) Refer to Fig. 4.4.6. We know that the integral represents the signed area of, the region between the graph of y = x 2 - 1 and the x axis. In other words, it is, , Copyright 1985 Springer-Verlag. All rights reserved.

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230, , Chapter 4 The integral, , the area of R, minus the area of R,. Evaluating,, , (b) For functions which are negative on part of an interval, we must recall,, from Section 4.3, that the integral represents the signed area between the graph, and the x axis. To get the ordinary area, we must integrate piece by piece., The area from x = 0 to x = 1 is Jhx3dx. The negative of the area from, x = - to x = 0 is J0 , / , x 3 d x . Thus the total area is, , +, , Finally, in this section, we consider the use of the fundamental theorem to, solve displacement problems. The following box summarizes the method,, which was justified earlier in this section., , Displacements and Velocity, If a particle on the x axis has velocity 6 = f(t) and position x = F ( t ) ,, then the displacement F ( b ) - F ( a ) between the times t = a and t = b is, obtained by integrating the velocity from r = a to 1 = h :, Displacement from, time t = a to t = b, , Example 10, , Solution, , = lh(velocity)dt., , An object moving in a straight line has velocity t. = 5t4, far does the object travel between t = 1 and t = 2?, , + 3 t 2 at time t . How, , The displacement equals the total distance travelled in this case, since z; > 0., Thus, the displacement is, , Thus, the object travels 38 units of length between t, , = 1 and, , t = 2. A, , We have seen that the geometric interpretation of integrals of functions that, can sometimes be negative requires the notion of signed area. Likewise, when, velocities are negative, we have to be careful with signs. The integral is always, the displacement; to get the actual distance travelled, we must change the sign, of the integral over the periods when the velocity is negative. See Fig. 4.4.7 for, a typical situation., , Figure 4.4.7. The total, distance travelled is, j:v dt - j:v dt; the, displacement is f2v dt., , > 0 lor, , starting, , u, , posltloll, , U</<<, , /, , =u, , Copyright 1985 Springer-Verlag. All rights reserved.

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4.4 The Fundamental Theorem of Calculus, , Example 11, , Solution, , 231, , An object on the x axis has velocity v = 2 t - t 2 at time t . If it starts out at, t - 0 , where is it at time t = 3? How far has it travelled?, x = -1, Let x = f ( t ) be the position at time t . Then, , +, , Since f ( 0 ) = - 1 , the object is again at x = 0 f ( 0 ) = - I at time t = 3., The object turns around when v changes sign, namely, at those t where, 2 t - t 2 = 0 or t = 0 , 2 . For O < t < 2 , v > 0 , and for 2 < t < 3 , v < O . The, total distance travelled is therefore, , Exercises for Section 4.4, Using the fundamental theorem of calculus, compute, the integrals in Exercises 1-4., , L3, , du., , 24., , J4, 2, , u-1, , du., , Calculate the areas of the regions in Exercises 25-28, (Figure 4.4.8)., , 4. i 8 ( 1 + 6 ) d x ., , 3. L 6 3 x d x ., , 23., , Compute the quantities in Exercises 5-8., 5. x3I41;., 6. ( x 2 z3&)I:., 7. (3x2 + 5)l:., 8. ( x 4 + x 2 + 2)1c2., Evaluate the integrals in Exercises 9-24., , +, , +, , 9. Lbs4l3ds., , II., , 17., , 1, , ".r"', , 10. JpZl(t4 8i)dt., , dr., , 1 2 x, l2, l2 5 ,, 1, , (i+413', , 14. J-:'l, , + x2 - x3)dx., , 18. l 7 $ 3, , + z2)dz., , 2, , 19., , dt., , 2, , 21., , dx., , 22., , -1, JP2, , (x2, , +x), , 2, , dx ., , Figure 4.4.8. Regions for, Exercises 25-28., , Copyright 1985 Springer-Verlag. All rights reserved.

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232, , Gkaplter 4 The Integral, , Interpret the integrals in Exercises 29 and 30 in terms of, areas, sketch, and evaluate., , 30. A 2 ( x 2- 3 ) d x ., In Exercises 31-40, find the area of the region between, the graph of each of the following functions and the x, axis on the given interval and sketch., 31. x 3 on [0,2]., 32. 1 / x 2 on [1,2]., 33. x 2 + 2 x + 3 on [I, 21., 34. x 3 + 3 x + 2 on [O,21., 35. x 4 + 2 on [ - l,l]., 36. 3x4 - 2 x 2 on [ - 1, I]., 37. x 4 + 3 x 2 + 1 ; - 2 < x < 1., 38. 8x6 + 3x4 - 2; 1 < x < 2., 39. ( I / x ~+) x ~ 1;< g 3., 40. ( 3 x + 5 ) / x 3 ; 1 < x < 2., , +, , 41. An object moving in a straight line has velocity, u = 6t4 + 3t2 at time t. How far does the object, travel between t = 1 and t = lo?, 42. An object moving in a straight line has velocity, u = 2t3 + t4 at time t. How far does the object, travel between t = 0 and t = 2?, 43. The velocity of an object on the x axis is u, = 4t - 2t2. I f it is at x = 1 at t = 0, where is it at, i = 4? Hclw far has it travelled?, , 44. The velocity of an object on the x axis is v, = t 2 - 3t + 2. If the object is at x = - 1 at t = 0,, where is it at = 2? How far has it travelled?, 45. The velocity of a stone dropped from a balloon is, 32t feet per second, where i is the time in seconds after release. How far does the stone travel, in the first 10 seconds?, 46. How far does the stone in Exercise 45 travel in, the second 10 seconds after its release? The third, 10 seconds?, *47. An object is thrown upwards from the earth's, surface with a velocity uo. (a) How far has it, travelled after it returns? (b) How far has it, travelled when its velocity is - 4uo?, *48. Suppose that F is continuous on [O,21, that F ' ( x ), < 2 for 0 $ x < f , and that F 1 ( x )< 1 whenever, f < x & 2. What can you say about the difference F(2) - F(O)?, *49. Prove that if h ( t ) is a step function on [ a ,b] such, that f ( t ) & h ( t ) for all t in the interval ( a ,b), then, h ( t )dt, where F is any antidei,, rivative for f on [a, b]., F ( b ) - F ( a ) ,<, , *50. Let ao, . . . , a, be a given set of numbers and, & = ~ , - a ~ -Let, ~ . bk=Cr=16i,dj=bi-br-l., Express the b's in terms of the a's and the d's in, terms of the 6's., , 4.5 Definite and, lndefinite Integrals, Integrals and sums have similar properties., When we studied antiderivatives in Section 2.5, we used the notation jf(x)dx, for an antiderivative off, and we called it an indefinite integral. This notation, and terminology are consistent with the fundamental theorem of calculus. We, can rewrite the fundamental theorem in terms of the indefinite integral in the, following way., , Notice that although the indefinite integral is a function involving an arbitrary, constant, the expression, , represents a well-defined number, since the constant cancels when we subtract, the value at a from the value at b., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.5 Definite and Indefinite Integrals, , 233, , An expression of the form j, tf(x)dx with the endpoints specified, which, we have been calling simply "an integral," is sometimes called a definite, integral to distinguish it from an indefinite integral., Note that a definite integral is a number, while an indefinite integral is a, function (determined up to an additive constant)., Remember that one may check an indefinite integral formula by differentiating., , indefinite lrrtegral Test, , +, , x)dx = F(x) C, differentiate the rightIf(, hand side and see if you get the integrand f(x)., , To check a given formula, , Example 9, Solution, , Check the formula, , J3x, , 'dx, , = x9/3, , + C., , We differentiate the right-hand side using the power rule:, , so the formula checks. A, The next example involves an integral that cannot be readily found with the, antidifferentiation rules., Example 2, , (a) Check the formula r x ( l, J, tempt to derive the formula.), (b) Find J2x(l, 0, , Solution, , + x)'dx, , = &(7x -, , 1)(1, , + x)' + C. (Do not, , at-, , + x16dx., , (a) We differentiate the right-hand side using the product rule and power of a, function rule:, d, l, 1 [7(l + x)' + (7x - 1)7(1 + x)'], - - (7x - l ) ( l, x)'] = -, , dx, , [ 56, , +, , 56, , = (1, , + x)'x., , Thus the formula checks., (b) By the fundamental theorem and the formula we just checked, we have, , 1, 2, , x(1, , + x)'dx, , 1, , = - (7x - l)(l, , 56, , + x)', , In the box on page 204 we listed five key properties of the summation process., In the following box we list the corresponding properties of the definite, integral., , Copyright 1985 Springer-Verlag. All rights reserved.

