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Unformatted text preview: 5E-05(pp 314-323) 1/17/06 3:37 PM Page 314 CHAPTER 5 To compute an area we approximate a region by rectangles and let the number of rectangles become large. The precise area is the limit of these sums of areas of rectangles. I ntegrals 5E-05(pp 314-323) 1/17/06 3:37 PM Page 315 In Chapter 2 we used the tangent and velocity problems to |||| Now is a good time to read (or reread) A Preview of Calculus (see page 2). It discusses the unifying ideas of calculus and helps put in perspective where we have been and where we are going. introduce the derivative, which is the central idea in differential calculus. In much the same way, this chapter starts with the area and distance problems and uses them to formulate the idea of a definite integral, which is the basic concept of integral calculus. We will see in Chapters 6 and 9 how to use the integral to solve problems concerning volumes, lengths of curves, population predictions, cardiac output, forces on a dam, work, consumer surplus, and baseball, among many others. There is a connection between integral calculus and differential calculus. The Fundamental Theorem of Calculus relates the integral to the derivative, and we will see in this chapter that it greatly simplifies the solution of many problems. |||| 5.1 Areas and Distances In this section we discover that in trying to find the area under a curve or the distance traveled by a car, we end up with the same special type of limit. The Area Problem We begin by attempting to solve the area problem: Find the area of the region S that lies under the curve y f x from a to b. This means that S, illustrated in Figure 1, is bounded 0], the vertical lines x a and by the graph of a continuous function f [where f x x b, and the x-axis. y y=ƒ x=a S FIGURE 1 0 S=s (x, y) | a¯x¯b, 0¯y¯ƒ d a x=b b x In trying to solve the area problem we have to ask ourselves: What is the meaning of the word area ? This question is easy to answer for regions with straight sides. For a rectangle, the area is defined as the product of the length and the width. The area of a triangle is half the base times the height. The area of a polygon is found by dividing it into triangles (as in Figure 2) and adding the areas of the triangles. A™ w h l FIGURE 2 A=lw A¡ A£ A¢ b 1 A= 2 bh A=A¡+A™+A£+A¢ 315 5E-05(pp 314-323) 316 ❙❙❙❙ 1/17/06 3:37 PM Page 316 CHAPTER 5 INTEGRALS However, it isn’t so easy to find the area of a region with curved sides. We all have an intuitive idea of what the area of a region is. But part of the area problem is to make this intuitive idea precise by giving an exact definition of area. Recall that in defining a tangent we first approximated the slope of the tangent line by slopes of secant lines and then we took the limit of these approximations. We pursue a similar idea for areas. We first approximate the region S by rectangles and then we take the limit of the areas of these rectangles as we increase the number of rectangles. The following example illustrates the procedure. x 2 from 0 to 1 EXAMPLE 1 Use rectangles to estimate the area under the parabola y Try placing rectangles to estimate the area. Resources / Module 6 / What Is Area? / Estimating Area under a Parabola (the parabolic region S illustrated in Figure 3). y (1, 1) y=≈ S 0 x 1 FIGURE 3 SOLUTION We first notice that the area of S must be somewhere between 0 and 1 because S is contained in a square with side length 1, but we can certainly do better than that. Suppose we divide S into four strips S1, S2 , S3, and S4 by drawing the vertical lines x 1 , 4 x 1 , and x 3 as in Figure 4(a). 2 4 y y (1, 1) (1, 1) y=≈ S¢ S™ S£ S¡ 0 1 4 1 2 FIGURE 4 x 1 3 4 0 1 4 1 2 (a) 3 4 1 x (b) We can approximate each strip by a rectangle whose base is the same as the strip and whose height is the same as the right edge of the strip [see Figure 4(b)]. In other words, the heights of these rectangles are the values of the function f x x 2 at the right end 1 11 13 3 points of the subintervals [0, 4 ], [ 4 , 2 ], [ 2 , 4 ], and [ 4 , 1]. Each rectangle has width 1 and the heights are ( 1 )2, ( 1 )2, ( 3 )2, and 12. If we let R 4 be 4 4 2 4 the sum of the areas of these approximating rectangles, we get y (1, 1) y=≈ R4 1 4 ( 1 )2 4 1 4 ( 1 )2 2 1 4 ( 3 )2 4 1 4 12 15 32 0.46875 From Figure 4(b) we see that the area A of S is less than R 4 , so A 0 1 4 FIGURE 5 1 2 3 4 1 x 0.46875 Instead of using the rectangles in Figure 4(b) we could use the smaller rectangles in Figure 5 whose heights are the values of f at the left endpoints of the subintervals. (The leftmost rectangle has collapsed because its height is 0.) The sum of the areas of these 5E-05(pp 314-323) 1/17/06 3:37 PM Page 317 S ECTION 5.1 AREAS AND DISTANCES ❙❙❙❙ 317 approximating rectangles is L4 1 4 02 1 4 ( 1 )2 4 1 4 ( 1 )2 2 1 4 ( 3 )2 4 7 32 0.21875 We see that the area of S is larger than L 4 , so we have lower and upper estimates for A: 0.21875 A 0.46875 We can repeat this procedure with a larger number of strips. Figure 6 shows what happens when we divide the region S into eight strips of equal width. y y (1, 1) (1, 1) y=≈ 0 F IGURE 6 Approximating S with eight rectangles 1 8 1 x 0 (a) Using left endpoints 1 8 1 x (b) Using right endpoints By computing the sum of the areas of the smaller rectangles L 8 and the sum of the areas of the larger rectangles R 8 , we obtain better lower and upper estimates for A: 0.2734375 n Ln Rn 10 20 30 50 100 1000 0.2850000 0.3087500 0.3168519 0.3234000 0.3283500 0.3328335 0.3850000 0.3587500 0.3501852 0.3434000 0.3383500 0.3338335 A So one possible answer to the question is to say that the true area of S lies somewhere between 0.2734375 and 0.3984375. We could obtain better estimates by increasing the number of strips. The table at the left shows the results of similar calculations (with a computer) using n rectangles whose heights are found with left endpoints L n or right endpoints R n . In particular, we see by using 50 strips that the area lies between 0.3234 and 0.3434. With 1000 strips we narrow it down even more: A lies between 0.3328335 and 0.3338335. A good estimate is obtained by averaging these numbers: A 0.3333335. From the values in the table in Example 1, it looks as if R n is approaching increases. We confirm this in the next example. y (1, 1) y=≈ lim R n 1 1 n FIGURE 7 1 3 as n EXAMPLE 2 For the region S in Example 1, show that the sum of the areas of the upper approximating rectangles approaches 1 , that is, 3 nl 0 0.3984375 1 3 x SOLUTION R n is the sum of the areas of the n rectangles in Figure 7. Each rectangle has width 1 n and the heights are the values of the function f x x 2 at the points 2 2 2 1 n, 2 n, 3 n, . . . , n n; that is, the heights are 1 n , 2 n , 3 n , . . . , n n 2. 5E-05(pp 314-323) 318 ❙❙❙❙ 1/17/06 3:37 PM Page 318 CHAPTER 5 INTEGRALS Thus 2 1 n 1 n 1 n Rn 1 n 12 1 n2 12 1 n3 2 2 n 22 22 1 n 2 3 n 32 1 n n n 2 n2 32 n2 Here we need the formula for the sum of the squares of the first n positive integers: The ideas in Examples 1 and 2 are explored in Module 5.1/5.2/8.7 for a variety of functions. 12 1 22 32 nn n2 1 2n 6 1 Perhaps you have seen this formula before. It is proved in Example 5 in Appendix E. Putting Formula 1 into our expression for R n , we get Rn 1 n3 nn 1 2n 6 1 n 1 2n 6n 2 1 Thus, we have |||| Here we are computing the limit of the sequence R n . Sequences were discussed in A Preview of Calculus and will be studied in detail in Chapter 12. Their limits are calculated in the same way as limits at infinity (Section 4.4). In particular, we know that 1 0 lim nl n lim R n nl n lim 1 2n 6n 2 nl lim 1 6 lim 1 6 nl nl 1 6 n 1 1 2n n 1 n 1 12 1 n 2 1 n 1 3 It can be shown that the lower approximating sums also approach 1 , that is, 3 lim L n nl 1 3 From Figures 8 and 9 it appears that, as n increases, both L n and R n become better and better approximations to the area of S. Therefore, we define the area A to be the limit of the y y n=10 R¡¸=0.385 0 F IGURE 8 y n=50 R∞¸=0.3434 n=30 R£¸Å0.3502 1 x 0 1 x 0 1 x 5E-05(pp 314-323) 1/17/06 3:37 PM Page 319 S ECTION 5.1 AREAS AND DISTANCES y y 0 1 FIGURE 9 The area is the number that is smaller than all upper sums and larger than all lower sums x 319 y n=50 L∞¸=0.3234 n=30 L£¸Å0.3169 n=10 L¡¸=0.285 ❙❙❙❙ 0 1 x 0 1 x sums of the areas of the approximating rectangles, that is, A lim R n 1 3 lim L n nl nl Let’s apply the idea of Examples 1 and 2 to the more general region S of Figure 1. We start by subdividing S into n strips S1, S2 , . . . , Sn of equal width as in Figure 10. The width of the interval a, b is b a, so the width of each of the n strips is x b a n These strips divide the interval [a, b] into n subintervals x0, x1 , where x 0 a and x n x1 x 1, x 2 , x2, x3 , ..., x n 1, x n b. The right endpoints of the subintervals are a x, x2 a 2 x, x3 a 3 x, ... y y=ƒ S¡ 0 a S™ ⁄ S£ ¤ Si ‹ . . . xi-1 Sn xi . . . xn-1 b x FIGURE 10 Let’s approximate the i th strip Si by a rectangle with width x and height f x i , which is the value of f at the right endpoint (see Figure 11). Then the area of the i th rectangle y Îx f(x i) 0 F IGURE 11 a ⁄ ¤ ‹ xi-1 xi b x 5E-05(pp 314-323) 320 ❙❙❙❙ 1/17/06 3:37 PM Page 320 CHAPTER 5 INTEGRALS y is f x i x. What we think of intuitively as the area of S is approximated by the sum of the areas of these rectangles, which is Rn 0 a ⁄ bx (a) n=2 f x1 x f x2 x f xn x Figure 12 shows this approximation for n 2, 4, 8, and 12. Notice that this approximation appears to become better and better as the number of strips increases, that is, as n l . Therefore, we define the area A of the region S in the following way. y 2 Definition The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles: A 0 a ⁄ ¤ ‹ b lim R n lim f x 1 nl x nl f x2 x f xn x x (b) n=4 y It can be proved that the limit in Definition 2 always exists, since we are assuming that f is continuous. It can also be shown that we get the same value if we use left endpoints: 3 0 a bx (c) n=8 y A lim L n a x nl f x1 x f xn 1 x In fact, instead of using left endpoints or right endpoints, we could take the height of the i th rectangle to be the value of f at any number x* in the i th subinterval x i 1, x i . We call i ** * the numbers x 1 , x 2 , . . . , x n the sample points. Figure 13 shows approximating rectangles when the sample points are not chosen to be endpoints. So a more general expression for the area of S is A 4 0 lim f x 0 nl * lim f x 1 * f x2 x nl * f xn x x y bx Îx (d) n=12 FIGURE 12 f(x*) i 0 a x* ¡ ⁄ ¤ ‹ * x™ xi-1 * x£ xi b xn-1 x* i x * xn FIGURE 13 This tells us to end with i=n. This tells us to add. This tells us to start with i=m. n μ f(xi) Îx i=m We often use sigma notation to write sums with many terms more compactly. For instance, n f xi i1 x f x1 x f x2 x f xn x 5E-05(pp 314-323) 1/17/06 3:37 PM Page 321 S ECTION 5.1 AREAS AND DISTANCES ❙❙❙❙ 321 So the expressions for area in Equations 2, 3, and 4 can be written as follows: |||| If you need practice with sigma notation, look at the examples and try some of the exercises in Appendix E. n A lim f xi nl x i1 n A lim f xi nl x 1 i1 n A f x* i lim nl x i1 We could also rewrite Formula 1 in the following way: n nn i2 1 2n 6 i1 1 EXAMPLE 3 Let A be the area of the region that lies under the graph of f x cos x between x 0 and x b, where 0 b 2. (a) Using right endpoints, find an expression for A as a limit. Do not evaluate the limit. (b) Estimate the area for the case b 2 by taking the sample points to be midpoints and using four subintervals. SOLUTION (a) Since a 0, the width of a subinterval is b x f x1 x ib n, and x n f x2 x cos x 1 x b n b n cos nb n. The sum of the areas of f xn cos x 2 cos b n n So x 1 b n, x 2 2b n, x 3 3b n, x i the approximating rectangles is Rn 0 x x cos x n 2b n b n cos 2b n cos x nb n b n According to Definition 2, the area is A lim R n nl lim nl b n cos b n cos 3b n cos nb n Using sigma notation we could write A lim nl b n n cos i1 ib n It is very difficult to evaluate this limit directly by hand, but with the aid of a computer algebra system it isn’t hard (see Exercise 25). In Section 5.3 we will be able to find A more easily using a different method. 5E-05(pp 314-323) 322 ❙❙❙❙ 1/17/06 3:37 PM Page 322 CHAPTER 5 INTEGRALS y (b) With n 4 and b 0, 8 , 8, 4 , vals are y=cos x 1 x* 1 0 π 8 FIGURE 14 π 4 3π 8 π 2 x 2 we have x 4, 3 8 , and 3 3 16 x* 2 16 24 8, so the subintervals are 2 . The midpoints of these subinter- 8, 5 16 x* 3 7 16 x* 4 and the sum of the areas of the four approximating rectangles (see Figure 14) is 4 f x* i x 16 M4 x i1 f cos 8 16 cos f3 16 cos 8 16 cos x 3 16 f5 3 16 cos 5 16 x cos 8 16 5 16 cos f7 16 cos 8 7 16 x 7 16 8 1.006 So an estimate for the area is A 1.006 The Distance Problem Now let’s consider the distance problem: Find the distance traveled by an object during a certain time period if the velocity of the object is known at all times. (In a sense this is the inverse problem of the velocity problem that we discussed in Section 2.1.) If the velocity remains constant, then the distance problem is easy to solve by means of the formula distance velocity time But if the velocity varies, it’s not so easy to find the distance traveled. We investigate the problem in the following example. EXAMPLE 4 Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30-second time interval. We take speedometer readings every five seconds and record them in the following table: Time (s) Velocity (mi h) 0 5 10 15 20 25 30 17 21 24 29 32 31 28 In order to have the time and the velocity in consistent units, let’s convert the velocity readings to feet per second (1 mi h 5280 3600 ft s): Time (s) Velocity (ft s) 0 5 10 15 20 25 30 25 31 35 43 47 46 41 During the first five seconds the velocity doesn’t change very much, so we can estimate the distance traveled during that time by assuming that the velocity is constant. If we take the velocity during that time interval to be the initial velocity (25 ft s), then we 5E-05(pp 314-323) 1/17/06 3:37 PM Page 323 S ECTION 5.1 AREAS AND DISTANCES ❙❙❙❙ 323 obtain the approximate distance traveled during the first five seconds: 25 ft s 5s 125 ft Similarly, during the second time interval the velocity is approximately constant and we take it to be the velocity when t 5 s. So our estimate for the distance traveled from t 5 s to t 10 s is 31 ft s 5 s 155 ft If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled: 25 5 31 5 35 5 43 5 47 5 46 5 1135 ft We could just as well have used the velocity at the end of each time period instead of the velocity at the beginning as our assumed constant velocity. Then our estimate becomes 31 5 35 5 43 5 47 5 46 5 41 5 1215 ft If we had wanted a more accurate estimate, we could have taken velocity readings every two seconds, or even every second. √ 40 20 0 FIGURE 15 10 20 30 t Perhaps the calculations in Example 4 remind you of the sums we used earlier to estimate areas. The similarity is explained when we sketch a graph of the velocity function of the car in Figure 15 and draw rectangles whose heights are the initial velocities for each time