Chapter 8 - 5E-08(pp 510-519) 1/17/06 5:19 PM Page 510...

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Unformatted text preview: 5E-08(pp 510-519) 1/17/06 5:19 PM Page 510 CHAPTER 8 The techniques of this chapter enable us to find the height of a rocket a minute after liftoff and to compute the escape velocity of the rocket. T echniques of Integration 5E-08(pp 510-519) 1/17/06 5:19 PM Page 511 Because of the Fundamental Theorem of Calculus, we can integrate a function if we know an antiderivative, that is, an indefinite integral. We summarize here the most important integrals that we have learned so far. yx n dx xn 1 n1 ye x dx ex 2 tan x y tan x d x dx 1 tan a C sin x cot x y cosh x d x C y cot x d x C x a C y csc x cot x d x C 1 C ax ln a dx 2 ln sec x a2 x ln x y csc x dx sec x cosh x 1 1 dx x y cos x d x C C y sinh x d x 2 y ya cos x y sec x tan x d x yx 1 C y sin x d x y sec x dx n C C y sa csc x sinh x x2 dx sin C C ln sin x 1 2 C C 1 x a C In this chapter we develop techniques for using these basic integration formulas to obtain indefinite integrals of more complicated functions. We learned the most important method of integration, the Substitution Rule, in Section 5.5. The other general technique, integration by parts, is presented in Section 8.1. Then we learn methods that are special to particular classes of functions such as trigonometric functions and rational functions. Integration is not as straightforward as differentiation; there are no rules that absolutely guarantee obtaining an indefinite integral of a function. Therefore, in Section 8.5 we discuss a strategy for integration. |||| 8.1 Integration by Parts Every differentiation rule has a corresponding integration rule. For instance, the Substitution Rule for integration corresponds to the Chain Rule for differentiation. The rule that corresponds to the Product Rule for differentiation is called the rule for integration by parts. The Product Rule states that if f and t are differentiable functions, then d f xtx dx f xt x txf x 511 5E-08(pp 510-519) 512 ❙❙❙❙ 1/17/06 5:19 PM Page 512 CHAPTER 8 TECHNIQUES OF INTEGRATION In the notation for indefinite integrals this equation becomes y t x f x dx yf or f xt x x t x dx ytxf f xtx x dx f xtx We can rearrange this equation as yf 1 x t x dx f xtx ytxf x dx Formula 1 is called the formula for integration by parts. It is perhaps easier to remember in the following notation. Let u f x and v t x . Then the differentials are du f x d x and dv t x d x, so, by the Substitution Rule, the formula for integration by parts becomes y u dv 2 EXAMPLE 1 Find y v du uv y x sin x d x. SOLUTION USING FORMULA 1 Suppose we choose f x and t x 1, we have x and t x sin x. Then f x 1 cos x. (For t we can choose any antiderivative of t .) Thus, using Formula y x sin x d x ytxf f xtx x cos x x dx y cos x d x x cos x y cos x d x x cos x sin x C It’s wise to check the answer by differentiating it. If we do so, we get x sin x, as expected. SOLUTION USING FORMULA 2 Let |||| It is helpful to use the pattern: u dv du v u Then and so x dv sin x d x du dx v cos x u d√ y x sin x dx y x sin x dx u x √ cos x x cos x y cos x d x x cos x sin x C √ y du cos x dx 5E-08(pp 510-519) 1/17/06 5:19 PM Page 513 S ECTION 8.1 INTEGRATION BY PARTS NOTE ❙❙❙❙ 513 Our aim in using integration by parts is to obtain a simpler integral than the one we started with. Thus, in Example 1 we started with x x sin x d x and expressed it in terms of the simpler integral x cos x d x. If we had chosen u sin x and dv x d x, then du cos x d x and v x 2 2, so integration by parts gives ■ y x sin x d x sin x x2 2 1 2 yx 2 cos x d x Although this is true, x x 2 cos x d x is a more difficult integral than the one we started with. In general, when deciding on a choice for u and dv, we usually try to choose u f x to be a function that becomes simpler when differentiated (or at least not more complicated) as long as dv t x d x can be readily integrated to give v. EXAMPLE 2 Evaluate y ln x d x. SOLUTION Here we don’t have much choice for u and dv. Let u Then du ln x dv 1 dx x dx x v Integrating by parts, we get y ln x d x dx x x ln x yx |||| It’s customary to write x 1 dx as x dx. x ln x y dx |||| Check the answer by differentiating it. x ln x x C Integration by parts is effective in this example because the derivative of the function fx ln x is simpler than f . EXAMPLE 3 Find 2t y t e dt. SOLUTION Notice that t 2 becomes simpler when differentiated (whereas e t is unchanged when differentiated or integrated), so we choose t2 u Then du dv 2 t dt v e t dt et Integration by parts gives 3 2t y t e dt t 2et 2 y t e t dt The integral that we obtained, x t e t dt, is simpler than the original integral but is still not obvious. Therefore, we use integration by parts a second time, this time with u t and 5E-08(pp 510-519) 514 ❙❙❙❙ 1/17/06 5:20 PM Page 514 CHAPTER 8 TECHNIQUES OF INTEGRATION e t dt. Then du dv e t, and d t, v t tet y e dt tet y t e dt t et C Putting this in Equation 3, we get t 2et 2t e dt 2 y t e t dt t 2et yt 2 tet 2t te EXAMPLE 4 Evaluate |||| An easier method, using complex numbers, is given in Exercise 48 in Appendix G. ye x 2 te et t 2e C t C1 where C1 2C sin x d x. SOLUTION Neither e x nor sin x becomes simpler when differentiated, but we try choosing u e x and dv parts gives sin x d x anyway. Then du ye 4 x e x dx and v e x cos x sin x d x ye x cos x, so integration by cos x d x The integral that we have obtained, x e x cos x d x, is no simpler than the original one, but at least it’s no more difficult. Having had success in the preceding example integrating by parts twice, we persevere and integrate by parts again. This time we use u e x and dv cos x d x. Then du e x dx, v sin x, and ye 5 x cos x d x e x sin x ye x sin x d x At first glance, it appears as if we have accomplished nothing because we have arrived at |||| Figure 1 illustrates Example 4 by e x sin x and showing the graphs of f x 1x cos x . As a visual check Fx 2 e sin x on our work, notice that f x 0 when F has a maximum or minimum. we get ye x sin x d x e x cos x e x sin x ye x sin x d x This can be regarded as an equation to be solved for the unknown integral. Adding 12 x e x sin x d x to both sides, we obtain F f _3 x e x sin x d x, which is where we started. However, if we put Equation 5 into Equation 4 2 y e x sin x d x e x cos x e x sin x 6 Dividing by 2 and adding the constant of integration, we get _4 FIGURE 1 ye x sin x d x 1 2 e x sin x cos x C If we combine the formula for integration by parts with Part 2 of the Fundamental Theorem of Calculus, we can evaluate definite integrals by parts. Evaluating both sides of Formula 1 between a and b, assuming f and t are continuous, and using the Fundamental 5E-08(pp 510-519) 1/17/06 5:20 PM Page 515 S ECTION 8.1 INTEGRATION BY PARTS ❙❙❙❙ 515 Theorem, we obtain y 6 b a EXAMPLE 5 Calculate y 1 0 f x t x dx f xtx b y a b a t x f x dx tan 1x dx. SOLUTION Let tan 1x u Then dv dx du x v x2 1 dx So Formula 6 gives y 1 0 1 tan 1x dx x tan 1x y 0 1 tan 1 1 |||| Since tan 1x 0 for x 0, the integral in Example 5 can be interpreted as the area of the region shown in Figure 2. 0 tan 1 0 x 1 0 dx x2 1 x 1 0 x2 1 dx dx x2 1 y To evaluate this integral we use the substitution t 1 x 2 (since u has another meaning in this example). Then dt 2 x dx, so x dx dt 2. When x 0, t 1; when x 1, t 2; so y y=tan–!x y 0 1 y 4 x 1 0 x x 1 0 x2 1 dx 1 2 y 1 2 FIGURE 2 y Therefore 1 0 tan 1x dx 4 2 1 dt t 1 2 ln 2 y 1 x 2 2 1 1 2 ln 1 x 1 0 ln t dx ln 2 ln 2 2 4 EXAMPLE 6 Prove the reduction formula |||| Equation 7 is called a reduction formula because the exponent n has been reduced to n 1 and n 2. where n n 1 n y sin n2 x dx 2 is an integer. SOLUTION Let Then 1 cos x sin n 1x n n y sin x dx 7 u sin n 1x du n dv v 1 sin n 2x cos x dx sin x d x cos x so integration by parts gives n y sin x d x cos x sin n 1x n 1 y sin n2 x cos 2x d x 5E-08(pp 510-519) ❙❙❙❙ 516 1/17/06 5:20 PM Page 516 CHAPTER 8 TECHNIQUES OF INTEGRATION Since cos 2x sin 2x, we have 1 n cos x sin n 1x y sin x dx n y sin 1 n2 x dx n n y sin x d x 1 As in Example 4, we solve this equation for the desired integral by taking the last term on the right side to the left side. Thus, we have n y sin n x d x cos x sin n 1x 1 cos x sin n 1x n n y sin x d x or n 1 n y sin 1 n n2 y sin x dx n2 x dx The reduction formula (7) is useful because by using it repeatedly we could eventually express x sin n x d x in terms of x sin x d x (if n is odd) or x sin x 0 d x x d x (if n is even). |||| 8.1 Exercises 1–2 |||| Evaluate the integral using integration by parts with the indicated choices of u and dv. y dy e2y 1 ■ 3–32 5. ■ ■ ■ ■ ■ ■ ■ ■ y x cos 5 x d x y re r2 4. 6. dr 2 yx 9. y ln 2 x sin x dx 1 dx y y cos x ln sin x 28. y y cos ln x dx 30. y y x 4 ln x 2 dx 32. y 12 0 x y xe cos 1x d x 8. y t sin 2 t d t 10. yx 2 2 1 cos m x d x ■ ■ ■ y arctan 4 t d t ln x 2 d x 13. y 15. ye 17. 12. 5 21. s3 1 ■ ■ arctan 1 x d x r3 1 t 0 ■ r2 s4 0 e s sin t ■ dr s ds ■ ■ ■ ■ |||| First make a substitution and then use integration by parts to evaluate the integral. x dx 33. y sin s x d x y 4 e sx d x 34. y 36. yx e 1 ln p d p s s 3 2 cos 2 d 5 x2 dx 3t 14. y t e dt sin 3 d 16. ye y y sinh y d y 18. y y cosh ay d y 2 ■ ■ ■ y 0 y 2 1 t sin 3 t d t ln x dx x2 22. y 0 y 4 1 x 2 1e ■ ■ ■ ■ ■ ■ ■ ■ |||| Evaluate the indefinite integral. Illustrate, and check that your answer is reasonable, by graphing both the function and its antiderivative (take C 0). x y x cos 39. 20. 1 ■ ; 37–40 cos 2 d 37. 19. x 5x dx 1 y sin yp 1 0 x csc 2x d x 33–36 35. 11. dx 4 dx ■ 7. 26. 2 y ■ Evaluate the integral. |||| 24. 0 sec 2 d , dv ■ x dx y 31. ln x, dv sec 2 d ; u y ■ 3. u 25. 29. 2. y x ln x d x; y 27. 1. 23. y 32 x dx 38. yx 3 e x dx 40. yx e ln x dx dx 2x 3 x2 dx st ln t dt ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E-08(pp 510-519) 1/17/06 5:21 PM Page 517 S ECTION 8.1 INTEGRATION BY PARTS 41. (a) Use the reduction formula in Example 6 to show that x 2 2 y sin x d x sin 2 x 4 ❙❙❙❙ 517 ; 53–54 |||| Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find (approximately) the area of the region bounded by the curves. C 53. y 54. y (b) Use part (a) and the reduction formula to evaluate x sin 4x d x. x sin x, arctan 3 x, ■ ■ y ■ x y ■ 2 2 x2 ■ ■ ■ ■ ■ ■ ■ ■ 42. (a) Prove the reduction formula 55–58 1 cos n 1x sin x n n y cos x d x n 1 n y cos |||| Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis. n2 x dx 55. y 2 y 0 n sin n x d x 1 y n 2 2 246 357 sin 2 n 1x d x 0 2 0 45–48 ln x n d x 45. y 46. yx e nx x ln x x ne x dx n n y ln x 2n 1 2n n1 x e, x ■ 1; about the y-axis 1, x 0, y ■ 1; about the y-axis x 0; about x 1 ; about the x-axis ■ ■ ■ ■ ■ ■ ■ ■ ■ x 2 ln x on the interval 1, 3 . decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is m, the fuel is consumed at rate r, and the exhaust gases are ejected with constant velocity ve (relative to the rocket). A model for the velocity of the rocket at time t is given by the equation vt tt m ve ln rt m 2 where t is the acceleration due to gravity and t is not too large. If t 9.8 m s 2, m 30,000 kg, r 160 kg s, and ve 3000 m s, find the height of the rocket one minute after liftoff. Use integration by parts to prove the reduction formula. |||| e ,y 0, 0 60. A rocket accelerates by burning its onboard fuel, so its mass 2n 2n 1 135 246 sin 2 nx d x 0, x x 2, y 59. Find the average value of f x 44. Prove that, for even powers of sine, y x ■ where n 2 is an integer. (b) Use part (a) to evaluate x0 2 sin 3x d x and x0 2 sin 5x d x. (c) Use part (a) to show that, for odd powers of sine, y e x, x 58. y sin n 2x d x 0 e x, y 57. y 43. (a) Use the reduction formula in Example 6 to show that cos 56. y (b) Use part (a) to evaluate x cos 2x d x. (c) Use parts (a) and (b) to evaluate x cos 4x d x. dx 61. A particle that moves along a straight line has velocity vt t 2e t meters per second after t seconds. How far will n y x n 1e x d x it travel during the first t seconds? 47. x2 y x x2 2n 48. ■ a2 1 n 2na 2 2n 1 tan x sec n 2x n1 n y sec x d x ■ 62. If f 0 a 2 n dx ■ ■ ■ y n n ■ ■ x 2 1 2 a 2n1 dx (n 1 2 ) y a 0 0 and f and t are continuous, show that t0 f x t x dx f at a y f ata a 0 f x t x dx 63. Suppose that f 1 y sec n2 x dx n 2, f 4 7, f 1 5, f 4 4 f is continuous. Find the value of x1 x f x d x. 1 3, and 64. (a) Use integration by parts to show that ■ ■ ■ ■ ■ yf 49. Use Exercise 45 to find x ln x 3 d x. 50. Use Exercise 46 to find x x e d x. x dx xf x y xf x dx 4x 51–52 |||| 51. y ■ Find the area of the region bounded by the given curves. xe 52. y 0.4 x , 5 ln x, ■ (b) If f and t are inverse functions and f is continuous, prove that ■ y y ■ 0, x 5 y x ln x ■ b a ■ ■ ■ ■ ■ ■ ■ f x dx bf b af a y fb fa t y dy [Hint: Use part (a) and make the substitution y f x .] 5E-08(pp 510-519) 518 ❙❙❙❙ 1/17/06 5:21 PM Page 518 CHAPTER 8 TECHNIQUES OF INTEGRATION (c) In the case where f and t are positive functions and b a 0, draw a diagram to give a geometric interpretation of part (b). (d) Use part (b) to evaluate x1e ln x d x. (c) Use parts (a) and (b) to show that 2n 2n x f x d x, by using cylindrical shells, but now we can use integration by parts to prove it using the slicing method of Section 6.2, at least for the case where f is one-to-one and therefore has an inverse function t. Use the figure to show that b 2d V a 2c y d ty c 2 lim nl x=g(y) 1 2 1 2 3 4 3 4 5 6 5 6 7 2n 2n 1 2n 2n 1 2 This formula is usually written as an infinite product: dy Make the substitution y f x and then use integration by parts on the resulting integral to prove that V xab 2 x f x d x. y I2 n 1 I2 n and deduce that lim n l I2 n 1 I2 n 1. (d) Use part (c) and Exercises 43 and 44 to show that xab 2 65. We arrived at Formula 6.3.2, V 1 2 2 1 2 2 3 4 3 4 5 6 5 6 7 and is called the Wallis product. (e) We construct rectangles as follows. Start with a square of area 1 and attach rectangles of area 1 alternately beside or on top of the previous rectangle (see the figure). Find the limit of the ratios of width to height of these rectangles. y=ƒ d x=b c x=a 0 66. Let In x0 2 a b x sin n x d x. (a) Show that I2 n 2 I2 n 1 I2 n. (b) Use Exercise 44 to show that I2 n 2 I2 n |||| 8.2 2n 2n 1 2 Trigonometric Integrals In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. EXAMPLE 1 Evaluate 3 y cos x d x. SOLUTION Simply substituting u cos x isn’t helpful, since then du sin x dx. In order to integrate powers of cosine, we would need an extra sin x factor. Similarly, a power of sine would require an extra cos x factor. Thus, here we can separate one cosine factor and convert the remaining cos2x factor to an expression involving sine using the identity sin 2x cos 2x 1: cos 3x cos 2x cos x 1 sin 2x cos x We can then evaluate the integral by substituting u sin x, so du 3 2 y cos x d x y cos x y 1 sin x y cos x d x u 2 du 1 3 sin 3x u C 1 1 3 u3 cos x d x and sin 2x cos x d x C 5E-08(pp 510-519) 1/17/06 5:22 PM Page 519 S ECTION 8.2 TRIGONOMETRIC INTEGRALS ❙❙❙❙ 519 In general, we try to write an integrand involving powers of sine and cosine in a form where we have only one sine factor (and the remainder of the expression in terms of cosine) or only one cosine factor (and the remainder of the expression in terms of sine). The identity sin 2x cos 2x 1 enables us to convert back and forth between even powers of sine and cosine. EXAMPLE 2 Find 5 2 y sin x cos x d x S OLUTION We could convert cos 2x to 1 sin 2x, but we would be left with an expression in terms of sin x with no extra cos x factor. Instead, we separate a single sine factor and rewrite the remaining sin 4x factor in terms of cos x : sin 5x cos 2x |||| Figure 1 shows the graphs of the integrand sin 5x cos 2x in Example 2 and its indefinite integral (with C 0). Which is which? Substituting u sin2x 2 cos 2x sin x cos x, we have du 5 2 cos 2x 2 cos 2x sin x 1 sin x dx and so sin 2x 2 cos 2x sin x d x y sin x cos x d x y 0.2 y _π cos 2x 2 cos 2x sin x d x y π 1 1 u2 2u2 u3 3 _ 0.2 1 3 FIGURE 1 2 cos 3x u5 5 2 5 u2 y du u7 7 u 6 du C 1 7 cos 5x 2u 4 cos 7x C In the preceding examples, an odd power of sine or cosine enabled us to separate a single factor and convert the remaining even power. If the integrand contains even powers of both sine and cosine, this strategy fails. In this case, we can take advantage of the following half-angle identities (see Equations 17b and 17a in Appendix D): 1 2 sin 2x |||| Example 3 shows that the area of the region shown in Figure 2 is 2. EXAMPLE 3 Evaluate y 0 1 cos 2 x 1 2 cos 2x and 1 cos 2 x sin 2x d x. SOLUTION If we write sin 2x 1 cos 2x, the integral is no simpler to evaluate. Using the half-angle formula for sin x, however, we have 2 1.5 y=sin@ x 0 0 _0.5 sin 2x d x 1 2 y 1 2 y ( 0 cos 2 x d x 1 1 2 [ (x ) 1 2 sin 2 1 2 (0 1 2 1 2 sin 2 x) sin 0) 0 1 2 π Notice that we mentally made the substitution u 2 x when integrating cos 2 x. Another method for evaluating this integral was given in Exercise 41 in Section 8.1. FIGURE 2 EXAMPLE 4 Find 4 y sin x d x. SOLUTION We could evaluate this integral using the reduction formula for x sin n x d x (Equation 8.1.7) together with Example 1 (as in Exercise 41 in Section 8.1), but a better 5E-08(pp 520-529) 520 ❙❙❙❙ 1/17/06 5:23 PM Page 520 CHAPTER 8 TECHNIQUES OF INTEGRATION method is to write sin 4x sin 2x 2 and use a half-angle formula: 4 sin 2x 2 d x y sin x d x y y 1 4 y 2 1 cos 2 x 2 1 2 cos 2 x dx cos 2 2 x d x Since cos 2 2 x occurs, we must use another half-angle formula 1 2 cos 2 2 x 1 cos 4 x This gives 1 4 y 1 4 4 y sin x d x y( ( 13 42 1 3 2 x 2 cos 2 x 1 2 2 cos 2 x 1 2 1 8 sin 2 x 1 cos 4 x d x cos 4 x) d x sin 4 x) C To summarize, we list guidelines to follow when evaluating integrals of the form 