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234, , Chapter 4 The Integral, , 1. i b [f(x), , 3. If a, , + g(x)] dx = i b f ( x ) d x + i b g ( x ) d x (sum rule)., , < b < c, then, , 5. If f(x), , <, , I c f ( x ) d x =Jbf(x)dx, a, +dcf(x)dx., , g(x) for all x satisfying a, , < x < b, then, , These properties hold for all functions f and g that have integrals. However,, while it is technically a bit less general, it is much easier to deduce the, properties from the antidifferentiation rules and the fundamental theorem of, calculus, assuming not only that f and g have integrals, but that they have, antiderivatives as well., Prove property 1 in the display above (assuming that f and g have antiderivatives)., , Example 3, Solullon, , Example 4, , Let F be an antiderivative for f and G be one for g. Then F, antiderivative for f g by the sum rule for antiderivatives. Thus,, , +, , +G, , is an, , Prove property 5., , Soiaati~n If f(x) < g(x) on (a, b), then ( F - G)'(x) = F'(x) - G1(x) = f(x) - g(x) < 0, for x in (a, b). Since a function with a negative derivative is decreasing, we get, , [ F ( b ) - ~ ( b )-][ ~ ( a -) G(a)] 0,, and so F(b) - F(a) < G(6) - G(a). By the fundamental theorem of calculus,, the last inequality can be written, , as required. A, Properties 2 and 3 can be proved in a yay similar to property 1. Note that, property 4 is obvious, since we know how to compute areas of rectangles., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.5 Definite and Indefinite integrals, , 235, , Example 5, , Explain property 3 in terms of (a) areas (assume that f is a positive function), and (b) distances and velocities., , Solution, , (a) Since r f ( x ) d x is the area under the graph of f from x = a to x = c,, property 3 merely states that the sum of the areas of regions A and B in Fig., 4.5.1 is the total area., , Figure 4.5.1. Illustrating, , the rule j:f(x) dx =, I:fcx> dx + Jif(x)dx., (b) Property 3 states that the displacement for a moving object between times, a and c equals the sum of the displacements between a and b and between b, and c. A, We have defined the integral J?(x)dx when a is less than b ; however, the, right-hand side of the equation, , makes sense even when a > h. Can we define j'tf(x)d.u for the case a, that this equation will still be true? The answer is simple:, !f b < a and f is iniegvahle on [ b . a ] , we dejine, , 6 , K ' e define jtf(x)d,u to he zero., Notice that if F' is integrable on [ / > . ( I ] , where h, preceding definition and the fundamental theorem., , !fa, , b so, , =, , < a., , then by the, , so the equation j ! : ~ ' ( x ) d x= F ( b ) - F ( a ) is still valid., , Example 6, , Find d 2 x ' d x ., , Solution j i x 3 d x = (x4/4)12 = i ( 1 6 - 1296) = -320. (Although the function f ( x ) = x', is positive, the integral is negative. To explain this, we remark that as x goes, from 6 to 2, " d x is negative.") A, , Copyright 1985 Springer-Verlag. All rights reserved.

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238, , Chapter 4 The Integral, , We have seen that the fundamental theorem of calculus enables us to compute, integrals by using antiderivatives. The relationship between integration and, differentiation is completed by an alternative version of the fundamental, theorem. Let us first state and prove it; its geometric meaning will be given, shortly., , Iff is continuous on [a,%], then, , We now justify the alternative version of the fundamental theorem. In, Exercises 49-53, it is shown that f has an antiderivative F. Let us accept this, fact here., The fundamental theorem applied to f on the interval [a, x] gives, , Differentiating both sides,, , d, dx, , = - F(x), , (since F(a) is constant), , =f(x>, , (since F is an antiderivative off j., , Thus the alternative version is proved., Notice that in the statement of the theorem we have changed the, (dummy) variable of integration to the letter "s" to avoid confusion with the, endpoint "x.", Example 7, Solution, , Verify the formula, , A, dx, , b, , xf(s)ds, , = f(x), , for f(x) = x., , The integral in question is, , (Ix, , f (s) ds, dx 0, so the formula holds. A, , Thus,, , Example 8, , Let F(x) =, , ds. Find F'(3)., jX, 1+s2+s3, 2, , Solullon, , Using the alternative version of the fundamental theorem, with f(s) =, 1/(I + s2+ s3), we have Fr(3) = f(3) = 1/(1 + 3' 33) = $. Notice that we, did not need to differentiate or integrate 1/(1 + s2 s3) to get the answer. A, , +, +, , At the top of the next page, we summarize the two forms of the fundamental, theorem., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.5 Definite and indefinite Integrals, , 237, , Fundamental Theorem of Calculus, Usual Version:, , ib, , F 1 ( x )dx, , =, , F(b) - F(a)., , Integrating the derivative of F gives the change in F., , A LXfw, , Alternative Version: -, , ds, , =f(x)., , Differentiating the integral off with respect to the upper limit gives f., , The alternative form of the fundamental theorem of calculus has an illuminating interpretation and explanation in terms of areas. Suppose that f ( x ) is, non-negative on [ a ,b]. Imagine uncovering the graph off by moving a screen, to the right, as in Fig. 4.5.2. When the screen is at x , the exposed area is, , A = ~ , f ( s ) d sThe, ., alternative eersion of the fundamental theorem can be, phrased as follows: as the screen moves to the right, the rate of change of, exposed area A with respect to x , d A / d x , equals f ( x ) . This same conclusion, can be seen graphically by investigating the difference quotient,, A ( x Ax) - A ( x ), Ax, The quantity A ( x Ax) - A ( x ) is the area under the graph of f between x, and x Ax. For Ax small, this area is approximately the area of the rectangle, , +, , +, , +, , Figure 4.5.3. The geometry, needed to explain why, d A / d x = f(x)., , with base Ax and height f ( x ) , as in Fig. 4.5.3. Therefore,, , and the approximation gets better as Ax becomes smaller. Thus, A(x, , + Ax) - A ( x ), , Ax, approaches f ( x ) as Ax +O, which means that d A / d x = f(x). I f f is continu-, , Copyright 1985 Springer-Verlag. All rights reserved.

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238, , Chapter 4 The Integral, , ous, this argument is the basis for a rigorous proof of the alternative version of, the fundamental theorem. See Exercises 49-53 for additional details., , Exercises Isr Section 4.5, In Exercises 1-4, check the integration formula by, differentiating the right-hand side., 1. 5 x 4 d x = x 5 + C., , i, , 19. Explain property 2 of integration in terms of (a), areas and (b) distances and velocities., 20. Explain property 5 of integration in terms of, (a) areas and (b) distances and velocities., ~f i ' f ( x ) d x = 3. i 2 f ( x ) dx = 4. and, , 6', , f(x)dx, , =, , -8., , calculate the quantities in Exercises 21-24, using the, properties of integration., , 5. (a) Check the following integral, Calculate the integrals in Exercises 25-28., ( b ) Evaluate, , i', , dl., , 3t2, , 6. ( a ) Check the following integration forniula:, , 27., , 1:, , dx., , Verify the formula, , ( b ) Evaluate, , 6', , [ ( s S +2s, , + 1)/(1, , 7. ( a ) Calculate the derivative of, I, , ( b ) Find, , ( 3 x 2 + .r4), , + x?)?, , (I, , 8. ( a ) Differentiate, ( b ) Find, , /*-( I, 3, , I, , + .r, , -, , .~)~]clr., , -, , x- l, , dx., , 'f(s)d.s = f ( x ) for the func-, , tions in Exercises 29 and 30., 29. /(.Y) = .u3 - 1 ., 30. /(.u) = u", .u2 + .u., 1, 31. Let F(r)=, , ( 4 - ,s)>, , i', , /_12, , + 8j3, , ds. Find F'(4)., , Y', -, , s 2 +I, , c1.r, and, , 28., , Evaluate the derivatives in Exercises 33-36., 3, d.r ., , 1, , + .r, , -, , I, , du in two ways., +x)', , Calculate the definite integrals in Exercises 9-18, , 37. Let c ( t ) be the velocity of a moving object. In, this context, interpret the formula, , 38. Interpret the alternative version of the fundamental theorem of calculus in the context of the, solar energy exaniple in the Supplement to Section 4.3., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.5 Definite and Indefinite Integrals, , 239, , 39. Suppose that, , :, :14+s2, &!, - . What is F1(x)?, , *45. Let F(x) =, , JX2, , s46. Calculate (a) Draw a graph off on the interval [O, 61., , s47. Find a formula for d, , (b) Find 16f(t)dt., , and explain, , 0, , your formula in terms of areas., , (c) Find b6f(x)dx., (d) Let F(t) = j b f(s)ds. Find the formula for, F(t) in [0,6] and draw a graph of F., (e) Find F1(t) for t in (0,6)., 40. (a) Give a formula for a function f whose graph, is the broken line segment ABCD in Fig., 4.5.4., (b) Find, , i f (t) dt ., lo, , 3, , (c) Find the area of quadrilateral ABCD by, means of geometry and compare the result, with the integral in part (b)., , 41. Let f be continuous on the interval I and let a ,, and a, be in I. Define the functions:, ), , I, , continuous on [a, b], then F(t) = ff(s) ds is an antiderivative off., *49. Prove property 3 for an integrable function f;, that is, if f is integrable on [a, b] and on [b, c],, then f is integrable on [a, c] and, , [Hint: Let I be the right-hand side. Show that, every number less than I is a lower sum for f on, [a, c] and, likewise, every number greater than I, is an upper sum. If S < I, show by a general fact, about inequalities that you can write S = SI, S,, where S , < j:f(x)dx and S2 < jZf(x)dx., Now piece together a lower sum corresponding, to S l with one for S,.], ~ 5 0 Prove, ., property 5 for integrable functions f and g., [Hint: Every lower sum for f is also one for g.], *51. Show that, , +, , Figure 4.5.4. Find a, formula for f., , F, , jr, , d, x, *48. Compute ds, I + X dx, ~., Exercises 49-53 outline a proof of this fact: if f is, , =f, , )d, aI, , and, , F,(t), , =, , itf(s) ds., "2, , (a) Show that Fl and F2 differ by a constant., (b) Express the constant F2- FI as an integral., 42. Develop a formula for j x ( l + x)" dx for n # - 1, or -2 by studying Example 2 [Hint: Guess the, answer (ax + b)(l + x)"+' and determine what a, and b have to be.], *43. (a) Combine the alternative version of the fundamental theorem of calculus with the chain, rule to prove that, , (b) Interpret (a) in terms of Fig. 4.5.5., , using property 3 of the integral., *52. Show that (I/h)J:+hf(s)ds lies between the maximum and minimum values of f on the interval, [t, t + h] (you may assume h > 0; a similar argument is needed for h < 0). Conclude that, , for some c between t and t + h, by the intermediate value theorem. (This result is sometimes, called the mean value theorem for integrals; we, will treat it again in Section 9.3.), *53. Use continuity off and the results from Exercises, 51 and 52 to show that F' = f., s54. Exercises 49-53 outlined a complete proof of the, alternative version of the fundamental theorem, of calculus. Use the "alternative version" to, prove the "main version" in Section 4.4., , Assume L L t f'=f, , wa%*anuaa~., , Figure 4.5.5. The rate of change of the exposed area, with respect to t is f(g(t)). g'(t)., , Copyright 1985 Springer-Verlag. All rights reserved.