interval. The area of the first rectangle is 25 5 125, which is also our estimate for the distance traveled in the first five seconds. In fact, the area of each rectangle can be interpreted as a distance because the height represents velocity and the width represents time. The sum of the areas of the rectangles in Figure 15 is L 6 1135, which is our initial estimate for the total distance traveled. In general, suppose an object moves with velocity v f t , where a t b and ft 0 (so the object always moves in the positive direction). We take velocity readings at times t0 a , t1, t2 , . . . , tn b so that the velocity is approximately constant on each subinterval. If these times are equally spaced, then the time between consecutive readings b a n. During the first time interval the velocity is approximately f t0 and so is t the distance traveled is approximately f t0 t. Similarly, the distance traveled during the second time interval is about f t1 t and the total distance traveled during the time interval a, b is approximately n f t0 t f t1 t f tn 1 t f ti t 1 i1 If we use the velocity at right endpoints instead of left endpoints, our estimate for the total distance becomes n f t1 t f t2 t f tn t f ti t i1 The more frequently we measure the velocity, the more accurate our estimates become, so it seems plausible that the exact distance d traveled is the limit of such expressions: n 5 d nl n f ti lim 1 t i1 We will see in Section 5.4 that this is indeed true. lim nl f ti i1 t 5E-05(pp 324-333) 324 ❙❙❙❙ 1/17/06 3:38 PM Page 324 CHAPTER 5 INTEGRALS Because Equation 5 has the same form as our expressions for area in Equations 2 and 3, it follows that the distance traveled is equal to the area under the graph of the velocity function. In Chapters 6 and 9 we will see that other quantities of interest in the natural and social sciences—such as the work done by a variable force or the cardiac output of the heart—can also be interpreted as the area under a curve. So when we compute areas in this chapter, bear in mind that they can be interpreted in a variety of practical ways. |||| 5.1 Exercises 1. (a) By reading values from the given graph of f , use five rect- right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or an overestimate? (b) Repeat part (a) using left endpoints. angles to find a lower estimate and an upper estimate for the area under the given graph of f from x 0 to x 10. In each case sketch the rectangles that you use. (b) Find new estimates using 10 rectangles in each case. 1 x 2 from x 1 to x 2 using three rectangles and right endpoints. Then improve your estimate by using six rectangles. Sketch the curve and the approximating rectangles. (b) Repeat part (a) using left endpoints. (c) Repeat part (a) using midpoints. (d) From your sketches in parts (a)–(c), which appears to be the best estimate? 5. (a) Estimate the area under the graph of f x y 5 y=ƒ 1 1 x 2 , 2 x 2. (b) Estimate the area under the graph of f using four approximating rectangles and taking the sample points to be (i) right endpoints (ii) midpoints In each case sketch the curve and the rectangles. (c) Improve your estimates in part (b) by using eight rectangles. ; 6. (a) Graph the function f x 0 10 x 5 2. (a) Use six rectangles to find estimates of each type for the area under the given graph of f from x 0 to x 12. (i) L 6 (sample points are left endpoints) (ii) R 6 (sample points are right endpoints) (iii) M6 (sample points are midpoints) (b) Is L 6 an underestimate or overestimate of the true area? (c) Is R 6 an underestimate or overestimate of the true area? (d) Which of the numbers L 6, R 6, or M6 gives the best estimate? Explain. 7–8 |||| With a programmable calculator (or a computer), it is possible to evaluate the expressions for the sums of areas of approximating rectangles, even for large values of n, using looping. (On a TI use the Is command or a For-EndFor loop, on a Casio use Isz, on an HP or in BASIC use a FOR-NEXT loop.) Compute the sum of the areas of approximating rectangles using equal subintervals and right endpoints for n 10, 30, and 50. Then guess the value of the exact area. y 8 7. The region under y sin x from 0 to 8. The region under y 1 x 2 from 1 to 2 y=ƒ 4 ■ CAS 0 4 8 3. (a) Estimate the area under the graph of f x 25 x 2 from 5 using five approximating rectangles and 4. (a) Estimate the area under the graph of f x x 0 to x ■ ■ ■ ■ ■ ■ ■ ■ ■ CAS ■ 9. Some computer algebra systems have commands that will draw approximating rectangles and evaluate the sums of their areas, at least if x* is a left or right endpoint. (For instance, in Maple i use leftbox, rightbox, leftsum, and rightsum.) (a) If f x s x, 1 x 4, find the left and right sums for n 10, 30, and 50. (b) Illustrate by graphing the rectangles in part (a). (c) Show that the exact area under f lies between 4.6 and 4.7. 12 x 1 x from x 1 to x 5 using four approximating rectangles and right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or an overestimate? (b) Repeat part (a) using left endpoints. ■ 10. (a) If f x 2, use the commands dissin sin x , 0 x cussed in Exercise 9 to find the left and right sums for n 10, 30, and 50. 5E-05(pp 324-333) 1/17/06 3:38 PM Page 325 S ECTION 5.1 AREAS AND DISTANCES (b) Illustrate by graphing the rectangles in part (a). (c) Show that the exact area under f lies between 0.87 and 0.91. ❙❙❙❙ 325 Use these data to estimate the height above the Earth’s surface of the space shuttle Endeavour, 62 seconds after liftoff. 15. The velocity graph of a braking car is shown. Use it to estimate the distance traveled by the car while the brakes are applied. 11. The speed of a runner increased steadily during the first three seconds of a race. Her speed at half-second intervals is given in the table. Find lower and upper estimates for the distance that she traveled during these three seconds. √ (ft /s) 60 40 t (s) 0 0.5 1.0 1.5 2.0 2.5 3.0 v (ft s) 0 6.2 10.8 14.9 18.1 19.4 20.2 20 12. Speedometer readings for a motorcycle at 12-second intervals 0 are given in the table. (a) Estimate the distance traveled by the motorcycle during this time period using the velocities at the beginning of the time intervals. (b) Give another estimate using the velocities at the end of the time periods. (c) Are your estimates in parts (a) and (b) upper and lower estimates? Explain. 4 2 6 t (seconds) 16. The velocity graph of a car accelerating from rest to a speed of 120 km h over a period of 30 seconds is shown. Estimate the distance traveled during this period. √ (km / h) 80 t (s) 0 12 24 36 48 60 v (ft s) 30 28 25 22 24 27 40 0 13. Oil leaked from a tank at a rate of r t liters per hour. The rate decreased as time passed and values of the rate at 2-hour time intervals are shown in the table. Find lower and upper estimates for the total amount of oil that leaked out. th r t (L h) 17–19 |||| Use Definition 2 to find an expression for the area under the graph of f as a limit. Do not evaluate the limit. 0 2 4 6 8 10 17. f x 4 sx, 1 8.7 7.6 6.8 6.2 5.7 5.3 18. f x 1 x 4, 2 19. f x x cos x, 0 14. When we estimate distances from velocity data, it is sometimes necessary to use times t0 , t1, t2 , t3 , . . . that are not equally spaced. We can still estimate distances using the time periods ti ti ti 1. For example, on May 7, 1992, the space shuttle Endeavour was launched on mission STS-49, the purpose of which was to install a new perigee kick motor in an Intelsat communications satellite. The table, provided by NASA, gives the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. ■ ■ ■ n 20. lim nl i1 0 185 319 447 742 1325 1445 4151 x 5 x ■ 2 ■ ■ ■ ■ ■ ■ ■ 2i n 5 i1 4n tan 10 i 4n Velocity (ft s) 0 10 15 20 32 59 62 125 2 n n ■ Launch Begin roll maneuver End roll maneuver Throttle to 89% Throttle to 67% Throttle to 104% Maximum dynamic pressure Solid rocket booster separation ■ 16 |||| Determine a region whose area is equal to the given limit. Do not evaluate the limit. nl Time (s) x 20–21 21. lim Event t 30 (seconds) 20 10 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 22. (a) Use Definition 2 to find an expression for the area under the curve y x 3 from 0 to 1 as a limit. (b) The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in part (a). 13 23 33 n3 nn 1 2 2 ■ 5E-05(pp 324-333) 326 CAS ❙❙❙❙ 1/17/06 3:38 PM Page 326 CHAPTER 5 INTEGRALS x 5 from 0 to 2 as a 23. (a) Express the area under the curve y CAS 25. Find the exact area under the cosine curve y cos x from x 0 to x b, where 0 b 2. (Use a computer algebra system both to evaluate the sum and compute the limit in Example 3.) In particular, what is the area if b 2? Compare your answer with the estimate obtained in Example 3(b). limit. (b) Use a computer algebra system to find the sum in your expression from part (a). (c) Evaluate the limit in part (a). CAS x4 24. (a) Express the area under the curve y 5x 2 x from 2 to 7 as a limit. (b) Use a computer algebra system to evaluate the sum in part (a). (c) Use a computer algebra system to find the exact area by evaluating the limit of the expression in part (b). |||| 5.2 26. (a) Let A n be the area of a polygon with n equal sides inscribed in a circle with radius r. By dividing the polygon into n congruent triangles with central angle 2 n, show that A n 1 nr 2 sin 2 n . 2 r 2. [Hint: Use Equation 3.5.2.] (b) Show that lim n l A n The Definite Integral We saw in Section 5.1 that a limit of the form n 1 f x* i lim nl * lim f x 1 x * f x2 x nl i1 * f xn x x arises when we compute an area. We also saw that it arises when we try to find the distance traveled by an object. It turns out that this same type of limit occurs in a wide variety of situations even when f is not necessarily a positive function. In Chapters 6 and 9 we will see that limits of the form (1) also arise in finding lengths of curves, volumes of solids, centers of mass, force due to water pressure, and work, as well as other quantities. We therefore give this type of limit a special name and notation. 2 Definition of a Definite Integral If f is a continuous function defined for x b, we divide the interval a, b into n subintervals of equal width x b a n. We let x 0 a , x 1, x 2 , . . . , x n ( b) be the endpoints of these ** subintervals and we let x 1 , x 2 , . . . , x * be any sample points in these subintervals, n so x * lies in the i th subinterval x i 1, x i . Then the definite integral of f from a i to b is a y b a n f x dx f x* i lim nl x i1 Because we have assumed that f is continuous, it can be proved that the limit in Definition 2 always exists and gives the same value no matter how we choose the sample points x *. (See Note 4 for a precise definition of this type of limit.) If we take the sample i points to be right endpoints, then x * x i and the definition of an integral becomes i 3 y b a n f x dx lim nl f xi i1 x 5E-05(pp 324-333) 1/17/06 3:38 PM Page 327 S ECTION 5.2 THE DEFINITE INTEGRAL If we choose the sample points to be left endpoints, then x * i becomes y b a xi 1 ❙❙❙❙ 327 and the definition n f x dx lim nl f xi 1 x i1 Alternatively, we could choose x * to be the midpoint of the subinterval or any other numi ber between x i 1 and x i . Although most of the functions that we encounter are continuous, the limit in Definition 2 also exists if f has a finite number of removable or jump discontinuities (but not infinite discontinuities). (See Section 2.5.) So we can also define the definite integral for such functions. The symbol x was introduced by Leibniz and is called an integral sign. It is an elongated S and was chosen because an integral is a limit of sums. In the notation xab f x d x, f x is called the integrand and a and b are called the limits of integration; a is the lower limit and b is the upper limit. The symbol dx has no official meaning by itself; xab f x d x is all one symbol. The procedure of calculating an integral is called integration. NOTE 1 ■ The definite integral xab f x d x is a number; it does not depend on x. In fact, we could use any letter in place of x without changing the value of the integral: NOTE 2 ■ y b a NOTE 3 ■ y f x dx b a y f t dt b a f r dr The sum n f x* i x i1 |||| Bernhard Riemann received his Ph.D. under the direction of the legendary Gauss at the University of Göttingen and remained there to teach. Gauss, who was not in the habit of praising other mathematicians, spoke of Riemann’s “creative, active, truly mathematical mind and gloriously fertile originality.” The definition (2) of an integral that we use is due to Riemann. He also made major contributions to the theory of functions of a complex variable, mathematical physics, number theory, and the foundations of geometry. Riemann’s broad concept of space and geometry turned out to be the right setting, 50 years later, for Einstein’s general relativity theory. Riemann’s health was poor throughout his life, and he died of tuberculosis at the age of 39. that occurs in Definition 2 is called a Riemann sum after the German mathematician Bernhard Riemann (1826–1866). We know that if f happens to be positive, then the Riemann sum can be interpreted as a sum of areas of approximating rectangles (see Figure 1). By comparing Definition 2 with the definition of area in Section 5.1, we see that the definite integral xab f x d x can be interpreted as the area under the curve y f x from a to b. (See Figure 2.) y y Îx 0 a x* i y=ƒ b x 0 a b x FIGURE 1 FIGURE 2 If ƒ˘0, the Riemann sum μ f(x*) Îx i is the sum of areas of rectangles. If ƒ˘0, the integral ja ƒ dx is the area under the curve y=ƒ from a to b. b 5E-05(pp 324-333) 328 ❙❙❙❙ 1/17/06 3:38 PM Page 328 CHAPTER 5 INTEGRALS y y=ƒ + + 0a b _ x If f takes on both positive and negative values, as in Figure 3, then the Riemann sum is the sum of the areas of the rectangles that lie above the x-axis and the negatives of the areas of the rectangles that lie below the x-axis (the areas of the gold rectangles minus the areas of the blue rectangles). When we take the limit of such Riemann sums, we get the situation illustrated in Figure 4. A definite integral can be interpreted as a net area, that is, a difference of areas: FIGURE 3 μ f(x*) Î x is an approximation to i the net area y + + _ bx b a f x dx In the spirit of the precise definition of the limit of a function in Section 2.4, we can write the precise meaning of the limit that defines the integral in Definition 2 as follows: 0 there is an integer N such that FIGURE 4 b a A2 ■ For every number j A1 where A 1 is the area of the region above the x-axis and below the graph of f , and A 2 is the area of the region below the x-axis and above the graph of f . NOTE 4 y=ƒ 0a y y ƒ dx is the net area b a n f x* i f x dx x i1 N and for every choice of x* in x i 1, x i . i for every integer n This means that a definite integral can be approximated to within any desired degree of accuracy by a Riemann sum. Although we have defined xab f x d x by dividing a, b into subintervals of equal width, there are situations in which it is advantageous to work with subintervals of unequal width. For instance, in Exercise 14 in Section 5.1 NASA provided velocity data at times that