0 and n 0 are integers. x sin mx cos nx dx, where m Strategy for Evaluating y sin m x cos nx dx (a) If the power of cosine is odd n 2 k 1 , save one cosine factor and use cos 2x 1 sin 2x to express the remaining factors in terms of sine: m x cos 2 k 1x d x y sin m x cos 2x k cos x d x y sin y sin m x1 sin 2x k cos x d x Then substitute u sin x. (b) If the power of sine is odd m 2 k 1 , save one sine factor and use sin 2x 1 cos 2x to express the remaining factors in terms of cosine: 2k 1 x cos n x d x y sin 2x k cos n x sin x d x y y sin 1 cos 2x k cos n x sin x d x Then substitute u cos x. [Note that if the powers of both sine and cosine are odd, either (a) or (b) can be used.] (c) If the powers of both sine and cosine are even, use the half-angle identities sin 2x 1 2 1 cos 2 x cos 2x It is sometimes helpful to use the identity sin x cos x 1 2 sin 2 x 1 2 1 cos 2 x 5E-08(pp 520-529) 1/17/06 5:24 PM Page 521 S ECTION 8.2 TRIGONOMETRIC INTEGRALS ❙❙❙❙ 521 We can use a similar strategy to evaluate integrals of the form x tan mx sec nx dx. Since d dx tan x sec 2x, we can separate a sec 2x factor and convert the remaining (even) power of secant to an expression involving tangent using the identity sec 2x 1 tan 2x. Or, since d dx sec x sec x tan x, we can separate a sec x tan x factor and convert the remaining (even) power of tangent to secant. EXAMPLE 5 Evaluate 6 4 y tan x sec x d x. SOLUTION If we separate one sec 2x factor, we can express the remaining sec 2x factor in terms of tangent using the identity sec 2x 1 by substituting u tan x with du sec 2x d x : 6 4 tan 2x. We can then evaluate the integral 6 2 2 y tan x sec x d x y tan x sec x sec x d x 6 y tan x yu 6 1 7 EXAMPLE 6 Find y tan 5 u 2 du 1 u7 7 tan 2x sec 2x d x 1 u9 9 u6 u 8 du C 1 9 tan 7x y tan 9x C sec 7 d . SOLUTION If we separate a sec 2 factor, as in the preceding example, we are left with a sec 5 factor, which isn’t easily converted to tangent. However, if we separate a sec tan factor, we can convert the remaining power of tangent to an expression sec 2 1. We can then evaluate the involving only secant using the identity tan 2 integral by substituting u sec , so du sec tan d : y tan 5 sec 7 d y tan 4 y sec 2 y sec 6 sec tan d u2 u 11 11 1 11 sec 11 1 2 sec 6 sec 1 2 u 6 du 2 u9 9 u7 7 2 9 sec 9 tan d u 10 y 2u 8 u 6 du C 1 7 sec 7 C The preceding examples demonstrate strategies for evaluating integrals of the form x tan mx sec nx dx for two cases, which we summarize here. 5E-08(pp 520-529) 522 ❙❙❙❙ 1/17/06 5:24 PM Page 522 CHAPTER 8 TECHNIQUES OF INTEGRATION Strategy for Evaluating y tan m x sec nx dx (a) If the power of secant is even n 2 k, k 2 , save a factor of sec 2x and use sec 2x 1 tan 2x to express the remaining factors in terms of tan x : m y tan x sec 2 kx d x m x sec 2x y tan y tan m x1 k1 sec 2x d x tan 2x k1 sec 2x d x Then substitute u tan x. (b) If the power of tangent is odd m 2 k 1 , save a factor of sec x tan x and use tan 2x sec 2x 1 to express the remaining factors in terms of sec x : 2k 1 Then substitute u x sec n x d x y tan 2x k sec n 1x sec x tan x d x y y tan sec 2x 1 k sec n 1x sec x tan x d x sec x. For other cases, the guidelines are not as clear-cut. We may need to use identities, integration by parts, and occasionally a little ingenuity. We will sometimes need to be able to integrate tan x by using the formula established in (5.5.5): y tan x d x ln sec x C We will also need the indefinite integral of secant: 1 y sec x d x ln sec x tan x C We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x tan x : sec x y sec x d x y sec x sec x y If we substitute u becomes x 1 u du sec x ln u tan x dx tan x sec 2x sec x tan x dx sec x tan x sec 2x d x, so the integral tan x, then du sec x tan x C. Thus, we have y sec x d x ln sec x tan x C 5E-08(pp 520-529) 1/17/06 5:25 PM Page 523 SECTION 8.2 TRIGONOMETRIC INTEGRALS E XAMPLE 7 Find ❙❙❙❙ 523 3 y tan x d x. SOLUTION Here only tan x occurs, so we use tan 2x sec 2x 1 to rewrite a tan 2x factor in 2 terms of sec x : 3 2 y tan x d x y tan x tan x d x y tan x sec 2x 1 dx 2 y tan x sec x d x y tan x d x tan 2x 2 ln sec x In the first integral we mentally substituted u C sec 2x d x. tan x so that du If an even power of tangent appears with an odd power of secant, it is helpful to express the integrand completely in terms of sec x. Powers of sec x may require integration by parts, as shown in the following example. EXAMPLE 8 Find 3 y sec x d x. SOLUTION Here we integrate by parts with u du sec x sec x tan x d x 3 tan x v 2 sec x tan x y sec x tan x d x sec x tan x y sec x sec x tan x Then y sec x d x sec 2x d x dv y sec x d x y sec x d x sec 2x 1 dx 3 Using Formula 1 and solving for the required integral, we get 3 y sec x d x 1 2 (sec x tan x ln sec x tan x ) C Integrals such as the one in the preceding example may seem very special but they occur frequently in applications of integration, as we will see in Chapter 9. Integrals of the form x cot m x csc n x d x can be found by similar methods because of the identity 1 cot 2x csc 2x. Finally, we can make use of another set of trigonometric identities: 2 To evaluate the integrals (a) x sin m x cos n x d x, (b) x sin m x sin n x d x, or (c) x cos m x cos n x d x, use the corresponding identity: |||| These product identities are discussed in Appendix D. (a) sin A cos B 1 2 sin A B sin A (b) sin A sin B 1 2 cos A B cos A B (c) cos A cos B 1 2 cos A B cos A B B 5E-08(pp 520-529) ❙❙❙❙ 524 1/17/06 5:25 PM Page 524 CHAPTER 8 TECHNIQUES OF INTEGRATION EXAMPLE 9 Evaluate y sin 4 x cos 5x d x. SOLUTION This integral could be evaluated using integration by parts, but it’s easier to use the identity in Equation 2(a) as follows: y sin 4 x cos 5x d x y 1 2 sin x sin 9x d x sin x sin 9x d x 1 2 1 2 |||| 8.2 1–47 (cos x 1 9 cos 9x C Exercises Evaluate the integral. |||| y 5 6 4 y 2 0 cos2 d 8. 4 9. y sin 3 t dt 11. y 1 13. 15. 0 y 4 0 d sin 4x cos 2x d x 3 y sin x scos x d x 2 3 17. y cos x tan x d x 19. y 1 sin x dx cos x 2 21. y sec x tan x d x 23. 25. 27. 29. y 2 0 y sin y 2 0 3 sin 2 2 6 12. y csc x d x 40. y y sin 5x sin 2 x d x 42. y sin 3x cos x d x y cos 7 44. y y 46. y cos x y t sec 2 6 cot 2x d x 2 4 3 csc 3 d 3 6 y x cos2x d x 14. 16. 0 y cos 2 0 sin 2x cos 2x d x cos5 sin y cos 5 y cot 20. y cos x sin 2 x d x 2 1 ■ sec 4 t 2 dt y tan x d x y sec 6 t dt 26. cos 5 d tan 2x dx sec 2x 4 2 x cot 6 x d x csc 3x d x cos x sin x dx sin 2 x dx 1 t 2 tan 4 t 2 dt ■ ■ 48. If x0 4 x0 2 4 d sin 4 d 18. 0 cot 3x d x d 24. 0 y csc 2 d y y tan x d x y 38. d 4 mx dx y 3 y tan 3 cos 5x dx 22. 2 36. ay dy 3 y sin x cos x dx 10. 2 cos 6. y tan x sec x d x 47. 5 y cos x sin x dx 4. 34. 45. sin 5x cos 3x dx y tan 43. 4 2 32. 41. 3 6 y cot 39. 7. y 2. y 37. 5. 2 y cos 35. 3. 3 y sin x cos x dx y tan x d x 33. 1. 31. 4 ■ ■ tan 6 x sec x d x ■ ■ ■ ■ ■ I , express the value of tan x sec x d x in terms of I. ; 49–52 |||| Evaluate the indefinite integral. Illustrate, and check that your answer is reasonable, by graphing both the integrand and its antiderivative (taking C 0 . 5 4 4 tan 5 x sec 4 x d x 3 y tan x sec x dx 28. 30. 0 49. sec 4 y tan 3 y sin x d x 50. y sin x cos x d x y sin 3x sin 6 x d x 52. y sec tan 4 d 0 3 4 2 x sec 5 2 x d x ■ y ■ 8 51. y 4 ■ tan 5x sec6x d x ■ ■ ■ ■ ■ ■ ■ 53. Find the average value of the function f x the interval , . x dx 2 ■ ■ 2 ■ 3 sin x cos x on ■ 5E-08(pp 520-529) 1/17/06 5:26 PM Page 525 S ECTION 8.3 TRIGONOMETRIC SUBSTITUTION 54. Evaluate x sin x cos x d x by four methods: (a) the substitution |||| current that varies from 155 V to 155 V with a frequency of 60 cycles per second (Hz). The voltage is thus given by the equation Et 155 sin 120 t Find the area of the region bounded by the given curves. 55. y sin x, y sin 3x, 56. y sin x, y 2 sin 2x, ■ ■ ■ ■ ■ x x 0, x 0, ■ where t is the time in seconds. Voltmeters read the RMS (rootmean-square) voltage, which is the square root of the average value of E t 2 over one cycle. (a) Calculate the RMS voltage of household current. (b) Many electric stoves require an RMS voltage of 220 V. Find the corresponding amplitude A needed for the voltage Et A sin 120 t . 2 x 2 ■ ■ ■ ■ ■ ■ ; 57–58 |||| Use a graph of the integrand to guess the value of the integral. Then use the methods of this section to prove that your guess is correct. 57. y 2 cos 3x d x 0 ■ ■ 58. ■ ■ ■ ■ ■ y 2 0 ■ 65–67 ■ ■ |||| Prove the formula, where m and n are positive integers. 65. sin 2 x cos 5 x d x ■ ■ y sin m x cos n x d x 66. y sin m x sin n x d x 67. y cos m x cos n x d x 0 0 59–62 |||| Find the volume obtained by rotating the region bounded by the given curves about the specified axis. 59. y sin x, x 60. y 2 tan x, y 0, x 0, x 4; about the x-axis 61. y cos x, y 0, x 0, x 2; about y 2, x ,y ■ cos x, y ■ ■ 0, x ■ ■ 0, x ■ 2; about y ■ ■ ■ ■ ■ n n ■ ■ ■ ■ ■ ■ ■ ■ N 1 ■ ■ a n sin n x fx n1 ■ a 1 sin x ■ a 2 sin 2 x a N sin Nx Show that the mth coefficient a m is given by the formula am 0. |||| 8.3 if m if m n n 68. A finite Fourier series is given by the sum 1 63. A particle moves on a straight line with velocity function vt sin t cos 2 t. Find its position function s f t if f0 0 if m if m 0; about the x-axis ■ 62. y 525 64. Household electricity is supplied in the form of alternating u cos x, (b) the substitution u sin x, (c) the identity sin 2 x 2 sin x cos x, and (d) integration by parts. Explain the different appearances of the answers. 55–56 ❙❙❙❙ 1 y f x sin m x d x Trigonometric Substitution In finding the area of a circle or an ellipse, an integral of the form x sa 2 x 2 d x arises, where a 0. If it were x x sa 2 x 2 d x, the substitution u a 2 x 2 would be effective but, as it stands, x sa 2 x 2 d x is more difficult. If we change the variable from x to by the substitution x a sin , then the identity 1 sin 2 cos 2 allows us to get rid of the root sign because sa 2 x2 sa 2 a 2 sin 2 sa 2 1 sin 2 sa 2 cos 2 a cos Notice the difference between the substitution u a 2 x 2 (in which the new variable is a function of the old one) and the substitution x a sin (the old variable is a function of the new one). In general we can make a substitution of the form x t t by using the Substitution Rule in reverse. To make our calculations simpler, we assume that t has an inverse function; that is, t is one-to-one. In this case, if we replace u by x and x by t in the Substitution Rule (Equation 5.5.4), we obtain yf x dx yf t t t t dt This kind of substitution is called inverse substitution. 5E-08(pp 520-529) 526 ❙❙❙❙ 1/17/06 5:27 PM Page 526 CHAPTER 8 TECHNIQUES OF INTEGRATION We can make the inverse substitution x a sin provided that it defines a one-to-one function. This can be accomplished by restricting to lie in the interval 2, 2 . In the following table we list trigonometric substitutions that are effective for the given radical expressions because of the specified trigonometric identities. In each case the restriction on is imposed to ensure that the function that defines the substitution is one-to-one. (These are the same intervals used in Appendix D in defining the inverse functions.) Table of Trigonometric Substitutions Expression Substitution sa 2 x2 x a sin , sa 2 x2 x a tan , sx 2 a2 x a sec , EXAMPLE 1 Evaluate SOLUTION Let x x2 s9 x 2 x2 sec 2 2 3 cos s9 cos 2 3 cos 2 dx y cos 2 d sin 2 csc 2 y cot œ„„„„„ 9-≈ d and 3 cos 2 d 1d cot ¨ tan 2 3 cos 3 cos d 9 sin 2 y x 1 2.) Thus, the Inverse Substitution Rule y 3 sec 2 d x. x2 x tan 2 3 2 or 2 2 s9 cos 2 2. Then d x 0 0 because y sin 2 1 2 9 sin 2 s9 1 2 2 3 sin , where s9 (Note that cos gives y 2 Identity C Since this is an indefinite integral, we must return to the original variable x. This can be done either by using trigonometric identities to express cot in terms of sin x 3 or by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right triangle. Since sin x 3, we label the opposite side and the hypotenuse as having lengths x and 3. Then the Pythagorean Theorem gives the length of the adjacent side as s9 x 2, so we can simply read the value of cot from the figure: FIGURE 1 (Although Since sin x2 s9 cot x sin ¨ = 3 x 0 in the diagram, this expression for cot x 3, we have sin 1 x 3 and so y x2 s9 x 2 dx x2 s9 x sin is valid even when 1 x 3 C 0.) 5E-08(pp 520-529) 1/17/06 5:27 PM Page 527 S ECTION 8.3 TRIGONOMETRIC SUBSTITUTION ❙❙❙❙ 527 EXAMPLE 2 Find the area enclosed by the ellipse x2 a2 (0, b) y2 b2 (a, 0) FIGURE 2 ¥ ≈ + =1 b@ a@ 1 SOLUTION Solving the equation of the ellipse for y, we get y 0 y2 b2 x x2 a2 1 a2 x2 a or 2 b sa 2 a y x2 Because the ellipse is symmetric with respect to both axes, the total area A is four times the area in the first quadrant (see Figure 2). The part of the ellipse in the first quadrant is given by the function b y 0xa sa 2 x 2 a 1 4 and so y A a 0 b sa 2 a x 2 dx To evaluate this integral we substitute x a sin . Then dx limits of integration we note that when x 0, sin 0, so sin 1, so 2. Also sa 2 x2 since 0 sa 2 a 2 sin 2 sa 2 cos 2 a cos d . To change the 0; when x a, a cos a cos 2. Therefore A 4 b a y a 0 4 ab y sa 2 2 0 [ cos 2 d 1 2 2 ab x 2 dx sin 2 4 b a 4 ab y 0 0 2 2 y 2 ab 0 21 2 a cos 1 2 a cos d cos 2 0 d 0 ab We have shown that the area of an ellipse with semiaxes a and b is a b. In particular, taking a b r, we have proved the famous formula that the area of a circle with radius r is r 2. NOTE Since the integral in Example 2 was a definite integral, we changed the limits of integration and did not have to convert back to the original variable x. ■ EXAMPLE 3 Find 1 y x sx SOLUTION Let x sx 2 2 2 4 2 tan , d x. 2 2. Then d x s4 tan 2 4 1 s4 sec 2 2 sec 2 d and 2 sec 2 sec Thus, we have dx y x sx 2 2 4 y 2 sec 2 d 4 tan 2 2 sec 1 4 sec y tan 2 d 5E-08(pp 520-529) 528 ❙❙❙❙ 1/17/06 5:28 PM Page 528 CHAPTER 8 TECHNIQUES OF INTEGRATION To evaluate this trigonometric integral we put everything in terms of sin sec tan 2 Therefore, making the substitution u y dx x 2s x 2 cos d sin 2 y 1 4 1 u x y x sx 2 EXAMPLE 4 Find y sx x 2 4 2 du u2 1 4 sin C 4 x and so sx 2 4 4x dx x 2 y C sx 2 We use Figure 3 to determine that csc FIGUR E 3 1 4 C csc 4 ¨ 2 cos sin 2 sin , we have 1 4 4 œ„„„„„ ≈+4 tan ¨ = cos 2 sin 2 1 cos and cos : 4 C d x. SOLUTION It would be possible to use the trigonometric substitution x 2 tan here (as in Example 3). But the direct substitution u x 2 4 is simpler, because then du 2 x d x and y NOTE x sx 2 4 1 2 dx du y su su C sx 2 4 C Example 4 illustrates the fact that even when trigonometric substitutions are possible, they may not give the easiest solution. You should look for a simpler method first. ■ EXAMPLE 5 Evaluate y sx dx 2 a2 , where a 0. SOLUTION 1 We let x dx a sec a sec , where 0 tan d and sx 2 a2 2 or sa 2 sec 2 sa 2 tan 2 1 3 2. Then a tan a tan Therefore y sx dx 2 a y 2 a sec tan a tan y sec x d d ln sec tan C ≈-a@ œ„„„„„ The triangle in Figure 4 gives tan ¨ a FIGU RE 4 x sec ¨ = a y sx dx 2 a2 sx 2 ln x a ln x a 2 a, so we have sx 2 a2 a sx 2 a2 C ln a C 5E-08(pp 520-529) 1/17/06 5:28 PM Page 529 S ECTION 8.3 TRIGONOMETRIC SUBSTITUTION Writing C1 C 529 ln a, we have y sx 1 SOLUTION 2 For x identity cosh 2 y dx 2 sx 2 ln x a2 a2 0 the hyperbolic substitution x sinh 2 y 1, we have sx 2 Since dx ❙❙❙❙ a2 sa 2 cosh 2 t C1 a cosh t can also be used. Using the sa 2 sinh 2 t 1 a sinh t a sinh t dt, we obtain y sx Since cosh t dx 2 a x a, we have t cosh 1 sx 2 a y dt t C x a and dx y 2 a sinh t dt a sinh t y 2 cosh 2 x a 1 C Although Formulas 1 and 2 look quite different, they are actually equivalent by Formula 7.6.4. NOTE As Example 5 illustrates, hyperbolic substitutions can be used in place of trigonometric substitutions and sometimes they lead to simpler answers. But we usually use trigonometric substitutions because trigonometric identities are more familiar than hyperbolic identities. ■ EXAMPLE 6 Find y x3 3 s3 2 0 4x 2 32 9 d x. 932 s4 x 2 9 )3 so trigonometric substitution is appropriate. Although s4 x 9 is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitution u 2 x. When we combine this with the tangent substitution, we have x 3 tan , 2 which gives d x 3 sec 2 d and 2 SOLUTION First we note that 4 x 2 2 s4 x 2 When x 0, tan y 0, so 4x 2 s9 tan 2 0; when x x3 3 s3 2 0 9 9 32 dx y 3 0 so that du y 3 16 cos 3 sec 3 s3 2, tan 3 16 Now we substitute u 3, u 1. 2 9 y 27 8 tan3 27 sec3 3 0 0 3 tan 3 d sec 1 3 2 s3, so 3. sec 2 d 3 16 y 3 0 sin3 d cos2 cos 2 sin d cos 2 sin d . When 0, u 1; when 5E-08(pp 530-539) 530 ❙❙❙❙ 1/17/06 6:03 PM Page 530 CHAPTER 8 TECHNIQUES OF INTEGRATION Therefore y x3 3 s3 2 4x 0 2 9 3 16 dx 32 3 16 x 2x y s3 EXAMPLE 7 Evaluate y u2 1 12 u 1 3 16 du y 12 1 1 u 2 du 12 1 u u 2 [( 1 2 3 16 2) 1 1 3 32 1 d x. x2 SOLUTION We can transform the integrand into a function for which trigonometric substitution is appropriate by first completing the square under the root sign: 2x 3 x2 x2 4 3 x 2x 3 x 2x We now substitute u |||| Figure 5 shows the graphs of the integrand in Example 7 and its indefinite integral (with C 0 ). Which is which? x 2 1. Then du y dx 2 sin , giving du x 2x x 2 cos u s4 x 2 y dx 1 d x and x 2 sin 2 cos 1 2 sin u 1 du u2 u2 d and s4 y y s3 2x 2 1 This suggests that we make the substitution u y s3 x2 1 2 cos , so 1d 2 cos d 3 2 _4 2 cos C s4 u2 sin s3 2x 1 x2 u 2 C _5 FIGURE 5 |||| 8.3 4–30 |||| Evaluate the integral using the indicated trigonometric substitution. Sketch and label the associated right triangle. 2 s3 2. 3. ■ yx ■ x x dx ; x3 2 9 ■ dx ; ■ x x y yx 9. dx ; y 7. 9 2 s9 y sx C x3 dx s16 x 2 4. 1 x 2 sx 2 3 1 2 Evaluate the integral. |||| 5. y x 1 Exercises 1–3 1. sin y sx 3 sec 3 sin 0 1 2 s2 3 t st 2 ■ ■ ■ ■ ■ ■ ■ 1 1 s25 3 tan ■ 2 x dx 2 16 2 dt 6. y dx 8. y 10. 