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240, , Chapter 4 The Integral, , 4.6 Applications of the, Integral, Areas between graphs can be calculated as integrals., We have seen that area under the graph of a function can be expressed as an, integral. After a word problem based on this fact, we will learn how to, calculate areas between graphs in the plane. Other applications of integration, concern recovering the total change in a quantity from its rate of change., Example 1, , A parabolic doorway with base 6 feet and height 8 feet is cut out of a wall., How many square feet of wall space are removed?, , Ssiutisn, , Place the coordinate system as shown in Fig. 4.6.1 and let the parabola have, equation y = ax2 + c. Sincey = 8 when x = 0, c = 8. Also, y = O when x = 3,, so O = a32 8, SO a = - 5 . Thus the parabola is y = - $ x 2 8. The area, under its graph is, , Y, , +, , +, , (, , dx= --x3+8x, 2"7, =(-$27+8-3, , -, , (7-: . 2 7 - 8 . 3, , so 32 square feet have been cut out. A, Now we turn to the problem of finding the area between the graphs of two, functions. Iff and g are two functions defined on [a,b], with f ( x ) < g ( x ) for, all x in [a,b],we define the region between the graphs off and g on [a,b ] to be, the set of those points ( x ,y ) such that a < x < b and f ( x ) < y < g ( x ) ., , Figure 4.6.1.Find the area, of the parabolic doorway., , Example 2, , Sketch and find the area of the region between the graphs of x 2 and x, [ - 1,lI., , Solution, =x, , + 3 on, , The region is shaded in Fig. 4.6.2. It is not quite of the form we have been, dealing with; however, we may note that if we add to it the region under the, graph of x 2 on [ - 1, I], we obt$in the region under the graph of x 3 on, [ - 1,1]. Denoting the area of the shaded region by A , we have, , +, , +, , J-1 1 x 2 d x + A =J-I 1 ( x + 3 ) d x, or, I, , 1, , A = S - ] ( x + 3 ) d ~ - J - ~ ~ x ~ d 1x( x=+~ 3- - x2)dx., Evaluating the integral yields A, , =, , (4x 2 + 3 x - f x 3 ) /1,= 5 f . A, , The method of Example 2 can be used to show that if 0 < f ( x ) < g ( x ) for x, in [a,b], then the area of the region between the graphs off and g on [ a ,b ] is, equal to J i g ( x )dx - J ~ ( X dx, ) = Jh, [ g ( x ) - f ( x ) ] dx., , Figure 4'62. The area, the shaded region is the, difference between two, areas under graphs., , Find the area between the graphs of y, , Example 3, Solution, , Since 0, , < x 3 < x 2 on [0,1], by, , =, , x 2 and y, , =, , x 3 for x between O and 1., , the principle just stated the area is, , Copyright 1985 Springer-Verlag. All rights reserved.

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4.6 Applications of the Integral, , 24'1, , The same method works even if f ( x ) can take negative values. It is only the, difference between f ( x ) and g ( x ) which matters., , Area Between Graphs, If f ( x ) < g ( x ) for all x in [ a ,b], and f and g are integrable on [ a ,b], then, the area between the graphs off and g on [ a ,b] equals, , There is a heuristic argument for this formula for the area between graphs, which gives a useful way of remembering the formula and deriving similar, ones. We can think of the region between the graphs as being composed of, infinitely many "infinitesimally wide" rectangles, of width d x , one for each x, in [ a ,b]. (See Fig. 4.6.3.), , Figure 4.6.3. We may think, of the shaded region as, being composed of, infinitely many rectangles,, each of infinitesimal width., , The total area is then the "continuous sum" of the areas of these, rectangles. The height of the rectangle over x is h ( x ) = g ( x ) - f ( x ) , the area, of the rectangle is [ g ( x ) - f ( x ) ] d x , and the continuous sum of these areas is, the integral J : [ g ( x )- f ( x ) ] d x . This kind of infinitesimal argument was used, frequently in the early days of calculus, when it was considered to be perfectly, acceptable. Nowadays, we usually take the viewpoint of Archimedes, who, used infinitesimals to discover results which he later proved by more rigorous,, but much more tedious, arguments., Example 4, Solution, 1, , Sketch and find the area of the region between the graphs of x and x 2, [ - 2921., , + 1 on, , The region is shaded in Fig. 4.6.4. By the formula for the area between curves,, the area is, 2, , A, Y=x?+ I, , J 2 2 [ ( ~ 2 + l ) - ( ~ ) ] d x =( x 2 + l - x ) d x =, -, , -2, , - - -82 - 3, , =-, , 42 ,, , )., , If the graphs o f f and g intersect, then the area of the region between them, must be found by breaking the region up into smaller pieces and applying the, preceding method to each piece., , I, , Figure 4.6.4. What is the, area of the shaded region?, , Copyright 1985 Springer-Verlag. All rights reserved.

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242, , Chapter 4 The Integral, , between x = a and x = b, first plot the graphs and locate points where, f ( x ) = g ( x ) . Suppose, for example, that f ( x ) > g ( x ) for a < x < c , f ( c ), = g ( c ) and f ( x ) < g ( x ) for c < x < b, as in Fig. 4.6.5. Then the area is, , Figure 4.6.5. The area of, the shaded region equals, , Find the area of the shaded region in Fig. 4.6.6., , Example 5, Solution, , First we locate the intersection points by setting x 2 = x . This has the solutions, x = 0 or 1. Between 0 and 1, x 2 < X , and between 1 and 2, x 2 > X , SO the area, is, 1, , A=j"(x-x2)dx+, , I;', , (x2-xpx, , Figure 4.6.6. Find the, shaded area., , Example 6, Solution, , Find the area between the graphs ofv,, and x = 2., , = x3, , and y,, , = 3 x 2 - 2x, , between x, , =0, , The graphs are plotted in Fig. 4.6.7. They intersect when x 3 = 3 x 2 - 2 x , i.e.,, x ( x 2 - 3 x + 2 ) = 0, i.e., x ( x - 2)(x - 1) = 0, which has solutions x = 0, 1 ,, and 2, as in the figure. The area is thus, , Figure 4.6.7. The graphs, needed to find the area, between y = x3 and, , y, , = 3x2 - 2x., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.6 Applications of the Integral, , 243, , In the next problem, the intersection points of two graphs determine the limits, of integration., Example 7, , +, , The curves x = y2 and x = 1 iy2(neither of which is the graph of a, function y = f(x)) divide the xy plane into five regions, only one of which is, bounded. Sketch and find the area of this bounded region., , Solution If we plot x as a function of y , we obtain the graphs and region shown in Fig., 4.6.8. We use our general rule for the area between graphs, which gives, A =l-:(l, , + $ y 2 - y2)dy, , =J-%(l- $ y2, = @ - ;(@)'-[-fi(-c, , )~], , 3, , Figwe 4.6.8. T h e bounded, region determined b y, x = y 2 a n d x = 1 +ty2., , Example 8, , Solution, , =2fi-$(fi), , =@(2-, , (Note that the roles of x and y have been reversed in this example.) A, Find the area between the graphs x, , = y2 - 2, , and y = x., , This is a good illustration of the fact that sometimes it is wise to pause and, think about various methods at our disposal rather than simply plunging, ahead., First of all, we sketch the graphs, as in Fig. 4.6.9. We can plot x = y 2 - 2, either by writing y = +., and graphing these two square-root functions, or, preferably, by regarding x as a function of y and drawing the corresponding parabola., , Figure 4.6.9. Find the, shaded area., , Copyright 1985 Springer-Verlag. All rights reserved.