were not equally spaced, but we were still able to estimate the distance traveled. And there are methods for numerical integration that take advantage of unequal subintervals. If the subinterval widths are x 1, x 2 , . . . , x n , we have to ensure that all these widths approach 0 in the limiting process. This happens if the largest width, max x i , approaches 0. So in this case the definition of a definite integral becomes NOTE 5 ■ y b a n f x dx EXAMPLE 1 Express lim max xi l 0 i 1 f x* i xi n x3 i lim nl as an integral on the interval 0, x i sin x i x i1 . SOLUTION Comparing the given limit with the limit in Definition 2, we see that they will be identical if we choose x3 fx x sin x x* i and xi (So the sample points are right endpoints and the given limit is of the form of Equation 3.) We are given that a 0 and b . Therefore, by Definition 2 or Equation 3, we have n x3 i lim nl i1 x i sin x i x y 0 x3 x sin x d x 5E-05(pp 324-333) 1/17/06 3:38 PM Page 329 S ECTION 5.2 THE DEFINITE INTEGRAL ❙❙❙❙ 329 Later, when we apply the definite integral to physical situations, it will be important to recognize limits of sums as integrals, as we did in Example 1. When Leibniz chose the notation for an integral, he chose the ingredients as reminders of the limiting process. In general, when we write n f x* i lim nl y x b f x dx a i1 by x, x * by x, and x by dx. i we replace lim Evaluating Integrals When we use the definition to evaluate a definite integral, we need to know how to work with sums. The following three equations give formulas for sums of powers of positive integers. Equation 4 may be familiar to you from a course in algebra. Equations 5 and 6 were discussed in Section 5.1 and are proved in Appendix E. n nn i 4 i1 n nn i2 5 1 2 1 2n 6 i1 n nn i3 6 1 2 1 2 i1 The remaining formulas are simple rules for working with sigma notation: n c 7 |||| Formulas 7–10 are proved by writing out each side in expanded form. The left side of Equation 8 is ca 1 The right side is c a1 ca 2 an n n ca i 8 c ai i1 ca n a2 nc i1 i1 n n ai 9 bi i1 These are equal by the distributive property. The other formulas are discussed in Appendix E. bi i1 n i1 n ai 10 n ai bi n ai i1 bi i1 i1 EXAMPLE 2 x3 (a) Evaluate the Riemann sum for f x endpoints and a 0, b 3, and n 6. 3 (b) Evaluate y x 3 0 Try more problems like this one. Resources / Module 6 / What Is Area? / Problems and Tests 6 x taking the sample points to be right 6 x d x. SOLUTION (a) With n 6 the interval width is x b a n 3 0 6 1 2 5E-05(pp 324-333) 330 ❙❙❙❙ 1/17/06 3:38 PM Page 330 CHAPTER 5 INTEGRALS and the right endpoints are x 1 0.5, x 2 x 6 3.0. So the Riemann sum is 1.0, x 3 1.5, x 4 2.0, x 5 2.5, and 6 R6 y f xi x i1 f 0.5 5 y=˛-6x 1 2 x 2.875 f 1.0 f 1.5 x 5.625 5 x f 2.0 4 0.625 x f 2.5 x f 3.0 x 9 3.9375 0 x 3 Notice that f is not a positive function and so the Riemann sum does not represent a sum of areas of rectangles. But it does represent the sum of the areas of the gold rectangles (above the x-axis) minus the sum of the areas of the blue rectangles (below the x-axis) in Figure 5. (b) With n subintervals we have FIGURE 5 b x a 3 n n Thus x 0 0, x 1 3 n, x 2 6 n, x 3 9 n, and, in general, x i using right endpoints, we can use Equation 3: y 3 n x3 6x dx n lim f xi nl 3 n nl lim nl lim nl 5 0 81 4 A™ 81 4 27 nl 3 FIGURE 6 3 81 n4 (˛-6x) dx=A¡-A™=_6.75 x (Equation 8 with c i1 i (Equations 10 and 8) nn 2 27 4 n i1 2 1 1 n 3 n) 18 i n 54 n2 i3 3 n 3i n 6 n 1 i1 3 27 3 i n3 i1 lim y=˛-6x A¡ 0 n 3i n f nl 3i n i1 81 n4 nl j 3 n n lim y lim i1 lim 0 x 3i n. Since we are 54 n n 1 n2 2 2 27 1 1 n 6.75 This integral can’t be interpreted as an area because f takes on both positive and negative values. But it can be interpreted as the difference of areas A 1 A 2 , where A 1 and A 2 are shown in Figure 6. Figure 7 illustrates the calculation by showing the positive and negative terms in the right Riemann sum R n for n 40. The values in the table show the Riemann sums approaching the exact value of the integral, 6.75, as n l . 5E-05(pp 324-333) 1/17/06 3:38 PM Page 331 S ECTION 5.2 THE DEFINITE INTEGRAL y n 5 0 x 3 331 Rn 40 100 500 1000 5000 y=˛-6x ❙❙❙❙ 6.3998 6.6130 6.7229 6.7365 6.7473 FIGURE 7 R¢¸Å_6.3998 A much simpler method for evaluating the integral in Example 2 will be given in Section 5.3. |||| Because f x x 4 is positive, the integral in Example 3 represents the area shown in Figure 8. EXAMPLE 3 (a) Set up an expression for x25 x 4 dx as a limit of sums. (b) Use a computer algebra system to evaluate the expression. y SOLUTION x 4, a (a) Here we have f x 2, b 5, and y=x$ So x0 0 b x 300 2 5 2, x 1 2 3 n, x 2 2 a 3 n n 6 n, x 3 2 9 n, and xi 2 3i n f xi x lim x FIGURE 8 From Equation 3, we get y 5 n x 4 dx n lim nl i1 lim 2 3 n nl n 2 i1 nl 3i n f2 i1 3i n 3 n 4 (b) If we ask a computer algebra system to evaluate the sum and simplify, we obtain n 3i n 2 i1 4 2062n 4 3045n 3 1170n 2 10n 3 27 Now we ask the CAS to evaluate the limit: y 5 2 x 4 dx lim nl 3 n n 2 i1 3 2062 10 3i n 3093 5 4 lim nl 3 2062n 4 3045n 3 1170n 2 10n 4 27 618.6 We will learn a much easier method for the evaluation of integrals in the next section. 5E-05(pp 324-333) 332 ❙❙❙❙ 1/17/06 3:38 PM Page 332 CHAPTER 5 INTEGRALS EXAMPLE 4 Evaluate the following integrals by interpreting each in terms of areas. (a) y 1 0 s1 x 2 dx (b) y 3 0 x 1 dx SOLUTION y (a) Since f x s1 x 2 0, we can interpret this integral as the area under the curve 2 from 0 to 1. But, since y 2 y s1 x 1 x 2, we get x 2 y 2 1, which shows that the graph of f is the quarter-circle with radius 1 in Figure 9. Therefore y= œ„„„„„ 1-≈ or ≈+¥=1 1 1 y s1 0 0 1 x 1 4 x 2 dx 1 2 4 (In Section 8.3 we will be able to prove that the area of a circle of radius r is r 2.) (b) The graph of y x 1 is the line with slope 1 shown in Figure 10. We compute the integral as the difference of the areas of the two triangles: FIGURE 9 y 3 0 x 1 dx A1 1 2 A2 22 y 1 2 11 1.5 (3, 2) y=x-1 A¡ 0 A™ 1 3 x _1 FIGURE 10 The Midpoint Rule We often choose the sample point x * to be the right endpoint of the i th subinterval because i it is convenient for computing the limit. But if the purpose is to find an approximation to an integral, it is usually better to choose x * to be the midpoint of the interval, which we i denote by x i . Any Riemann sum is an approximation to an integral, but if we use midpoints we get the following approximation. Module 5.1/5.2/8.7 shows how the Midpoint Rule estimates improve as n increases. Midpoint Rule y b a n f x dx f xi x x f x1 f xn i1 where x and xi b a n 1 2 xi 1 xi EXAMPLE 5 Use the Midpoint Rule with n midpoint of x i 1, x i 5 to approximate y 2 1 1 dx. x 5E-05(pp 324-333) 1/17/06 3:38 PM Page 333 S ECTION 5.2 THE DEFINITE INTEGRAL ❙❙❙❙ 333 SOLUTION The endpoints of the five subintervals are 1, 1.2, 1.4, 1.6, 1.8, and 2.0, so the midpoints are 1.1, 1.3, 1.5, 1.7, and 1.9. The width of the subintervals is x 2 1 5 1 , so the Midpoint Rule gives 5 y 1 y= x y 2 1 1 dx x x f 1.1 1 5 1 1.1 f 1.3 1 1.3 f 1.5 1 1.5 f 1.7 1 1.7 f 1.9 1 1.9 0.691908 0 1 2 x FIGURE 11 Since f x 1 x 0 for 1 x 2, the integral represents an area, and the approximation given by the Midpoint Rule is the sum of the areas of the rectangles shown in Figure 11. At the moment we don’t know how accurate the approximation in Example 5 is, but in Section 8.7 we will learn a method for estimating the error involved in using the Midpoint Rule. At that time we will discuss other methods for approximating definite integrals. If we apply the Midpoint Rule to the integral in Example 2, we get the picture in Figure 12. The approximation M40 6.7563 is much closer to the true value 6.75 than the right endpoint approximation, R 40 6.3998, shown in Figure 7. y 5 y=˛-6x 0 3 x FIGURE 12 M¢¸Å_6.7563 Properties of the Definite Integral When we defined the definite integral xab f x dx, we implicitly assumed that a b. But the definition as a limit of Riemann sums makes sense even if a b. Notice that if we reverse a and b, then x changes from b a n to a b n. Therefore y a b If a b, then x f x dx y b a f x dx 0 and so y a a f x dx 0 We now develop some basic properties of integrals that will help us to evaluate integrals in a simple manner. We assume that f and t are continuous functions. 5E-05(pp 334-343) 334 ❙❙❙❙ 1/17/06 3:39 PM Page 334 CHAPTER 5 INTEGRALS Properties of the Integral 1. 2. area=c(b-a) 0 a b b y f+g g b a b a b a c dx cb fx a, t x dx where c is any constant y b a b c y f x dx, c f x dx t x dx y b a b t x dx a where c is any constant a fx y f x dx y f x dx b t x dx a Property 1 says that the integral of a constant function f x c is the constant times the length of the interval. If c 0 and a b, this is to be expected because c b a is the area of the shaded rectangle in Figure 13. Property 2 says that the integral of a sum is the sum of the integrals. For positive functions it says that the area under f t is the area under f plus the area under t. Figure 14 helps us understand why this is true: In view of how graphical addition works, the corresponding vertical line segments have equal height. In general, Property 2 follows from Equation 3 and the fact that the limit of a sum is the sum of the limits: c dx=c(b-a) a y b a x F IG URE 13 j y 4. y=c c y 3. y y y f b a n fx t x dx lim nl f xi t xi x i1 n 0 bx a lim nl x t xi i1 n lim b nl [ ƒ+©] dx= a j b a b ƒ dx+ja © dx |||| Property 3 seems intuitively reasonable because we know that multiplying a function by a positive number c stretches or shrinks its graph vertically by a factor of c. So it stretches or shrinks each approximating rectangle by a factor c and therefore it has the effect of multiplying the area by c. y b a x i1 n FIGURE 14 j n f xi f xi x lim nl i1 y f x dx b a t xi x i1 t x dx Property 3 can be proved in a similar manner and says that the integral of a constant times a function is the constant times the integral of the function. In other words, a constant (but only a constant) can be taken in front of an integral sign. Property 4 is proved by writing f t f t and using Properties 2 and 3 with c 1. EXAMPLE 6 Use the properties of integrals to evaluate y 1 4 0 3x 2 dx. SOLUTION Using Properties 2 and 3 of integrals, we have y 1 0 4 3x 2 d x y 1 0 1 3x 2 dx 4 dx y 4 dx 41 0 y We know from Property 1 that y 1 0 0 1 0 4 4 dx 1 3 y x 2 dx 0 5E-05(pp 334-343) 1/17/06 3:39 PM Page 335 S ECTION 5.2 THE DEFINITE INTEGRAL 1 1 3 and we found in Example 2 in Section 5.1 that y x 2 dx 0 y 1 3x 2 d x 4 0 y 1 0 3 335 . So 1 3 y x 2 dx 4 dx 4 ❙❙❙❙ 0 1 3 5 The next property tells us how to combine integrals of the same function over adjacent intervals: y y=ƒ y 5. 0 a c b x FIGURE 15 c a y f x dx b c f x dx y b a f x dx This is not easy to prove in general, but for the case where f x 0 and a c b Property 5 can be seen from the geometric interpretation in Figure 15: The area under y f x from a to c plus the area from c to b is equal to the total area from a to b. EXAMPLE 7 If it is known that x010 f 17 and x08 f x dx x dx 12, find x810 f x dx. SOLUTION By Property 5, we have y 8 0 y so 10 8 f x dx y f x dx 10 0 y 10 8 f x dx y f x dx 8 0 y 10 f x dx 0 f x dx 17 12 5 Notice that Properties 1–5 are true whether a b, a b, or a b. The following properties, in which we compare sizes of functions and sizes of integrals, are true only if a b. Comparison Properties of the Integral 6. If f x 0 for a 7. If f x t x for a 8. If m fx b b, then y f x dx x a M for a mb y b b, then y f x dx x a 0. y b a t x dx. b, then x a y b a f x dx Mb a M y=ƒ m 0 a FIGURE 16 b x If f x 0, then xab f x d x represents the area under the graph of f , so the geometric interpretation of Property 6 is simply that areas are positive. But the property can be proved from the definition of an integral (Exercise 64). Property 7 says that a bigger function has a bigger integral. It follows from Properties 6 and 4 because f t 0. Property 8 is illustrated by Figure 16 for the case where f x 0. If f is continuous we could take m and M to be the absolute minimum and maximum values of f on the inter- 5E-05(pp 334-343) 336 ❙❙❙ 1/17/06 3:39 PM Page 336 CHAPTER 5 INTEGRALS val a, b . In this case Property 8 says that the area under the graph of f is greater than the area of the rectangle with height m and less than the area of the rectangle with height M . Proof of Property 8 Since m M , Property 7 gives fx y b a m dx y b a y f x dx b a M dx Using Property 1 to evaluate the integrals on the left and right sides, we obtain mb y a b a f x dx Mb a Property 8 is useful when all we want is a rough estimate of the size of an integral without going to the bother of using the Midpoint Rule. EXAMPLE 8 Use Property 8 to estimate y 4 1 s x d x. S OLUTION Since f x s x is an increasing function, its absolute minimum on 1, 4 is m f1 1 and its absolute maximum on 1, 4 is M f 4 s4 2. Thus, Property 8 gives y y=œ„ x 14 2 1 y 4 1 sx dx 24 sx dx 1 6 or 1 3 0 1 4 4 1 x The result of Example 8 is illustrated in Figure 17. The area under y s x from 1 to 4 is greater than the area of the lower rectangle and less than the area of the large rectangle. FIGURE 17 |||| 5.2 y Exercises 2 x 2, 0 x 2, with four subintervals, taking the sample points to be right endpoints. Explain, with the aid of a diagram, what the Riemann sum represents. 1. Evaluate the Riemann sum for f x 2. If f x 3x 7, 0 x 3, evaluate the Riemann sum with n 6, taking the sample points to be left endpoints. What does the Riemann sum represent? Illustrate with a diagram. 3. If f x sx 2, 1 x 6, find the Riemann sum with n 5 correct to six decimal places, taking the sample points to be midpoints. What does the Riemann sum represent? Illustrate with a diagram. 4. (a) Find the Riemann sum for f x x 2 sin 2 x, 0 x 3, with six terms, taking the sample points to be right endpoints. (Give your answer correct to six decimal places.) Explain what the Riemann sum represents with the aid of a sketch. (b) Repeat part (a) with midpoints as the sample points. 8 5. The graph of a function f is given. Estimate x0 f x d x using four subintervals with (a) right endpoints, (b) left endpoints, and (c) midpoints. y f 1 0 1 x 5E-05(pp 334-343) 1/17/06 3:40 PM Page 337 ❙❙❙❙ S ECTION 5.2 THE DEFINITE INTEGRAL 6. The graph of t is shown. Estimate x 3 3 t x dx with six sub- 15. Use a calculator or computer to make a table of values of right Riemann sums R n for the integral x0 sin x d x with n 5, 10, 50, and 100. What value do these numbers appear to be approaching? intervals using (a) right endpoints, (b) left endpoints, and (c) midpoints. y 16. Use a calculator or computer to make a table of values of g left and right Riemann sums L n and R n for the integral x02 s1 x 4 d x with n 5, 10, 50, and 100. Between what two numbers must the value of the integral lie? Can you make a similar statement for the integral x 2 1 s1 x 4 d x ? Explain. 