2 0 x 3 sx 2 4 dx sx 2 a 2 dx x4 y st t5 2 2 dt 1, so 5E-08(pp 530-539) 1/17/06 6:03 PM Page 531 S ECTION 8.3 TRIGONOMETRIC SUBSTITUTION 11. y s1 13. y 15. y 17. y sx 12. y sx 2 9 dx x3 14. y u s5 16. yx 18. y a x2 x2 2 x 21. y 23. y s5 25. y s9x 27. y 29. dx 7 s1 y y x s1 0 ■ 2 9x 2 d x 22. y x 2 dx 24. y st 26. 28. y 6x 8 dx 2x 2 dx 2 30. 4 x dx ■ ■ ■ ■ 9 b2 t y s4 x 1 2 1 t2 sx 2 0 5 2 y 0 ■ ¨ 32 O dt dt 6t 2 y dx 4x x 2 dx 2 ln ( x a2 a2) sx 2 2 38. A charged rod of length L produces an electric field at point 52 P a, b given by cos t dt s1 sin 2 t ■ dx x 4 sx 2 ; 37. Use a graph to approximate the roots of the equation x 2 s4 x 2 2 x. Then approximate the area bounded by the curve y x 2 s4 x 2 and the line y 2 x. dx ■ x Graph the integrand and its indefinite integral on the same screen and check that your answer is reasonable. 13 x2 R ; 36. Evaluate the integral EP ■ y b La a 4 0 x2 b2 32 dx ■ where is the charge density per unit length on the rod and 0 is the free space permittivity (see the figure). Evaluate the integral to determine an expression for the electric field E P . 31. (a) Use trigonometric substitution to show that y sx Q 1 dx x2 ■ P dx 2 ax y s 25 4x y u2 dx s16 x 2 531 equation x 2 y 2 r 2. Then A is the sum of the area of the triangle POQ and the area of the region PQR in the figure.] 4 dx du 20. dx x 3s4 x2 x sx 2 0 2 x 23 dx 32 x 2 19. ■ 1 4x 2 dx ❙❙❙❙ y C P (a, b) (b) Use the hyperbolic substitution x a sinh t to show that 0 y sx dx 2 sinh a2 x a 1 L x C 39. Find the area of the crescent-shaped region (called a lune) These formulas are connected by Formula 7.6.3. bounded by arcs of circles with radii r and R. (See the figure.) 32. Evaluate y x 2 x2 a2 32 dx (a) by trigonometric substitution. (b) by the hyperbolic substitution x 33. Find the average value of f x sx 2 r R a sinh t. 1 x, 1 x 7. 34. Find the area of the region bounded by the hyperbola 9x 2 4y 2 36 and the line x 1 2 3. r 2 for the area of a sector of a circle 2 with radius r and central angle . [Hint: Assume 0 and place the center of the circle at the origin so it has the 35. Prove the formula A 40. A water storage tank has the shape of a cylinder with diameter 10 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 7 ft, what percentage of the total capacity is being used? 41. A torus is generated by rotating the circle x 2 y R2 about the x-axis. Find the volume enclosed by the torus. r2 5E-08(pp 530-539) 532 ❙❙❙❙ 1/17/06 6:03 PM Page 532 CHAPTER 8 TECHNIQUES OF INTEGRATION |||| 8.4 Integration of Rational Functions by Partial Fractions In this section we show how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate. To illustrate the method, observe that by taking the fractions 2 x 1 and 1 x 2 to a common denominator we obtain 2 x 1 1 2x x 2 2 x1 x 1x 2 x x 2 5 x 2 If we now reverse the procedure, we see how to integrate the function on the right side of this equation: yx x 5 2 x 2 dx 2 y x 1 1 2 ln x x 1 2 ln x dx 2 C To see how the method of partial fractions works in general, let’s consider a rational function Px Qx fx where P and Q are polynomials. It’s possible to express f as a sum of simpler fractions provided that the degree of P is less than the degree of Q. Such a rational function is called proper. Recall that if Px an x n an 1x n 1 a1 x a0 where a n 0, then the degree of P is n and we write deg P n. If f is improper, that is, deg P deg Q , then we must take the preliminary step of dividing Q into P (by long division) until a remainder R x is obtained such that deg R deg Q . The division statement is Px Qx fx 1 Sx Rx Qx where S and R are also polynomials. As the following example illustrates, sometimes this preliminary step is all that is required. EXAMPLE 1 Find ≈+x +2 x-1 ) ˛ +x ˛-≈ ≈+x ≈-x 2x 2x-2 2 y x3 x x d x. 1 SOLUTION Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write y x3 x x dx 1 y x3 3 x2 x x2 2 2 2x 2 x 1 2 ln x dx 1 C 5E-08(pp 530-539) 1/17/06 6:03 PM Page 533 S ECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ❙❙❙❙ 533 The next step is to factor the denominator Q x as far as possible. It can be shown that any polynomial Q can be factored as a product of linear factors (of the form ax b) and irreducible quadratic factors (of the form ax 2 b x c, where b 2 4 ac 0). For instance, if Q x x 4 16, we could factor it as x2 Qx 4 x2 4 x 2 x2 2x 4 The third step is to express the proper rational function R x Q x (from Equation 1) as a sum of partial fractions of the form A ax b Ax or i ax 2 B bx c j A theorem in algebra guarantees that it is always possible to do this. We explain the details for the four cases that occur. CASE I ■ The denominator Q x is a product of distinct linear factors. This means that we can write Qx a1 x b1 a 2 x b2 ak x bk where no factor is repeated (and no factor is a constant multiple of another). In this case the partial fraction theorem states that there exist constants A1, A2 , . . . , Ak such that Rx Qx 2 A1 a 1 x b1 A2 a2 x b2 Ak a k x bk These constants can be determined as in the following example. EXAMPLE 2 Evaluate x2 y 2x 3 2x 3x 2 1 d x. 2x SOLUTION Since the degree of the numerator is less than the degree of the denominator, we don’t need to divide. We factor the denominator as 2x 3 3x 2 2x x 2x 2 3x 2 x 2x 1x 2 Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand (2) has the form x2 x 2x 3 |||| Another method for finding A, B, and C is given in the note after this example. 2x 1 1x 2 A x B 2x C 1 x 2 To determine the values of A, B, and C, we multiply both sides of this equation by the product of the denominators, x 2 x 1 x 2 , obtaining 4 x2 2x 1 A 2x 1x 2 Bx x 2 Cx 2 x 1 Expanding the right side of Equation 4 and writing it in the standard form for polynomials, we get 5 x2 2x 1 2A B 2C x 2 3A 2B Cx 2A 5E-08(pp 530-539) 534 ❙❙❙❙ 1/17/06 6:04 PM Page 534 CHAPTER 8 TECHNIQUES OF INTEGRATION |||| Figure 1 shows the graphs of the integrand in Example 2 and its indefinite integral (with K 0). Which is which? 2 The polynomials in Equation 5 are identical, so their coefficients must be equal. The coefficient of x 2 on the right side, 2 A B 2C, must equal the coefficient of x 2 on the left side—namely, 1. Likewise, the coefficients of x are equal and the constant terms are equal. This gives the following system of equations for A, B, and C: 2A 2C 1 3A 2B C 2 2A _3 B 2B 2C 3 1 _2 FIGURE 1 1 2 Solving, we get A y x2 2x 3 1 5 ,B 2x 3x 2 1 dx 2x y 1 2 |||| We could check our work by taking the terms to a common denominator and adding them. 1 10 , and C , and so 11 2x 1 1 5 2x 1 1 10 ln x ln 2 x 1 1 10 x 2 1 10 1 ln x In integrating the middle term we have made the mental substitution u gives du 2 d x and dx du 2. NOTE dx 2 2x K 1, which We can use an alternative method to find the coefficients A, B, and C in Example 2. Equation 4 is an identity; it is true for every value of x. Let’s choose values of x that simplify the equation. If we put x 0 in Equation 4, then the second and third terms on the right side vanish and the equation then becomes 2 A 1, or A 1 . Likewise, 2 1 1 1 1 x 2 gives 5B 4 4 and x 2 gives 10C 1, so B 5 and C 10 . (You may object that Equation 3 is not valid for x 0, 1 , or 2, so why should Equation 4 be valid for those 2 values? In fact, Equation 4 is true for all values of x, even x 0, 1 , and 2. See Exercise 67 2 for the reason.) ■ EXAMPLE 3 Find yx dx 2 a2 , where a 0. SOLUTION The method of partial fractions gives 1 x2 a2 x 1 ax A a x B a x a and therefore Ax a Bx a 1 Using the method of the preceding note, we put x a in this equation and get A 2a 1, so A 1 2a . If we put x a, we get B 2a 1, so B 1 2a . Thus y dx x2 a2 1 2a y 1 1 x a x a 1 (ln x 2a a ln x dx a ) C 5E-08(pp 530-539) 1/17/06 6:04 PM Page 535 S ECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS Since ln x ❙❙❙❙ 535 ln x y , we can write the integral as ln y dx y 6 x 2 a 1 x ln 2a x 2 a a C See Exercises 53–54 for ways of using Formula 6. CASE II ■ Q x is a product of linear factors, some of which are repeated. Suppose the first linear factor a 1 x b1 is repeated r times; that is, a 1 x b1 r occurs in the factorization of Q x . Then instead of the single term A1 a 1 x b1 in Equation 2, we would use A1 a 1 x b1 7 A2 Ar a1 x b1 2 a1 x r b1 By way of illustration, we could write x3 x 1 x2 x 1 3 A x B x2 C x D 1 E x 1 2 x 1 3 but we prefer to work out in detail a simpler example. EXAMPLE 4 Find y x 4 2x 2 x3 x2 4x 1 d x. x1 SOLUTION The first step is to divide. The result of long division is x 4 2x 2 x3 x2 4x 1 x1 x x3 The second step is to factor the denominator Q x we know that x 1 is a factor and we obtain x3 x2 Since the linear factor x x 1 x 1 x2 2 4x 1. Since Q 1 x x x 1x 1x 1 1 occurs twice, the partial fraction decomposition is 4x 12x Ax A |||| Another method for finding the coefficients: Put x 1 in (8): B 2. Put x 1: C 1. Put x 0: A B C 1. x2 1 1 A 1 x B 1 x Multiplying by the least common denominator, x 8 1 1 x x 4x x2 x x3 1 1x 1 Bx C x2 B C 1 1 1 2C x 2 2 A B C 0 A B 2C 4 A B C 0 1 1 , we get x Cx A Now we equate coefficients: x 1 B 2 C 0, 5E-08(pp 530-539) 536 ❙❙❙❙ 1/17/06 6:04 PM Page 536 CHAPTER 8 TECHNIQUES OF INTEGRATION Solving, we obtain A y x 4 2x 2 x3 x2 1, B 2, and C 4x 1 dx x1 y x x x2 2 ■ 1 1 x2 2 CASE III 1, so x x 2 1 ln x x x 2 1 x 2 x 1 2 1 ln 1 x x ln x 1 1 1 dx 1 1 K K Q x contains irreducible quadratic factors, none of which is repeated. If Q x has the factor ax 2 b x c, where b 2 4 ac 0, then, in addition to the partial fractions in Equations 2 and 7, the expression for R x Q x will have a term of the form Ax 9 ax 2 B bx c where A and B are constants to be determined. For instance, the function given by fx x x 2 x 2 1 x 2 4 has a partial fraction decomposition of the form x 2 x2 x A 1 x2 4 x Bx x2 2 C 1 Dx x2 E 4 The term given in (9) can be integrated by completing the square and using the formula dx y 10 2x 2 x3 EXAMPLE 5 Evaluate y SOLUTION Since x 3 4x 2 x a 1 tan a 2 x a C x4 d x. 4x x x2 4 can’t be factored further, we write 2x 2 x 4 x x2 4 Multiplying by x x 2 1 A x Bx x2 C 4 4 , we have 2x 2 4 A x2 4 Bx Cx A x B x2 Cx 4A Equating coefficients, we obtain A Thus A 1, B B 1, and C y 2x 2 x3 2 C 1 4A 4 1 and so x4 dx 4x y 1 x x x2 1 4 dx 5E-08(pp 530-539) 1/17/06 6:05 PM Page 537 S ECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ❙❙❙❙ 537 In order to integrate the second term we split it into two parts: y x x2 1 dx 4 x y 2 x yx dx 4 1 2 dx 4 We make the substitution u x 2 4 in the first of these integrals so that du We evaluate the second integral by means of Formula 10 with a 2: y 2x 2 x 4 dx x x2 4 y 1 dx x y 1 2 ln x EXAMPLE 6 Evaluate 4x 2 y 4x 3x 4x 2 x 4 ln x 2 1 y dx x2 4 1 2 x2 1 tan 2 x d x. dx 4 x2 K 2 d x. 3 SOLUTION Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain 4x 2 4x 2 3x 4x 2 3 x 1 1 4x 4x 2 3 Notice that the quadratic 4 x 2 4 x 3 is irreducible because its discriminant is b 2 4ac 32 0. This means it can’t be factored, so we don’t need to use the partial fraction technique. To integrate the given function we complete the square in the denominator: 4x 2 4x 3 2x This suggests that we make the substitution u x u 1 2, so 4x 2 4x 2 3x 4x 2 dx 3 x y 1 4 yu 2 2 x 1 8 ln u 2 2 x ■ 1 2 x NOTE 1 1 8 ln 4 x 2 4x 1 2 2 u 2x 1 4x y x y 1 u2 1 1 4 4x 2 d x and dx du 1 4 du 2 1. Then, du 3 2 u 2 1 y 1 4 x 1 u 2 1 s2 3 2 tan 1 4 s2 u yu 2 1 du 2 du u 1 C s2 tan 1 2x s2 1 C Example 6 illustrates the general procedure for integrating a partial fraction of the form Ax B ax 2 bx c where b 2 4 ac 0 5E-08(pp 530-539) 538 ❙❙❙❙ 1/17/06 6:05 PM Page 538 CHAPTER 8 TECHNIQUES OF INTEGRATION We complete the square in the denominator and then make a substitution that brings the integral into the form y Cu u2 D du a2 Cy u u2 Dy du a2 1 u2 a2 du Then the first integral is a logarithm and the second is expressed in terms of tan 1. CASE IV ■ Q x contains a repeated irreducible quadratic factor. If Q x has the factor a x 2 partial fraction (9), the sum A1 x B1 ax 2 bx c 11 c r, where b 2 bx A2 x ax 2 4ac 0, then instead of the single B2 bx Ar x c 2 ax Br 2 bx r c occurs in the partial fraction decomposition of R x Q x . Each of the terms in (11) can be integrated by first completing the square. |||| It would be extremely tedious to work out by hand the numerical values of the coefficients in Example 7. Most computer algebra systems, however, can find the numerical values very quickly. For instance, the Maple command convert f, parfrac, x EXAMPLE 7 Write out the form of the partial fraction decomposition of the function xx Apart[f] gives the following values: E 1, 15 8 1 8 B , 1 8 F I 1 2 , x3 1 x2 xx , C , D G J x2 x 1 1 x2 1 Ex x2 F 1 Gx x2 3 SOLUTION or the Mathematica command A x3 1 x2 H x2 x 1 1 x2 1 3 1, 3 4 A x , 1 2 B x Cx D x2 x 1 1 EXAMPLE 8 Evaluate y 1 x 2x 2 x x2 1 x3 H 12 Ix x2 J 13 d x. 2 SOLUTION The form of the partial fraction decomposition is 1 x 2x 2 x x2 1 Multiplying by x x 2 x3 2x 2 x3 Bx x2 C 1 Bx C x x2 Dx x2 E 12 1 2, we have 1 A x2 1 A x4 x A x 2 2x 2 A 2 1 B x4 x2 Cx 3 B x4 1 2A B Dx C x3 Ex Dx 2 x D x2 C Ex Ex A If we equate coefficients, we get the system A B 0 C which has the solution A 1 2A 1, B B 1, C D 1, D 2 C 1, and E E 1 0. Thus A 1 5E-08(pp 530-539) 1/17/06 6:05 PM Page 539 S ECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 1 x 2x 2 x x2 1 x3 y dx 2 |||| In the second and fourth terms we made the mental substitution u x 2 1. 1 x x x2 y y dx x yx 1 2 ln x 1 1 x x2 x 2 dx dx yx 1 1 tan 1x ln x 2 539 dx 2 1 ❙❙❙❙ 2 y 1 x dx x2 1 1 2x 2 1 2 K We note that sometimes partial fractions can be avoided when integrating a rational function. For instance, although the integral x2 x x2 y 1 dx 3 could be evaluated by the method of Case III, it’s much easier to observe that if u x x2 3 x 3 3x, then du 3x 2 3 d x and so y x2 x x2 1 dx 3 1 3 ln x 3 3x C Rationalizing Substitutions Some nonrational functions can be changed into rational functions by means of appropriate substitutions. In particular, when an integrand contains an expression of the form n n st x , then the substitution u st x may be effective. Other instances appear in the exercises. EXAMPLE 9 Evaluate SOLUTION Let u y sx 4 x d x. 4. Then u 2 sx x u2 4, so x 4 and dx 2u du. Therefore y sx 4 x dx yu 2 u 2 y 4 1 4 u 2 sx 4 x dx 2 y du 2u 2 sx 8y 22 4 4 du 4 as u 2u 2 and using du u2 1 8 u2 du 4 We can evaluate this integral either by factoring u 2 partial fractions or by using Formula 6 with a 2: y u2 2y 2u du ln 2 ln 4 u u sx sx 2 2 4 4 C 2 2 C 5E-08(pp 540-549) ❙❙❙❙ 540 1/17/06 5:30 PM Page 540 CHAPTER 8 TECHNIQUES OF INTEGRATION |||| 8.4 Exercises 1–6 |||| Write out the form of the partial fraction decomposition of the function (as in Example 7). Do not determine the numerical values of the coefficients. 1. (a) 2x 3 3x x x 2. (a) 3 x 3. (a) x (b) 1 x (b) 3 x 1 x2 2 3x 2 x3 (b) 4 4. (a) x 6. (a) 4 x x4 x x2 3 ■ 7. y 11. y ■ x yx 9. x ■ ■ 37. 2 x x 1 x2 x 2x 1 1 3 x2 t4 t x t2 1 t2 2 6 x9 5x x 1 3 x 2 1 1 4 dx dx 1 x 13. 15. yx y r2 yr 10. y y 0 17. y 19. y 21. y 23. 25. 27. 29. 1 y y y 4y yy 14. dx bx 2x x 1 2 2 3 dx 12 2 16. 7y 12 dy 2y 3 1 x 5 2 x 1 dx 5x 2 3x 2 dx x 3 2x 2 x 3 1 y y sx 45. y sx dt 46. y sx dx 47. ye ■ 0 x 1 2 3x 2 x 10 1 x2 9 dx x 3 x 2 2x 1 dx x2 1 x2 2 yx x 2 1 y y x 1 0 ax 10 dx 6 x2 1 y 20. y 22. ys 4 2x 5 dx 26. 28. 30. y y y y dx b x 3 4x x2 x 2x x3 x x2 3x x dx 2 2 dx s x x 1 dx 3 x2 x 6 dx x 3 3x x2 x x 3 2x 1 dx 1 2 x2 1 2 x4 ■ ■ yx dx 1 2 x 3 5x dx x 5x 2 4 4 0 ■ dx x4 x2 1 1 ■ dx 2 ■ ■ ■ ■ x 40. y 44. dx 4 y 1 sx yx 42. dx 1 1 3 sx 1 0 2 dx dx ■ 3 x3 2 1 1 3 sx 1 3 4 sx dx dx [Hint: Substitute u dx [Hint: Substitute u e 2x 3e x 2x ■ 3 13 ■ 2 ■ 48. dx ■ ■ ■ sx dx x2 x 6 sx .] sx .] 12 cos x dx sin x sin x y 2 ■ ■ ■ ■ ■ 49–50 |||| Use integration by parts, together with the techniques of this section, to evaluate the integral. 49. y ln x 2 ■ x ■ 50. 2 dx ■ ■ ■ ■ y x tan ■ 1 x dx ■ ■ ■ ■ 1 x 2 2 x 3 to decide whether ; 51. Use a graph of f x 2 x0 f x d x is positive or negative. Use the graph to give a rough estimate of the value of the integral and then use partial fractions to find the exact value. 2 3 1 1 sx 16 9 ■ ds 2 ■ y x sx 1 x 1 18. 24. dx 1 ■ 39. 2 ■ 1 4t t 4 dx 1 x3 3 13 |||| Make a substitution to express the integrand as a rational function and then evaluate the integral. dr 4 2 x 2x 2 y 38. 2 3 2 ■ ■ ax x 41. ■ 4 yx x2 0 39–48 2 4 34. dx dx 4 x ■ x3 ■ 12. 2 y 3x 2 y 36. 2x 3 x 2 1 x6 ■ 8. dx yx x2 5 x 4x 1 32. dx 1 43. (b) 3 35. Evaluate the integral. |||| 2 (b) 1 ■ 7–38 (b) 3 x4 5. (a) ■ x 4x x2 y 1 3 x 1 x 3 yx 33. 1 2x 2 31. 2x x1 dx 5x 2 4 1 x3 ; 52. Graph both y same screen. 2 x 2 and an antiderivative on the 53–54 |||| Evaluate the integral by completing the square and using Formula 6. 53. ■ yx ■ dx 2 54. 