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244, , Chapter 4 The Integral, , The intersection points are obtained by setting y2 - 2 = y , which gives, , y, , =, , - 1 and 2., , Method 1, , Write y = lr J&%??, , Method 2, , Regard x as a function of y and obtain the area as, , for the first graph and obtain the area in two pieces as, , Method 2, while requiring the slight trick of regarding x as a function of y , is, simpler. A, The velocity-displacement relationship holds for all rates of change. If a, quantity Q depends on x and has a rate of change r, then by the fundamental, theorem,, , If the rate of change of Q with respect to x for a < x < b is given by, Q is obtained by integrating:, , r = f(x), then the change in, , This relationshi:, can be used in a variety of ways, depending on the interpretation of Q, r, and x. For example, we can view the box on page 230 as a, special instance with Q the position, r the velocity, and x the time., Example 9, , Solution, , Water is flowing into a tub at 3t2 + 6t liters per minute at time t, between, t = 0 and t = 2. How many liters enter the tub during this period?, Let Q(t) denote the number of liters at time t. Then Qt(t) = 3t2, , + 6t, so, , ., Thus, 20 liters enter the tub during the 3-minute interval. A, , The final example comes from economics. We recall from the discussion on, page 106 that the marginal revenue at production level x is Rt(x), where R is, the revenue., , Copyright 1985 Springer-Verlag. All rights reserved.

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4.6 Applications of the Integral, , Example 10, , T h e marginal revenue for a company a t production level x is given b y, 15 - 0 . 1 ~ If. R ( x ) denotes the revenue a n d R(0) = 0, find R(100)., , Ssiullon, , By the fundamental theorem, R(100) - R ( 0 ) = J,!,OO~'(x)dx.But R(0) = 0,, a n d the marginal revenue R ' ( x )is 15 - 0.1 x . Thus,, , Exercises for Section 4.6, 1. A parabolic arch with base 8 meters and height, 10 meters is erected. How much area does it, enclose?, 2. A parabolic arch with base 10 meters and height, 12 meters is erected. How much area does it, enclose?, 3. A swimming pool has the shape of the region, bounded by y = x 2 and y = 2. A swimming pool, cover is estimated to cost $2.00 per square foot., If one unit along each of the x and v, axes is 50, feet. then how much shotlld the cover cost?, 4. An artificial 1ake.with two bays has the shape of, the region above the curve .v = x4 - x 2 and below the line y = 8 ( x and y are measured in, kilometers). If the lake is 10 meters deep, how, many cubic meters of water does it hold?, 5. Find the area of the shaded region in Fig. 4.6.10., , area of the shaded region?, , ', , V ' ], , 15., , 16., , 3, , +, , x4+x2, , x59, , + x4, , 4(x6 - 1), x6+ 1, , and, , and, , +, +, , 7+- x4 x 2, x59 x4, , (3x6 - 1)(x6 - I ), x6(x6, , + 1), , on [ I , 21., 17. Find the area between the graphs of y = x3 and, y = 5x2 6x between x = 0 and x = 3., 18. Find the area between the graphs of y = x3 + 1, a n d y = x 2 - 1 betweenx= - 1 a n d x = 1., 19. The curves y = x3 and y = x divide the plane, into six regions, only two of which are bounded., Find the areas of the bounded regions., 20. The lines y = x and y = 2x and the curve y, = 2/x2 together divide the plane into several, regions, one of which is bounded. (a) How many, regions are there? (b) Find the area of the, bounded region., , +, , A x 2, 4, , 6. Find the area of the shaded region in Fig. 4.6.11., Find ihe area between the graphs on the designated, intervals in Exercises 7-10., 7. y = (2/x2) x4 and y = 1 between x = 1 and, x = 2., 8. y = x4 and y = x3 between x = - 1 and x = 0., 9. y = f ia n d y = x b e t w e e n x = O a n d x = 1., 10. y = 3fia n d y = 1/x2 betweenx = 8 a n d x = 27., In Exercises 11-16, find the area between the graphs of, each pair of functions on the given interval:, 11. x and x4 on [O, I]., 12. x2 and 4x4 on [2,3]., 13. 3x2 and x4 2 on [ - f ,f]., 14. x4 1 and 1/x2 on [ I , 21., , +, , +, , 245, , +, , Figure 4.6.11. What is the, area of the shaded region?, , Copyright 1985 Springer-Verlag. All rights reserved.

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Chapter 4 The integral, , 246, , In Exercises 21-24, find the area between the given, graphs., 21. x = y 2 - 3 a n d x = 2 y, 22. x = y 2 + 8 a n d x = -6y, 23. x = y 3 a n d y = 2 x, 24. x = y 4 - 2 a n d x = , v 2, 25. Water is flowing out of a reservoir at 300t2 liters, per second for t between 0 and 5. How many, liters are released in this period?, 26. Air is escaping from a balloon at 3t2 + 2t cubic, centimeters per second for t between 1 and 3., How much air escapes during this period?, 27. Suppose that the marginal revenue of a company, at production level x is given by 30 - 0 . 0 2 ~0.0001x2. If R(0) = 0, find R(300)., 28. (a) Find the total revenue R(x) from selling x, units of a product, if the marginal revenue is, 36 - 0 . 0 1 ~+ 0.00015x2 and R(0) = 0., (b) How much revenue is produced when the, items from x = 50 to x = 100 are sold?, 29. (a) Use calculus to find the area of the triangle, whose vertices are (0,0), (a, h), and (b,O)., (Assume 0 < a < b and 0 < h . ) Compare, your result with a formula from geometry., (b) Repeat part (a) for the case 0 < b < a ., 30. In Example 2 of Section 4.5, it was shown that, JX(1 + x ) * ~ =, x 9(7r - I)(I + x)' + C. Use, this result to find the area under the graph of, 1 + x(l + x)* between x = - 1 and x = 1., 31. Fill in the blank, referring to the Supplement to, Section 4.3: Light meter is to energy-storage dial, as -is to odometer., 32. Fill in the blank: Marginal revenue is to, , 34. A small gold mine in northern Nevada was reopened in January 1979, producing 500,000 tons, of ore in the first year. Let A(t) be the number of, tons of ore produced, in thousands, t years after, 1979. Productivity A'(t) is expected to decline by, 20,000 tons per year until 1990., (a) Find a formula for Af(t), assuming that the, production decline is constant., (b) How much ore is mined, approximately, T, years after 1979 during a time period At?, (c) Find, by definite integration, the predicted, number of tons of ore to be mined from, 1981 through 1986., 35. Let W ( t )be the number of words learned after t, minutes are spent memorizing a French vocab), ulary list. Typically, W ( 0 )= 0 and W 1 ( t =, 4(t/ 100) - 3(t/ 100)'., (a) Apply the fundamental theorem of calculus, to show that, W ( t )= Jt[4(x/100), , -, , 3(x/100)'] dx., , 0, , (b) Evaluate the integral in part (a)., (c) How many words are learned after 1 hour, and 40 minutes of study?, 36. The region under the graph y = l / x 2 on [ l , 41 is, to be divided into two parts of equal area by a, vertical line. Where should the line be drawn?, +37. Where would you draw a horizontal line to divide the region in the preceding exercise into two, parts of equal area?, +38. Find the area in square centimeters, correct to, within I square centimeter, of the region in Fig., 4.6.13., , as /(XI is to LXf(s)ds., 33. A circus tent is equipped with four exhaust fans, at one end, each capable of moving 5500 cubic, feet of air per minute. The rectangular base of, the tent is 80 feet by 180 feet. Each corner is, supported by a 20-foot-high post and the roof is, supported by a center beam which is 32 feet off, the ground and runs down the center of the tent, for its 180-foot length. Canvas drapes in a parabolic shape from the center beam to the sides 20, feet off the ground. (See Fig. 4.6.12.) Determine, the elapsed time for a complete change of air in, the tent enclosure., , Figure 4.6.13. Find the area, of the "blob.", , Exhaust fans, , 3, , Figure 4.6.12. The circus, tent in Exercise 33., , Copyright 1985 Springer-Verlag. All rights reserved.