1 0 337 x 1 17–20 Express the limit as a definite integral on the given |||| interval. n 17. lim nl 7. A table of values of an increasing function f is shown. Use the 25 0 table to find lower and upper estimates for x f x d x. x i sin x i x, 0, i1 n xi 18. lim nl 1 i1 x, xi 1, 5 n x 0 5 10 15 20 25 nl fx 42 37 25 6 15 s 2 x* i 19. lim x* i 2 x, 1, 8] i1 36 n 20. lim 8. The table gives the values of a function obtained from an nl 6 0 experiment. Use them to estimate x f x d x using three equal subintervals with (a) right endpoints, (b) left endpoints, and (c) midpoints. If the function is known to be a decreasing function, can you say whether your estimates are less than or greater than the exact value of the integral? x 0 1 2 3 4 fx 9.3 9.0 8.3 6.5 5 2.3 11. ■ CAS y 10 2 y 1 0 1 dx, sin x 2 dx, ■ ■ n n ■ 10. 4 12. 5 ■ ■ y 0 y 5 1 ■ sec x 3 dx, n y y 25. y ■ ■ 5 1 1 sin x 2 d x 0, 2 ■ ■ ■ ■ 2 2 1 y 24. x 2 dx 2 0 22. y 4 1 5 0 x2 ■ 1 dx, 1 ■ n ■ ■ 1 2x ■ ■ 5 dx 2x 3 dx x 3 dx ■ ■ ■ ■ ■ ■ ■ ■ 26. (a) Find an approximation to the integral x x x ■ ■ ■ ■ ■ b2 b 27. Prove that y x dx a2 2 a b3 b 28. Prove that y x 2 dx . a3 3 a . 29–30 |||| Express the integral as a limit of Riemann sums. Do not evaluate the limit. 0.315 Deduce that the approximation using the Midpoint Rule with n 5 in Exercise 11 is accurate to two decimal places. 2 x 3x d x using a Riemann sum with right endpoints and n 8. (b) Draw a diagram like Figure 3 to illustrate the approximation in part (a). (c) Use Equation 3 to evaluate x04 x 2 3x d x. (d) Interpret the integral in part (c) as a difference of areas and illustrate with a diagram like Figure 4. 4 tions for Exercise 7 in Section 5.1), compute the left and right sin x 2 on the interval Riemann sums for the function f x 0, 1 with n 100. Explain why these estimates show that 0 ■ 3x dx 14. With a programmable calculator or computer (see the instruc- y x, 4 0 and graphs the corresponding rectangles (use middlesum and middlebox commands in Maple), check the answer to Exercise 11 and illustrate with a graph. Then repeat with n 10 and n 20. 0.306 5 6 13. If you have a CAS that evaluates midpoint approximations 1 6 x* i |||| Use the form of the definition of the integral given in Equation 3 to evaluate the integral. ■ sx 3 2 21–25 23. 10.5 9–12 |||| Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places. 9. ■ 21. 6 7.6 ■ 3 x* i 4 i1 29. ■ y x 6 1 2 ■ x5 ■ 30. dx ■ ■ ■ ■ y 2 0 x 2 sin x dx ■ ■ ■ ■ ■ 5E-05(pp 334-343) 338 CAS ❙❙❙❙ 1/17/06 3:40 PM Page 338 CHAPTER 5 INTEGRALS 31–32 |||| Express the integral as a limit of sums. Then evaluate, using a computer algebra system to find both the sum and the limit. 31. y ■ ■ y 32. sin 5x dx 0 ■ ■ ■ ■ 10 2 ■ evaluate x25 1 integrals to evaluate x14 2 x 2 ■ ■ ■ (c) 2 0 y 7 5 (b) f x dx (d) f x dx y 5 f x dx 0 y 9 b 47. Write as a single integral in the form xa f x d x : f x dx 0 y y=ƒ 2 2 2 4 6 x 9 0 x 8 2f x y f x dx 5 48. If x1 f x dx 49. If x09 f x d x 0 2 cos x dx 1 (from Exercise 25 in Section 5.1), together with the properties of integrals, to evaluate x0 2 2 cos x 5x dx. y 2 1 dx. 46. Use the result of Exercise 27 and the fact that x0 it in terms of areas. y 3x ■ 33. The graph of f is shown. Evaluate each integral by interpreting (a) 3 x 4 dx. 45. Use the results of Exercises 27 and 28 and the properties of x 6 dx ■ 44. Use the properties of integrals and the result of Example 3 to 5 2 12 and x45 f x dx 1 y f x dx 2 f x dx 3.6, find x14 f x dx. 37 and x09 t x d x 3 t x dx. 16, find 5 50. Find x0 f x dx if 3 for x x for x fx 3 3 34. The graph of t consists of two straight lines and a semicircle. Use it to evaluate each integral. (a) y 2 0 (b) t x dx y 6 2 (c) t x dx y 7 0 51–54 |||| Use the properties of integrals to verify the inequality without evaluating the integrals. t x dx 51. y=© 2 y 4 y 52. y 4 0 2 s5 1 7x 4 54. 35. |||| 31 2 0 y( x y 36. 1 dx 39. ■ 0 y (1 s9 3 y 2 1 ■ 0 y ■ 2 sx x 2 dx s1 6 2 1 sin 2 x dx 2 s2 sin x dx ■ ■ 1 dx 3 ■ ■ ■ ■ ■ ■ ■ ■ Evaluate the integral by interpreting it in terms of areas. 55–60 2 2 x 2 ) dx y 38. y 40. x dx ■ ■ ■ 9 41. Given that y s x d x 4 ■ 38 3 3 1 10 0 ■ 55. 3 y 59. y 1 dx x 2 y Use Property 8 to estimate the value of the integral. 1 56. y 58. y 60. y 2 0 sx 3 1 dx 2x dx x ■ |||| x 2 dx s4 57. 37. 1 y 6 ■ 35–40 1 4 y x dx y 53. 2 0 sin 3 x dx 3 4 tan x dx 2 x3 0 3x 3 dx 5 dx ■ ■ ■ 4 ■ 1 1 ■ , what is y st dt ? x 4 dx s1 ■ ■ ■ ■ ■ 3 ■ 4 4 ■ sin2x dx ■ ■ ■ ■ 9 61–62 |||| Use properties of integrals, together with Exercises 27 and 28, to prove the inequality. 1 42. Evaluate y x 2 cos x dx. 1 1 43. In Example 2 in Section 5.1 we showed that x0 x 2 dx Use this fact and the properties of integrals to evaluate x01 5 6 x 2 dx. 1 3 . 61. ■ y 3 sx 4 1 ■ 1 dx ■ ■ 26 3 ■ 62. ■ ■ 2 y 0 ■ 2 x sin x dx ■ ■ 8 ■ ■ 5E-05(pp 334-343) 1/17/06 3:40 PM Page 339 D ISCOVERY PROJECT AREA FUNCTIONS 63. Prove Property 3 of integrals. 67–68 i1 65. If f is continuous on a, b , show that y b a [Hint: fx y f x dx fx 68. lim b nl f x dx a ■ f x .] 2 0 f x sin 2 x dx 1 n i1 ■ xi y 2 0 [Hint: Consider f x n 1 in 1 ■ 69. Find x12 x 66. Use the result of Exercise 65 to show that y i4 n5 67. lim nl 1 ■ x 4.] 2 ■ ■ ■ ■ ■ ■ ■ 2 dx. Hint: Choose x* to be the geometric mean of i and x i (that is, x* sx i 1 x i ) and use the identity i 1 mm 1 f x dx 1 m 1 m 1 DISCOVERY PROJECT Area Functions 1. (a) Draw the line y 2 t 1 and use geometry to find the area under this line, above the t-axis, and between the vertical lines t 1 and t 3. (b) If x 1, let A x be the area of the region that lies under the line y 2 t 1 between t 1 and t x. Sketch this region and use geometry to find an expression for A x . (c) Differentiate the area function A x . What do you notice? 2. (a) If x 339 Express the limit as a definite integral. |||| n 64. Prove Property 6 of integrals. ❙❙❙❙ 1, let y Ax x 1 1 t 2 dt A x represents the area of a region. Sketch that region. (b) Use the result of Exercise 28 in Section 5.2 to find an expression for A x . (c) Find A x . What do you notice? (d) If x 1 and h is a small positive number, then A x h A x represents the area of a region. Describe and sketch the region. (e) Draw a rectangle that approximates the region in part (d). By comparing the areas of these two regions, show that Ax h h Ax 1 x2 (f) Use part (e) to give an intuitive explanation for the result of part (c). ; 3. (a) Draw the graph of the function f x cos x 2 in the viewing rectangle 0, 2 by 1.25, 1.25 . (b) If we define a new function t by tx y x 0 cos t 2 dt then t x is the area under the graph of f from 0 to x [until f x becomes negative, at which point t x becomes a difference of areas]. Use part (a) to determine the value of x at which t x starts to decrease. [Unlike the integral in Problem 2, it is impossible to evaluate the integral defining t to obtain an explicit expression for t x .] (c) Use the integration command on your calculator or computer to estimate t(0.2), t(0.4), t(0.6), . . . , t(1.8), t(2). Then use these values to sketch a graph of t. (d) Use your graph of t from part (c) to sketch the graph of t using the interpretation of t x as the slope of a tangent line. How does the graph of t compare with the graph of f ? ■ 5E-05(pp 334-343) 340 ❙❙❙❙ 1/17/06 3:40 PM Page 340 CHAPTER 5 INTEGRALS 4. Suppose f is a continuous function on the interval a, b and we define a new function t by the equation y tx x a f t dt Based on your results in Problems 1–3, conjecture an expression for t x . |||| 5.3 The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus is appropriately named because it establishes a connection between the two branches of calculus: differential calculus and integral calculus. Differential calculus arose from the tangent problem, whereas integral calculus arose from a seemingly unrelated problem, the area problem. Newton’s teacher at Cambridge, Isaac Barrow (1630–1677), discovered that these two problems are actually closely related. In fact, he realized that differentiation and integration are inverse processes. The Fundamental Theorem of Calculus gives the precise inverse relationship between the derivative and the integral. It was Newton and Leibniz who exploited this relationship and used it to develop calculus into a systematic mathematical method. In particular, they saw that the Fundamental Theorem enabled them to compute areas and integrals very easily without having to compute them as limits of sums as we did in Sections 5.1 and 5.2. The first part of the Fundamental Theorem deals with functions defined by an equation of the form Investigate the area function interactively. Resources / Module 6 / Areas and Derivatives / Area as a Function y y tx 1 x a f t dt y=f(t) area=© 0 a x b FIGURE 1 t where f is a continuous function on a, b and x varies between a and b. Observe that t depends only on x, which appears as the variable upper limit in the integral. If x is a fixed number, then the integral xax f t dt is a definite number. If we then let x vary, the number xax f t dt also varies and defines a function of x denoted by t x . If f happens to be a positive function, then t x can be interpreted as the area under the graph of f from a to x, where x can vary from a to b. (Think of t as the “area so far” function; see Figure 1.) x0x f EXAMPLE 1 If f is the function whose graph is shown in Figure 2 and t x t dt, find the values of t 0 , t 1 , t 2 , t 3 , t 4 , and t 5 . Then sketch a rough graph of t. x00 f SOLUTION First we notice that t 0 y 2 t dt 0. From Figure 3 we see that t 1 is the area of a triangle: y=f(t) 1 0 1 2 4 y t1 t 1 0 1 2 f t dt 12 1 f t dt 1 To find t 2 we add to t 1 the area of a rectangle: FIGURE 2 t2 y 2 0 f t dt y 1 0 f t dt y 2 1 12 3 5E-05(pp 334-343) 1/17/06 3:40 PM Page 341 SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ❙❙❙❙ 341 We estimate that the area under f from 2 to 3 is about 1.3, so t3 y t2 3 2 f t dt 3 1.3 4.3 y 2 y 2 y 2 y 2 y 2 1 1 1 1 1 0 1 t 0 1 g(1)=1 t 2 0 g(2)=3 1 2 3 t 0 1 2 4 t 0 1 2 4 t g(3)Å4.3 FIGURE 3 g(4)Å3 For t g(5)Å1.7 3, f t is negative and so we start subtracting areas: y t4 g t3 y t5 4 t4 y 4 3 f t dt 4.3 f t dt 1.3 3 3.0 3 2 5 4 1.3 1.7 1 0 1 2 3 4 We use these values to sketch the graph of t in Figure 4. Notice that, because f t is positive for t 3, we keep adding area for t 3 and so t is increasing up to x 3, where it attains a maximum value. For x 3, t decreases because f t is negative. 5x FIGURE 4 x ©=j f(t) dt If we take f t a t and a 0, then, using Exercise 27 in Section 5.2, we have y tx h ƒ a FIGURE 5 x 0 t dt x2 2 Notice that t x x, that is, t f . In other words, if t is defined as the integral of f by Equation 1, then t turns out to be an antiderivative of f , at least in this case. And if we sketch the derivative of the function t shown in Figure 4 by estimating slopes of tangents, f in Example 1 too. we get a graph like that of f in Figure 2. So we suspect that t To see why this might be generally true we consider any continuous function f with fx 0. Then t x xax f t dt can be interpreted as the area under the graph of f from a to x, as in Figure 1. In order to compute t x from the definition of derivative we first observe that, t x is obtained by subtracting areas, so it is the area under the for h 0, t x h graph of f from x to x h (the gold area in Figure 5). For small h you can see from the figure that this area is approximately equal to the area of the rectangle with height f x and width h : y 0 x x+h b tx t so h tx tx h h tx hf x fx 5E-05(pp 334-343) 342 ❙❙❙❙ 1/17/06 3:41 PM Page 342 CHAPTER 5 INTEGRALS Intuitively, we therefore expect that tx tx lim h h hl0 tx fx The fact that this is true, even when f is not necessarily positive, is the first part of the Fundamental Theorem of Calculus. The Fundamental Theorem of Calculus, Part 1 If f is continuous on a, b , then the func|||| We abbreviate the name of this theorem as FTC1. In words, it says that the derivative of a definite integral with respect to its upper limit is the integrand evaluated at the upper limit. tion t defined by y tx x f t dt a a x b is continuous on a, b and differentiable on a, b , and t x Proof If x and x tx h are in a, b , then h tx y xh x a y xh x 2 y m y x f t dt a xh x y f t dt x a f t dt (by Property 5) f t dt 0, tx h h tx 1 h y xh x f t dt M mh that is, F IGURE 6 y For now let us assume that h 0. Since f is continuous on x, x h , the Extreme Value Theorem says that there are numbers u and v in x, x h such that f u m and fv M , where m and M are the absolute minimum and maximum values of f on x, x h . (See Figure 6.) By Property 8 of integrals, we have y=ƒ 0 f t dt f t dt a y and so, for h f x. xu √=x+h y f uh y xh x x Since h xh x f t dt Mh f t dt f vh 0, we can divide this inequality by h : fu 1 h y xh x f t dt fv Now we use Equation 2 to replace the middle part of this inequality: 3 fu tx h h tx fv Inequality 3 can be proved in a similar manner for the case h 0. (See Exercise 55.) 5E-05(pp 334-343) 1/17/06 3:41 PM Page 343 SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS Module 5.3 provides visual evidence for FTC1. ❙❙❙❙ Now we let h l 0. Then u l x and v l x, since u and v lie between x and x Therefore lim f u lim f u lim f v h. fx lim f v 343 fx hl0 u lx and hl0 v lx because f is continuous at x. We conclude, from (3) and the Squeeze Theorem, that tx 4 lim tx h h hl0 tx fx If x a or b, then Equation 4 can be interpreted as a one-sided limit. Then Theorem 3.2.4 (modified for one-sided limits) shows that t is continuous on a, b . Using Leibniz notation for derivatives, we can write FTC1 as d dx 5 y x a f t dt fx when f is continuous. Roughly speaking, Equation 5 says that if we first integrate f and then differentiate the result, we get back to the original function f . y EXAMPLE 2 Find the derivative of the function t x SOLUTION Since f t s1 x 0 s1 t 2 dt. t 2 is continuous, Part 1 of the Fundamental Theorem of Calculus gives tx s1 EXAMPLE 3 Although a formula of the form t x x2 xax f t dt may seem like a strange way of defining a function, books on physics, chemistry, and statistics are full of such functions. For instance, the Fresnel function Sx y x 0 sin t 2 2 dt is named after the French physicist Augustin Fresnel (1788–1827), who is famous for his works in optics. This function first appeared in Fresnel’s theory of the diffraction of light waves, but more recently it has been applied to the design of highways. Part 1 of the Fundamental Theorem tells us how to differentiate the Fresnel function: Sx sin x2 2 This means that we can apply all the methods of differential calculus to analyze S (see Exercise 49). 