2x ■ ■ ■ ■ ■ y 4x ■ 2x 2 ■ 1 12 x ■ 7 dx ■ 55. The German mathematician Karl Weierstrass (1815–1897) noticed that the substitution t tan x 2 will convert any ■ 5E-08(pp 540-549) 1/17/06 5:31 PM Page 541 S ECTION 8.5 STRATEGY FOR INTEGRATION x 2 1 and t2 s1 sin x 2 t t2 (b) Show that t2 t2 1 1 cos x and 2t sin x 2 56–59 t2 1 dt CAS 58. y dx 5 sin x ■ 1 sin x 2 3 1 ■ 60–61 ■ |||| 57. cos x ■ ■ 59. dx ■ y 3 sin x y 2 sin x ■ ■ ■ 1 sin 2 x ■ fx dx 4 cos x 1 dx ■ ■ Find the area of the region under the given curve from x x x 61. y ■ 1 6x 2 ■ 1 , 1 ■ 8 a a 5, b 2, b CAS 4 x 3 27x 2 5 x 32 13x 4 50 x 3 286 x 2 299x 100 x 6 12 x 5 7x 3 13x 2 8 80 x 116 x 4 80 x 3 41x 2 5 20 x 4 (b) Use part (a) to find x f x d x and graph f and its indefinite integral on the same screen. (c) Use the graph of f to discover the main features of the graph of x f x d x. 10 67. Suppose that F, G, and Q are polynomials and ■ ■ ■ ■ ■ ■ ■ ■ ■ Fx Qx 62. Find the volume of the resulting solid if the region under the curve y 1 x 2 3x 2 from x 0 to x about (a) the x-axis and (b) the y-axis. 1 is rotated 63. One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number of sterile males that mate with fertile females but produce no offspring. If P represents the number of female insects in a population, S the number of sterile males introduced each generation, and r the population’s natural growth rate, then the |||| 8.5 70 66. (a) Find the partial fraction decomposition of the function fx 3 , 30 x 5 (b) Use part (a) to find x f x d x (by hand) and compare with the result of using the CAS to integrate f directly. Comment on any discrepancy. a to b. 60. y dP decomposition of the function Use the substitution in Exercise 55 to transform the integrand into a rational function of t and then evaluate the integral. y3 S 65. (a) Use a computer algebra system to find the partial fraction |||| 56. S 1P 1 as a difference of squares by first adding and subtracting the same quantity. Use this factorization to evaluate x 1 x 4 1 d x. (c) Show that dx r 64. Factor x 4 t2 1 P yP Suppose an insect population with 10,000 females grows at a rate of r 0.10 and 900 sterile males are added. Evaluate the integral to give an equation relating the female population to time. (Note that the resulting equation can’t be solved explicitly for P.) t s1 541 female population is related to time t by rational function of sin x and cos x into an ordinary rational function of t. x (a) If t tan x 2 , , sketch a right triangle or use trigonometric identities to show that cos ❙❙❙❙ Gx Qx 0. Prove that F x for all x except when Q x all x. [Hint: Use continuity.] 68. If f is a quadratic function such that f 0 y fx xx 1 2 3 G x for 1 and dx is a rational function, find the value of f 0 . Strategy for Integration As we have seen, integration is more challenging than differentiation. In finding the derivative of a function it is obvious which differentiation formula we should apply. But it may not be obvious which technique we should use to integrate a given function. Until now individual techniques have been applied in each section. For instance, we usually used substitution in Exercises 5.5, integration by parts in Exercises 8.1, and partial fractions in Exercises 8.4. But in this section we present a collection of miscellaneous integrals in random order and the main challenge is to recognize which technique or formula to use. No hard and fast rules can be given as to which method applies in a given situation, but we give some advice on strategy that you may find useful. 5E-08(pp 540-549) 542 ❙❙❙❙ 1/17/06 5:32 PM Page 542 CHAPTER 8 TECHNIQUES OF INTEGRATION A prerequisite for strategy selection is a knowledge of the basic integration formulas. In the following table we have collected the integrals from our previous list together with several additional formulas that we have learned in this chapter. Most of them should be memorized. It is useful to know them all, but the ones marked with an asterisk need not be memorized since they are easily derived. Formula 19 can be avoided by using partial fractions, and trigonometric substitutions can be used in place of Formula 20. Table of Integration Formulas Constants of integration have been omitted. 1. yx n dx xn 1 n1 3. ye x dx ex 5. y sin x d x 7. y sec x d x 9. y sec x tan x d x 2 1 dx x 2. tan x sec x y cos x d x 8. cos x ya 6. 1 y 4. n y csc x d x x ln x ax ln a dx sin x 2 cot x 10. y csc x cot x d x 12. y csc x d x ln csc x ln sin x 11. y sec x d x ln sec x 13. y tan x d x ln sec x 14. y cot x d x 15. y sinh x d x cosh x 16. y cosh x d x 17. y 18. y *19. y *20. y dx x 2 a 2 dx x2 a2 1 tan a tan x x a 1 1 x ln 2a x a a sa x cot x sinh x dx 2 csc x 2 dx sx 2 a 2 sin 1 ln x x a sx 2 a2 Once you are armed with these basic integration formulas, if you don’t immediately see how to attack a given integral, you might try the following four-step strategy. 1. Simplify the Integrand if Possible Sometimes the use of algebraic manipulation or trigonometric identities will simplify the integrand and make the method of integration obvious. Here are some examples: y s x (1 tan y sec 2 d sx ) dx sin y cos y sin y (s x x) d x cos2 d cos d 1 2 y sin 2 d 5E-08(pp 540-549) 1/17/06 5:32 PM Page 543 SECTION 8.5 STRATEGY FOR INTEGRATION sin x cos x 2 d x y sin 2x y y 1 ❙❙❙❙ 543 cos 2x d x 2 sin x cos x 2 sin x cos x d x 2. Look for an Obvious Substitution Try to find some function u t x in the intet x d x also occurs, apart from a constant factor. grand whose differential du For instance, in the integral y x x 2 1 dx we notice that if u x 2 1, then du 2 x d x. Therefore, we use the substitution u x 2 1 instead of the method of partial fractions. 3. Classify the Integrand According to Its Form If Steps 1 and 2 have not led to the solution, then we take a look at the form of the integrand f x . (a) Trigonometric functions. If f x is a product of powers of sin x and cos x, of tan x and sec x, or of cot x and csc x, then we use the substitutions recommended in Section 8.2. (b) Rational functions. If f is a rational function, we use the procedure of Section 8.4 involving partial fractions. (c) Integration by parts. If f x is a product of a power of x (or a polynomial) and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then we try integration by parts, choosing u and dv according to the advice given in Section 8.1. If you look at the functions in Exercises 8.1, you will see that most of them are the type just described. (d) Radicals. Particular kinds of substitutions are recommended when certain radicals appear. (i) If s x 2 a 2 occurs, we use a trigonometric substitution according to the table in Section 8.3. n n (ii) If sax b occurs, we use the rationalizing substitution u sax b. n More generally, this sometimes works for st x . 4. Try Again If the first three steps have not produced the answer, remember that there are basically only two methods of integration: substitution and parts. (a) Try substitution. Even if no substitution is obvious (Step 2), some inspiration or ingenuity (or even desperation) may suggest an appropriate substitution. (b) Try parts. Although integration by parts is used most of the time on products of the form described in Step 3(c), it is sometimes effective on single functions. Looking at Section 8.1, we see that it works on tan 1x, sin 1x, and ln x, and these are all inverse functions. (c) Manipulate the integrand. Algebraic manipulations (perhaps rationalizing the denominator or using trigonometric identities) may be useful in transforming the integral into an easier form. These manipulations may be more substantial than in Step 1 and may involve some ingenuity. Here is an example: y1 dx cos x y y 1 1 1 cos x 1 1 cos x dx sin 2x cos x dx cos x y csc 2x y 1 1 cos x sin 2x cos x dx cos 2x dx 5E-08(pp 540-549) 544 ❙❙❙❙ 1/17/06 5:32 PM Page 544 CHAPTER 8 TECHNIQUES OF INTEGRATION (d) Relate the problem to previous problems. When you have built up some experience in integration, you may be able to use a method on a given integral that is similar to a method you have already used on a previous integral. Or you may even be able to express the given integral in terms of a previous one. For instance, x tan 2x sec x d x is a challenging integral, but if we make use of the identity tan 2x sec 2x 1, we can write 2 3 y tan x sec x d x y sec x d x y sec x d x and if x sec 3x d x has previously been evaluated (see Example 8 in Section 8.2), then that calculation can be used in the present problem. (e) Use several methods. Sometimes two or three methods are required to evaluate an integral. The evaluation could involve several successive substitutions of different types, or it might combine integration by parts with one or more substitutions. In the following examples we indicate a method of attack but do not fully work out the integral. EXAMPLE 1 tan 3x y cos x d x 3 In Step 1 we rewrite the integral: y tan 3x dx cos 3x 3 3 y tan x sec x d x The integral is now of the form x tan m x sec n x d x with m odd, so we can use the advice in Section 7.2. Alternatively, if in Step 1 we had written y tan 3x dx cos 3x sin 3x y cos x 3 sin 3x 1 dx cos 3x y cos x d x 6 then we could have continued as follows with the substitution u y sin 3x dx cos 6x y y EXAMPLE 2 ye sx 1 cos 2x sin x d x cos 6x u2 1 u 6 du y u y 4 u2 1 u u cos x : 6 6 du du dx According to Step 3(d)(ii) we substitute u ye sx dx s x. Then x u 2, so dx 2u du and 2 y ue u du The integrand is now a product of u and the transcendental function e u so it can be integrated by parts. 5E-08(pp 540-549) 1/17/06 5:33 PM Page 545 SECTION 8.5 STRATEGY FOR INTEGRATION ❙❙❙❙ 545 x5 1 dx 3 3x 2 10x No algebraic simplification or substitution is obvious, so Steps 1 and 2 don’t apply here. The integrand is a rational function so we apply the procedure of Section 8.4, remembering that the first step is to divide. E XAMPLE 3 yx EXAMPLE 4 y x sln x dx Here Step 2 is all that is needed. We substitute u du d x x, which occurs in the integral. ln x because its differential is 1x dx 1x Although the rationalizing substitution EXAMPLE 5 y 1 1 u x x works here [Step 3(d)(ii)], it leads to a very complicated rational function. An easier method is to do some algebraic manipulation [either as Step 1 or as Step 4(c)]. Multiplying numerator and denominator by s1 x, we have y 1 1 x dx x 1 x dx x2 y s1 y s1 1 x2 sin 1x dx s1 y s1 x2 x x2 dx C Can We Integrate All Continuous Functions? The question arises: Will our strategy for integration enable us to find the integral of every 2 continuous function? For example, can we use it to evaluate x e x d x ? The answer is no, at least not in terms of the functions that we are familiar with. The functions that we have been dealing with in this book are called elementary functions. These are the polynomials, rational functions, power functions x a , exponential functions a x , logarithmic functions, trigonometric and inverse trigonometric functions, hyperbolic and inverse hyperbolic functions, and all functions that can be obtained from these by the five operations of addition, subtraction, multiplication, division, and composition. For instance, the function fx x2 x 3 1 2x ln cosh x 1 xe sin 2 x is an elementary function. If f is an elementary function, then f is an elementary function but x f x d x need not 2 be an elementary function. Consider f x e x . Since f is continuous, its integral exists, and if we define the function F by Fx y x 0 2 e t dt 5E-08(pp 540-549) ❙❙❙❙ 546 1/17/06 5:34 PM Page 546 CHAPTER 8 TECHNIQUES OF INTEGRATION then we know from Part 1 of the Fundamental Theorem of Calculus that ex Fx 2 2 Thus, f x e x has an antiderivative F , but it has been proved that F is not an elementary function. This means that no matter how hard we try, we will never succeed in evalu2 ating x e x d x in terms of the functions we know. (In Chapter 11, however, we will see how 2 to express x e x d x as an infinite series.) The same can be said of the following integrals: y ex dx x y sx 3 y sin x y cos e dx 1 dx ln x y 1 dx 2 y x dx sin x dx x In fact, the majority of elementary functions don’t have elementary antiderivatives. You may be assured, though, that the integrals in the following exercises are all elementary functions. |||| 8.5 1–80 1. 3. 5. Evaluate the integral. |||| sin x sec x dx tan x y y 2t 2 t 0 y 1 3 2 9. yx 8. r ln r dr x 2 1 3 x x2 16. y 18. y1 20. ye dt 22. y x sin s x )8 d x 24. y ln x 17. y x sin x d x 19. ye 21. yt e 23. y (1 32 x s1 x 2 dx 2 1 5 y y 0 4x x ex dx 4 s1 3 dx 1 x2 s2 2 0 x 2 dx e2t dt e4t sx 31. y 33. dx s1 ln x dx x ln x 15. 2t x2 14. y 3 1 y sin x cos cos x dx 1 x2 yx 28. y sin sat dt 1 dw 2 30. y x dx x 32. y y s3 34. y 35. y 36. y sin 4 x cos 3x dx 37. y 38. y 39. y 40. y s4y 41. y 42. yx y e s1 44. y s1 45. yx e dx 46. y1 dx 2 47. yx a dx a2 48. y 3w 5 w 0 cot x d x x 4 0 y 43. x4 y x csc x y 29. 3x 2 2 dx 2x 8 26. 3 dx dx 12. 5 13. 0 x y s3 y cot x ln sin x 2 d yx dx cos 5 d 12 3 10. 4x y sin y tan yx 27. 2. 6. 4 y 1 3 3x 2 2 dx 2x 8 25. 4. dt e arctan y dy 1 y2 1 7. 11. Exercises 1 x dx 1 dx dx 1 1 1 x 2 dx 2x x 8 sin x dx 1 4 0 cos2 tan2 d x x2 1 s1 tan2 d x 5 x 2 e x dx x3 x2 dx 2 x2 2 4x dx s2 x 2x 1 4 2 4 4 0 2 1 dx 3 4 cot x dx cot x tan 5 sec 3 d 1 2 4y tan 1x d x e x dx ex dx ex 1 x x4 a4 dx 3 dy 5E-08(pp 540-549) 1/17/06 5:35 PM Page 547 ❙❙❙❙ S ECTION 8.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS 49. 1 y x s4 x 1 1 51. y x s4 x 53. yx 55. yx 57. y x sx 59. ye 3x 61. yx 10 63. y sx e 65. 2 50. dx 1 s4 x 54. y 56. y sx c dx 58. yx dx 60. yx dx 62. y 64. y dx 1 4 sx 3 1 ex x4 16 1 dx 1 1 |||| 8.6 sx dx dx 66. sin x 2 d x x x ln x y 2 2 1 ln 1 4 3 2 dx x dx 1 3 sx x3 x1 3 dx dx 10 dx ln tan x dx sin x cos x 3 u u3 arctan st dt st 3 68. y1 e 2x dx ex 70. y 72. y1 74. ye 76. y 78. y sin x 80. y sin 1 du u2 67. y 69. 1 dx x4 1 yx 1 4 y sx 2 52. 2 sinh mx dx sx yx y1 71. yx 73. y 75. y sin x sin 2 x sin 3x d x 77. y1 79. y x sin ■ 1 x 4x 2 4 3 1 2 x2 x dx 4 dx sx dx x3 2 x cos x d x ■ ■ ■ ■ ■ 2 e x and y antiderivatives, but y 2x 2 2 x2 x 2 x 1 e d x. 81. The functions y 1 2e x 547 e x dx ln x 1 dx x2 st dt 3 st x dx e x2 x bx sin 2 x dx sec x cos 2 x dx sec x sin x cos x dx 4 x cos 4 x ■ ■ ■ ■ ■ ■ 2 x 2e x don’t have elementary 2 1 e x does. Evaluate Integration Using Tables and Computer Algebra Systems In this section we describe how to use tables and computer algebra systems to integrate functions that have elementary antiderivatives. You should bear in mind, though, that even the most powerful computer algebra systems can’t find explicit formulas for the antideriv2 atives of functions like e x or the other functions described at the end of Section 8.5. Tables of Integrals Tables of indefinite integrals are very useful when we are confronted by an integral that is difficult to evaluate by hand and we don’t have access to a computer algebra system. A relatively brief table of 120 integrals, categorized by form, is provided on the Reference Pages at the back of the book. More extensive tables are available in CRC Standard Mathematical Tables and Formulae, 30th ed. by Daniel Zwillinger (Boca Raton, FL: CRC Press, 1995) (581 entries) or in Gradshteyn and Ryzhik’s Table of Integrals, Series, and Products, 6e (New York: Academic Press, 2000), which contains hundreds of pages of integrals. It should be remembered, however, that integrals do not often occur in exactly the form listed in a table. Usually we need to use substitution or algebraic manipulation to transform a given integral into one of the forms in the table. EXAMPLE 1 The region bounded by the curves y arctan x, y about the y-axis. Find the volume of the resulting solid. 0, and x SOLUTION Using the method of cylindrical shells, we see that the volume is V y 1 0 2 x arctan x d x 1 is rotated 5E-08(pp 540-549) 548 ❙❙❙❙ 1/17/06 5:35 PM Page 548 CHAPTER 8 TECHNIQUES OF INTEGRATION |||| The Table of Integrals appears on the Reference Pages at the back of the book. In the section of the Table of Integrals entitled Inverse Trigonometric Forms we locate Formula 92: 1 y u tan u2 u du 1 u 2 tan 1u 2 C Thus, the volume is V y 2 1 0 x x tan 1x d x 2 2 1 1 tan x 4 x2 2 x 1 2 1 EXAMPLE 2 Use the Table of Integrals to find 1 2 1 0 x 2 tan 1x 2 tan 1 1 1 0 1 2 y x2 d x. 4x 2 s5 SOLUTION If we look at the section of the table entitled Forms involving sa 2 u 2, we see that the closest entry is number 34: u2 y sa 2 u2 u sa 2 2 du a2 sin 2 u2 u a 1 C This is not exactly what we have, but we will be able to use it if we first make the substitution u 2 x : y x2 4x 2 s5 y dx 5 (so a Then we use Formula 34 with a 2 y s5 x2 4x 2 1 8 dx y s5 u2 x s5 8 u 2 2 du s5 u 2 2 u2 4x 2 u2 du s5 u2 u2 5 sin 2 u s5 2 5 sin 16 EXAMPLE 3 Use the Table of Integrals to find y s5 ): 1 8 du 1 8 yx 3 1 2x s5 1 u s5 C C sin x d x. SOLUTION If we look in the section called Trigonometric Forms, we see that none of the entries explicitly includes a u 3 factor. However, we can use the reduction formula in entry 84 with n 3: yx 85. yu n cos u du u n sin u n y un 1 sin u du 3 sin x d x x 3 cos x 3 y x 2 cos x d x We now need to evaluate x x 2 cos x dx. We can use the reduction formula in entry 85 with n 2, followed by entry 82: yx 2 cos x d x x 2 sin x 2 y x sin x d x x 2 sin x 2 sin x x cos x K 5E-08(pp 540-549) 1/17/06 5:36 PM Page 549 SECTION 8.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS ❙❙❙❙ 549 Combining these calculations, we get yx where C 3 x 3 cos x sin x d x 3x 2 sin x 6 x cos x y x sx 2 2x SOLUTION Since the table gives forms involving sa 2 not sax 2 bx If we make the substitution u pattern sa 2 u 2 : y x sx 2 2x x 2x 4 x 2, sa 2 y u su u sa 2 2 u2 a2 ln (u 2 sa 2 u2) a 2, but u 1 (so x y 4 dx 2 1 1), the integrand will involve the 2 1 2 3 du 2 1 2 y st dt y su 2 u su 2 2 3 du 3 du 3 du t 2 3 du 3: 1 3 u2 3 32 s3 : 3 2 3 y su u2 2 32 3 For the second integral we use Formula 21 with a C 3 1 su 2 u The first integral is evaluated using the substitution t u 2 du x 2, and s x 2 x y u su 2 4 d x. c , we first complete the square: x2 y sa C 3K . EXAMPLE 4 Use the Table of Integrals to find 21. 