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Review Exercises for Chapter 4, , 247, , Review Exercises for Chapter 4, O 21. (a) Find upper and lower sums for, , Compute the sums in Exercises 1-8., 4, , 1., , 2 i2., i= l, 3, , 2., , 2 j3., , J=, , within 0.2 of one another. (b) Look at the average of these sums. Can you guess what the exact, integral is?, , 1, , 22. Find upper and lower sums for, , & of one another., , L3:, , - dx within, , 500, , 5. c ( 3 i + 7 ) ., i= l, , n+3 .2, 6., , 2, i=,, , 1+, , (b) Find the area under the graph of the function x / ( l x212,from x = 0 to x = 1 ., 24. Find ( d / d x ) [ x 3 / ( l+ x3)].( b ) Find the area under the graph of x 2 / ( 1 x3)' from x = 1 to, x = 2., 25. Find the area under the graph of y = mx b, from x = a , to x = a2 and verify your answer by, using plane geometry. Assume that m x + b > 0, on [al,a21., 26. Find the area under the graph of the function, y = ( 1 / x 2 ) + x + 1 fromx= 1 tox=2., 27. Find the area under the graph of y = x 2 + 1, from - 1 to 2 and sketch the region., 28. Find the area under the graph of, , +, , 1 (n is a non-negative integer)., 1, , +, , 10, , 7., , 2 [ ( i+, , - i4]., , i=O, , +, , 9. Let f be defined on [O,11 by, , from x = - 1 to x = 1 and sketch., 29. (a) Verify the integration formula, , Find b 1 f ( x j d x ., , 10. Let f be defined by, , Find J P 2 ~ ( xdx., ), 11. Interpret the integral in Exercise 9 in terms of, distances and velocities., 12. Interpret the integral in Exercise 10 in terms of, distances and velocities., Evaluate the definite integrals in Exercises 13-16., , 2s2, , ds, , 16., , l2, , x2, , 3x, x+l, , +, , +, , ( b ) Find the area under the graph of y =, x 2 / ( x 3+ 6)2 between x = 0 and x = 2., 30. Find the area between the graphs of y = x 3 and, y = 5 x 2 + 2 x between x = 0 and x = 2. Sketch., 3 1. The curves y = x6 - 3 and y = - x 2 - 1 divide, the plane into five regions, one of which is, bounded. Find its area., 32. Find the area of each of the numbered regions in, Fig. 4.R. 1., , dx., , Find the area under the graphs of the functions between the indicated limits in Exercises 17-20., 17. y = x 3 + x 2 , 0 < x < 1., , 18. y = x 2 + 2 x + 1 , 1 < ~ < 2 ., x4, , Figure 4.R.1. Find the area, of the numbered regions., , Copyright 1985 Springer-Verlag. All rights reserved.

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248, , Chapter 4 The Integral, , 33. Find the area of the region bounded by the, graphs x = y 2 - 6 and x = -v., 34. Find the area of the region bounded by the, graphs = x2 - 2 and y = 2 - x2., 35. An object is thrown at t = 0 from an airplane,, and it has vertical velocity u = - 10 - 321 feet, per second at time t . If the object is still falling, after 10 seconds, what can you say about the, altitude of the plane at t = O?, 36. Suppose the velocity of an object at position x is, x n , where n is some integer +=1. Find the time, required to travel from x = & to x = 1, 37. (a) At time t = 0, a container has 1 liter of, water in it. Water is poured in at the rate of, 3t2 - 2t 3 liters per minute (t = time in, minutes). If the container has a leak which, can drain 2 liters per minute, how much, water is in the container at the end of 3, minutes?, (b) What if the leak is 4 liters per minute?, (c) What if the leak is 8 liters per minute?, [Hint: What happens if the tank is empty, for a while?], 38. Water is poured into a container at a rate of t, liters per minute. At the same time, water is, leaking out at the rate of t2 liters per minute., Assume that the container is empty at t = 0., (a) When does the amount of water in the container reach its maximum?, (b) When is the container empty again?, 39. Suppose that a supply curve p = S(x) and a demand curve p = D(x) are graphed and that there, is a unique point (a, b) at which supply equals, demand ( p = price/unit in dollars, x = number, of units). The (signed) area enclosed by x = 0,, x = a , p = b, and p = D(x) is called the consumer's surplus or the consumer's loss depending on, whether the sign is positive or negative. Similarly,, the (signed) area enclosed by x = 0, x = a, p = b,, and p = S(x) is called the producer's surplus or, the producer's loss depending on whether the sign, is positive or negative., (a) Let D(x) > b. Explain why the consumer's, , +, , sutplus is, , 1 4D (x), , -, , b] dx., , (b) Let S(x) 9 b. Explain why the producer's, surplus is jo"[b - Six)] dx., (c) "If the price stabilizes at $6 per unit, then, some people are still willing to pay a higher, price, but benefit by paying the lower price, of $6 per unit. The total of these benefits, over [O,a] is the consumer's surplus." Explain this in the language of integration., (d) Find the consumer's and producer's surplus, for the supply curve p = x2/8 and the de1., mand curve p = - (x/4), 40. The demand for wood products in 1975 was, about 12.6 billion cubic feet. By measuring order, , +, , increases, it was determined that x years after, 1975, the demand increased by 9x/1000; that is,, D1(x) = 9x/1000, where D(x) is the demand x, years after 1975, in billions of cubic feet., (a) Use the fundamental theorem of calculus to, show that, , (b) Find D(x)., (c) Find the demand for wood in 1982., 4 1. Suppose that an object on the x axis has velocity, v = t2 - 4t - 5. How far does it travel between, t = 0 and t = 6?, 42. Show that the actual distance travelled by a bus, with velocity v = f(t) is Ji(f(t)( dt. What condition on v = f(t) means that the bus made a, round trip befween t = a and t = b?, 43. A rock is dropped off a bridge over a gorge. The, sound of the splash is heard 5.6 seconds after the, rock was dropped. (Assume the rock falls with, velocity 321 feet per second and sound travels at, 1080 feet per second.), (a) Show by integration that the rock falls 16t2, feet after t seconds, and that the sound of, the splash travels 1080t feet in t seconds., (b) The time T required for the rock to hit the, water must satisfy 1 6 ~ =, ' 1080i5.6 - T),, because the rock and the sound wave travel, equal distances. Find T., (c) Find the height of the bridge., (d) Find the number of seconds required for the, sound of the splash to travel from the water, to the bridge., 44. The current I(t) and charge Q(t) at time t (in, amperes and coulombs, respectively) in a circuit, are related by the equation I(t) = Qf(t)., (a) Given Q(0) = 1, use the fundamental theorem of calculus to justify the formula Q(t), = 1 + jbI(r)dr., (b) The voltage drop V (in volts) across a resistor of resistance R ohms is related to the, current I (in amperes) by the formula, V = RI. Suppose that in a simple circuit, with a resistor made of nichrome wire, V, = 4.36, R = 1, and Q(0) = 1. Find Q(t)., (c) Repeat (b) for a circuit with a 12-volt battery and 4-ohm resistance., 45. A ruptured sewer line causes lake contamination, near a ski resort. The concentration C(t) of bacteria (number per cubic centimeter) after t days, is given by C1(t) = 103(t - 7), 0 < t 9 6, after, treatment of the lake at t = 0., (a) An inspector will be sent out after the bacteria concentration has dropped to half its, original value C(0). On which day should, the inspector be sent if C(0) = 40,000?, (b) What is the total change in the concentration from the fourth day to the sixth day?, , Copyright 1985 Springer-Verlag. All rights reserved.

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Review Exercises for Chapter 4, , 46. A baseball is thrown vertically upwards from the, ground with an initial upward velocity of 50 feet, per second. How far has it travelled when it, strikes the ground?, 47. Let, , 48. Let, , Compute L 2 y ( r )dt., d J x -Ldr., 49. Calculate dx 0 I + S ~, d JX, 50. Calculate dt., dx 2 1 - t 4, 5 1 . Find the area between the graph of the function, in Exercise 47 and the x axis., d, +52. Find dx. (Hint: See Exercise, , 249, , +54. Find l b ( 3 x 2 + x) dx "by hand.", *55. Let f be defined on [0, 11 by, , (a) Show that there are no upper sums for f on, [O, I], and hence that f is not integrable., (b) Show that every number less than 2 is a, lower sum. [Nint: Use step functions which, are zero on an interval [O,6) and approximate f very closely on [ e , 11. Take 6 small, and use the integrability off on [ E , I].], (c) Show that no number greater than or equal, to 2 is a lower sum. [Hint: Show ~ f ( f (+~ ), j'y(x)dx < 2 for all E in (0, I).], (d) If you had to assign a value to Jhf(x)dx,, what value would you assign?, +56. Modeling your discussion after the preceding, exercise, find the upper or lower sums for each of, the following functions on [0, 1 1 :, , 43, Section 4.5.), , Copyright 1985 Springer-Verlag. All rights reserved.