5E-05(pp 344-353) 344 ❙❙❙❙ 1/17/06 3:44 PM Page 344 CHAPTER 5 INTEGRALS Figure 7 shows the graphs of f x sin x 2 2 and the Fresnel function x Sx x0 f t dt. A computer was used to graph S by computing the value of this integral for many values of x. It does indeed look as if S x is the area under the graph of f from 0 to x [until x 1.4 when S x becomes a difference of areas]. Figure 8 shows a larger part of the graph of S. y y 1 f 0 0.5 S x 1 x 1 FIGURE 7 FIGURE 8 x The Fresnel function S(x)=j0 sin (πt@/2) dt ƒ=sin (π≈/2) x S(x)=j sin (πt@/2) dt 0 If we now start with the graph of S in Figure 7 and think about what its derivative should look like, it seems reasonable that S x f x . [For instance, S is increasing when f x 0 and decreasing when f x 0.] So this gives a visual confirmation of Part 1 of the Fundamental Theorem of Calculus. EXAMPLE 4 Find d dx y x4 1 sec t dt. SOLUTION Here we have to be careful to use the Chain Rule in conjunction with FTC1. Let u x 4. Then d dx y x4 1 sec t dt d dx y u 1 sec t dt d du y sec u du dx u 1 sec t dt sec x 4 du dx (by the Chain Rule) (by FTC1) 4x 3 In Section 5.2 we computed integrals from the definition as a limit of Riemann sums and we saw that this procedure is sometimes long and difficult. The second part of the Fundamental Theorem of Calculus, which follows easily from the first part, provides us with a much simpler method for the evaluation of integrals. The Fundamental Theorem of Calculus, Part 2 If f is continuous on a, b , then |||| We abbreviate this theorem as FTC2. y b a f x dx Fb Fa where F is any antiderivative of f , that is, a function such that F f. 5E-05(pp 344-353) 1/17/06 3:44 PM Page 345 S ECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS Proof Let t x ❙❙❙❙ 345 xax f t dt. We know from Part 1 that t x f x ; that is, t is an antiderivative of f . If F is any other antiderivative of f on a, b , then we know from Corollary 4.2.7 that F and t differ by a constant: Fx 6 tx C for a x b. But both F and t are continuous on a, b and so, by taking limits of both sides of Equation 6 (as x l a and x l b ), we see that it also holds when x a and x b. If we put x a in the formula for t x , we get y ta So, using Equation 6 with x Fb b and x Fa a a f t dt 0 a, we have tb C tb ta ta C y tb b a f t dt Part 2 of the Fundamental Theorem states that if we know an antiderivative F of f , then we can evaluate xab f x d x simply by subtracting the values of F at the endpoints of the interval a, b . It’s very surprising that xab f x d x, which was defined by a complicated procedure involving all of the values of f x for a x b, can be found by knowing the values of F x at only two points, a and b. Although the theorem may be surprising at first glance, it becomes plausible if we interpret it in physical terms. If v t is the velocity of an object and s t is its position at time t, s t , so s is an antiderivative of v. In Section 5.1 we considered an object that then v t always moves in the positive direction and made the guess that the area under the velocity curve is equal to the distance traveled. In symbols: y b a v t dt sb sa That is exactly what FTC2 says in this context. EXAMPLE 5 Evaluate the integral y 1 2 x 3 dx. x 3 is continuous on 2, 1 and we know from Sec14 tion 4.10 that an antiderivative is F x 4 x , so Part 2 of the Fundamental Theorem gives SOLUTION The function f x y 1 2 x 3 dx F1 F 2 1 4 1 4 1 4 2 4 15 4 Notice that FTC2 says we can use any antiderivative F of f. So we may as well use 14 14 7 or 1 x 4 C. the simplest one, namely F x 4 x , instead of 4 x 4 We often use the notation Fx b a Fb Fa 5E-05(pp 344-353) 346 ❙❙❙❙ 1/17/06 3:44 PM Page 346 CHAPTER 5 INTEGRALS So the equation of FTC2 can be written as y b f x dx a Fx b a Other common notations are F x b where a b a and F x x 2 is F x using Part 2 of the Fundamental Theorem: SOLUTION An antiderivative of f x y A 1 0 1 x3 3 2 x dx f . x 2 from 0 to 1. EXAMPLE 6 Find the area under the parabola y |||| In applying the Fundamental Theorem we use a particular antiderivative F of f . It is not necessary to use the most general antiderivative. F 0 1 3 x 3. The required area A is found 13 3 03 3 1 3 If you compare the calculation in Example 6 with the one in Example 2 in Section 5.1, you will see that the Fundamental Theorem gives a much shorter method. y y=cos x 1 EXAMPLE 7 Find the area under the cosine curve from 0 to b, where 0 SOLUTION Since an antiderivative of f x cos x is F x b 2. sin x, we have area=1 0 π 2 x A y b 0 cos x dx In particular, taking b from 0 to 2 is sin 2 FIGURE 9 sin x b 0 sin b sin 0 sin b 2, we have proved that the area under the cosine curve 1. (See Figure 9.) When the French mathematician Gilles de Roberval first found the area under the sine and cosine curves in 1635, this was a very challenging problem that required a great deal of ingenuity. If we didn’t have the benefit of the Fundamental Theorem, we would have to compute a difficult limit of sums using obscure trigonometric identities (or a computer algebra system as in Exercise 25 in Section 5.1). It was even more difficult for Roberval because the apparatus of limits had not been invented in 1635. But in the 1660s and 1670s, when the Fundamental Theorem was discovered by Barrow and exploited by Newton and Leibniz, such problems became very easy, as you can see from Example 7. EXAMPLE 8 What is wrong with the following calculation? | y 3 1 1 dx x2 1 x 1 3 1 1 3 1 4 3 SOLUTION To start, we notice that this calculation must be wrong because the answer is negative but f x 1 x 2 0 and Property 6 of integrals says that xab f x d x 0 when f 0. The Fundamental Theorem of Calculus applies to continuous functions. It can’t be applied here because f x 1 x 2 is not continuous on 1, 3 . In fact, f has an infinite discontinuity at x 0, so y 3 1 1 dx x2 does not exist 5E-05(pp 344-353) 1/17/06 3:44 PM Page 347 SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ❙❙❙❙ 347 D ifferentiation and Integration as Inverse Processes We end this section by bringing together the two parts of the Fundamental Theorem. The Fundamental Theorem of Calculus Suppose f is continuous on a, b . 1. If t x 2. y b a y x a f x dx f t dt, then t x Fb f x. F a , where F is any antiderivative of f , that is, F f. We noted that Part 1 can be rewritten as d dx y x a f t dt fx which says that if f is integrated and then the result is differentiated, we arrive back at the f x , Part 2 can be rewritten as original function f . Since F x y b a F x dx Fb Fa This version says that if we take a function F , first differentiate it, and then integrate the F a . Taken result, we arrive back at the original function F , but in the form F b together, the two parts of the Fundamental Theorem of Calculus say that differentiation and integration are inverse processes. Each undoes what the other does. The Fundamental Theorem of Calculus is unquestionably the most important theorem in calculus and, indeed, it ranks as one of the great accomplishments of the human mind. Before it was discovered, from the time of Eudoxus and Archimedes to the time of Galileo and Fermat, problems of finding areas, volumes, and lengths of curves were so difficult that only a genius could meet the challenge. But now, armed with the systematic method that Newton and Leibniz fashioned out of the Fundamental Theorem, we will see in the chapters to come that these challenging problems are accessible to all of us. |||| 5.3 Exercises 1. Explain exactly what is meant by the statement that “differenti- ation and integration are inverse processes.” 2. Let t x x0x f t dt, where f is the function whose graph is shown. (a) Evaluate t x for x 0, 1, 2, 3, 4, 5, and 6. (b) Estimate t 7 . (c) Where does t have a maximum value? Where does it have a minimum value? (d) Sketch a rough graph of t. y 3 2 1 0 1 4 6 t 5E-05(pp 344-353) 348 ❙❙❙❙ 1/17/06 3:45 PM Page 348 CHAPTER 5 INTEGRALS x0x f 3. Let t x t dt, where f is the function whose graph 12. F x 13. h x is shown. (a) Evaluate t 0 , t 1 , t 2 , t 3 , and t 6 . (b) On what interval is t increasing? (c) Where does t have a maximum value? (d) Sketch a rough graph of t. y y 15. y y 17. y y t 1 ■ y 21. y 23. y 25. y 27. y 29. y 31. y 33. y 35. y 36. 3 1 0 y t 1 Sketch the area represented by t x . Then find t x in two ways: (a) by using Part 1 of the Fundamental Theorem and (b) by evaluating the integral using Part 2 and then differentiating. |||| ■ ■ x 1 t 2 dt ■ 6. t x ■ ■ ■ ■ ■ x y (1 9. t y 11. F x y x 0 y y 2 y 2 x 2 s1 ■ ■ y ■ 10. t u t sin t dt x 1 y ■ t 4 5 dt 2 y x 2 cos t 2 dt ■ ■ t sin t dt sin3t dt ■ ■ 22. y x 4 5 dx 24. y 3 dt t4 26. y 28. y 30. y 32. y (3 34. y 8 4x 2 1 0 2 1 5 2 3 dx 2 dx x3 5 x5 dx x2 0 4 0 2 sec 2 t dt csc 2 d 2 f x dx 0 ■ x x2 5 y x4 x5 where f x f x dx ■ ■ ■ 2 ■ ■ ■ 4 6 dx 1 0 ■ ■ sx dx 3 2 2 4 1 x 5 dx cos d 1 dx sx 1 x sx ) dx 0 6 0 if 0 if 1 ■ y 2 dy 3y 83 1 csc cot d x x x if sin x if 0 where f x ■ 1 2 x 0 x ■ ■ ■ ■ ; 37–40 |||| Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area. 37. y 1 u 3 cos t 2 dt ■ x 5 dx 3 sx, 0 x 27 38. y x 4, 1 x 6 39. y 2 Hint: y cos t 2 dt x 8. t x 2t dt st ) dt 0 7–18 |||| Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. 7. t x 0 1 x2 20. 1 f t dt, where f is the function whose graph is f y y cos x 1 r 3 dr s1 |||| Use Part 2 of the Fundamental Theorem of Calculus to evaluate the integral, or explain why it does not exist. y 5. t x y 18. y du x2 0 19–36 shown. (a) Evaluate t 3 and t 3 . (b) Estimate t 2 , t 1 , and t 0 . (c) On what interval is t increasing? (d) Where does t have a maximum value? (e) Sketch a rough graph of t. (f) Use the graph in part (e) to sketch the graph of t x . Compare with the graph of f . 5–6 u2 1 1 3x y 14. h x 16. y u3 1 ■ 19. 3 sin4t dt cos t dt t sx 3 ■ 0 x 1x f 1 4. Let t x tan d 2 y x 10 x sin x, 0 x 40. y sec2x, 0 x dx ■ ■ ■ ■ 3 ■ ■ ■ ■ ■ ■ ■ ■ 5E-05(pp 344-353) 1/17/06 3:45 PM Page 349 S ECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS 41– 42 |||| Evaluate the integral and interpret it as a difference of areas. Illustrate with a sketch. 41. y 2 ■ x 3 dx 1 ■ y 42. ■ ■ ■ ■ 5 2 4 ■ y ■ ■ y 43. t x 3x 2x u2 u2 1 du 1 3x Hint: y f u du y 2x y 44. t x s2 tan x y f u du 3x 0 f u du |||| 51. 1 x2 0 2x t 4 x0x f Let t x y 3 dt 2 y 45. y ■ x3 sx ■ 47. If F x ■ ■ y x 1 y 46. y st sin t dt ■ ■ ■ cos x ■ y f t dt, where f t 5x t2 u4 s1 u 1 cos u 2 du ■ 1 t dt, where f is the function whose graph is shown. (a) At what values of x do the local maximum and minimum values of t occur? (b) Where does t attain its absolute maximum value? (c) On what intervals is t concave downward? (d) Sketch the graph of t. Find the derivative of the function. |||| sin t dt t ■ 51–52 43–46 x 0 ■ 349 (d) Does this function have horizontal asymptotes? (e) Solve the following equation correct to one decimal place: sin x dx ■ ❙❙❙❙ ■ f 1 ■ 0 _1 ■ 2 4 6 t 8 _2 du, find F 2 . 52. 48. Find the interval on which the curve y y 1 t x 0 1 y f 0.4 t2 dt 0.2 is concave upward. 0 49. The Fresnel function S was defined in Example 3 and graphed CAS in Figures 7 and 8. (a) At what values of x does this function have local maximum values? (b) On what intervals is the function concave upward? (c) Use a graph to solve the following equation correct to two decimal places: y 0 CAS sin 2 t 2 dt 0.2 50. The sine integral function ■ ■ ■ 53. lim nl 54. lim y x 0 sin t dt t is important in electrical engineering. [The integrand ft sin t t is not defined when t 0, but we know that its 1 and this makes f limit is 1 when t l 0. So we define f 0 a continuous function everywhere.] (a) Draw the graph of Si. (b) At what values of x does this function have local maximum values? (c) Find the coordinates of the first inflection point to the right of the origin. 5 7 t 9 ■ ■ ■ ■ ■ ■ ■ ■ ■ 53–54 |||| Evaluate the limit by first recognizing the sum as a Riemann sum for a function defined on 0, 1 . nl Si x 3 _0.2 n x 1 ■ ■ i1 i3 n4 1 n 1 n ■ 2 n ■ 3 n ■ ■ n n ■ 55. Justify (3) for the case h ■ ■ ■ ■ ■ 0. 56. If f is continuous and t and h are differentiable functions, find a formula for d dx 57. (a) Show that 1 (b) Show that 1 y hx tx f t dt s1 x 3 1 x01 s1 x 3 dx x 3 for x 1.25. 0. 5E-05(pp 344-353) 350 ❙❙❙❙ 1/17/06 3:45 PM Page 350 CHAPTER 5 INTEGRALS 62. A high-tech company purchases a new computing system 58. Let 0 x 2 0 fx and if if x if if y tx x 0 1 x x 0 x x 2 whose initial value is V . The system will depreciate at the rate f f t and will accumulate maintenance costs at the rate t t t , where t is the time measured in months. The company wants to determine the optimal time to replace the system. (a) Let 1 2 0 59. Find a function f and a number a such that 6 for all x y a and 2, h 1 2, 13, and h is continu- ■ ■ 1 t 63–68 63. A y t 0 y y f s ds What does C represent and why would the company want to minimize C? (c) Show that C has a minimum value at the numbers t T where C T fT. 30 30 t 0 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ The following exercises are intended only for those who have already covered Chapter 7. 65. Ct |||| 5.4 Vt 2 12,900 t Determine the length of time T for the total depreciation x0t f s ds to equal the initial value V . Dt (c) Determine the absolute minimum of C on 0, T . (d) Sketch the graphs of C and f t in the same coordinate system, and verify the result in part (a) in this case. 61. A manufacturing company owns a major piece of equipment that depreciates at the (continuous) rate f f t , where t is the time measured in months since its last overhaul. Because a fixed cost A is incurred each time the machine is overhauled, the company wants to determine the optimal time T (in months) between overhauls. (a) Explain why x0t f s ds represents the loss in value of the machine over the period of time t since the last overhaul. (b) Let C C t be given by t s ds V t if 0 450 if t 0. h1 3, h 2 6, h 2 5, h 2 ous everywhere. Evaluate x12 h u du. fs tt 2 sx 60. Suppose h is a function such that h 1 t 0 V 15 0 ft ft dt t2 y Show that the critical numbers of C occur at the numbers t where C t ft tt. (b) Suppose that (a) Find an expression for t x similar to the one for f x . (b) Sketch the graphs of f and t. (c) Where is f differentiable? Where is t differentiable? x 1 t Ct f t dt |||| 1 dx 2x 9 1 ■ y 1 1 64. s1 eu ■ 1 66. dt t2 du ■ ■ ■ ■ y y 68. 6 s3 2 12 67. Evaluate the integral. y ■ 1 0 10 x dx 4 1 t2 0 2 1 u2 4 u 1 ■ 3 ■ dt du ■ ■ ■ Indefinite Integrals and the Net Change Theorem We saw in Section 5.3 that the second part of the Fundamental Theorem of Calculus provides a very powerful method for evaluating the definite integral of a function, assuming that we can find an antiderivative of the function. In this section we introduce a notation for antiderivatives, review the formulas for antiderivatives, and use them to evaluate definite integrals. We also reformulate FTC2 in a way that makes it easier to apply to science and engineering problems. 