6 sin x ln(u su 2 3) Thus y x sx 2 2x 1 3 x2 4 dx 2x 4 32 x 1 2 sx 2 2x 4 3 2 ln( x 1 sx 2 2x 4) C Computer Algebra Systems We have seen that the use of tables involves matching the form of the given integrand with the forms of the integrands in the tables. Computers are particularly good at matching patterns. And just as we used substitutions in conjunction with tables, a CAS can perform substitutions that transform a given integral into one that occurs in its stored formulas. So it isn’t surprising that computer algebra systems excel at integration. That doesn’t mean that integration by hand is an obsolete skill. We will see that a hand computation sometimes produces an indefinite integral in a form that is more convenient than a machine a n s w e r . To begin, let’s see what happens when we ask a machine to integrate the relatively simple function y 1 3x 2 . Using the substitution u 3x 2, an easy calculation by hand gives 1 y 3x 2 d x 13 ln 3x 2 C 5E-08(pp 550-559) 550 ❙❙❙❙ 1/17/06 6:00 PM Page 550 CHAPTER 8 TECHNIQUES OF INTEGRATION whereas Derive, Mathematica, and Maple all return the answer 1 3 ln 3x 2 The first thing to notice is that computer algebra systems omit the constant of integration. In other words, they produce a particular antiderivative, not the most general one. Therefore, when making use of a machine integration, we might have to add a constant. Second, the absolute value signs are omitted in the machine answer. That is fine if our problem is concerned only with values of x greater than 2 . But if we are interested in other 3 values of x, then we need to insert the absolute value symbol. In the next example we reconsider the integral of Example 4, but this time we ask a machine for the answer. y x sx EXAMPLE 5 Use a computer algebra system to find 2 2x 4 d x. SOLUTION Maple responds with the answer 1 3 x2 2x 4 1 4 32 2 sx 2 2x 2x 3 s3 arcsinh 1 2 3 4 x This looks different from the answer we found in Example 4, but it is equivalent because the third term can be rewritten using the identity ln( x sx 2 1) s3 1 3 arcsinh x |||| This is Equation 7.6.3. x s1 1 3 Thus arcsinh s3 1 3 x ln ln 1 1 s3 ln 1 s3 [ x | x s1 ln( x 1 2 x 2 1 3 sx 2 4) 2x The resulting extra term 3 ln(1 s3 ) can be absorbed into the constant of integration. 2 Mathematica gives the answer 5 6 x2 3 x 6 sx 2 2x 4 3 1x arcsinh 2 s3 Mathematica combined the first two terms of Example 4 (and the Maple result) into a single term by factoring. Derive gives the answer 1 6 sx 2 2x 4 2x 2 x 5 3 2 ln(sx 2 2x 4 x 1) The first term is like the first term in the Mathematica answer, and the second term is identical to the last term in Example 4. EXAMPLE 6 Use a CAS to evaluate yx x2 5 8 d x. SOLUTION Maple and Mathematica give the same answer: 1 18 x 18 5 2 x 16 50 x 14 1750 3 x 12 4375x 10 21875 x 8 218750 3 x6 156250 x 4 390625 2 x2 5E-08(pp 550-559) 1/17/06 6:00 PM Page 551 ❙❙❙❙ S ECTION 8.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS 551 It’s clear that both systems must have expanded x 2 5 8 by the Binomial Theorem and then integrated each term. If we integrate by hand instead, using the substitution u x 2 5, we get x2 yx |||| Derive and the TI-89/92 also give this answer. 1 18 5 8 dx x2 5 9 C For most purposes, this is a more convenient form of the answer. EXAMPLE 7 Use a CAS to find 5 2 y sin x cos x d x. SOLUTION In Example 2 in Section 8.2 we found that 5 1 3 2 y sin x cos x dx 1 2 5 cos 3x 1 7 cos7x 8 105 cos 5x cos 3x C Derive and Maple report the answer 1 7 sin 2x cos 3x whereas Mathematica produces 10 5 64 5 0 cos x 1 192 3 320 cos 3x cos 5x 1 448 cos 7x We suspect that there are trigonometric identities which show these three answers are equivalent. Indeed, if we ask Derive, Maple, and Mathematica to simplify their expressions using trigonometric identities, they ultimately produce the same form of the answer as in Equation 1. F EXAMPLE 8 If f x FIGURE 1 Graph F for 0 x x 60 sin 4x cos 5x, find the antiderivative F of f such that F 0 5. Where does F have extreme values and inflection points? 0. SOLUTION The antiderivative of f produced by Maple is 10 fª F x2 20 3 _7 FIGURE 2 |||| Use the indicated entry in the Table of Integrals on the Reference Pages to evaluate the integral. 2x 2 s7 x sin x cos 6x 4 7 cos 4x sin x 16 21 cos 2x sin x 32 21 sin x Exercises 1–4 y 20 7 sin 3x cos 6x and we note that F 0 0. This expression could probably be simplified, but there’s no need to do so because a computer algebra system can graph this version of F as easily as any other version. A graph of F is shown in Figure 1. To locate the extreme values of F, we graph its derivative F f in Figure 2 and observe that F has a local maximum when x 2.3 and a local minimum when x 2.5. The graph of F f in Figure 2 shows that F has inflection points when x 0.7, 1.3, 1.8, 2.4, 3.3, and 3.9. 5 |||| 8.6 1 2 Fx f 0 1. 4 35 sin 4x cos 3x 2 dx ; entry 33 2. y s3 2x y sec 4. 3x 3. ye 2 3 x dx ; entry 71 sin 3 d ; entry 98 dx ; entry 55 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E-08(pp 550-559) 552 ❙❙❙❙ 1/17/06 6:01 PM Page 552 CHAPTER 8 TECHNIQUES OF INTEGRATION 5–30 |||| Use the Table of Integrals on the Reference Pages to evaluate the integral. 5. y 1 2 x cos 1x dx 0 y 6. 1 x 2 s4 x 2 3 2 7 CAS 35–42 |||| Use a computer algebra system to evaluate the integral. Compare the answer with the result of using tables. If the answers are not the same, show that they are equivalent. dx 9. 11. 3x ye dx x 2s4x 2 y y 0 1 t 2e t dt y x2 1 x 3 4 dx y sin x cos 2x d x 38. y tan x 2 sec 4x d x 39. 3 y x s1 2x dx 40. y sin 4x d x y tan5x d x 42. y x 5s x 3 dy x 2 cos 3x d x y 12. 36. x 2 dx s2y 2 y2 y 10. 9 3 x 2 dx 41. y csc 8. cos 4 x d x y x 2s5 37. 7. 35. 0 2 1 dx ■ ■ 3 tan 1 z dz z2 13. y 15. ye 17. y y s6 19. y sin x cos x ln sin x 21. 23. x ex dx 3 e2x 5 25. y 27. y se 29. y sx 0 ■ ■ ■ CAS cos 3 x 2 d x x s4 x 6 x 4e y ye t ■ x s4 x 2, 0 x 32. The region under the curve y 2 x 1 dx 2x 1 dx instead. Why do you think it was successful with this form of the integrand? CAS 44. Try to evaluate y ■ 1 x ln x 2 dx ln x s1 with a computer algebra system. If it doesn’t return an answer, make a substitution that changes the integral into one that the CAS can evaluate. 3 dt CAS ■ ■ 45–48 |||| Use a CAS to find an antiderivative F of f such that F 0 0. Graph f and F and locate approximately the x-coordinates of the extreme points and inflection points of F . 45. f x x2 x4 2 bu. 46. f x 47. f x xe 48. f x ■ ■ 1 x2 x 1 sin x, 5 x sin x cos x, 0 x 4 3 34. Verify Formula 31 (a) by differentiation and (b) by substituting a sin . ■ x dx 33. Verify Formula 53 in the Table of Integrals (a) by differentiation u ■ 4 2, is rotated about the a ■ 43. Computer algebra systems sometimes need a helping hand dx sin t ■ ■ y 2 x s2 2x dx x ■ If it doesn’t return an answer, ask it to try tan x from 0 to 4 is rotated about the x-axis. Find the volume of the resulting solid. and (b) by using the substitution t ■ y 2 x s4 31. Find the volume of the solid obtained when the region under the curve y y-axis. ■ dx sec 2 tan 2 d tan 2 ■ ■ from human beings. Ask your CAS to evaluate y s9 0 ■ 2e x 3 30. 2 ■ ■ dx 1 y sin 1 2 s2 28. 1 dx 10 y 2 sx d x x5 26. dx x ■ x 2 x 4 dx ■ ye 24. ln x 2x 2 22. y sec x d x s4 yx 20. dx y x sin x 18. 4y2 dy 4y y sin 16. sech e x d x 2 y 1 14. x x6 ■ 6 5 x 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E-08(pp 550-559) 1/17/06 6:01 PM Page 553 D ISCOVERY PROJECT PATTERNS IN INTEGRALS ❙❙❙❙ 553 DISCOVERY PROJECT CAS Patterns in Integrals In this project a computer algebra system is used to investigate indefinite integrals of families of functions. By observing the patterns that occur in the integrals of several members of the family, you will first guess, and then prove, a general formula for the integral of any member of the family. 1. (a) Use a computer algebra system to evaluate the following integrals. (i) y (iii) y x 1 2x 3 x 1 2x 5 dx (ii) y dx (iv) y 1 1x x 1 x 2 2 5 dx dx (b) Based on the pattern of your responses in part (a), guess the value of the integral y x 1 ax b dx if a b. What if a b? (c) Check your guess by asking your CAS to evaluate the integral in part (b). Then prove it using partial fractions. 2. (a) Use a computer algebra system to evaluate the following integrals. (i) y sin x cos 2 x dx (ii) y sin 3 x cos 7x dx (iii) y sin 8 x cos 3 x dx (b) Based on the pattern of your responses in part (a), guess the value of the integral y sin ax cos bx dx (c) Check your guess with a CAS. Then prove it using the techniques of Section 7.2. For what values of a and b is it valid? 3. (a) Use a computer algebra system to evaluate the following integrals. (i) (iv) y ln x dx (ii) y x ln x dx y x 3 ln x dx (v) y x 2 ln x dx y x 7 ln x dx (iii) (b) Based on the pattern of your responses in part (a), guess the value of y x n ln x dx (c) Use integration by parts to prove the conjecture that you made in part (b). For what values of n is it valid? 4. (a) Use a computer algebra system to evaluate the following integrals. (i) y xe x dx (ii) y x 2e x dx (iv) y x 4e x dx (v) y x 5e x dx (iii) y x 3e x dx (b) Based on the pattern of your responses in part (a), guess the value of x x 6e x d x. Then use your CAS to check your guess. (c) Based on the patterns in parts (a) and (b), make a conjecture as to the value of the integral nx yx e dx when n is a positive integer. (d) Use mathematical induction to prove the conjecture you made in part (c). 5E-08(pp 550-559) 554 ❙❙❙❙ 1/17/06 6:01 PM Page 554 CHAPTER 8 TECHNIQUES OF INTEGRATION |||| 8.7 Approximate Integration There are two situations in which it is impossible to find the exact value of a definite integral. The first situation arises from the fact that in order to evaluate xab f x d x using the Fundamental Theorem of Calculus we need to know an antiderivative of f . Sometimes, however, it is difficult, or even impossible, to find an antiderivative (see Section 8.5). For example, it is impossible to evaluate the following integrals exactly: y 1 2 e x dx 0 y 1 x 3 dx s1 1 The second situation arises when the function is determined from a scientific experiment through instrument readings or collected data. There may be no formula for the function (see Example 5). In both cases we need to find approximate values of definite integrals. We already know one such method. Recall that the definite integral is defined as a limit of Riemann sums, so any Riemann sum could be used as an approximation to the integral: If we divide a, b into n subintervals of equal length x b a n, then we have y y b a n f x* i f x dx x i1 where x * is any point in the ith subinterval x i 1, x i . If x * is chosen to be the left endpoint i i of the interval, then x * x i 1 and we have i 0 x¸ ⁄ x™ x£ x¢ x y 1 f x dx a (a) Left endpoint approximation Ln f xi 1 x i1 If f x 0, then the integral represents an area and (1) represents an approximation of this area by the rectangles shown in Figure 1(a). If we choose x * to be the right endpoint, then i x * x i and we have i y y 2 0 n b x¸ ⁄ x™ x£ x¢ x (b) Right endpoint approximation y b a n f x dx Rn f xi x i1 [See Figure 1(b).] The approximations L n and Rn defined by Equations 1 and 2 are called the left endpoint approximation and right endpoint approximation, respectively. In Section 5.2 we also considered the case where x * is chosen to be the midpoint xi of i the subinterval x i 1, x i . Figure 1(c) shows the midpoint approximation Mn , which appears to be better than either L n or Rn. Midpoint Rule y b a f x dx where 0 x –¡ x™ – – x£ – x¢ (c) Midpoint approximation FIGURE 1 x and xi Mn b x f x1 f x2 a n x 1 2 xi 1 xi midpoint of x i 1, x i f xn 5E-08(pp 550-559) 1/17/06 6:01 PM Page 555 S ECTION 8.7 APPROXIMATE INTEGRATION ❙ ❙❙❙ 555 Another approximation, called the Trapezoidal Rule, results from averaging the approximations in Equations 1 and 2: y b a n 1 2 f x dx n f xi x 1 f xi i1 x 2 i1 f x0 f x1 2 f x1 x f x0 2 f x1 n x 2 x 2 f x2 f xi f xi 1 i1 f x2 f xn 2 f xn 1 1 f xn f xn y Trapezoidal Rule y b a f x dx where x 0 x¸ ⁄ x™ x£ x¢ x x f x0 2 Tn b a n and xi a 2 f x2 2 f xn 1 f xn i x. The reason for the name Trapezoidal Rule can be seen from Figure 2, which illustrates the case f x 0. The area of the trapezoid that lies above the ith subinterval is FIGURE 2 Trapezoidal approximation y= 2 f x1 x f xi f xi 1 x f xi 2 2 f xi 1 and if we add the areas of all these trapezoids, we get the right side of the Trapezoidal Rule. 1 x EXAMPLE 1 Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with n approximate the integral x12 1 x d x. 5 to SOLUTION (a) With n 5, a zoidal Rule gives y 1 2 F IGURE 3 2 1 1 dx x T5 0.1 1, and b 2, we have x 0.2 f1 2 1 1 2 f 1.2 2 1.2 2 1.4 2 2 f 1.4 2 1.6 15 2 f 1.6 2 1.8 0.2, and so the Trape- 2 f 1.8 f2 1 2 0.695635 y= This approximation is illustrated in Figure 3. (b) The midpoints of the five subintervals are 1.1, 1.3, 1.5, 1.7, and 1.9, so the Midpoint Rule gives 21 y1 x d x x f 1.1 f 1.3 f 1.5 f 1.7 f 1.9 1 x 1 5 1 FIGURE 4 2 1 1.1 1 1.3 0.691908 This approximation is illustrated in Figure 4. 1 1.5 1 1.7 1 1.9 5E-08(pp 550-559) ❙❙❙❙ 556 1/17/06 6:01 PM Page 556 CHAPTER 8 TECHNIQUES OF INTEGRATION In Example 1 we deliberately chose an integral whose value can be computed explicitly so that we can see how accurate the Trapezoidal and Midpoint Rules are. By the Fundamental Theorem of Calculus, y 2 1 y b a f x dx approximation error 1 dx x ln x]1 2 ln 2 0.693147 . . . The error in using an approximation is defined to be the amount that needs to be added to the approximation to make it exact. From the values in Example 1 we see that the errors in the Trapezoidal and Midpoint Rule approximations for n 5 are ET 0.002488 and EM 0.001239 In general, we have ET Module 5.1/5.2/8.7 allows you to compare approximation methods. Approximations to y 2 y b a f x dx Tn and y EM b a f x dx Mn The following tables show the results of calculations similar to those in Example 1, but for n 5, 10, and 20 and for the left and right endpoint approximations as well as the Trapezoidal and Midpoint Rules. 1 dx x n Rn Tn Mn 0.745635 0.718771 0.705803 0.645635 0.668771 0.680803 0.695635 0.693771 0.693303 0.691908 0.692835 0.693069 n Corresponding errors Ln 5 10 20 1 EL ER ET EM 5 10 20 0.052488 0.025624 0.012656 0.047512 0.024376 0.012344 0.002488 0.000624 0.000156 0.001239 0.000312 0.000078 We can make several observations from these tables: 1. In all of the methods we get more accurate approximations when we increase the 2. |||| It turns out that these observations are true in most cases. 3. 4. 5. value of n. (But very large values of n result in so many arithmetic operations that we have to beware of accumulated round-off error.) The errors in the left and right endpoint approximations are opposite in sign and appear to decrease by a factor of about 2 when we double the value of n. The Trapezoidal and Midpoint Rules are much more accurate than the endpoint approximations. The errors in the Trapezoidal and Midpoint Rules are opposite in sign and appear to decrease by a factor of about 4 when we double the value of n. The size of the error in the Midpoint Rule is about half the size of the error in the Trapezoidal Rule. Figure 5 shows why we can usually expect the Midpoint Rule to be more accurate than the Trapezoidal Rule. The area of a typical rectangle in the Midpoint Rule is the same as the trapezoid ABCD whose upper side is tangent to the graph at P. The area of this trapezoid is closer to the area under the graph than is the area of the trapezoid AQRD used in 5E-08(pp 550-559) 1/17/06 6:01 PM Page 557 SECTION 8.7 APPROXIMATE INTEGRATION C 557 the Trapezoidal Rule. [The midpoint error (shaded red) is smaller than the trapezoidal error (shaded blue).] These observations are corroborated in the following error estimates, which are proved in books on numerical analysis. Notice that Observation 4 corresponds to the n 2 in each denominator because 2 n 2 4n 2. The fact that the estimates depend on the size of the second derivative is not surprising if you look at Figure 5, because f x measures how much the graph is curved. [Recall that f x measures how fast the slope of y f x changes.] P B A ❙❙❙❙ D x i-1 xi – xi K for a 3 Error Bounds Suppose f x in the Trapezoidal and Midpoint Rules, then C R Kb a 12n 2 ET P x b. If ET and EM are the errors 3 and Kb a 24n 2 EM 3 B Let’s apply this error estimate to the Trapezoidal Rule approximation in Example 1. If fx 1 x, then f x 1 x 2 and f x 2 x 3. Since 1 x 2, we have 1 x 1, so Q A D 2 x3 fx 2 13 2 FIGURE 5 Therefore, taking K |||| K can be any number larger than all the values of f x , but smaller values of K give better error bounds. 2, a ET 1, b 2, and n 22 1 12 5 2 5 in the error estimate (3), we see that 3 1 150 0.006667 Comparing this error estimate of 0.006667 with the actual error of about 0.002488, we see that it can happen that the actual error is substantially less than the upper bound for the error given by (3). EXAMPLE 2 How large should we take n in order to guarantee that the Trapezoidal and Midpoint Rule approximations for x12 1 x d x are accurate to within 0.0001? SOLUTION We saw in the preceding calculation that f x 2 for 1 x 2, so we can take K 2, a 1, and b 2 in (3). Accuracy to within 0.0001 means that the size of the error should be less than 0.0001. Therefore, we choose n so that 213 12n 2 0.0001 Solving the inequality for n, we get n2 |||| It’s quite possible that a lower value for n would suffice, but 41 is the smallest value for which the error bound formula can guarantee us accuracy to within 0.0001. or Thus, n 2 12 0.0001 n 1 s0.0006 41 will ensure the desired accuracy. 40.8 5E-08(pp 550-559) 558 ❙❙❙❙ 1/17/06 6:01 PM Page 558 CHAPTER 8 TECHNIQUES OF INTEGRATION For the same accuracy with the Midpoint Rule we choose n so that 213 24n 2 which gives y 0.0001 1 s0.0012 n 29 EXAMPLE 3 2 (a) Use the Midpoint Rule with n 10 to approximate the integral x01 e x dx. (b) Give an upper bound for the error involved in this approximation. y=e x SOLUTION 2 (a) Since a 0, b y 1 0 1, and n 2 e x dx 10, the Midpoint Rule gives x f 0.05 0.1 e 0.0025 f 0.15 e 0.0225 e 0.4225 0 1 x f 0.85 e 0.0625 e 0.5625 e 0.1225 e 0.7225 f 0.95 e 0.2025 e 0.3025 e 0.9025 1.460393 Figure 6 illustrates this approximation. 