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Chapter 9, , Many physical and geometric quantities can be expressed as integrals., Our applications of integration in Chapter 4 were limited to area, distancevelocity, and rate problems. In this chapter, we will see how to use integrals to, set up problems involving volumes, averages, centers of mass, work, energy,, and power. The techniques developed in Chapter 7 make it possible to solve, many of these problems completely., , 9.1 Volumes by the Slice, Method, The volume of a solid region is an integral of its cross-sectional areas., By thinking of a region in space as being composed of "infinitesimally thin, slices," we shall obtain a formula for volumes in terms of the areas of slices. In, this section, we apply the formula in a variety of special cases. Further, methods for calculating volumes will appear when we study multiple integration in Chapter 17., , Figure 9.1.1. The area of, the shaded region is, J t l ( x )dx., , We will develop the slice method for volumes by analogy with the, computation of areas by integration. Iff and g are functions with f(x) & g(x), on [a, b], then the area between the graphs off and g is [ g(x) - f(x)] dx (see, Section 4.6). We recall the infinitesimal argument for this formula. Think of, the region as being composed of infinitesimally thin strips obtained by cutting, with lines perpendicular to the x axis. Denote the vertical line through x by, Lx; the intersection of Lx with the region between the graphs has length, l(x) = g(x) -:f(x), and the corresponding "infinitesimal rectangle" with thickness dx has area I(x) dx (= height X width) (see Fig. 9.1.1). The area of the, entire region, obtained by "summing" the infinitesimal areas, is, , Given a region surrounded by a closed curve, we can often use, the same formula Jb,l(x)dx to find its area. To implement this, we position it conveniently with respect to the axes and .determine a and b, by noting where the ends of the region are. We determine I(x) by using the geometry of the situation at hand. This is done for a disk of radius r in, , Copyright 1985 Springer-Verlag. All rights reserved.

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420, , Chapter 9 Applications of Integration, , Fig. 9.1.2. We may evaluate the integral 1'_,2 I/dx by using integral, tables to obtain the answer vr2, in agreement with elementary geometry. (One, can also readily evaluate integrals of this type by using the substitution, x = r cos 8.), , Figure 9.1.2. Area of the, disk = j ? , 2 J Z 7, , dx., , To find the volume of a solid region, we imagine it sliced by a family of, parallel planes: The plane P, is perpendicular to a fixed x axis in space at a, distance x from a reference point (Fig. 9.1.3)., The plane P, cuts the solid in a plane. region R,; the corresponding, "infinitesimal piece" of the solid is a slab whose base is a region R, and whose, thickness is dx (Fig. 9.1.4). The volume of such a cylinder is equal to the area, , Figure 9.1.3. The plane P,, is at distance x horn Po., , Figure 9.1.4. An infinitesimally thin slice of a solid., , of the base R , times the thickness dx. If we denote the area of R, by A(x),, then this volume is A (x) dx. Thus the volume of the entire solid, obtained by, summing, is the integral ~ b , ~ ( x ) dwhere, x,, the limits a and b are determined, by the ends of the solid., , The Slice Method, Let S be a solid and P, be a family of parallel planes such that:, 1. S lies between Pa and P,;, 2. the area of the slice of S cut by P, is A(x)., , Then the volume of S is equal to, , lb~, (x) dx., , Copyright 1985 Springer-Verlag. All rights reserved.

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9.1 Volumes by the Slice Method, , 421, , The slice method can also be justified using step functions. We shall see how, to do this below., In simple cases, the areas A (x) can be computed by elementary geometry., For more complicated problems, it may be necessary to do a preliminary, integration to find the A(x)'s themselves., Example 11, Solution, , Find the volume of a ball' of radius r., Draw the ball above the x axis as in Fig. 9.1.5., , region R, of area A ( x ), , Figure 9.1.5. The area of, the slice at x of a ball of, radius r is, A ( x ) = n-(r2 - x2)., , Y, , Let the plane Po pass through the center of the ball. The ball lies, The area of, between P-, and P,,and the slice R, is a disk of radius, the slice is s x (radiu~)~;, i.e., A(x) = ~(d-)~, = n(r2 - x2). Thus the, volume is, , .-d, , Example 2, , Find the volume of the conical solid in Fig. 9.1.6. (The base is a circle.), , Figure 9.1.6. Find the, volume of this oblique, circular cone., , Solution, , Figure9.1.7. IDEI/IABI, = .lGEI/lGBI = IG~IIIGCI, by similar triangles. But, lABl= r, lGCl= h, and, IGFI = h - x , andso, 1 D E 1 = [(h - x ) / h ] r ., , We let the x axis be vertical and choose the family P, of planes such that Po, contains the base of the cone and P, is at distance x above Po.Then the cone, lies between Po and P,, and the plane section by P, is a disk with radius, [(h - x)/h]r and area n[(h - x)/h12r2 (see Fig. 9.1.7). By the slice method,, ' A sphere is the set of points in space at a fixed distance from a point. A ball is the solid region, enclosed by a sphere, just as a disk is the plane region enclosed by a circle., , Copyright 1985 Springer-Verlag. All rights reserved.

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422, , Chapter 9 Applications of Integration, , the volume is, , Example 3, , Find the volume of the solid W shown in Fig. 9.1.8. It can be thought of as a, wedge-shaped piece of a cylindrical tree of radius r obtained by making, two saw cuts to the tree's center, one horizontally and one at an, angle 0., X, , Figure 9.1.8. Find the, volume of the wedge W., , Solution, , With the setup in Fig. 9.1.8, we slice W by planes to produce triangles R, of, and its height, area A(x) as shown. The base b of the triangle is b =, is h = b tan0 = $-tan$., Thus, A (x) = $ bh = $(r2- x2)tan0. Hence,, the volume is, , d, m, ,, , Notice that even though we started with a region with a circular boundary, n, does not occur in the answer! A, Example 4, , A ball of radius r is cut into three pieces by parallel planes at a distance of r/3, on each side of the center. Find the volume of each piece., , Solution, , The middle piece lies between the planes P_,/, and P,/, of Example 1, and, the area function is A(x) = n(r2 - x2) as before, so the volume of the middle, piece is, , This leaves a volume of (4 - E)nr3 = nr3 to be divided between the two, outside pieces. Since they are congruent, each of them has volume nr3. (YOU, may check this by computing JiI3n(r2 - x2)dx.) A, ., One way to construct a solid is to take a plane region R,,as shown in Fig. 9.1.9, and revolve it around the x axis so that it sweeps out a solid region S. Such, solids are common in woodworking shops (lathe-tooled table legs), in pottery, , Copyright 1985 Springer-Verlag. All rights reserved.

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9.1 Volumes by the Slice Method, , 423, , Revolve, , 3, , Figure 9.1.9. S is the solid, of revolution obtained by, revolving the plane region, R about the x axis., , studios (wheel-thrown pots), and in nature (unicellular organisms).' They are, called solids of revolution and are said to have axial symmetry., Suppose that region R is bounded by the lines x = a, x = b, and y = 0,, and by the graph of the functiony = f(x). To compute the volume of S by the, slice method, we use the family of planes perpendicular to the x axis, with Po, passing through the origin. The plane section of S by P, is a circular disk of, radius f(x) (see Fig. 9.1. lo), so its area A (x) is m [ f(x)12. By the basic formula, of the slice method, the volume of S is, , Figure 9.1.10. The volume, of a solid of revolution, obtained by the disk, method., , I, , k x A, We use the term "disk method" for this special case of the slice method since, the slices are disks., , Disk Method, The volume of the solid of revolution obtained by revolving the region, under the graph of a (non-negative) function f(x) on [a,b] about the x, , Exarnple 5, , The region under the graph of x2 on [0, I] is revolved about the x axis. Sketch, the resulting solid and find its volume., , Solution, , The solid, which is shaped something like a trumpet, is sketched in Fig. 9.1.1 1., , Figure 9.1.11. The volume, of this solid of revolution is, nJ:(x2)' dx., See D'Arcy Thompson, On Growth and Form, abridged edition, Cambridge University Press, , (1969)., , Copyright 1985 Springer-Verlag. All rights reserved.

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424, , Chapter 9 Applications of Integration, , According to the disk method, its volume is, , Example 6, , The region between the graphs of sinx and x on [0, n/2] is revolved about the, x axis. Sketch the resulting solid and find its volume., , Solution, , The solid is sketched in Fig. 9.1.12. It has the form of a hollowed-out cone., , Figure 9.1.12. The region, between the graphs of sin x, and x is revolved about the, x axis., , The volume is that of the cone minus that of the hole. The cone is obtained by, revolving the region under the graph of x on [O, 11 about the axis, so its, volume is, , The hole is obtained by revolving the region under the graph of sinx on, [O, n/2] about the x axis, so its volume is, nc'2sin2x dx = ni'/2, , -, , 2x dx, , (since cos 2x, , =, , 1 - 2 sin2x), , Thus the volume of our solid is r4/24 - r 2 / 4 w 1.59. A, The volume of the solid obtained by rotating the region between the graphs of, two functionsf and g (with f(x) < g(x) on [a, b])can be done as in Example 6, or by the washer method which proceeds as follows. In Fig. 9.1.13, the volume, of the shaded region (the "washer") is the area x thickness. The area of the, washer is the area of the complete disk minus that of the hole. Thus, the, washer's volume is, , r.(, , mi2- n [ f ( x ) ~ 2dx., ), , Thus, the total volume is, , The reader should notice that this method gives the same answer as one finds, by using the method of Example 6., , Copyright 1985 Springer-Verlag. All rights reserved.