5E-05(pp 344-353) 1/17/06 3:46 PM Page 351 SECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM ❙❙❙❙ 351 I ndefinite Integrals Both parts of the Fundamental Theorem establish connections between antiderivatives and definite integrals. Part 1 says that if f is continuous, then xax f t dt is an antiderivative of f. F a , where F is an antiPart 2 says that xab f x dx can be found by evaluating F b derivative of f. We need a convenient notation for antiderivatives that makes them easy to work with. Because of the relation given by the Fundamental Theorem between antiderivatives and integrals, the notation x f x d x is traditionally used for an antiderivative of f and is called an indefinite integral. Thus yf x dx Fx Fx means fx For example, we can write yx 2 x3 3 dx C x3 3 d dx because x2 C So we can regard an indefinite integral as representing an entire family of functions (one antiderivative for each value of the constant C ). | You should distinguish carefully between definite and indefinite integrals. A definite integral xab f x d x is a number, whereas an indefinite integral x f x d x is a function (or family of functions). The connection between them is given by Part 2 of the Fundamental Theorem. If f is continuous on a, b , then y b a yf f x dx x dx b a The effectiveness of the Fundamental Theorem depends on having a supply of antiderivatives of functions. We therefore restate the Table of Antidifferentiation Formulas from Section 4.10, together with a few others, in the notation of indefinite integrals. Any formula can be verified by differentiating the function on the right side and obtaining the integrand. For instance 1 y cf C d tan x dx C y tan x because fx t x dx yf yx 2 y sec x dx n xn 1 n1 sec2x Table of Indefinite Integrals x dx y k dx kx y sin x dx 2 y sec x dx c y f x dx C cos x tan x y sec x tan x dx y cos x dx C 2 y csc x dx C sec x dx C sin x cot x y csc x cot x dx C ytx x dx n 1 C C csc x C dx 5E-05(pp 344-353) 352 ❙❙❙❙ 1/17/06 3:46 PM Page 352 CHAPTER 5 INTEGRALS Recall from Theorem 4.10.1 that the most general antiderivative on a given interval is obtained by adding a constant to a particular antiderivative. We adopt the convention that when a formula for a general indefinite integral is given, it is valid only on an interval. Thus, we write 1 1 C y x 2 dx x with the understanding that it is valid on the interval 0, or on the interval , 0 . This is true despite the fact that the general antiderivative of the function f x 1 x 2, x 0, is 1 x 1 x Fx if x 0 C2 if x 0 EXAMPLE 1 Find the general indefinite integral |||| The indefinite integral in Example 1 is graphed in Figure 1 for several values of C. The value of C is the y-intercept. y 4 10 x 4 2 sec 2x d x SOLUTION Using our convention and Table 1, we have 10 x 4 y _1.5 C1 2 sec2x d x 10 y x 4 dx 1.5 10 _4 FIGURE 1 2 y sec2x dx x5 5 2x 5 2 tan x 2 tan x C C You should check this answer by differentiating it. EXAMPLE 2 Evaluate cos d. sin2 y SOLUTION This indefinite integral isn’t immediately apparent in Table 1, so we use trigonometric identities to rewrite the function before integrating: y cos d sin2 y 1 sin y csc EXAMPLE 3 Evaluate y 3 0 x3 cos sin cot d d csc C 6 x d x. SOLUTION Using FTC2 and Table 1, we have y 3 0 x 3 6x dx x4 4 ( 1 4 81 4 3 x2 6 2 3 4 27 0 3 32 ) 0 0 (1 4 04 6.75 Compare this calculation with Example 2(b) in Section 5.2. 3 02 ) 5E-05(pp 344-353) 1/17/06 3:46 PM Page 353 SECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM |||| Figure 2 shows the graph of the integrand in Example 4. We know from Section 5.2 that the value of the integral can be interpreted as the sum of the areas labeled with a plus sign minus the areas labeled with a minus sign. E XAMPLE 4 Find y 12 SOLUTION The Fundamental Theorem gives y y 12 x 0 12 x2 2 12 sin x dx y=x-12 sin x 1 2 + 12 cos x 0 12 2 12 cos 12 0 72 + _ _ 12 cos 12 60 10 353 12 sin x dx. x 0 ❙❙❙❙ cos 0 12 12 cos 12 12 x This is the exact value of the integral. If a decimal approximation is desired, we can use a calculator to approximate cos 12. Doing so, we get y FIGURE 2 12 x 0 EXAMPLE 5 Evaluate y 9 2t 2 12 sin x dx t 2 st t2 1 1 70.1262 dt. SOLUTION First we need to write the integrand in a simpler form by carrying out the division: y 9 2t 2 1 t 2 st t2 1 y dt 9 1 18 t 3 2 18 9 2t 2 1 t 2 32 3 t 1 1 9 32 1 9 dt 9 1 1 2 3 9 2 t t3 2 2t [2 t1 2 2 (2 2 3 1 2 3 1 32 1 9 1 32 1 1 ) 4 9 Applications Part 2 of the Fundamental Theorem says that if f is continuous on a, b , then y b a f x dx Fb where F is any antiderivative of f. This means that F ten as y b a F x dx Fb Fa f , so the equation can be rewrit- Fa We know that F x represents the rate of change of y F x with respect to x and Fb F a is the change in y when x changes from a to b. [Note that y could, for instance, increase, then decrease, then increase again. Although y might change in both directions, Fb F a represents the net change in y.] So we can reformulate FTC2 in words as follows. 5E-05(pp 354-363) 354 ❙❙❙❙ 1/17/06 3:47 PM Page 354 CHAPTER 5 INTEGRALS The Net Change Theorem The integral of a rate of change is the net change: y b a F x dx Fb Fa This principle can be applied to all of the rates of change in the natural and social sciences that we discussed in Section 3.4. Here are a few instances of this idea: ■ If V t is the volume of water in a reservoir at time t, then its derivative V t is the rate at which water flows into the reservoir at time t. So y t2 t1 V t dt V t2 V t1 is the change in the amount of water in the reservoir between time t1 and time t2 . ■ If C t is the concentration of the product of a chemical reaction at time t, then the rate of reaction is the derivative d C dt. So y dC dt dt t2 t1 C t2 C t1 is the change in the concentration of C from time t1 to time t2 . ■ If the mass of a rod measured from the left end to a point x is m x , then the linear density is x m x . So y b x dx a mb ma is the mass of the segment of the rod that lies between x ■ a and x b. If the rate of growth of a population is dn dt, then y t2 t1 dn dt dt n t2 n t1 is the net change in population during the time period from t1 to t2 . (The population increases when births happen and decreases when deaths occur. The net change takes into account both births and deaths.) ■ If C x is the cost of producing x units of a commodity, then the marginal cost is the derivative C x . So y x2 x1 C x dx C x2 C x1 is the increase in cost when production is increased from x1 units to x2 units. ■ If an object moves along a straight line with position function s t , then its velocity is v t s t , so 2 y t2 t1 v t dt s t2 s t1 is the net change of position, or displacement, of the particle during the time period from t1 to t2 . In Section 5.1 we guessed that this was true for the case where the object moves in the positive direction, but now we have proved that it is always true. 5E-05(pp 354-363) 1/17/06 3:47 PM Page 355 S ECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM ■ ❙❙❙❙ If we want to calculate the distance traveled during the time interval, we have to consider the intervals when v t 0 (the particle moves to the right) and also the intervals when v t 0 (the particle moves to the left). In both cases the distance is computed by integrating v t , the speed. Therefore y 3 t2 t1 vt dt total distance traveled Figure 3 shows how both displacement and distance traveled can be interpreted in terms of areas under a velocity curve. √ √(t) t™ displacement=j √(t) dt=A¡-A™+A£ t¡ A¡ t™ A£ 0 t¡ t t™ A™ distance=j | √(t)| dt=A¡+A™+A£ t¡ F IGURE 3 ■ The acceleration of the object is a t y t2 t1 v t , so a t dt v t2 v t1 is the change in velocity from time t1 to time t2 . Resources / Module 7 / Physics and Engineering / Start of Spy Tracks EXAMPLE 6 A particle moves along a line so that its velocity at time t is vt t 2 t 6 (measured in meters per second). (a) Find the displacement of the particle during the time period 1 (b) Find the distance traveled during this time period. t 4. SOLUTION (a) By Equation 2, the displacement is s4 y s1 4 1 y v t dt t3 3 4 1 t2 t 4 t2 2 6 dt 9 2 6t 1 This means that the particle moved 4.5 m toward the left. (b) Note that v t t2 t 6 t 3 t 2 and so v t 0 on the interval 1, 3 and v t 0 on 3, 4 . Thus, from Equation 3, the distance traveled is |||| To integrate the absolute value of v t , we use Property 5 of integrals from Section 5.2 to split the integral into two parts, one where vt 0 and one where v t 0. y 4 1 vt dt y 3 1 y 3 1 v t dt t2 t3 3 61 6 y 4 3 v t dt t 6 dt t2 2 6t 10.17 m 3 1 y 4 3 t2 t3 3 t 6 dt t2 2 6t 4 3 355 5E-05(pp 354-363) 356 ❙❙❙❙ 1/17/06 3:48 PM Page 356 CHAPTER 5 INTEGRALS EXAMPLE 7 Figure 4 shows the power consumption in the city of San Francisco for a day in September (P is measured in megawatts; t is measured in hours starting at midnight). Estimate the energy used on that day. P 800 600 400 200 0 FIGURE 4 3 6 9 12 15 18 t 21 Pacific Gas & Electric SOLUTION Power is the rate of change of energy: P t Theorem, y 24 0 y P t dt 24 0 E t dt E t . So, by the Net Change E 24 E0 is the total amount of energy used that day. We approximate the value of the integral using the Midpoint Rule with 12 subintervals and t 2: y 24 P t dt P1 P3 440 400 420 620 790 840 840 0 P5 P 21 P 23 810 690 670 t 550 2 850 15,840 The energy used was approximately 15,840 megawatt-hours. How did we know what units to use for energy in Example 7? The integral x024 P t dt is defined as the limit of sums of terms of the form P t * t. Now P t * is measured in i i megawatts and t is measured in hours, so their product is measured in megawatt-hours. The same is true of the limit. In general, the unit of measurement for xab f x dx is the product of the unit for f x and the unit for x. |||| A note on units |||| 5.4 1– 4 1. 2. |||| Exercises Verify by differentiation that the formula is correct. y sx x 2 1 dx sx 2 1 4. yx x sin x cos x ■ 5–14 C ■ |||| 5. 3. y sa 1 2 x2 3 dx x a 2 sa 2 sx 2 a sx 2 a 2 a 2x dx 2 C C ■ y x cos x dx 1 2 x2 C yx 7. y ■ ■ ■ ■ ■ ■ ■ Find the general indefinite integral. 34 x3 6. dx 6x 1 dx y sx dx 8. yx1 3 2 x 4 dx ■ ■ 5E-05(pp 354-363) 1/17/06 3:48 PM Page 357 S ECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM 9. y 1 10. t 2 dt t2 y u2 11. y (2 s x ) 2 dx 12. y sin 13. y1 sin x dx sin2x 14. y 357 43. The area of the region that lies to the right of the y-axis and to 1 u2 1 ❙❙❙❙ du the left of the parabola x 2y y 2 (the shaded region in the figure) is given by the integral x02 2y y 2 d y. (Turn your head clockwise and think of the region as lying below the curve x 2y y 2 from y 0 to y 2.) Find the area of the region. sin 2 x dx sin x 3 cos d y 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ x=2y-¥ ■ ; 15–16 |||| Find the general indefinite integral. Illustrate by graphing several members of the family on the same screen. 15. y x sx dx 16. y cos x 0 2 sin x d x x 1 ■ ■ ■ 17– 40 17. y 19. y 21. y 23. y 25. 27. 6x 0 0 2 6y 2 y 1 2 dy y3 4y 3 2 26. sx ) d x 3 28. 4 5 dx x 4 y 4 sin 0 64 1 1 1 39. y (x 4 y 2 ■ ■ ■ ■ ■ 2 2v 9 5y7 y y y 1 dv 0 dx x 46. The current in a wire is defined as the derivative of the charge: It xab I t sec sin Q t . (See Example 3 in Section 3.4.) What does dt represent? 47. If oil leaks from a tank at a rate of r t gallons per minute at tan d sin sec2 3 1 45. If w t is the rate of growth of a child in pounds per year, what does x510 w t dt represent? 2 time t, what does x0120 r t dt represent? tan2 48. A honeybee population starts with 100 bees and increases d at a rate of n t bees per week. What does 100 represent? x015 n t dt x 2 3 dx 1 49. In Section 4.8 we defined the marginal revenue function R x 8 1 as the derivative of the revenue function R x , where x is the 5000 number of units sold. What does x1000 R x d x represent? x1 dx 3 sx 2 3 0 ■ 2 50. If f x is the slope of a trail at a distance of x miles from the sin x d x ■ start of the trail, what does x35 f x d x represent? ■ ■ ■ y x x 2 x 4. Then use this information to estimate the area of the region that lies under the curve and above the x-axis. 2x 3x 4 51. If x is measured in meters and f x is measured in newtons, what are the units for x0100 f x d x ? ; 41. Use a graph to estimate the x-intercepts of the curve ; 42. Repeat Exercise 41 for the curve y y=1 y=$ „ œx 3x 2 dx sx 3 1 y dy 1 x x 4 0 dx u 2 du 5 3v 0 y 4 y 1, and the curve y sx . Find the area of this region by writing x as a function of y and integrating with respect to y (as in Exercise 43). s 2 t dt 2 y ■ u3 y3 y 40. ■ 9 4x 3 1 4 1 38. 2 x ) dx 1 u5 2 1 y 36. 5 sx 4 ) dx 5 0 ■ ■ 2x 1 3 sx dx sx y (sx y 32. d 0 0 34. 3 cos 1 0 30. cos2 d cos2 1 4 37. ■ y 3 1 24. 1 y 20. 14 d y y t dt 0 33. 18. 5 dx y 0 ■ 44. The boundaries of the shaded region are the y-axis, the line 22. y 35. 4x y x (sx 31. ■ 1 2 du 1 y ■ st 1 4 1 29. ■ 3u 2 y 2 5y 4 3 ■ Evaluate the integral. |||| 2 ■ 6 2x . 52. If the units for x are feet and the units for a x are pounds per foot, what are the units for da d x? What units does x28 a x d x have? 5E-05(pp 354-363) 358 ❙❙❙❙ 1/17/06 3:48 PM Page 358 CHAPTER 5 INTEGRALS 62. Water flows in and out of a storage tank. A graph of the rate of 53–54 |||| The velocity function (in meters per second) is given for a particle moving along a line. Find (a) the displacement and (b) the distance traveled by the particle during the given time interval. 53. v t 3t 5, 54. v t 2 2t ■ t ■ ■ 0 t ■ 3 1 8, t change r t of the volume of water in the tank, in liters per day, is shown. If the amount of water in the tank at time t 0 is 25,000 L, use the Midpoint Rule to estimate the amount of water four days later. ■ r 2000 6 ■ ■ ■ ■ ■ ■ ■ The acceleration function (in m s2 ) and the initial velocity are given for a particle moving along a line. Find (a) the velocity at time t and (b) the distance traveled during the given time interval. 55–56 55. a t t 56. a t 2t ■ 1000 |||| ■ 4, 5, v0 0 3, v 0 ■ ■ t 4, 0 ■ ■ 0 t 2 4t 3 _1000 10 ■ 1 3 ■ ■ ■ ■ ■ 57. The linear density of a rod of length 4 m is given by x 9 2 s x measured in kilograms per meter, where x is measured in meters from one end of the rod. Find the total mass of the rod. 58. Water flows from the bottom of a storage tank at a rate of rt 200 4 t liters per minute, where 0 t 50. Find the amount of water that flows from the tank during the first 10 minutes. 59. The velocity of a car was read from its speedometer at 10-second intervals and recorded in the table. Use the Midpoint Rule to estimate the distance traveled by the car. t (s) v (mi h) t (s) v (mi h) 0 10 20 30 40 50 0 38 52 58 55 51 60 70 80 90 100 56 53 50 47 45 63. Economists use a cumulative distribution called a Lorenz curve to describe the distribution of income between households in a given country. Typically, a Lorenz curve is defined on 0, 1 with endpoints 0, 0 and 1, 1 , and is continuous, increasing, and concave upward. The points on this curve are determined by ranking all households by income and then computing the percentage of households whose income is less than or equal to a given percentage of the total income of the country. For example, the point a 100, b 100 is on the Lorenz curve if the bottom a% of the households receive less than or equal to b% of the total income. Absolute equality of income distribution would occur if the bottom a% of the households receive a% of the income, in which case the Lorenz curve would be the line y x. The area between the Lorenz curve and the line y x measures how much the income distribution differs from absolute equality. The coefficient of inequality is the ratio of the area between the Lorenz curve and the line y x to the area under y x. y (1, 1) 1 y=x y=L (x) 60. Suppose that a volcano is erupting and readings of the rate r t at which solid materials are spewed into the atmosphere are given in the table. The time t is measured in seconds and the units for r t are tonnes (metric tons) per second. t 0 1 2 3 4 5 6 rt 2 10 24 36 46 54 60 0 (a) Show that the coefficient of inequality is twice the area between the Lorenz curve and the line y x, that is, show that coefficient of inequality (a) Give upper and lower estimates for the quantity Q 6 of erupted materials after 6 seconds. (b) Use the Midpoint Rule to estimate Q 6 . 