2 2 (b) Since f x e x , we have f x 2 xe x and f x 0 x 1, we have x 2 1 and so FIGURE 6 0 Taking K 6e, a 0, b bound for the error is |||| Error estimates are upper bounds for the error. They give theoretical, worst-case scenarios. The actual error in this case turns out to be about 0.0023. fx 1, and n 6e 1 3 24 10 2 2 4x 2 e x 2 2 2 4 x 2 e x . Also, since 6e 10 in the error estimate (3), we see that an upper e 400 0.007 Simpson ’s Rule Another rule for approximate integration results from using parabolas instead of straight line segments to approximate a curve. As before, we divide a, b into n subintervals of equal length h x b a n, but this time we assume that n is an even number. Then on each consecutive pair of intervals we approximate the curve y f x 0 by a parabola as shown in Figure 7. If yi f x i , then Pi x i , yi is the point on the curve lying above x i. A typical parabola passes through three consecutive points Pi , Pi 1 , and Pi 2 . y y P¸ P¡ P∞ P™ P£ 0 a=x¸ FIGURE 7 ⁄ x™ x£ P¸ (_h, y¸) Pß P¡ (0, › ) P™ (h, fi) P¢ x¢ x∞ xß=b x _h FIGURE 8 0 h x 5E-08(pp 550-559) 1/17/06 6:01 PM Page 559 SECTION 8.7 APPROXIMATE INTEGRATION ❙❙❙❙ 559 To simplify our calculations, we first consider the case where x 0 h, x 1 0, and x 2 h. (See Figure 8.) We know that the equation of the parabola through P0 , P1 , and P2 is of the form y Ax 2 Bx C and so the area under the parabola from x h to x h is |||| Here we have used Theorem 5.5.6. Notice that Ax 2 C is even and Bx is odd. y h h Ax 2 Bx h 2 y Ax 2 C dx C dx 0 h x3 2A 3 Cx h3 3 Ch 2A But, since the parabola passes through P0 y0 A h y1 Ah 2 y0 and therefore B h 2 Ah 2 3 6C h, y0 , P1 0, y1 , and P2 h, y2 , we have Bh h Ah 2 C y2 2 0 C Bh C C 4y1 2 Ah 2 y2 6C Thus, we can rewrite the area under the parabola as h y0 3 4y1 y2 Now, by shifting this parabola horizontally we do not change the area under it. This means that the area under the parabola through P0 , P1 , and P2 from x x 0 to x x 2 in Figure 7 is still h y0 4y1 y2 3 Similarly, the area under the parabola through P2 , P3 , and P4 from x h y2 3 4y3 x 2 to x x 4 is y4 If we compute the areas under all the parabolas in this manner and add the results, we get b a f x dx h y0 3 4y1 y2 h y2 3 h y0 3 y 4y1 2y2 4y3 4y3 2y4 h yn 3 y4 2yn 2 4yn 1 4yn 2 1 yn yn Although we have derived this approximation for the case in which f x 0, it is a reasonable approximation for any continuous function f and is called Simpson’s Rule after the English mathematician Thomas Simpson (1710–1761). Note the pattern of coefficients: 1, 4, 2, 4, 2, 4, 2, . . . , 4, 2, 4, 1. 5E-08(pp 560-569) 560 ❙❙❙❙ 1/17/06 5:57 PM Page 560 CHAPTER 8 TECHNIQUES OF INTEGRATION Simpson’s Rule |||| Thomas Simpson was a weaver who taught himself mathematics and went on to become one of the best English mathematicians of the 18th century. What we call Simpson’s Rule was actually known to Cavalieri and Gregory in the 17th century, but Simpson popularized it in his best-selling calculus textbook, entitled A New Treatise of Fluxions. y b a f x dx x f x0 3 Sn 4 f x1 2 f xn where n is even and x b y 2 1 1 dx x 1 x, n 4 f x3 4 f xn 2 f xn 1 a n. 10 to approximate x12 1 x d x. EXAMPLE 4 Use Simpson’s Rule with n SOLUTION Putting f x 2 f x2 10, and x 0.1 in Simpson’s Rule, we obtain S10 x f1 3 0.1 3 4 f 1.1 1 1 4 1.1 2 f 1.2 2 1.2 4 1.3 4 f 1.3 2 1.4 2 f 1.8 4 1.5 2 1.6 4 1.7 4 f 1.9 2 1.8 f2 4 1.9 1 2 0.693150 Notice that, in Example 4, Simpson’s Rule gives us a much better approximation S10 0.693150 to the true value of the integral ln 2 0.693147. . . than does the Trapezoidal Rule T10 0.693771 or the Midpoint Rule M10 0.692835 . It turns out (see Exercise 48) that the approximations in Simpson’s Rule are weighted averages of those in the Trapezoidal and Midpoint Rules: 1 3 S2 n Tn 2 3 Mn (Recall that ET and EM usually have opposite signs and EM is about half the size of ET .) In many applications of calculus we need to evaluate an integral even if no explicit formula is known for y as a function of x. A function may be given graphically or as a table of values of collected data. If there is evidence that the values are not changing rapidly, then the Trapezoidal Rule or Simpson’s Rule can still be used to find an approximate value for xab y d x, the integral of y with respect to x. E XAMPLE 5 Figure 9 shows data traffic on the link from the United States to SWITCH, the Swiss academic and research network, on February 10, 1998. D t is the data throughput, measured in megabits per second Mb s . Use Simpson’s Rule to estimate the total amount of data transmitted on the link up to noon on that day. D 8 6 4 2 FIGURE 9 0 3 6 9 12 15 18 21 24 t (hours) 5E-08(pp 560-569) 1/17/06 5:57 PM Page 561 S ECTION 8.7 APPROXIMATE INTEGRATION ❙❙❙❙ 561 SOLUTION Because we want the units to be consistent and D t is measured in megabits per second, we convert the units for t from hours to seconds. If we let A t be the amount of data (in megabits) transmitted by time t, where t is measured in seconds, then At D t . So, by the Net Change Theorem (see Section 5.4), the total amount of data transmitted by noon (when t 12 60 2 43,200) is A 43,200 y 43,200 0 D t dt We estimate the values of D t at hourly intervals from the graph and compile them in the table. t hours t seconds Dt t hours t seconds Dt 0 1 2 3 4 5 6 0 3,600 7,200 10,800 14,400 18,000 21,600 3.2 2.7 1.9 1.7 1.3 1.0 1.1 7 8 9 10 11 12 25,200 28,800 32,400 36,000 39,600 43,200 1.3 2.8 5.7 7.1 7.7 7.9 Then we use Simpson’s Rule with n y 43,200 0 A t dt t 3 D0 12 and t 4D 3600 3600 to estimate the integral: 2D 7200 4D 39,600 D 43,200 3600 3.2 4 2.7 2 1.9 4 1.7 2 1.3 4 1.0 3 2 1.1 4 1.3 2 2.8 4 5.7 2 7.1 4 7.7 7.9 143,880 Thus, the total amount of data transmitted up to noon is about 144,000 megabits, or 144 gigabits. In Exercises 27 and 28 you are asked to demonstrate, in particular cases, that the error in Simpson’s Rule decreases by a factor of about 16 when n is doubled. That is consistent with the appearance of n 4 in the denominator of the following error estimate for Simpson’s Rule. It is similar to the estimates given in (3) for the Trapezoidal and Midpoint Rules, but it uses the fourth derivative of f . f4 x K for a If ES is the error involved in using Simpson’s Rule, then 4 Error Bound for Simpson’s Rule Suppose that ES Kb a 180 n 4 5 x b. 5E-08(pp 560-569) 562 ❙❙❙❙ 1/17/06 5:57 PM Page 562 CHAPTER 8 TECHNIQUES OF INTEGRATION EXAMPLE 6 How large should we take n in order to guarantee that the Simpson’s Rule approximation for x12 1 x d x is accurate to within 0.0001? SOLUTION If f x 4 1 x, then f 24 x 5. Since x x 4 f |||| Many calculators and computer algebra systems have a built-in algorithm that computes an approximation of a definite integral. Some of these machines use Simpson’s Rule; others use more sophisticated techniques such as adaptive numerical integration. This means that if a function fluctuates much more on a certain part of the interval than it does elsewhere, then that part gets divided into more subintervals. This strategy reduces the number of calculations required to achieve a prescribed accuracy. Therefore, we can take K choose n so that 1, we have 1 x 24 x5 x 1 and so 24 24 in (4). Thus, for an error less than 0.0001 we should 24 1 5 180n 4 0.0001 n4 or 24 180 0.0001 n This gives 1 4 s0.00075 6.04 Therefore, n 8 (n must be even) gives the desired accuracy. (Compare this with Example 2, where we obtained n 41 for the Trapezoidal Rule and n 29 for the Midpoint Rule.) E XAMPLE 7 2 (a) Use Simpson’s Rule with n 10 to approximate the integral x01 e x d x. (b) Estimate the error involved in this approximation. SOLUTION (a) If n |||| Figure 10 illustrates the calculation in Example 7. Notice that the parabolic arcs are 2 so close to the graph of y e x that they are practically indistinguishable from it. y 1 0 10, then x 2 e x dx 0.1 and Simpson’s Rule gives x f0 3 4 f 0.1 0.1 0 e 3 y 4e 0.01 4e 0.49 2 f 0.2 2e 0.04 2e 0.64 2 f 0.8 4e 0.09 4e 0.81 2e 0.16 4e 0.25 4 f 0.9 f1 2e 0.36 e1 1.462681 y=e ≈ f and so, since 0 x FIGURE 10 1 x Therefore, putting K most 4 x 12 48 x 2 16 x 4 e x 2 1, we have 0 0 2 e x is (b) The fourth derivative of f x f 76e, a 4 x 12 0, b 48 1, and n 76e 1 5 180 10 4 16 e 1 76e 10 in (4), we see that the error is at 0.000115 (Compare this with Example 3.) Thus, correct to three decimal places, we have y 1 0 2 e x dx 1.463 5E-08(pp 560-569) 1/17/06 5:57 PM Page 563 ❙❙❙❙ S ECTION 8.7 APPROXIMATE INTEGRATION |||| 8.7 1. Let I Exercises x04 f (Round your answers to six decimal places.) Compare your results to the actual value to determine the error in each approximation. x d x, where f is the function whose graph is shown. (a) Use the graph to find L 2 , R2, and M2. (b) Are these underestimates or overestimates of I ? (c) Use the graph to find T2. How does it compare with I ? (d) For any value of n, list the numbers L n , Rn , Mn , Tn , and I in increasing order. y 3 5. 0 ■ ■ ■ ■ ■ y 1 0 ■ sx e ■ dx, n ■ 6 ■ ■ ■ f y 1 9. y 11. y 13. y 15. y 17. y 1 2 3 4x 2. The left, right, Trapezoidal, and Midpoint Rule approximations were used to estimate x02 f x d x, where f is the function whose graph is shown. The estimates were 0.7811, 0.8675, 0.8632, and 0.9540, and the same number of subintervals were used in each case. (a) Which rule produced which estimate? (b) Between which two approximations does the true value of x02 f x d x lie? y 4 s1 x 2 d x, ln x d x, 1x 2 1 12 0 2 e 1 x d x, 1 n cos x d x, x 5 1 1 3 1 0 n ■ y 5 ■ 8. 10. 12. y y 18. 8 n ■ y 16. 8 4 dy, y y 8 10 n y 14. n sin e t 2 dt, n x02 e 6 ■ ■ ■ 12 0 sin x 2 dx, n dt t2 3 4 0 4 0 6 4 4 2 ■ s1 , n 6 sx d x, 1 0 4 n 8 t4 s x sin x d x, n ln x 3 2 d x, ex d x, x n ■ 8 n 10 10 ■ ■ ■ x2 d x. (b) Estimate the errors in the approximations of part (a). (c) How large do we have to choose n so that the approximations Tn and Mn to the integral in part (a) are accurate to within 0.00001? y=ƒ 0 2 0 19. (a) Find the approximations T10 and M10 for the integral 1 2 1 20. (a) Find the approximations T8 and M8 for x0 cos x 2 d x. x (b) Estimate the errors involved in the approximations of part (a). (c) How large do we have to choose n so that the approximations Tn and Mn to the integral in part (a) are accurate to within 0.00001? 1 2 ; 3. Estimate x0 cos x d x using (a) the Trapezoidal Rule and (b) the Midpoint Rule, each with n 4. From a graph of the integrand, decide whether your answers are underestimates or overestimates. What can you conclude about the true value of the integral? 1 21. (a) Find the approximations T10 and S10 for x0 e x d x and the corresponding errors ET and ES. (b) Compare the actual errors in part (a) with the error estimates given by (3) and (4). (c) How large do we have to choose n so that the approximations Tn , Mn , and Sn to the integral in part (a) are accurate to within 0.00001? sin x 2 2 in the viewing rectangle 0, 1 by 0, 0.5 and let I x01 f x d x. (a) Use the graph to decide whether L 2 , R2 , M2, and T2 underestimate or overestimate I . (b) For any value of n, list the numbers L n , Rn , Mn , Tn , and I in increasing order. (c) Compute L 5 , R5 , M5, and T5. From the graph, which do you think gives the best estimate of I ? ; 4. Draw the graph of f x 22. How large should n be to guarantee that the Simpson’s Rule 2 approximation to x01 e x dx is accurate to within 0.00001? CAS Use (a) the Midpoint Rule and (b) Simpson’s Rule to approximate the given integral with the specified value of n. |||| ■ 6. 8 7–18 |||| Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.) 7. 0 x 2 sin x dx, n y 2 5–6 563 23. The trouble with the error estimates is that it is often very diffi- cult to compute four derivatives and obtain a good upper bound K for f 4 x by hand. But computer algebra systems have no 5E-08(pp 560-569) 564 ❙❙❙❙ 1/17/06 5:57 PM Page 564 CHAPTER 8 TECHNIQUES OF INTEGRATION problem computing f 4 and graphing it, so we can easily find a value for K from a machine graph. This exercise deals with approximations to the integral I x02 f x d x, where fx e cos x. (a) Use a graph to get a good upper bound for f x . (b) Use M10 to approximate I . (c) Use part (a) to estimate the error in part (b). (d) Use the built-in numerical integration capability of your CAS to approximate I . (e) How does the actual error compare with the error estimate in part (c)? (f) Use a graph to get a good upper bound for f 4 x . (g) Use S10 to approximate I . (h) Use part (f) to estimate the error in part (g). (i) How does the actual error compare with the error estimate in part (h)? (j) How large should n be to guarantee that the size of the error in using Sn is less than 0.0001? CAS 1 24. Repeat Exercise 23 for the integral y s4 Use Simpson’s Rule to estimate the area of the pool. 6.2 x 1 x 3 dx 0 ■ ■ y 26. ■ ■ ■ ■ 2 ■ ■ ■ ■ ■ 27–28 |||| Find the approximations Tn , Mn , and Sn for n 6 and 12. Then compute the corresponding errors ET, EM, and ES. (Round your answers to six decimal places. You may wish to use the sum command on a computer algebra system.) What observations can you make? In particular, what happens to the errors when n is doubled? 27. ■ y 4 1 ■ ■ y 28. sx dx ■ ■ ■ ■ 2 1 ■ fx 6.8 6.5 6.3 6.4 6.9 2.0 2.4 2.8 3.2 7.6 8.4 8.8 9.0 32. A radar gun was used to record the speed of a runner during the first 5 seconds of a race (see the table). Use Simpson’s Rule to estimate the distance the runner covered during those 5 seconds. t (s) v (m s) t (s) v (m s) 0 0.5 1.0 1.5 2.0 2.5 0 4.67 7.34 8.86 9.73 10.22 3.0 3.5 4.0 4.5 5.0 10.51 10.67 10.76 10.81 10.81 33. The graph of the acceleration a t of a car measured in ft s2 is shown. Use Simpson’s Rule to estimate the increase in the velocity of the car during the 6-second time interval. a 12 xe x d x ■ x 1 for all x, estimate the (b) If it is known that 4 f x error involved in the approximation in part (a). e x dx 0 ■ fx 0.0 0.4 0.8 1.2 1.6 |||| Find the approximations L n , Rn , Tn , and Mn for n 4, 8, and 16. Then compute the corresponding errors EL , ER, ET , and EM. (Round your answers to six decimal places. You may wish to use the sum command on a computer algebra system.) What observations can you make? In particular, what happens to the errors when n is doubled? y 4.8 value of the integral x03.2 f x d x. 25–26 25. 5.6 5.0 4.8 31. (a) Use the Midpoint Rule and the given data to estimate the x 3 d x. 1 7 .2 6.8 ■ ■ ■ 29. Estimate the area under the graph in the figure by using (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule, each with n 4. 8 4 0 2 4 6t y 34. Water leaked from a tank at a rate of r t liters per hour, where the graph of r is as shown. Use Simpson’s Rule to estimate the total amount of water that leaked out during the first six hours. r 4 1 0 1 2 3 4x 30. The widths (in meters) of a kidney-shaped swimming pool were measured at 2-meter intervals as indicated in the figure. 2 0 2 4 6t 5E-08(pp 560-569) 1/17/06 5:57 PM Page 565 S ECTION 8.7 APPROXIMATE INTEGRATION 35. The table (supplied by San Diego Gas and Electric) gives the power consumption in megawatts in San Diego County from midnight to 6:00 A.M. on December 8, 1999. Use Simpson’s Rule to estimate the energy used during that time period. (Use the fact that power is the derivative of energy.) t P t 1814 1735 1686 1646 1637 1609 1604 3:30 4:00 4:30 5:00 5:30 6:00 565 39. The region bounded by the curves y 3 s1 x 3, y 0, x 0, and x 2 is rotated about the x-axis. Use Simpson’s Rule with n 10 to estimate the volume of the resulting solid. CAS 40. The figure shows a pendulum with length L that makes a maxi- mum angle 0 with the vertical. Using Newton’s Second Law it can be shown that the period T (the time for one complete swing) is given by P 0:00 0:30 1:00 1:30 2:00 2:30 3:00 ❙❙❙❙ 1611 1621 1666 1745 1886 2052 T where k sin( 1 2 If L 1 m and find the period. 4 0 L t y 2 0 s1 dx k 2 sin 2x ) and t is the acceleration due to gravity. 42 , use Simpson’s Rule with n 0 10 to 36. Shown is the graph of traffic on an Internet service provider’s T1 data line from midnight to 8:00 A.M. D is the data throughput, measured in megabits per second. Use Simpson’s Rule to estimate the total amount of data transmitted during that time period. ¨¸ D 0.8 41. The intensity of light with wavelength traveling through a diffraction grating with N slits at an angle is given by I N 2 sin 2k k 2, where k Nd sin and d is the distance between adjacent slits. A helium-neon laser with wavelength 632.8 10 9 m is emitting a narrow band of light, given by 10 6 10 6, through a grating with 4 10,000 slits spaced 10 m apart. Use the Midpoint Rule with n 10 to estimate the total light intensity x10 0 I d emerg1 ing from the grating. 0.4 6 0 2 4 6 8 t (hours) 6 37. If the region shown in the figure is rotated about the y-axis to form a solid, use Simpson’s Rule with n volume of the solid. 