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9.1 Volumes by the Slice Method, , 425, , Figure 9.1.13. The washer, method., , 9.1.14. A "stepwise, cylindrical" solid., , Our formula for volumes by the slice method was introduced via infinitesimal~.A more rigorous argument for the formula is based on the use of upper, and lower sums.3To present this argument, we first look at the case where S is, composed of n cylinders, as in Fig. 9.1.14., If the ith cylinder Ci lies between the planes P , - , and Pxi and has, cross-sectional area. k,, then the function A (x) is a step function on the, interval [x,, x,]; in fact, A(x) = k, for x in (xi-, ,xi). TI, :volume of C, is the, product of its base area k, by its height Ax, = xi - xi-,, so the volume of the, total figure is C?, ,k,Ax,; but this is just the integral pOA(x)dx of the step, function A(x). We conclude that if S is a "stepwise cylindrical" solid between, the planes Pa and P, , then, volume s =, , l, , b, , A (x) dx., , If S is a reasonably "smooth" solid region, we expect that it can be squeezed, arbitrarily closely between stepwise cylindrical regions on the inside and, outside. Specifically, for every positive number E , there should be a stepwise, cylindrical region S, inside S and another such region Sooutside S such that, (volume So) - (volume S,) < E . If Ai(x) and A,(x) are the corresponding, functions, then A, and A, are step functions, and we have the inequality, A,(x) < A (x) < A,(x), so, volume S, =, , Lb~,, , (x) dx <, , (x) dx <, , Ib~,, , (x) dx = volume So., , Since S encloses Si and So encloses S, volume Si < volume S < volume So., Thus the numbers (volume S ) and, (x) dx both belong to the same interval, [(volume S,), (volume So)], which has length less than E . It follows that the, differencebetween (volume S ) and Jb,~(x)dxis less than any positive number, E ; the only way this can be so is if the two numbers are equal., , Supplement to Section 9.1:, Gavalieri's Delicatessen, The idea behind the slice method goes back, beyond the invention of calculus,, to Francesco Bonaventura Cavalieri (1598-1647), a student of Galileo and, then professor at the University of Bologna. An accurate report of the events, leading to Cavalieri's discovery is not available, so we have taken the liberty of, inventing one., Even this justification, as we present it, is not yet completely satisfactory. For example, do we, get the same answer if we slice the solid a different way? The answer is yes, but the proof uses, multiple integrals (see Chapter 17)., , Copyright 1985 Springer-Verlag. All rights reserved.

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426, , Chapter 9 Applications of Integration, , Cavalieri's delicatessen usually produced bologna in cylindrical form, so, ., length. One day, the, that the volume would be computed as 1 ~ radius2., casings were a bit weak, and the bologna came out with odd bulges. The scale, was not working that day, either, so the only way to compute the price of the, bologna was in terms of its volume., Cavalieri took his best knife and sliced the bologna into n very thin slices,, each of thickness Ax, and measured the radii r , ,r,, . . . , r,, of the slices, (fortunately, they were round). He then estimated the volume to be, C'i'=,~r?Ax~,, the sum of the volumes of the slices., Cavalieri was moonlighting from his regular job as a professor at the, University of Bologna. That afternoon, he went back to his desk and began, the book "Geometria indivisibilium continuorum nova quandum ratione, promota" ("Geometry shows the continuous indivisibility between new rations, and getting promoted"), in which he stated what is now known4 as Cavalieri's, principle :, , If two solids are sliced by a family of parallel planes in such a way that, corresponding sections have equal areas, then the two solids have the same, volume., The book was such a success that Cavalieri sold his delicatessen and, retired to a life of occasional teaching and eternal glory., Honest!, , Exercises for Section 9.1, In Exercises 1-4, use the slice method to find the, volume of the indicated solid., , a, , Figure 9.1.15. The solids, for Exercises 1-4., , 1. The solid in Fig. 9.1.15(a); each plane section is, a circle of radius I., 2. The parallelepiped in Fig. 9.1.15(b); the base is a, rectangle with sides a and b., 3. The solid in Fig. 9.1.15(c); the base is a figure of, area A and the figure at height x has area, A, = [(h - x ) / h I 2 ~ ., 4. The solid in Fig. 9.1.15(d); the base is a right, triangle with sides b and I., 5. Find the volume of the tent in Fig. 9.1.16. The, plane section at height x above the base is a, square of side i(6 - x ) ~ i. The height of the, tent is 5 feet., , 6. What would the volume of the tent in the previous exercise be if the base and cross sections, were equilateral triangles instead of squares (with, the same side lengths)?, 7. The base of a solid S is the disk in the xy plane, with radius 1 and center (0,O). Each section of S, cut by a plane perpendicular to the x axis is an, equilateral triangle. Find the volume of S ., , Copyright 1985 Springer-Verlag. All rights reserved.

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9.1 Volumes by the Slice Method, , 8. A plastic container is to have the shape of a, truncated pyramid with upper and lower bases, being squares of side length 10 and 6 centimeters, respectively. How high should the container, be to hold exactly one liter (= 1000 cubic centimeters)?, 9. The conical solid in Fig. 9.1.6 is to be cut by, horizontal planes into four pieces of equal vorume. Where should the cuts be made? [Hint:, What is the volume of the portion of the cone, above the plane P,?], 10. The tent in Exercise 5 is to be cut into two pieces, of equal volume by a plane parallel to the base., Where should the cut be made?, (a) Express your answer as the root of a fifthdegree polynomial., 8 (b) Find an approximate solution using the, method of bisection., 11. A wedge is cut in a tree of radius 0.5 meter by, making two cuts to the tree's center, one horizontal and another at an angle of 15' to the first., Find the volume of the wedge., 12. A wedge is cut in a tree of radius 2 feet by, making two cuts to the tree's center, one horizontal and another at an angle of 20" to the first., Find the volume of the wedge., 13. Find the volume of the solid in Fig. 9.1.17(a)., 14. Find the volume of the solid in Fig. 9.1.17(b)., , 20. The region under the graph of J, G, on2, lo, 11., 21. The semicircular region with center (a,O) and, radius r (assume that 0 < r < a, y > 0)., and, 22. The region between the graphs of, 5 + x on [0, I]. (Evaluate the integral using geometry or the tables.), 23. The square region with vertices (4,6), (5,6),, (5,7), and (4,7)., 24. The region in Exercise 23 moved 2 units upward., 25. The region in Exercise 23 rotated by 45" around, its center., 26. The triangular region with vertices (1, I), (2,2),, and (3,l)., *27. A vase with axial symmetry has the cross section, shown in Fig. 9.1.18 when it is cut by a plane, through its axis of symmetry. Find the volume of, the vase to the nearest cubic centimeter., , *28. A right circular cone of base radius r and height, 14 is to be cut into three equal pieces by parallel, planes which are parallel to the base. Where, should the cuts be made?, *29. Find the formula for the volume of a doughnut, with outside radius R and a hole of radius r., *30. Use the fact that the area of a disk of radius r is, ?rr2= Y-JJdx to compute the area inside the ellipse y2/4 x2 = r2., *3 1. Prove Cavalieri's principle., *32. Using Cavalieri's principle, without integration,, find a relation between the volumes of:, (a) a hemisphere of radius 1;, (b) a right circular cone of base radius 1 and, height 1;, (c) a right circular cylinder of base radius 1 and, height 1., [Hint: Consider two of the solids side by side as, a single solid. The sum of two volumes will equal, the third.], , +, , Figure 9.1.17. Find the, volumes of these solids., In Exercises 15-26, find the volume of the solid obtained by revolving each of the given regions about the, x axis and sketch the region., 15. The region under the graph of 3x + 1 on [O, 21., 16. The region under the graph of 2 - (x - 1)' on, [O, 21., 17. The region under the graph of cos x 1 on, [O, 2 4 ., 18. The region under the graph of cos 2x on [0, ?r/4]., 19. The region under the graph of x(x - 1)' on, [1,21., , +, , 427, , Copyright 1985 Springer-Verlag. All rights reserved.