61. The marginal cost of manufacturing x yards of a certain fabric 2 is C x 3 0.01x 0.000006 x (in dollars per yard). Find the increase in cost if the production level is raised from 2000 yards to 4000 yards. x 1 1 2y x 0 L x dx (b) The income distribution for a certain country is represented by the Lorenz curve defined by the equation Lx 5 12 x2 7 12 x What is the percentage of total income received by the bottom 50% of the households? Find the coefficient of inequality. 5E-05(pp 354-363) 1/17/06 3:48 PM Page 359 ❙❙❙❙ WRITING PROJECT NEWTON, LEIBNIZ, AND THE INVENTION OF CALCULUS ; 64. On May 7, 1992, the space shuttle Endeavour was launched on mission STS-49, the purpose of which was to install a new perigee kick motor in an Intelsat communications satellite. The table gives the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. Event Time (s) Launch Velocity (ft s) 0 10 185 End roll maneuver 15 319 Throttle to 89% 20 447 Throttle to 67% 32 742 Throttle to 104% 59 1325 Maximum dynamic pressure 62 1445 125 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ The following exercises are intended only for those who have already covered Chapter 7. 65–67 65. y 67. |||| x e y Evaluate the integral. 2 x x 1 1 66. dx 0 Begin roll maneuver ■ 359 y 9 1 sx sx 4 2 dx 4151 Solid rocket booster separation (a) Use a graphing calculator or computer to model these data by a third-degree polynomial. (b) Use the model in part (a) to estimate the height reached by the Endeavour, 125 seconds after liftoff. ■ x2 1 1 ■ x2 ■ ■ 1 dx ■ ■ ■ ■ ■ ■ ■ ■ 68. The area labeled B is three times the area labeled A. Express b in terms of a. y y y=´ y=´ B A 0 a x 0 b x WRITING PROJECT Newton, Leibniz, and the Invention of Calculus We sometimes read that the inventors of calculus were Sir Isaac Newton (1642–1727) and Gottfried Wilhelm Leibniz (1646–1716). But we know that the basic ideas behind integration were investigated 2500 years ago by ancient Greeks such as Eudoxus and Archimedes, and methods for finding tangents were pioneered by Pierre Fermat (1601–1665), Isaac Barrow (1630–1677), and others. Barrow, Newton’s teacher at Cambridge, was the first to understand the inverse relationship between differentiation and integration. What Newton and Leibniz did was to use this relationship, in the form of the Fundamental Theorem of Calculus, in order to develop calculus into a systematic mathematical discipline. It is in this sense that Newton and Leibniz are credited with the invention of calculus. Read about the contributions of these men in one or more of the given references and write a report on one of the following three topics. You can include biographical details, but the main thrust of your report should be a description, in some detail, of their methods and notations. In particular, you should consult one of the sourcebooks, which give excerpts from the original publications of Newton and Leibniz, translated from Latin to English. The Role of Newton in the Development of Calculus The Role of Leibniz in the Development of Calculus The Controversy between the Followers of Newton and Leibniz over Priority in the Invention of Calculus References 1. Carl Boyer and Uta Merzbach, A History of Mathematics (New York: Wiley, 1987), Chapter 19. 5E-05(pp 354-363) 360 ❙❙❙❙ 1/17/06 3:49 PM Page 360 CHAPTER 5 INTEGRALS 2. Carl Boyer, The History of the Calculus and Its Conceptual Development (New York: Dover, 1959), Chapter V. 3. C. H. Edwards, The Historical Development of the Calculus (New York: Springer-Verlag, 1979), Chapters 8 and 9. 4. Howard Eves, An Introduction to the History of Mathematics, 6th ed. (New York: Saunders, 1990), Chapter 11. 5. C. C. Gillispie, ed., Dictionary of Scientific Biography (New York: Scribner’s, 1974). See the article on Leibniz by Joseph Hofmann in Volume VIII and the article on Newton by I. B. Cohen in Volume X. 6. Victor Katz, A History of Mathematics: An Introduction (New York: HarperCollins, 1993), Chapter 12. 7. Morris Kline, Mathematical Thought from Ancient to Modern Times (New York: Oxford University Press, 1972), Chapter 17. Sourcebooks 1. John Fauvel and Jeremy Gray, eds., The History of Mathematics: A Reader (London: MacMillan Press, 1987), Chapters 12 and 13. 2. D. E. Smith, ed., A Sourcebook in Mathematics (New York: Dover, 1959), Chapter V. 3. D. J. Struik, ed., A Sourcebook in Mathematics, 1200–1800 (Princeton, N.J.: Princeton University Press, 1969), Chapter V. |||| 5.5 The Substitution Rule Because of the Fundamental Theorem, it’s important to be able to find antiderivatives. But our antidifferentiation formulas don’t tell us how to evaluate integrals such as y 2 x s1 1 |||| Differentials were defined in Section 3.10. If u f x , then du f x dx x 2 dx To find this integral we use the problem-solving strategy of introducing something extra. Here the “something extra” is a new variable; we change from the variable x to a new variable u. Suppose that we let u be the quantity under the root sign in (1), u 1 x 2. Then the differential of u is du 2 x dx. Notice that if the dx in the notation for an integral were to be interpreted as a differential, then the differential 2 x dx would occur in (1) and, so, formally, without justifying our calculation, we could write y 2 x s1 2 x 2 dx y s1 2 3 x 2 2 x dx u3 2 2 3 C y su du x2 1 32 C But now we can check that we have the correct answer by using the Chain Rule to differentiate the final function of Equation 2: d dx [2 3 x2 1 32 C] 2 3 3 2 x2 1 12 2x 2x sx 2 1 In general, this method works whenever we have an integral that we can write in the form x f t x t x d x. Observe that if F f , then 3 yF t x t x dx Ftx C 5E-05(pp 354-363) 1/17/06 3:49 PM Page 361 S ECTION 5.5 THE SUBSTITUTION RULE ❙❙❙❙ 361 because, by the Chain Rule, d Ftx dx F tx t x If we make the “change of variable” or “substitution” u we have yF or, writing F t x t x dx Ftx C Fu t x , then from Equation 3 C yF u du f , we get yf t x t x dx yf u du Thus, we have proved the following rule. t x is a differentiable function whose range is an 4 The Substitution Rule If u interval I and f is continuous on I , then yf t x t x dx yf u du Notice that the Substitution Rule for integration was proved using the Chain Rule for differentiation. Notice also that if u t x , then du t x d x, so a way to remember the Substitution Rule is to think of dx and du in (4) as differentials. Thus, the Substitution Rule says: It is permissible to operate with dx and du after integral signs as if they were differentials. EXAMPLE 1 Find yx 3 cos x 4 2 d x. x 4 2 because its differential is du 4 x 3 dx, which, apart from the constant factor 4, occurs in the integral. Thus, using x 3 dx du 4 and the Substitution Rule, we have SOLUTION We make the substitution u |||| Check the answer by differentiating it. cos x 4 2 dx 1 4 sin u C 1 4 3 y cos u 1 4 yx sin x 4 du 2 1 4 y cos u du C Notice that at the final stage we had to return to the original variable x. The idea behind the Substitution Rule is to replace a relatively complicated integral by a simpler integral. This is accomplished by changing from the original variable x to a new variable u that is a function of x. Thus, in Example 1 we replaced the integral x x 3 cos x 4 2 d x by the simpler integral 1 x cos u du. 4 The main challenge in using the Substitution Rule is to think of an appropriate substitution. You should try to choose u to be some function in the integrand whose differential also occurs (except for a constant factor). This was the case in Example 1. If that is not 5E-05(pp 354-363) 362 ❙❙❙❙ 1/17/06 3:49 PM Page 362 CHAPTER 5 INTEGRALS possible, try choosing u to be some complicated part of the integrand. Finding the right substitution is a bit of an art. It’s not unusual to guess wrong; if your first guess doesn’t work, try another substitution. EXAMPLE 2 Evaluate SOLUTION 1 Let u y s2 x 2x 1 d x. 1. Then du 2 dx, so dx du 2. Thus, the Substitution Rule gives y s2 x du 2 y su 1 dx u3 2 32 1 2 1 3 2x 1 SOLUTION 2 Another possible substitution is u dx s2 x 1 du (Or observe that u 2 2x 1 y s1 SOLUTION Let u 1 4x 2 x s1 4x 2 dx ©= ƒ dx _1 FIGURE 1 ƒ= x 1-4≈ œ„„„„„„ 1 ©=j ƒ dx=_ 4 œ„„„„„„ 1-4≈ C C 1 . Then s2 x u du C 1 du yu 1 3 2 u du du 2x 1 32 1 8 8 x dx, so x dx du 1 8 y su 1 8 y u3 2 C d x. 4 x 2. Then du 1 f _1 x 32 du 2 dx.) Therefore u3 3 EXAMPLE 3 Find 1 3 C dx yu 1 dx 12 yu s2 x so 1, so 2u du y s2 x 1 2 1 8 (2 su ) yu 12 1 4 C du and du 4x 2 s1 C The answer to Example 3 could be checked by differentiation, but instead let’s check it with a graph. In Figure 1 we have used a computer to graph both the integrand 1 fx x s1 4 x 2 and its indefinite integral t x 4 x 2 (we take the case 4 s1 C 0). Notice that t x decreases when f x is negative, increases when f x is positive, and has its minimum value when f x 0. So it seems reasonable, from the graphical evidence, that t is an antiderivative of f . EXAMPLE 4 Calculate SOLUTION If we let u y cos 5x dx. 5x, then du y cos 5x dx 1 5 5 dx, so dx y cos u du 1 5 1 5 sin u du. Therefore C 1 5 sin 5x C 5E-05(pp 354-363) 1/17/06 3:50 PM Page 363 SECTION 5.5 THE SUBSTITUTION RULE E XAMPLE 5 Find y s1 1 x 2. Then du y s1 363 x 2 x 5 dx. SOLUTION An appropriate substitution becomes more obvious if we factor x 5 as x 4 u ❙❙❙❙ du 2. Also x 2 2 x dx, so x dx y s1 x 2 x 5 dx y su 1 2 ( 12 27 1 7 u 1 du 2 2 1 2 2u 3 2 u7 2 2 5 2 x2 1 u x. Let 1 2: x 2 x 4 x dx u5 2 y 1, so x 4 u u2 2u 1 du u 1 2 du 2 3 u5 2 2 5 72 y su u3 2) x2 1 C 1 3 52 1 x2 32 C Definite Integrals When evaluating a definite integral by substitution, two methods are possible. One method is to evaluate the indefinite integral first and then use the Fundamental Theorem. For instance, using the result of Example 2, we have y 4 1 dx 1 dx 1 3 9 4 y s2 x 1 3 0 s2 x 32 1 1 3 0 1 3 32 2x 1 27 4 32 1 0 26 3 Another method, which is usually preferable, is to change the limits of integration when the variable is changed. 5 |||| This rule says that when using a substitution in a definite integral, we must put everything in terms of the new variable u, not only x and dx but also the limits of integration. The new limits of integration are the values of u that correspond to x a and x b. The Substitution Rule for Definite Integrals If t is continuous on a, b and f is continuous on the range of u y b a t x , then y f t x t x dx tb ta f u du Proof Let F be an antiderivative of f . Then, by (3), F t x is an antiderivative of f t x t x , so by Part 2 of the Fundamental Theorem, we have y b a f t x t x dx Ftx b a F tb F ta But, applying FTC2 a second time, we also have y tb ta EXAMPLE 6 Evaluate y 4 0 f u du s2 x Fu tb ta F tb F ta 1 d x using (5). SOLUTION Using the substitution from Solution 1 of Example 2, we have u dx du 2. To find the new limits of integration we note that when x 0, u 1 and when x 4, u 9 2x 1 and 5E-05(pp 364-371) 364 ❙❙❙❙ 1/17/06 3:53 PM Page 364 CHAPTER 5 INTEGRALS y Therefore 4 0 s2 x 91 2 1 y 1 dx |||| The geometric interpretation of Example 6 is shown in Figure 2. The substitution u 2 x 1 stretches the interval 0, 4 by a factor of 2 and translates it to the right by 1 unit. The Substitution Rule shows that the two areas are equal. su du 2 3 1 3 32 9 9 1 2 u3 2 1 26 3 13 2 Observe that when using (5) we do not return to the variable x after integrating. We simply evaluate the expression in u between the appropriate values of u. y y 3 3 y=œ„„„„„ 2x+1 2 2 1 œ„ u 2 1 0 0 x 4 y= 1 9 u FIGURE 2 dx . 5x 2 2 EXAMPLE 7 Evaluate |||| The integral given in Example 7 is an abbreviation for y 1 2 1 3 5x 2 y SOLUTION Let u 5x. Then du 7. Thus when x 3 2, u 3 1 dx y dx 3 5x 2 1 5 dx, so dx 1 5 2 y 7 2 1 5 du 5. When x 2 and du u2 7 1 u 1 5 1, u 1 7 1 5u 2 1 2 7 2 1 14 Symmetry The next theorem uses the Substitution Rule for Definite Integrals (5) to simplify the calculation of integrals of functions that possess symmetry properties. 6 Integrals of Symmetric Functions Suppose f is continuous on (a) If f is even f (b) If f is odd f a f x , then x x a f x dx 2 x f x dx. f x , then xa a f x dx x a, a . a 0 0. Proof We split the integral in two: 7 y a a f x dx y 0 a f x dx y a 0 f x dx y 0 a f x dx In the first integral on the far right side we make the substitution u y a 0 f x dx x. Then 5E-05(pp 364-371) 1/17/06 3:53 PM Page 365 S ECTION 5.5 THE SUBSTITUTION RULE du d x and when x a, u a y 0 ❙❙❙❙ 365 a. Therefore y f x dx a f 0 u f y du u du a 0 f u du and so Equation 7 becomes (a) If f is even, then f y a a 0 a a x _a a 0 x6 EXAMPLE 8 Since f x 0 y a y 2 2 x6 0 f x dx 0 tan x f x , it is even and so 1 dx 0 x2 1 2 x 2(128 7 0 x 4 satisfies f 2) 284 7 x f x , it is odd and so tan x dx x2 x4 1 1 1 0 Exercises sin sx dx , sx x 3 u 2 10 dx, 1 dx, 4. y y 6. yx4 sx x [ y y cos 3x dx, yx a 0 2 5. 3. y 2 y x6 1 dx Evaluate the integral by making the given substitution. 2 a 2 y f x dx f x dx f u du 1 satisfies f FIGURE 3 2. a 0 2 1 x7 7 EXAMPLE 9 Since f x a _a 1. a 0 x (b) ƒ odd, j ƒ dx=0 |||| f x dx f u and so Equation 8 gives 0 |||| 5.5 y f u du f x dx a 0 Theorem 6 is illustrated by Figure 3. For the case where f is positive and even, part (a) says that the area under y f x from a to a is twice the area from 0 to a because of symmetry. Recall that an integral xab f x d x can be expressed as the area above the x-axis and below y f x minus the area below the axis and above the curve. Thus, part (b) says the integral is 0 because the areas cancel. y 1–6 0 a y a (a) ƒ even, j ƒ dx=2 j ƒ dx _a y u a a f u so Equation 8 gives f x dx y _a a y f x dx u (b) If f is odd, then f y a y 8 y cos 3x u u 4 x 3 x 4 1 2 sx dx, u 1 sin d , u cos 2x 4 u 3 2x 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E-05(pp 364-371) ❙❙❙❙ 366 7–32 1/17/06 3:54 PM Page 366 CHAPTER 5 INTEGRALS Evaluate the indefinite integral. |||| 9. 11. 13. y 2x x y 2 3 4 dx 3x 20 2 1 y s1 4x x 2y 15. 2x y sin 19. y 21. y cos 2 12. dx 14. dy 5 1 y s4 17. 10. dx 3 y 8. yx y 45. x 2 1 5t 4 3 y sec 2 20. y sx sin 1 22. cos st dt st yy 51. y 53. y y dt 2.7 4 18. t dt tan3 d 44. y sin d cos2 46. 2 48. y 1 dx 50. y 3 52. y 54. y 12 y 6 s2y 3 0 1 dy tan 2 d x3 2 dx 1 5 tan dx 13 3 s1 0 2 2x x sx 1 dx 4 x 0 a 2 x sx 2 0 ■ sin6 d y 6 dx 2 1 16. t dt y 42. csc t cot t dt 16 dx 2 2x 0 x 6 dx x y y 0 5 9 dx 2 y y 47. x3 y 49. 2 sec 2 t 4 dt y 43. 7. 41. ■ a 2 dx ■ ■ a 0 ■ ■ ■ 2 2 2 0 3 2 x 2 sin x dx 1 x6 cos x sin sin x d x x 4 s1 0 a 2x x sa 2 0 a a ■ x sx 2 ■ dx x 2 dx a 2 dx ■ ■ ■ sec 2 d ; 55–56 23. 25. y s1 z2 3 24. dz z3 2 y scot x csc x dx y sax ax b 2bx 2 c |||| Use a graph to give a rough estimate of the area of the region that lies under the given curve. Then find the exact area. dx 55. y 56. y cos x dx x2 26. y 28. y sx s2 x 2 sin x ■ 3 27. y sec x tan x dx 29. yx 30. y sin t sec 31. y sx a cx a sb x 4 ■ 2 ■ 2 1 dx c 0, a 3 3 1 x dx 35. 3x 3x 2 1 ■ ■ x2 y s1 ■ ■ ■ ■ ■ y 2 0 y 1 0 ■ |||| x 34. dx y sin x cos x dx 37–54 39. 4 ■ ■ ■ ■ ■ ■ ■ ■ 59. Breathing is cyclic and a full respiratory cycle from the begin- dx x ■ ■ y sx 36. 1 3 ■ 37. 1 ■ x 4 dx by making a substitution and interpreting the resulting integral in terms of an area. 32. 