8 to estimate the 42. Use the Trapezoidal Rule with n 10 to approximate x dx. Compare your result to the actual value. Can you explain the discrepancy? x020 cos 43. Sketch the graph of a continuous function on 0, 2 for which y 4 the Trapezoidal Rule with n Midpoint Rule. 2 is more accurate than the 44. Sketch the graph of a continuous function on 0, 2 for which 2 the right endpoint approximation with n than Simpson’s Rule. 0 2 4 6 10 x 8 45. If f is a positive function and f x 2 is more accurate 0 for a x b, show that 38. The table shows values of a force function f x where x is measured in meters and f x in newtons. Use Simpson’s Rule to estimate the work done by the force in moving an object a distance of 18 m. Tn y b a f x dx Mn 46. Show that if f is a polynomial of degree 3 or lower, then b Simpson’s Rule gives the exact value of xa f x d x. 1 2 x 0 3 6 9 12 15 18 47. Show that fx 9.8 9.1 8.5 8.0 7.7 7.5 7.4 48. Show that 3 Tn 1 Tn Mn 2 3 Mn T2 n. S2 n. 5E-08(pp 560-569) 566 ❙❙❙❙ 1/17/06 5:57 PM Page 566 CHAPTER 8 TECHNIQUES OF INTEGRATION |||| 8.8 Improper Integrals In defining a definite integral xab f x d x we dealt with a function f defined on a finite interval a, b and we assumed that f does not have an infinite discontinuity (see Section 5.2). In this section we extend the concept of a definite integral to the case where the interval is infinite and also to the case where f has an infinite discontinuity in a, b . In either case the integral is called an improper integral. One of the most important applications of this idea, probability distributions, will be studied in Section 9.5. Type I: Infinite Intervals Consider the infinite region S that lies under the curve y 1 x 2, above the x-axis, and to the right of the line x 1. You might think that, since S is infinite in extent, its area must be infinite, but let’s take a closer look. The area of the part of S that lies to the left of the line x t (shaded in Figure 1) is Try painting a fence that never ends. Resources / Module 6 / How To Calculate / Start of Improper Integrals y At Notice that A t t 1 1 dx x2 1 x t 1 t 1 1 1 no matter how large t is chosen. y y= 1 ≈ area=1 - x=1 0 FIGURE 1 1 t t 1 x We also observe that lim A t 1 t lim 1 tl tl The area of the shaded region approaches 1 as t l of the infinite region S is equal to 1 and we write y 1 y 1 2 x t 1 1 3 x (see Figure 2), so we say that the area 1 dx x2 1 y area= 4 5 area= 2 3 0 lim y tl y y 1 area= 2 0 1 dx x2 1 0 1 area=1 5x 0 1 x FIGURE 2 Using this example as a guide, we define the integral of f (not necessarily a positive function) over an infinite interval as the limit of integrals over finite intervals. 5E-08(pp 560-569) 1/17/06 5:57 PM Page 567 SECTION 8.8 IMPROPER INTEGRALS 1 ❙❙❙❙ 567 Definition of an Improper Integral of Type 1 (a) If xat f x d x exists for every number t y a a, then t lim y f x d x f x dx tl a provided this limit exists (as a finite number). (b) If xtb f x d x exists for every number t b, then y b f x dx lim tl y b t f x dx provided this limit exists (as a finite number). The improper integrals xa f x d x and xb f x d x are called convergent if the corresponding limit exists and divergent if the limit does not exist. (c) If both xa f x d x and xa f x d x are convergent, then we define y f x dx y a f x dx y a f x dx In part (c) any real number a can be used (see Exercise 74). Any of the improper integrals in Definition 1 can be interpreted as an area provided that f is a positive function. For instance, in case (a) if f x 0 and the integral xa f x d x is convergent, then we define the area of the region S x, y x a, 0 y f x in Figure 3 to be y AS a f x dx y y=ƒ S FIGURE 3 0 a x This is appropriate because xa f x d x is the limit as t l f from a to t. EXAMPLE 1 Determine whether the integral x1 of the area under the graph of 1 x d x is convergent or divergent. SOLUTION According to part (a) of Definition 1, we have y 1 1 dx x lim y tl t 1 1 dx x lim ln t tl lim ln x tl ln 1 t 1 lim ln t tl The limit does not exist as a finite number and so the improper integral x1 1 x d x is divergent. 5E-08(pp 560-569) 568 ❙❙❙❙ 1/17/06 5:57 PM Page 568 CHAPTER 8 TECHNIQUES OF INTEGRATION y y= Let’s compare the result of Example 1 with the example given at the beginning of this section: 1 ≈ 1 d x converges x2 y 1 finite area 0 x 1 FIGURE 4 y= 1 x 1 Geometrically, this says that although the curves y 1 x 2 and y 1 x look very similar for x 0, the region under y 1 x 2 to the right of x 1 (the shaded region in Figure 4) has finite area whereas the corresponding region under y 1 x (in Figure 5) has infinite area. Note that both 1 x 2 and 1 x approach 0 as x l but 1 x 2 approaches 0 faster than 1 x. The values of 1 x don’t decrease fast enough for its integral to have a finite value. EXAMPLE 2 Evaluate y y 0 xe x d x. SOLUTION Using part (b) of Definition 1, we have y 0 xe x d x We integrate by parts with u 1 y lim tl infinite area 0 1 d x diverges x y 0 xe x d x t e x d x so that du x, dv e x: d x, v x y 0 xe x d x xe x 0 FIGURE 5 We know that e t l 0 as t l t y te t t 0 1 t e x dx et , and by l’Hospital’s Rule we have lim te t lim tl tl t et et lim tl 1 e lim tl t 0 Therefore y 0 xe x d x te t lim tl 0 EXAMPLE 3 Evaluate 1 y x2 1 1 1 x 2 0 1 d x. SOLUTION It’s convenient to choose a y 1 et 1 dx 0 in Definition 1(c): y 1 0 1 x 2 dx y 0 1 x2 1 dx We must now evaluate the integrals on the right side separately: y 0 1 1 x 2 dx lim y tl dx t 0 1 lim tan 1t tl x 2 lim tan 1x tl tan 1 0 t 0 lim tan 1t tl 2 5E-08(pp 560-569) 1/17/06 5:58 PM Page 569 SECTION 8.8 IMPROPER INTEGRALS y 1 0 1 x2 dx dx 0 y lim tl tl lim tan 1 0 569 0 t tan 1t tl 0 lim tan 1x x2 1 t ❙❙❙❙ 2 2 Since both of these integrals are convergent, the given integral is convergent and 1 y= 1+≈ y 0 FIGURE 6 1 y area=π x dx x2 1 2 2 0, the given improper integral can be interpreted as the area of Since 1 1 x 2 the infinite region that lies under the curve y 1 1 x 2 and above the x-axis (see Figure 6). E XAMPLE 4 For what values of p is the integral y 1 convergent? 1 dx xp 1, then the integral is divergent, so let’s SOLUTION We know from Example 1 that if p assume that p 1. Then 1 t im y1 x p d x tll y1 x p d x tl lim tl If p 1, then p 1 0, so as t l , t p y 1 1 dx xp 1 tp 1 t1 1 1 1 pt l 1 if p 1 1 1, then p l 1 p1 and 1 t p 1 p x1 1 p and so the integral converges. But if p xt xp1 p1 lim 1 l 0. Therefore 0 and so as t l and the integral diverges. We summarize the result of Example 4 for future reference: 2 y 1 1 d x is convergent if p xp 1 and divergent if p 1. Type 2: Discontinuous Integrands Suppose that f is a positive continuous function defined on a finite interval a, b but has a vertical asymptote at b. Let S be the unbounded region under the graph of f and above the x-axis between a and b. (For Type 1 integrals, the regions extended indefinitely in a 5E-08(pp 570-579) 570 ❙❙❙❙ 1/17/06 5:40 PM Page 570 CHAPTER 8 TECHNIQUES OF INTEGRATION y horizontal direction. Here the region is infinite in a vertical direction.) The area of the part of S between a and t (the shaded region in Figure 7) is y=ƒ 0 x=b a x tb y At t a f x dx If it happens that A t approaches a definite number A as t l b , then we say that the area of the region S is A and we write FIGURE 7 y b a f x dx lim y t a tlb f x dx We use this equation to define an improper integral of Type 2 even when f is not a positive function, no matter what type of discontinuity f has at b. 3 Definition of an Improper Integral of Type 2 (a) If f is continuous on a, b and is discontinuous at b, then |||| Parts (b) and (c) of Definition 3 are illustrated in Figures 8 and 9 for the case where f x 0 and f has vertical asymptotes at a and c, respectively. y b a f x dx lim y t a tlb f x dx if this limit exists (as a finite number). (b) If f is continuous on a, b and is discontinuous at a, then y y b a f x dx lim tla y b t f x dx if this limit exists (as a finite number). 0 at b The improper integral xab f x d x is called convergent if the corresponding limit exists and divergent if the limit does not exist. x y y b a EXAMPLE 5 Find 0 b, and both xac f x d x and (c) If f has a discontinuity at c, where a c xcb f x d x are convergent, then we define FIGURE 8 a bx c y 1 5 2 sx y c a f x dx y b c f x dx d x. 2 SOLUTION We note first that the given integral is improper because f x 1 sx 2 has the vertical asymptote x 2. Since the infinite discontinuity occurs at the left endpoint of 2, 5 , we use part (b) of Definition 3: FIGURE 9 y 5 2 y y= f x dx 1 œ„„„„ x-2 dx sx 2 lim t l2 y t 5 dx sx 2 5 lim 2 s x 2 lim 2(s3 st t l2 t l2 3 area=2œ„ 0 1 FIGURE 10 2 3 4 t 2) 2 s3 5 x Thus, the given improper integral is convergent and, since the integrand is positive, we can interpret the value of the integral as the area of the shaded region in Figure 10. 5E-08(pp 570-579) 1/17/06 5:41 PM Page 571 S ECTION 8.8 IMPROPER INTEGRALS y EXAMPLE 6 Determine whether 0 /2 /2 0 sec x d x y lim tl /2 t 0 tan x lim and tan t l E XAMPLE 7 Evaluate y dx 3 x 0 tan t /2 tl ln sec t /2 as t l . Using sec x d x lim ln sec x tl because sec t l divergent. 571 sec x d x converges or diverges. SOLUTION Note that the given integral is improper because lim x l / 2 sec x part (a) of Definition 3 and Formula 14 from the Table of Integrals, we have y ❙❙❙❙ t 0 ln 1 2 . Thus, the given improper integral is if possible. 1 1 is a vertical asymptote of the integrand. Since it SOLUTION Observe that the line x occurs in the middle of the interval 0, 3 , we must use part (c) of Definition 3 with c 1: 3 dx 1 dx 3 dx y0 x 1 y0 x 1 y1 x 1 y where dx 1 0 x lim 1 t l1 y dx t 0 x lim ln x 1 lim (ln t 1 t l1 lim ln 1 1 t l1 ln 1 t 0 ) t t l1 because 1 t l 0 as t l 1 . Thus, x01 d x x 1 is divergent. This implies that x03 d x x 1 is divergent. [We do not need to evaluate x13 d x x 1 .] | WARNING If we had not noticed the asymptote x 1 in Example 7 and had instead confused the integral with an ordinary integral, then we might have made the following erroneous calculation: ■ y dx 3 0 x ln x 1 1 3 0 ln 2 ln 1 ln 2 This is wrong because the integral is improper and must be calculated in terms of limits. From now on, whenever you meet the symbol xab f x d x you must decide, by looking at the function f on a, b , whether it is an ordinary definite integral or an improper integral. EXAMPLE 8 Evaluate y 1 0 ln x d x. SOLUTION We know that the function f x lim x l 0 ln x ln x has a vertical asymptote at 0 since . Thus, the given integral is improper and we have y 1 0 ln x d x lim tl0 y t 1 ln x d x 5E-08(pp 570-579) 572 ❙❙❙❙ 1/17/06 5:42 PM Page 572 CHAPTER 8 TECHNIQUES OF INTEGRATION Now we integrate by parts with u y 1 t ln x, dv ln x d x d x, du x ln x 1 y t 1 t ln t t ln t x: dx t 1 ln 1 d x x, and v 1 1 t t To find the limit of the first term we use l’Hospital’s Rule: lim t ln t ln t 1t lim tl0 tl0 y 1t 1 t2 lim tl0 0 1 lim x t tl0 area=1 y Therefore 1 0 ln x d x lim t ln t tl0 0 y=ln x 0 1 1 0 t 1 Figure 11 shows the geometric interpretation of this result. The area of the shaded region above y ln x and below the x-axis is 1. FIGURE 11 A C omparison Test for Improper Integrals Sometimes it is impossible to find the exact value of an improper integral and yet it is important to know whether it is convergent or divergent. In such cases the following theorem is useful. Although we state it for Type 1 integrals, a similar theorem is true for Type 2 integrals. Comparison Theorem Suppose that f and t are continuous functions with fx tx 0 for x a. (a) If xa f x d x is convergent, then xa t x d x is convergent. (b) If xa t x d x is divergent, then xa f x d x is divergent. y f g 0 a FIGURE 12 x We omit the proof of the Comparison Theorem, but Figure 12 makes it seem plausible. If the area under the top curve y f x is finite, then so is the area under the bottom curve y t x . And if the area under y t x is infinite, then so is the area under y f x . [Note that the reverse is not necessarily true: If xa t x dx is convergent, xa f x dx may or may not be convergent, and if xa f x dx is divergent, xa t x dx may or may not be divergent.] EXAMPLE 9 Show that y 0 e x2 d x is convergent. SOLUTION We can’t evaluate the integral directly because the antiderivative of e an elementary function (as explained in Section 8.5). We write y 0 e x2 dx y 1 0 e x2 dx y 1 e x2 dx x2 is not 5E-08(pp 570-579) 1/17/06 5:42 PM Page 573 S ECTION 8.8 IMPROPER INTEGRALS y y=e 573 and observe that the first integral on the right-hand side is just an ordinary definite integral. In the second integral we use the fact that for x 1 we have x 2 x, so x 2 x 2 and therefore e x e x. (See Figure 13.) The integral of e x is easy to evaluate: _x 2 y=e_x t e x dx y 1 0 ❙❙❙❙ lim y e x d x tl x 1 1 lim e 1 e tl FIGURE 13 t e 1 2 Thus, taking f x e x and t x e x in the Comparison Theorem, we see that 2 x2 x1 e d x is convergent. It follows that x0 e x d x is convergent. 2 In Example 9 we showed that x0 e x d x is convergent without computing its value. In Exercise 70 we indicate how to show that its value is approximately 0.8862. In probability theory it is important to know the exact value of this improper integral, as we will see in Section 9.5; using the methods of multivariable calculus it can be shown that the exact value is s 2. Table 1 illustrates the definition of an improper integral by showing how 2 the (computer-generated) values of x0t e x d x approach s 2 as t becomes large. In fact, 2 these values converge quite quickly because e x l 0 very rapidly as x l . TABLE 1 x0t e t 1 2 3 4 5 6 x2 dx 0.7468241328 0.8820813908 0.8862073483 0.8862269118 0.8862269255 0.8862269255 EXAMPLE 10 The integral because y 1 1 e x d x is divergent by the Comparison Theorem x 1 e x x 1 x and x1 1 x d x is divergent by Example 1 [or by (2) with p 1]. Table 2 illustrates the divergence of the integral in Example 10. It appears that the values are not approaching any fixed number. TABLE 2 x1t t 2 5 10 100 1000 10000 |||| 8.8 y (c) y 1 x 4e 2 0 e x x dx 0.8636306042 1.8276735512 2.5219648704 4.8245541204 7.1271392134 9.4297243064 Exercises 1. Explain why each of the following integrals is improper. (a) 1 x2 x 4 (b) dx x 5x 6 dx y (d) y 2 0 sec x dx 1 0 x2 5 dx 2. Which of the following integrals are improper? Why? (a) y (c) y 2 1 1 dx 2x 1 sin x dx 1 x2 (b) y (d) y 1 1 0 2 1 2x ln x 1 dx 1 dx 5E-08(pp 570-579) ❙❙❙❙ 574 1/17/06 5:44 PM Page 574 CHAPTER 8 TECHNIQUES OF INTEGRATION 1 x 3 from x 1 to x t and evaluate it for t 10, 100, and 1000. Then find the total area under this curve for x 1. 3. Find the area under the curve y 1 x 1.1 and t x 1 x 0.9 in the viewing rectangles 0, 10 by 0, 1 and 0, 100 by 0, 1 . (b) Find the areas under the graphs of f and t from x 1 to x t and evaluate for t 10, 100, 10 4, 10 6, 10 10, and 10 20. (c) Find the total area under each curve for x 1, if it exists. ; 4. (a) Graph the functions f x 39. y 2 z 2 ln z dz 0 ■ ■ 41–46 ■ |||| 40. ■ ■ ■ y 1 0 ■ ln x dx sx ■ ■ 5. 1 y 3x 1 1 2 1 1 7. y 9. y 11. y 13. y xe 15. y 17. dx 6. y 1 0 2x 5 y 10. y 12. y 2 14. y x 2e sin d 16. y y x x2 1 dx 2x 18. y 19. y se ds 20. y 21. y ln x dx x 22. y 23. s2 x 1 0 1 29. 31. 33. y 35. y 0 1 3 2 33 y 37. dx x6 24. dx 0 0 1 dx x4 x 32. 1 15 ex e x 1 dx x, y x 0, 0 y x, y 0 x 2, 0 y sec 2x 0, 0 y 1 sx ■ ■ dx dz 3z 2 dx y 1 53. y 0 y 9 1 y y y 38. y 2 0 9 ■ 2} ■ ■ ■ ■ cos 2x dx 1 x2 1 50. y dx e 2x 52. y dx x sin x 54. y x 1 /2 0 ■ ■ ■ ■ ■ ■ 2 e ■ x dx x 1 x x6 s1 1 1 0 ■ dx ex dx sx ■ ■ ■ 55. The integral y 0 1 dy 4y 1 x2 1 x x 2x 3 dx 3 4 0 dx x2 s1 1 0 9 dx 1 0 34. 3 sx x x2 |||| Use the Comparison Theorem to determine whether the integral is convergent or divergent. 51. 3 ■ 9 49–54 1 dx x sx 0 2 x2 x x /2 e 1 (s x 1), use your calculator or computer to make a table of approximate values of x2t t x d x for t 5, 10, 100, 1000, and 10,000. Does it appear that x2 t x d x is convergent or divergent? (b) Use the Comparison Theorem with f x 1 s x to show that x2 t x d x is divergent. (c) Illustrate part (b) by graphing f and t on the same screen for 2 x 20. Use your graph to explain intuitively why x2 t x d x is divergent. re r 3 dr x 2 ■ y ; 48. (a) If t x cos 2 d e 2, 0 sin 2 x x 2, use your calculator or computer to make a table of approximate values of x1t t x d x for t 2, 5, 10, 100, 1000, and 10,000. Does it appear that x1 t x d x is convergent? (b) Use the Comparison Theorem with f x 1 x 2 to show that x1 t x d x is convergent. (c) Illustrate part (b) by graphing f and t on the same screen for 1 x 10. Use your graph to explain intuitively why x1 t x d x is convergent. v 4 dv z2 ■ y ; 47. (a) If t x dt x3 1, 0 { x, y ■ y 36. dx ; 44. S 49. y 30. y x arctan x dx 1 x2 2 1 28. 1 dx x2 x, y 0 y y 1 1 6 x, y x ; 46. S dx 2 2 2t e 42. S ln x dx x3 26. sec x d x 0 1 0 1 dx sx 3 0 y 5s dx x2 0 ln x dx x2 1 y x2 9 y y x2 x2 y 27. dy 1 2 25. y2 e 4 w x, y x ■ x 8. dw dx ■ ex 41. S ; 45. S Determine whether each integral is convergent or divergent. Evaluate those that are convergent. |||| ■ Sketch the region and find its area (if the area is finite). ; 43. S 5–40 ■ 6 dx 1 sx 1 x dx is improper for two reasons: The interval 0, is infinite and the integrand has an infinite discontinuity at 0. Evaluate it by expressing it as a sum of improper integrals of Type 2 and Type 1 as follows: 1 1 1 1 y0 s x 1 x d x y0 s x 1 x d x y1 s x 1 x d x ■ 5E-08(pp 570-579) 1/17/06 5:45 PM Page 575 S ECTION 8.8 IMPROPER INTEGRALS 56. Evaluate 1 x sx 2 y 2 by the same method as in Exercise 55. 57–59 |||| Find the values of p for which the integral converges and evaluate the integral for those values of p. 57. y 59. y ■ 1 dx xp 1 0 1 1 x ln x y 58. e p ■ ■ y ys R s 2r x r dr sr 2 s 2 If the actual density of stars in a cluster is x r find the perceived density y s . dx 1 2 R r 2, 67. A manufacturer of lightbulbs wants to produce bulbs that last x p ln x d x 0 575 a photograph. Suppose that in a spherical cluster of radius R the density of stars depends only on the distance r from the center of the cluster. If the perceived star density is given by y s , where s is the observed planar distance from the center of the cluster, and x r is the actual density, it can be shown that dx 4 ❙❙❙❙ ■ ■ ■ ■ ■ ■ ■ ■ 60. (a) Evaluate the integral x0 x ne x d x for n 0, 1, 2, and 3. (b) Guess the value of x0 x ne x d x when n is an arbitrary positive integer. (c) Prove your guess using mathematical induction. 