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9.3 Average Values and the Mean Value Theorem for Integrals, , 16. Find the volume in Exercise 15 by the slice, method., 17. Find the volume of the solid torus obtained by, rotating the disk (x - 3)' y 2 < 4 about the y, axis., 18. Find the volume of the solid torus obtained by, rotating the disk x2 (y - 5)' < 9 about the x, axis., 19. A spherical shell of radius r and thickness h is,, by definition, the region between two concentric, spheres of radius r - h/2 and r + h/2., (a) Find a formula for the volume V(r, h) of a, spherical shell of radius r and thickness h., (b) For fixed r, what is (d/dh)V(r,h) when, h = O? Interpret your result in terms of the, surface area of the sphere., 20. In Exercise 19, find (d/dr) V(r, h) when h is held, fixed. Give a geometric interpretation of your, answer., , +, , +, , 433, , *21. (a) Find the volume of the solid torus T,,,obtained by rotating the disk with radius a and, center (b, 0) about they axis, 0 < a < b., (b) What is the volume of the region between, the solid tori Ta,band T,,,,,, assuming, O<a+h<b?, (c) Using the result in (b), guess a formula for, the area of the torus which is the surface of, T,,,. (Compare Exercise 19)., *22. Let f(x) and g(y) be inverse functions with f(a), = a , f(b) = p, 0 < a < b, 0 < a < p. Show that, , Interpret this statement geometrically., *23. Use Exercise 22 to compute the volume of the, solid obtained by revolving the graph y =, cos-lx, 0 < x < 1, about the x axis., , 9.3 Average Values and, the Mean Value Theorem for Integrals, The average height of a region under a graph is its area divided by the length of, the base., The average value of a function on an interval will be defined in terms of an, integral, just as the average or mean of a list a , , . . . , a, of n numbers is, defined in terms of a sum as (l/n)Cr= ,ai., If a grain dealer buys wheat from n farmers, buying bi bushels from the, ith farmer at the price of pi dollars per bushel, the average price is determined, not by taking the simple average of the pi's, but rather by the "weighted, average":, , -, , Paverage -, , L'r=I pib; - total dollars, C?=,bi total bushels, , ', , If a cyclist changes speed intermittently, travelling at v, miles per hour, from to to t,, v, miles per hour from tl to t,, and so on up to time t,, then the, average speed for the trip is, "average, , -, , Cr=10i(~i- ti- I) - total miles, total hours, Cr=,(ti - ti- ,), , If, in either of the last two examples, the b;s or (ti - ti- ,)'s are all equal,, then the average value is simply the usual average of the pi's or the vi's., I f f is a step function on [a, b] and we have a partition (xo,x,, . . . , x,), with f(x) = ki on (xi-, ,xi), then the average value off on the interval [a, b] is, defined to be, , In other words, each interval is weighted by its length., How can we define the average value of a function which is not a step, function? For instance, it is common to talk of the average temperature at a, , Copyright 1985 Springer-Verlag. All rights reserved.

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434, , Chapter 9 Applications of Integration, , place on earth, although the temperature is not a step function. We may, rewrite (1) as, , and this leads us to adopt formula (2) as the definition of the average value for, any integrable function f, not just a step function., , Example 1, Solution, , Example 2, , Find the average value of f(x) = x2 on [0,2]., By definition, we have, , Show that if v = f(t) is the velocity of a moving object, then the definition of, agrees with the usual notion of average velocity., , ij[a,b], , Solution, , By the definition,, , but J:vdt is the distance travelled between t = a and t = b, so Ula,b.l =, (distance travelled)/(time of travel), which is the usual definition of average, velocity. A, Example 3, Solution, , Find the average value of, , Solution, , s on [ -, , 1,1]., , Jsl , d l dx)/2;, , but, , = 1, which is, , f n,, , By the formula for average values,, = (1:, I!., dx is the area of the upper semicircle of x2, , ,Js, , so, Example 4, , d, , +, , - ,,,] = 7r/4 a 0.785., , A, , Find, x2sin(x3), , =, , n, , J'x2sin x3dx, 0, , du, sin u 3, , r3, , (substituting u, , = x3), , Copyright 1985 Springer-Verlag. All rights reserved.

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9.3 Average Values and the Mean Value Theorem for Integrals, , 435, , We may rewrite the definition of the average value in the form, , and the right-hand side can be interpreted as the integral of a constant, function:, , Geometrically, the average value is the height of the rectangle with base, [a, b] which has the same area as the region under the graph off (see Fig., 9.3.1). Physically, if the graph off is a picture of the surface of wavy water in, a narrow channel, then the average value off is the height of the water when it, settles., An important property of average values is given in the following statement:, , f(.~)[O,b,, , Figure 9.3.1. The average, value is defined so that the, area of the rectangle equals, the area under the graph., The dots on the x &is, indicate places where the, average value is attained., , If m, , < f(x) < M for all x, , in [a, b], then m, , -, , < f ( ~ ) [ ~ ,<b ]M., , Indeed, the integrals J:m dx and J b , ~ d xare lower and upper sums for f on, [a, b], so, , <, , m(b - a), , Ib, , f(x)dx< M(b - a)., , Dividing by (b - a) gives the desired result., By the extreme value theorem (Section 3.5), f(x) attains a minimum value, m and a maximum value M on [a, b]. Then m < f(x) < M for x in [a, b], so, lies between m and M, by the preceding proposition. By the first, version of the intermediate value theorem (Section 3.1), applied to the interval, between the points where f(x) = m and f(x) = M, we conclude, - that there is a, x, in this interval (and thus in [a, b]), such that f(x,) = f ( ~ ) [ ~ , b ] ., In other words, we have proved that the average value of a continuous, function on an interval is always attained somewhere on the interval. This, result is known as the mean value theorem for integrals., , fola,bl, , Mean Value Theorem for Integrals, Let f be continuous on [a, b]. Then there is a point x, in (a, b) such that, , Notice that in Fig. 9.3.1, the mean value is attained at three different points., Example 5, , Give another proof of the mean value theorem for integrals by using the, fundamental theorem of calculus and the mean value theorem for derivatives., , Solution, , Let f be continuous on [a, b], and define F(x) = pJ(s)ds. By the fundamental, theorem of calculus (alternative version), Ff(x) = f(x) for x in (a, b). (Exercise, 29 asks you to verify that F is continuous at a and b-we accept it here.) By, the mean value theorem for derivatives, there is some x, in (a, b) such that, Ff(xo)=, , F(b> - F(a), b-a, , Copyright 1985 Springer-Verlag. All rights reserved.

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Chapter 9 Applications of Integration, , 436, , Substituting for F a n d F' in terms off, we have, , f ( ~ 0 =), , J ~ ( x dx, ) - pJ(x) dx, , b-a, , - Jbef(x) dx, b-a, , = f ( ~ [a,b], ), , which establishes the mean value theorem for integrals., , 9, , A, , Exercises for Section 9.3, In Exercises 1-4, find the average value of the given, function on the given interval., 2. x2 + 1 on [I, 21, 1. x3 on [0, 11, 4. cos2xsinx on [O,a/2], 3. x/(x2 1) on [I, 21, Calculate, each of the average values in Exercises 5-16., 6. z3 + z2 1[1,21, 5. x310,21, . ., , +, , +, , 7. 1/(1 + t2)[-1,Il, , 8' [(x', , +, , -, , 22. Show that if a, , < b < c, then, b-a, , 7GJ[a,c1 = ( =), , -, , f ( t ) [a,bl, , c-b, , +, , -, , ( Gf), ( 0 [b,cl ., , *23. How is the average of f(x) on [a, b] related to, that of f(x) + k for a constant k? Explain the, + l ) I [ - ~ . ~ ~, answer in terms of a graph., , *24. If f(x) = g(x) + h(x) on [a, b], show that the, average off on [a, b] is the sum of the averages, 12. (x2 + x - l ) ~ i n x [ ~ , , / ~, 11. sin x cos 2x[o,,/21, of g and h on [a, b]., *25. Suppose that f' exists and is continuous on [a, b]., 13. x3, 14. ~ ~ [ o , I I, Prove the mean value theorem for derivatives, from the mean value theorem for integrals., 16. Inx[,,,], 15. ~ i n ~ x , ~ , , ], *26. Let f be defined on the real line and let, 44 = f ( x ) ~ o , ~ l ., 17. What was the average temperature in Goose, (a) Derive the formula, Brow on June 13, 1857? (See Fig. 9.3.2)., a'(x> = (1 /x)[f(x> - a(x>l., (b) Interpret the formula in the cases f(x), = a(x), f(x) < a(x), and f(x) > a(x)., (c) When baseball players strike out, it lowers, their batting average more at the beginning, of the season than at the end. Explain why., *27. The geometric mean of the positive numbers, al;. - ,a, is the nth root of the product, a , - . a,. Define the geometric mean of a positive function f(x) on [a,b]. [Hint:Use logat, rithms.], Midnighc, *28. (a) Use the idea of Exercise 27 to prove the, arithmetic-geometric mean inequality (see ExFigure 9.3.2. Temperature, in Goose Brow on June 13,, ample 12, Section 3.5). [Hint: Use the fact that, e x is concave upwards.] (b) Generalize from, 1857., numbers to functions., *29. Iff is continuous on [a, b] and F(x) = yJ(s)ds,, 18. Find the average temperature in Goose Brow, verify directly using the definition of continuity, (Fig. 9.3.2) during the periods midnight to P.M., in Section 11.1 that F is continuous on [a, b]., and P.M. to midnight. How is the average over, *30. (a) At what point of the interval [O,a] is the, the whole day related to these numbers?, average value of e x achieved?, 19. (a) Find t2 + 3t 2[o,xl as a function of x., @J, (b) Denote the expression found in part (a) by, (b) Evaluate this function of x for x = 0.1, 0.01,, p(a). Evaluate p(a) for a = 1, 10, 100, 1000, 0.0001., Try, to, explain, what, is, happening., and a = 0.1, 0.01, 0.0001, and 0.000001. Be, 20. Find C O S ~ [ ~ , ~ +as@ a] function of 6 and evaluate, sure that your answers are reasonable., (c) Guess the limits lim,,,p(a)/a, and, the limit as 6 +0., = 0 then f(b) = f(a)., 21. Show that if, lima+,p(a)/a., 10. sin- 'x[ -, , 9., , +m[l,31, , -, , +, , fo[a,bl, , Copyright 1985 Springer-Verlag. All rights reserved.