2x ■ x 3 s4 x 2 dx by writing it as a sum of two integrals and interpreting one of those integrals in terms of an area. ■ ■ |||| Evaluate the indefinite integral. Illustrate and check that your answer is reasonable by graphing both the function and its antiderivative (take C 0). y sin 2 x, 0 57. Evaluate x 2 2 x ; 33–36 33. 1 1 58. Evaluate x0 x s1 dx ■ ■ x 5 cos t dt ■ ■ 1, 0 2 2 y tan ■ 1 2 ■ manufacture a new calculator. The rate of production of these calculators after t weeks is d ■ ■ 60. Alabama Instruments Company has set up a production line to dx sec ning of inhalation to the end of exhalation takes about 5 s. The maximum rate of air flow into the lungs is about 0.5 L s. This 1 explains, in part, why the function f t 2 sin 2 t 5 has often been used to model the rate of air flow into the lungs. Use this model to find the volume of inhaled air in the lungs at time t. ■ ■ dx dt 5000 1 t 100 10 2 calculators week Evaluate the definite integral, if it exists. x x2 1 1 25 38. dx 2x 3 5 dx y 40. y 7 0 s 0 s4 3x dx x cos x 2 dx (Notice that production approaches 5000 per week as time goes on, but the initial production is lower because of the workers’ unfamiliarity with the new techniques.) Find the number of calculators produced from the beginning of the third week to the end of the fourth week. 5E-05(pp 364-371) 1/17/06 3:55 PM Page 367 S ECTION 5.5 THE SUBSTITUTION RULE 4 2 61. If f is continuous and y f x d x 10, find y f 2 x d x. 0 0 9 3 4, find y x f x 2 d x. 62. If f is continuous and y f x d x 0 63. Suppose f is continuous on 0 . y a f x dx y a b y a fx y c dx bc ac f x dx For the case where f x 0, draw a diagram to interpret this equation geometrically as an equality of areas. 64. Show that the area under the graph of y sin sx from 0 to 4 is the same as the area under the graph of y 2 x sin x from 0 to 2. 65. If a and b are positive numbers, show that y 1 0 xa 1 x b dx y 1 0 xb 1 y 0 x f sin x d x x a dx x to show that 2 y 0 ■ ■ ■ ■ ■ ■ f sin x d x ■ ■ 367 ■ The following exercises are intended only for those who have already covered Chapter 7. 67–82 |||| Evaluate the integral. dx y5 69. y ln x x 71. ye x 73. x 68. y 70. y1 72. ye y x ln x 74. ye 75. y cot x d x 76. y 77. y1 78. y y e1 x dx x2 80. y dx x sln x 82. 81. ■ 66. Use the substitution u ■ 79. f x dx 0 and 0 a b, draw a diagram For the case where f x to interpret this equation geometrically as an equality of areas. (b) Prove that b ■ 67. (a) Prove that b ■ ❙❙❙❙ 3x 2 dx e x dx s1 dx 1 2 1 y e4 e ■ x2 x dx x2 ■ ■ ■ y 0 1 cos t sin t dt ex x 1 dx sin x dx 1 cos2x x x4 1 1 xe 12 0 ■ ■ 83. Use Exercise 66 to evaluate the integral y dx tan 1x dx x2 0 ■ 1 x sin x dx cos2x x2 dx dx sin 1x dx s1 x 2 ■ ■ ■ ■ 5E-05(pp 364-371) ❙❙❙❙ 368 |||| 1/17/06 3:55 PM Page 368 CHAPTER 5 INTEGRALS 5 Review ■ CONCEPT CHECK 1. (a) Write an expression for a Riemann sum of a function f . Explain the meaning of the notation that you use. (b) If f x 0, what is the geometric interpretation of a Riemann sum? Illustrate with a diagram. (c) If f x takes on both positive and negative values, what is the geometric interpretation of a Riemann sum? Illustrate with a diagram. (b) If r t is the rate at which water flows into a reservoir, what does xtt r t d t represent? 2 1 5. Suppose a particle moves back and forth along a straight line with velocity v t , measured in feet per second, and accelera- tion a t . 120 (a) What is the meaning of x60 v t dt ? 120 (b) What is the meaning of x60 v t dt ? 2. (a) Write the definition of the definite integral of a continuous function from a to b. (b) What is the geometric interpretation of xab f x d x if fx 0? (c) What is the geometric interpretation of xab f x d x if f x takes on both positive and negative values? Illustrate with a diagram. 120 (c) What is the meaning of x60 a t dt ? 6. (a) Explain the meaning of the indefinite integral x f x d x. (b) What is the connection between the definite integral xab f x d x and the indefinite integral x f x d x ? 7. Explain exactly what is meant by the statement that “differen- tiation and integration are inverse processes.” 3. State both parts of the Fundamental Theorem of Calculus. 4. (a) State the Net Change Theorem. 8. State the Substitution Rule. In practice, how do you use it? ■ TRUE-FALSE QUIZ Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. b a fx y t x dx b a y f x dx b a y a f x t x dx y b a f x dx b a 5 f x dx b a x f x dx b a sf x d x y a 1 y 10. y 11. b a y a 12. 0, then a b a t x for a t x for a x b, then x t x for a x b, b. sin x 1 x4 6x 9 1 5 1 2 ax 2 1 dx x4 bx c dx dx 2 5 2 y ax 2 0 0 c dx 3 8 x x 3 d x represents the area under the curve y from 0 to 2. 13. All continuous functions have derivatives. f x dx f 1. t x dx x02 x5 5 b x y f x dx b y f x dx f3 t x dx 5 y f x dx y b a 9. 5. If f is continuous on a, b and f x y y then f x 4. If f is continuous on a, b , then y 1 8. If f and t are differentiable and f x b 3. If f is continuous on a, b , then y 3 6. If f is continuous on 1, 3 , then y f v dv t x dx 2. If f and t are continuous on a, b , then b ■ 7. If f and t are continuous and f x 1. If f and t are continuous on a, b , then y ■ 14. All continuous functions have antiderivatives. x x3 5E-05(pp 364-371) 1/17/06 3:56 PM Page 369 C HAPTER 5 REVIEW ■ EXERCISES ❙❙❙❙ 369 ■ 7. The figure shows the graphs of f, f , and x0x f t dt. 1. Use the given graph of f to find the Riemann sum with six subintervals. Take the sample points to be (a) left endpoints and (b) midpoints. In each case draw a diagram and explain what the Riemann sum represents. Identify each graph, and explain your choices. y b y c y=ƒ 2 x a 0 2 6 x 8. Evaluate: x2 fx 0 x x with four subintervals, taking the sample points to be right endpoints. Explain, with the aid of a diagram, what the Riemann sum represents. (b) Use the definition of a definite integral (with right endpoints) to calculate the value of the integral y 2 0 x2 (b) 9–28 s1 11. y 13. y 15. y 17. y 19. x 2 ) dx by interpreting it in terms of areas. y 0 y x 8x 3 y y 21. y sx 23. y sin 25. y 27. y 3x 2 dx T x4 8x 1 1 x 9 dx 10. y 12. y du 14. y (su 1 5 dy 16. y 2 18. y 20. y 2 dx 4x 22. y csc t cos t dt 24. y sin x cos cos x sec 2 tan 2 d 26. y x2 1 28. y 1 9 x 9 dx su 0 2u2 u 1 3. Evaluate y0 ( x 2 x x cos dx 2 3 2 x x sin cos d x 2 3 t 2 t sin cos dt 2 3 sin Evaluate the integral, if it exists. |||| 9. x dx (c) Use the Fundamental Theorem to check your answer to part (b). (d) Draw a diagram to explain the geometric meaning of the integral in part (b). 1 0 d dx d (c) dx 2 d dx 2 y (a) 2. (a) Evaluate the Riemann sum for 1 y y2 0 dt 5 t 1 4 0 1 0 1 2 y 2s1 1 n sin xi x nl i1 as a definite integral on the interval 0, the integral. 6 5. If x0 f x d x CAS 5 6. (a) Write x1 x 10 and x04 f x d x and then evaluate 7, find x46 f x d x. 2 x 5 dx as a limit of Riemann sums, taking the sample points to be right endpoints. Use a computer algebra system to evaluate the sum and to compute the limit. (b) Use the Fundamental Theorem to check your answer to part (a). ■ 1 v 2 cos v 3 dv 0 x 8 0 3 0 ■ 2 4 dx ■ ■ ■ ■ ■ y 3 dy sin 3 t dt 0 4. Express lim 1 2 du 4 0 0 7 dx sin x dx 1 x2 1 1 2 4 0 4 0 ■ 3 t dt dx tan t 3 sec2t dt 1 sx 1 dx ■ ■ ■ ■ 5E-05(pp 364-371) ❙❙❙❙ 370 1/17/06 3:56 PM Page 370 CHAPTER 5 INTEGRALS 43. Use the Midpoint Rule with n ; 29–30 |||| Evaluate the indefinite integral. Illustrate and check that your answer is reasonable by graphing both the function and its antiderivative (take C 0). x01 s1 44. A particle moves along a line with velocity function vt t 2 t, where v is measured in meters per second. cos x dx s1 sin x 29. y 30. y sx ■ x3 2 ■ Find (a) the displacement and (b) the distance traveled by the particle during the time interval 0, 5 . dx 1 ■ 5 to approximate x 3 d x. 45. Let r t be the rate at which the world’s oil is consumed, where ■ ■ ■ ■ ■ ■ ■ ■ ■ ; 31. Use a graph to give a rough estimate of the area of the region that lies under the curve y exact area. x s x, 0 x 4. Then find the t is measured in years starting at t 0 on January 1, 2000, and r t is measured in barrels per year. What does x03 r t d t represent? 46. A radar gun was used to record the speed of a runner at the times given in the table. Use the Midpoint Rule to estimate the distance the runner covered during those 5 seconds. cos2 x sin3 x and use the graph to guess the value of the integral x02 f x d x. Then evaluate the integral to confirm your guess. ; 32. Graph the function f x t (s) 33–38 33. F x y 34. F x y 35. t x y x x y tan s 2 ds t x3 0 36. t x s1 y x cos sx y 38. y 3x 1 2x ■ ■ 39– 40 |||| d y 40. y 3 ■ ■ ■ ■ ■ ■ ■ ■ 8000 ■ 4000 ■ x ■ 41– 42 41. y 42. y ■ 3 dx 1 5 3 2 4 ■ ■ ■ ■ ■ ■ ■ ■ 8 12 16 20 t 24 (weeks) ■ fx Use the properties of integrals to verify the inequality. 1 3 sin x dx x ■ 4 48. Let ■ x 2 cos x dx 0 0 dx ■ |||| 1 1 10.51 10.67 10.76 10.81 10.81 r Use Property 8 of integrals to estimate the value of the sx 2 1 3.0 3.5 4.0 4.5 5.0 12000 integral. 39. 0 4.67 7.34 8.86 9.73 10.22 week, where the graph of r is as shown. Use the Midpoint Rule with six subintervals to estimate the increase in the bee population during the first 24 weeks. sin t 4 dt ■ v (m s) 47. A population of honeybees increased at a rate of r t bees per t 2 dt s1 t (s) dt t3 cos x 3 1 37. y t 4 dt s1 1 v (m s) 0 0.5 1.0 1.5 2.0 2.5 Find the derivative of the function. |||| if 3 x 0 if 0 x 1 Evaluate x1 3 f x d x by interpreting the integral as a difference of areas. s2 2 ■ x1 s1 x 2 49. Find an antiderivative F of f x ■ ■ ■ ■ ■ ■ ■ ■ F1 0. x 2 sin x 2 such that 5E-05(pp 364-371) 1/17/06 3:57 PM Page 371 C HAPTER 5 REVIEW x0x sin( 1 t 2) dt was introduced in 2 Section 5.3. Fresnel also used the function 50. The Fresnel function S x Cx CAS y x 0 52. Find a function f and a value of the constant a such that x 2 y f t dt cos ( 1 t 2) dt 2 b 2 y f x f x dx x CAS cos ( 1 t 2) dt 2 0.7 54. Find lim 1 h y 2h 2 y 0 f t dt x sin x y 0 y 1 0 51. If f is a continuous function such that fa 2 t 3 dt. y f x dx 1 0 f1 x dx 56. Evaluate ft x s1 2 55. If f is continuous on 0, 1 , prove that (d) Plot the graphs of C and S on the same screen. How are these graphs related? x fb a hl0 0 1 53. If f is continuous on a, b , show that in his theory of the diffraction of light waves. (a) On what intervals is C increasing? (b) On what intervals is C concave upward? (c) Use a graph to solve the following equation correct to two decimal places: y 2 sin x a 1 for all x, find an explicit formula for f x . t2 dt lim nl 1 n 1 n 9 2 n 9 3 n 9 n n 9 ❙❙❙❙ 371 5E-05(pp 372-373) 1/17/06 3:58 PM PROBLEMS PLUS Page 372 Before you look at the solution of the following example, cover it up and first try to solve the problem yourself. x EXAMPLE 1 Evaluate lim y 3 x x l3 x 3 sin t dt . t SOLUTION Let’s start by having a preliminary look at the ingredients of the function. What happens to the first factor, x x 3 , when x approaches 3? The numerator approaches 3 and the denominator approaches 0, so we have x x |||| The principles of problem solving are discussed on page 58. 3 l as x and xl3 x 3 as l xl3 The second factor approaches x33 sin t t dt, which is 0. It’s not clear what happens to the function as a whole. (One factor is becoming large while the other is becoming small.) So how do we proceed? One of the principles of problem solving is recognizing something familiar. Is there a part of the function that reminds us of something we’ve seen before? Well, the integral y x 3 sin t dt t has x as its upper limit of integration and that type of integral occurs in Part 1 of the Fundamental Theorem of Calculus: d dx y x a f t dt fx This suggests that differentiation might be involved. Once we start thinking about differentiation, the denominator x 3 reminds us of something else that should be familiar: One of the forms of the definition of the derivative in Chapter 3 is Fx Fa Fa lim xla xa and with a 3 this becomes F3 lim x l3 Fx x F3 3 So what is the function F in our situation? Notice that if we define Fx y x 3 sin t dt t then F 3 0. What about the factor x in the numerator? That’s just a red herring, so let’s factor it out and put together the calculation: lim x l3 |||| Another approach is to use l’Hospital’s Rule. x x 3 y x 3 sin t dt t y lim x lim x l3 3 lim x l3 x l3 Fx x 3F 3 sin 3 3 3 sin 3 372 x 3 sin t dt t x3 F3 3 (FTC1) 5E-05(pp 372-373) 1/17/06 3:59 PM Page 373 P RO B L E M S 1. If x sin y x x2 0 f t dt, where f is a continuous function, find f 4 . 2. In this problem we approximate the sine function on the interval 0, ; by three quadratic functions, each of which has the same zeros as the sine function on this interval. (a) Find a quadratic function f such that f 0 0 and which has the same maximum f value as sin on 0, . (b) Find a quadratic function t such that t 0 t 0 and which has the same rate of change as the sine function at 0 and . (c) Find a quadratic function h such that h 0 h 0 and the area under h from 0 to is the same as for the sine function. (d) Illustrate by graphing f, t, h, and the sine function in the same viewing rectangle 0, by 0, 1 . Identify which graph belongs to each function. 3. Show that 1 17 y 1 2 1 1 4. Suppose the curve y x4 dx 7 . 24 f x passes through the origin and the point 1, 1 . Find the value of the integral x01 f x d x. 5. Find a function f such that f 1 1, f 4 7, and f x 3 for all x, or prove that such a function cannot exist. 2cx x 2 c 3 for c 0 and look at the regions enclosed by these curves and the x-axis. Make a conjecture about how the areas of these regions are related. Prove your conjecture in part (a). Take another look at the graphs in part (a) and use them to sketch the curve traced out by the vertices (highest points) of the family of functions. Can you guess what kind of curve this is? Find the equation of the curve you sketched in part (c). tx 1 cos x x y0 s1 t 3 dt, where t x y0 1 sin t 2 dt, find f 2 . ; 6. (a) Graph several members of the family of functions f x (b) (c) (d) 7. If f 8. If f x x0x x 2 sin t 2 dt, find f x . b 9. Find the interval a, b for which the value of the integral xa 2 x 2 d x is a maximum. x 10000 10. Use an integral to estimate the sum si. i1 n 11. (a) Evaluate x0 x d x, where n is a positive integer. (b) Evaluate xab x d x, where a and b are real numbers with 0 12. Find d2 dx 2 x yy 0 sin t 1 s1 b. u 4 du dt. x 13. If f is a differentiable function such that y f t dt 2 a 0 fx 2 for all x, find f . 14. A circular disk of radius r is used in an evaporator and is rotated in a vertical plane. If it is to 2 2 be partially submerged in the liquid so as to maximize the exposed wetted area of the disk, 2 above the surface show that the center of the disk should be positioned at a height r s1 of the liquid. x 15. Prove that if f is continuous, then y f u x 0 u du x yy 0 u 0 f t dt du. 16. The figure shows a region consisting of all points inside a square that are closer to the center 2 FIGURE FOR PROBLEM 16 than to the sides of the square. Find the area of the region. 17. Evaluate lim nl 1 sn sn 1 1 sn sn 2 1 sn sn n . 373 ...
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