61. (a) Show that x x d x is divergent. about 700 hours but, of course, some bulbs burn out faster than others. Let F t be the fraction of the company’s bulbs that burn out before t hours, so F t always lies between 0 and 1. (a) Make a rough sketch of what you think the graph of F might look like. (b) What is the meaning of the derivative r t F t? (c) What is the value of x0 r t dt ? Why? 68. As we will see in Section 10.4, a radioactive substance decays (b) Show that t lim y x d x tl ■ exponentially: The mass at time t is m t m 0 e kt, where m 0 is the initial mass and k is a negative constant. The mean life M of an atom in the substance is 0 t This shows that we can’t define k y te kt dt M y f x dx lim y f x d x tl t 62. The average speed of molecules in an ideal gas is v 4 s M 2RT 0 t 32 y 0 v 3e Mv 2 2RT dv where M is the molecular weight of the gas, R is the gas constant, T is the gas temperature, and v is the molecular speed. Show that For the radioactive carbon isotope, 14 C, used in radiocarbon dating, the value of k is 0.000121. Find the mean life of a 14 C atom. 69. Determine how large the number a has to be so that y 1 a x2 1 v 63. We know from Example 1 that the region x, y x that by rotating finite volume. 1, 0 y 1 x has infinite area. Show about the x-axis we obtain a solid with 65. Find the escape velocity v0 that is needed to propel a rocket of mass m out of the gravitational field of a planet with mass M and radius R. Use Newton’s Law of Gravitation (see Exercise 29 in Section 6.4) and the fact that the initial kinetic energy of 1 mv 2 supplies the needed work. 0 2 66. Astronomers use a technique called stellar stereography to determine the density of stars in a star cluster from the observed (two-dimensional) density that can be analyzed from x2 d x by writing it as the sum of x04 e x d x and x4 e x d x. Approximate the first integral by using Simpson’s Rule with n 8 and show that the second integral is smaller than x4 e 4 x d x, which is less than 0.0000001. 2 71. If f t is continuous for t 0, the Laplace transform of f is the function F defined by 64. Use the information and data in Exercises 29 and 30 of Sec- tion 6.4 to find the work required to propel a 1000-kg satellite out of Earth’s gravitational field. 0.001 70. Estimate the numerical value of x0 e 2 8RT M dx Fs y 0 f te st dt and the domain of F is the set consisting of all numbers s for which the integral converges. Find the Laplace transforms of the following functions. (a) f t (b) f t (c) f t 1 et t ft Me at for t 0, where M and a are constants, then the Laplace transform F s exists for s a. 72. Show that if 0 ft Me at and 0 f t Ke at for t 0, where f is continuous. If the Laplace transform of f t is F s 73. Suppose that 0 5E-08(pp 570-579) ❙❙❙❙ 576 1/17/06 5:46 PM Page 576 CHAPTER 8 TECHNIQUES OF INTEGRATION 77. Find the value of the constant C for which the integral and the Laplace transform of f t is G s , show that Gs sF s f0 s a y 74. If x f x d x is convergent and a and b are real numbers, show that y a y f x dx a x2 75. Show that x0 x 2e 76. Show that x0 e 1 2 dx x2 dx integrals as areas. |||| y f x dx x0 x01 s e b x2 y f x dx b f x dx y 0 ln y d y by interpreting the ■ CONCEPT CHECK 2. How do you evaluate x sin mx cos nx dx if m is odd? What if n is odd? What if m and n are both even? x 2 occurs in an integral, what substitution might you try? What if sa 2 x 2 occurs? What if sx 2 a 2 occurs? 4. What is the form of the partial fraction expansion of a rational function P x Q x if the degree of P is less than the degree of Q and Q x has only distinct linear factors? What if a linear factor is repeated? What if Q x has an irreducible quadratic factor (not repeated)? What if the quadratic factor is repeated? ■ 4 A can be put in the form 4 x2 B x x x x2 4 A can be put in the form 4 x x x2 xx 4 A can be put in the form 2 4 x x 4. x2 x x2 A 4 can be put in the form 4 x x2 5. y 3. 6. 2 0 y x 4 1 x 2 1 dx 1 2 dx x x2 C 1 3x dx 1 ■ b 5. State the rules for approximating the definite integral xa f x dx with the Midpoint Rule, the Trapezoidal Rule, and Simpson’s Rule. Which would you expect to give the best estimate? How do you approximate the error for each rule? 2 B 4 x 2 . . B 4 a f x dx (b) y b f x dx (c) y f x dx b 7. Define the improper integral xa f x d x for each of the follow- ing cases. (a) f has an infinite discontinuity at a. (b) f has an infinite discontinuity at b. (c) f has an infinite discontinuity at c, where a c b. 8. State the Comparison Theorem for improper integrals. ■ 8. The Midpoint Rule is always more accurate than the Trapezoidal Rule. (b) Every elementary function has an elementary antiderivative. C 2 y 9. (a) Every elementary function has an elementary derivative. . B (a) 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. 2. 2 6. Define the following improper integrals. 3. If the expression sa 2 2 x converges. Evaluate the integral for this value of C. use it? xx x2 C 4 78. Find the value of the constant C for which the integral 1. State the rule for integration by parts. In practice, how do you 1. 1 sx 2 converges. Evaluate the integral for this value of C. d x. 8 Review 2 0 10. If f is continuous on 0, x0 and x1 f x d x is convergent, then f x d x is convergent. 11. If f is a continuous, decreasing function on 1, lim x l f x . and 0 , then x1 f x d x is convergent. 12. If xa f x d x and xa t x d x are both convergent, then xa fx t x d x is convergent. ln 15 13. If xa f x d x and xa t x dx are both divergent, then xa 1 d x is convergent. x s2 7. If f is continuous, then x f x dx lim t l xt t f x d x. fx 14. If f x diverges. t x d x is divergent. t x and x0 t x dx diverges, then x0 f x dx also 5E-08(pp 570-579) 1/17/06 5:48 PM Page 577 CHAPTER 8 REVIEW ■ EXERCISES Note: Additional practice in techniques of integration is provided in Exercises 7.5. 1–40 1. 3. y x 5 x 0 10 cos 1 sin 2 y 0 7 4. d 3 5. y tan x sec x d x 7. sin ln t dt t 9. y y 4 1 6. 8. x 3 2 ln x d x sx 2 x 2 11. y 13. yx 15. y sin 1 1 10. cos 5 d 17. y x sec x tan x d x 19. y 9x x 2 21. y sx 23. y csc 25. y 2 2 4 yy y sin x dx 1 x2 1 y y x2 x y sec 6 d tan 2 1 20. y sin t y ye 26. y1 y 31. y dx ex 33. y 35. y sx 37. y 0 y e s1 32. y dx dx cos x 32 sin x 2 cos 2 x d x y 48. y 50. 4x ■ ■ ■ 0 ■ ■ 1 dt 1 y 6 sy 2 1 1 2 0 1 1 dy 2 3x dx x1 dx 3 sx 4 tan 1x dx x2 1 ■ ■ ■ 52. 2 dx ■ ■ ■ ■ y sx x3 2 1 ■ dx ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ |||| Use the Table of Integrals on the Reference Pages to evaluate the integral. x x sin x dx cos 3 x 57. y sx 34. y arcsin x 2 d x 36. y1 38. y x tan 1 2x ■ 0 tan tan 2 ■ ye 1 5 ■ ■ 55. 4 dx 4x 2 ■ 2x e 2 ■ y ln x dx x t2 t2 1 y dx x x2 dx 3 55–58 ■ 32 1 x 3 46. ■ ■ 1 dx 1 30. e xse x 1 dx ex 8 x2 4 x2 sx ■ 2x d x by hand? (Don’t actually carry out the integration.) (b) How would you evaluate x x 5e 2 x d x using tables? (Don’t actually do it.) (c) Use a CAS to evaluate x x 5e 2 x d x. (d) Graph the integrand and the indefinite integral on the same screen. cos x dx y sx ■ 54. (a) How would you evaluate x x 5e dx x 28. x 5 sec x d x ln 10 10 y 1 3 0 ■ 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. CAS 1 44. ln x dx sx 4 ■ ; 53. Graph the function f x cos 2t 3 x y dx x ln x 0 ■ dt x 24. cos 3x sin 2 x d x 51. 2 ■ Evaluate the integral or show that it is divergent. 2x 2 ■ ■ stan d sin 2 |||| Evaluate the indefinite integral. Illustrate and check that your answer is reasonable by graphing both the function and its antiderivative (take C 0). x 2 8x 3 dx x 3 3x 2 y y ■ 4 ; 51–52 2 dx 2 18. 22. x 2 6x 4 dx 1 x2 2 49. sarctan x dx 1 x2 1 0 y ■ 3 y 42. 1 dy dx x 2s1 x 2 y y 47. 12 40. dx ■ 1 y 45. |||| y 43. 1 4y 2 ■ 3 1 2 ■ 41. 2t 1 xe 2x 1 2x 12 0 41–50 dy dt 4 3 29. 1 dx 4x dx y 1 5 4x 27. 0 1 6x dx 3x 3 x2 y 0.6 y ye 16. x 2 5 0 14. dx 3 y 12. dx y ■ 2. dx 577 ■ 39. Evaluate the integral. |||| ❙❙❙❙ ■ 2 x ■ ■ ■ y csc t dt 58. 1 dx ■ 5 56. e 2x dx s1 y s1 ■ cot x dx 2 sin x ■ ■ ■ 59. Verify Formula 33 in the Table of Integrals (a) by differentia- tion and (b) by using a trigonometric substitution. d x 2 dx ■ 60. Verify Formula 62 in the Table of Integrals. 61. Is it possible to find a number n such that x0 x n d x is convergent? ■ 5E-08(pp 570-579) 578 ❙❙❙❙ 1/17/06 5:49 PM Page 578 CHAPTER 8 TECHNIQUES OF INTEGRATION 62. For what values of a is x0 e ax cos x dx convergent? Evaluate the integral for those values of a. point to be 53 cm. The circumference 7 cm from each end is 45 cm. Use Simpson’s Rule to make your estimate. 63–64 |||| Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule with n 10 to approximate the given integral. Round your answers to six decimal places. 63. y 1 64. x 4 dx s1 0 2 y 0 ssin x d x 28 cm ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 65. Estimate the errors involved in Exercise 63, parts (a) and (b). How large should n be in each case to guarantee an error of less than 0.00001? 71. Use the Comparison Theorem to determine whether the integral x3 y x is convergent or divergent. 1 66. Use Simpson’s Rule with n e x x from x curve y 6 to estimate the area under the 1 to x 4. 67. The speedometer reading (v) on a car was observed at 1-minute intervals and recorded in the chart. Use Simpson’s Rule to estimate the distance traveled by the car. t (min) v (mi h) t (min) v (mi h) 0 1 2 3 4 5 40 42 45 49 52 54 6 7 8 9 10 56 57 57 55 56 week, where the graph of r is as shown. Use Simpson’s Rule with six subintervals to estimate the increase in the bee population during the first 24 weeks. r 12000 4000 C AS dx y2 x2 1 and the line y 3. 73. Find the area bounded by the curves y between x 0 and x cos x and y cos 2x . 74. Find the area of the region bounded by the curves y 1 (2 s x ), y 1 (2 s x ), and x 75. The region under the curve y 1. 2 cos x, 0 x 2, is rotated about the x-axis. Find the volume of the resulting solid. 76. The region in Exercise 75 is rotated about the y-axis. Find the volume of the resulting solid. y and lim x l f x f x dx 0 0, show that f0 78. We can extend our definition of average value of a continuous function to an infinite interval by defining the average value of to be f on the interval a, 1 t lim y f x dx tl t aa (a) Find the average value of y tan 1x on the interval 0, . (b) If f x 0 and xa f x dx is divergent, show that the average value of f on the interval a, is lim x l f x , if this limit exists. (c) If xa f x dx is convergent, what is the average value of f on the interval a, ? (d) Find the average value of y sin x on the interval 0, . 8000 0 2 72. Find the area of the region bounded by the hyperbola 77. If f is continuous on 0, 68. A population of honeybees increased at a rate of r t bees per 5 4 8 12 16 20 t 24 (weeks) 69. (a) If f x sin sin x , use a graph to find an upper bound for f 4 x . (b) Use Simpson’s Rule with n 10 to approximate x0 f x d x and use part (a) to estimate the error. (c) How large should n be to guarantee that the size of the error in using Sn is less than 0.00001? 70. Suppose you are asked to estimate the volume of a football. You measure and find that a football is 28 cm long. You use a piece of string and measure the circumference at its widest 79. Use the substitution u y 0 1 x to show that 1 ln x dx x2 0 80. The magnitude of the repulsive force between two point charges with the same sign, one of size 1 and the other of size q, is q F 4 0r 2 where r is the distance between the charges and 0 is a constant. The potential V at a point P due to the charge q is defined to be the work expended in bringing a unit charge to P from infinity along the straight line that joins q and P. Find a formula for V . 5E-08(pp 570-579) 1/17/06 5:49 PM Page 579 PROBLEMS PLUS Cover up the solution to the example and try it yourself first. EXAMPLE (a) Prove that if f is a continuous function, then a y 0 y f x dx a 0 fa x dx (b) Use part (a) to show that 2 y 0 sin n x dx sin x cos n x n 4 for all positive numbers n. SOLUTION |||| The principles of problem solving are discussed on page 58. (a) At first sight, the given equation may appear somewhat baffling. How is it possible to connect the left side to the right side? Connections can often be made through one of the principles of problem solving: introduce something extra. Here the extra ingredient is a new variable. We often think of introducing a new variable when we use the Substitution Rule to integrate a specific function. But that technique is still useful in the present circumstance in which we have a general function f . Once we think of making a substitution, the form of the right side suggests that it should be u a x. Then du d x. When x 0, u a; when x a, u 0. So y a 0 fa y x dx 0 a y f u du a 0 f u du But this integral on the right side is just another way of writing x0a f x d x. So the given equation is proved. (b) If we let the given integral be I and apply part (a) with a 2, we get I |||| The computer graphs in Figure 1 make it seem plausible that all of the integrals in the example have the same value. The graph of each integrand is labeled with the corresponding value of n. 1 3 4 2 y 0 2 sin n x dx sin n x cos n x FIGURE 1 0 sin n sin n 2x A well-known trigonometric identity tells us that sin cos 2x sin x, so we get I y 0 2 2 2x cos n x 2 x dx cos x and cos n x dx cos n x sin n x Notice that the two expressions for I are very similar. In fact, the integrands have the same denominator. This suggests that we should add the two expressions. If we do so, we get 1 2I 0 y 2 π 2 Therefore, I y 0 2 sin n x sin n x cos n x dx cos n x y 0 2 1 dx 2 4. 579 5E-08(pp 580-581) 1/17/06 5:38 PM Page 580 ; 1. Three mathematics students have ordered a 14-inch pizza. Instead of slicing it in the tradi- P RO B L E M S tional way, they decide to slice it by parallel cuts, as shown in the figure. Being mathematics majors, they are able to determine where to slice so that each gets the same amount of pizza. Where are the cuts made? 1 d x. x7 x The straightforward approach would be to start with partial fractions, but that would be brutal. Try a substitution. 2. Evaluate y 1 3 3. Evaluate y (s1 0 x 3 ) dx. 7 s1 x7 4. A man initially standing at the point O walks along a pier pulling a rowboat by a rope of 14 in length L. The man keeps the rope straight and taut. The path followed by the boat is a curve called a tractrix and it has the property that the rope is always tangent to the curve (see the figure). (a) Show that if the path followed by the boat is the graph of the function y f x , then FIGURE FOR PROBLEM 1 y pier (b) Determine the function y L sL 2 x dy dx fx x2 f x. (x, y) 5. A function f is defined by (L, 0) O FIGURE FOR PROBLEM 4 y fx x cos t cos x 0 0 t dt x 2 Find the minimum value of f . 6. If n is a positive integer, prove that y 1 0 ln x n d x 1 n n! 7. Show that y 1 0 x 2 n dx 1 2 2 n n! 2 2n 1 ! Hint: Start by showing that if In denotes the integral, then Ik 2k 2k 1 2 Ik 3 ; 8. Suppose that f is a positive function such that f is continuous. (a) How is the graph of y f x sin n x related to the graph of y as n l ? (b) Make a guess as to the value of the limit lim nl y 1 0 f x ? What happens f x sin nx dx based on graphs of the integrand. (c) Using integration by parts, confirm the guess that you made in part (b). [Use the fact that, since f is continuous, there is a constant M such that f x M for 0 x 1.] 9. If 0 580 a b, find lim tl0 y 1 0 1t bx a1 x t dx . 5E-08(pp 580-581) 1/17/06 5:39 PM Page 581 sin e x and use the graph to estimate the value of t such that xtt maximum. Then find the exact value of t that maximizes this integral. ; 10. Graph f x y 11. The circle with radius 1 shown in the figure touches the curve y 1 f x d x is a 2 x twice. Find the area of the region that lies between the two curves. 12. A rocket is fired straight up, burning fuel at the constant rate of b kilograms per second. Let v v t be the velocity of the rocket at time t and suppose that the velocity u of the exhaust gas is constant. Let M M t be the mass of the rocket at time t and note that M decreases as the fuel burns. If we neglect air resistance, it follows from Newton’s Second Law that y=| 2x | 0 F IGURE FOR PROBLEM 11 x F where the force F M dv dt ub M t. Thus 1 M dv dt ub Mt Let M1 be the mass of the rocket without fuel, M2 the initial mass of the fuel, and M0 M1 M2 . Then, until the fuel runs out at time t M2 b, the mass is M M0 bt. (a) Substitute M M0 b t into Equation 1 and solve the resulting equation for v. Use the initial condition v 0 0 to evaluate the constant. (b) Determine the velocity of the rocket at time t M2 b. This is called the burnout velocity. (c) Determine the height of the rocket y y t at the burnout time. (d) Find the height of the rocket at any time t. 13. Use integration by parts to show that, for all x 0 y 0 ln 1 sin t x 0, t dt 2 ln 1 x ; 14. The Chebyshev polynomials Tn are defined by Tn x cos n arccos x 0, 1, 2, 3, . . . n (a) What are the domain and range of these functions? (b) We know that T0 x 1 and T1 x x. Express T2 explicitly as a quadratic polynomial and T3 as a cubic polynomial. (c) Show that, for n 1, Tn 1 x 2 x Tn x Tn 1 x (d) Use part (c) to show that Tn is a polynomial of degree n. (e) Use parts (b) and (c) to express T4 , T5 , T6 , and T7 explicitly as polynomials. (f) What are the zeros of Tn ? At what numbers does Tn have local maximum and minimum values? (g) Graph T2 , T3 , T4 , and T5 on a common screen. (h) Graph T5 , T6 , and T7 on a common screen. (i) Based on your observations from parts (g) and (h), how are the zeros of Tn related to the zeros of Tn 1 ? What about the x-coordinates of the maximum and minimum values? (j) Based on your graphs in parts (g) and (h), what can you say about x1 1 Tn x d x when n is odd and when n is even? (k) Use the substitution u arccos x to evaluate the integral in part (j). (l) The family of functions f x cos c arccos x are defined even when c is not an integer (but then f is not a polynomial). Describe how the graph of f changes as c increases. 581 ...
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