05 - CHAPTER Integration 5 Section 5.1 Antiderivatives and...

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Unformatted text preview: CHAPTER Integration 5 Section 5.1 Antiderivatives and Indefinite Integration . . . . . . . . . 395 Section 5.2 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 Section 5.3 Reimann Sums and Definite Integrals . . . . . . . . . . . 420 Section 5.4 The Fundamental Theorem of Calculus . . . . . . . . . . 429 Section 5.5 Integration by Substitution . . . . . . . . . . . . . . . . . 441 Section 5.6 Numerical Integration Section 5.7 The Natural Logarithmic Function: Integration . . . . . . 470 Section 5.8 Inverse Trigonometric Functions: Integration . . . . . . . 480 Section 5.9 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . 489 Review Exercises . . . . . . . . . . . . . . . . . . . 460 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510 CHAPTER Integration Section 5.1 1. d3 dx x 3 3. d 13 x dx 3 5 Antiderivatives and Indefinite Integration d 3x dx 3 C C x2 4x C 9x 4 x 4 9 x4 2. d4 x dx 2 4. d 2 x2 3 dx 3x 2x 1 x 1 x2 4x3 C d 23 x dx 3 C x1 dy dt 3t2 y 5. t3 dy dx x3 2 y 7. 25 x 5 Check: 10. 11. 3 2 2 dx x x2 13. 1 dx 2x3 3 dx 1 dx 3x 2 d d 2 x 32 dx x3 dx 3x dx C x2 1 x3 2 2x 2 C 2 Check: C 34 x 4 C 1 x 1 d dx x 2 1 x2 C C 2x 3 Simplify x4 3 C 3 2x Integrate x1 3 dx x 12. 14. x3 C Rewrite 1 dx x2 xx dy dx y 2 32 12 C Check: 3t2 C d 25 x dx 5 x dx 1 C 8. Given 9. r C d3 t dt x 2x dr d 6. Check: 2 2 43 1 x 1 12 x 12 x4 4 3 3 dx 1x2 22 C 1 x 9 2 dx 1x1 91 C C C x C 1 x 2 C 2 C x2 2 3 14 x 4 32 x 2 1 4x2 1 9x C C C 395 396 15. Chapter 5 x Integration x2 2 3 dx 3x 16. C Check: 17. d x2 dx 2 3x C x x3 5 dx 14 x 4 5x 19. x3 2 2x Check: 21. Check: 24. x2 x2 1 2x x4 3 d dx x 2 2x1 1 x 1 3x 2 x3 x 2 2 2x 12 x3 3 3x x x 4 2 2x y2 y dy Check: 29. d 27 y dy 7 dx 1 dx Check: d x dx 2 x2 2 2x C 3x 2 x 1 3x 27 y 7 y5 2 dy C x C C 1 2 2 23 x 3 x1 2 x x 2 C y5 2 1 2x 1 3 3x 2 1 dx x4 x x2 12 2 3x 2 4 x2 2 x x 4 2x 3 4 1 dx 1 x2 6x 2 4 4 1 3x3 4 3x2 5x 15 C 1 x3 1 C C C 3 2t2 1 2 dt 4t4 4t2 d 45 t dt 5 43 t 3 43 t 3 t 1 dt t C 4t 4 1 Check: 3 dt Check: 3t t2 dt t2 d 13 t dt 3 34 t 4 3t d 3t dt 3t3 dt C C C 3 t2 4t2 2t2 C 2 30. x3 1 3 Check: y2 y C 1 x4 45 t 5 28. 1 x 1 C C 1 x C 47 x 7 3 3 2 C 2 x3 C x dx x 4x 3 C x 1 3x3 x x x3 21 x 15 C 3 2x x4 d dx 4 5 x4 1 dx d47 x dx 7 2x1 26. 2 dx x 27. 25 x 5 dx 2x x2 2 d3 x dx x3 C 2x 3 Check: 22. dx x3 Check: 4 1 C C C 3x 2 2 dx 2 2x 1 x3 1 x2 d4 x dx C 1 dx Check: x1 x 6x 2 20. 1 x3 2 dx C 1 2x2 23 x 3 2 x C x3 dx d 25 x dx 5 x2 C C x2 2 Check: 5 2 2 d 5x dx 18. 2 x x dx 1 2x2 Check: 25. 2 3 d dx x x 25 x 5 x2 2 5x 4x3 3 x3 C 1 dx x x2 5x d 25 x dx 5 1 dx x3 Check: 23. d 14 x dx 4 x dx Check: C Check: 5 1 13 t 3 34 t 4 3t3 1 1 2 C 3t t2 S ection 5.1 31. 2 sin x d dx Check: 33. 1 3 cos x dx 2 cos x d t dt d dx Check: 1 sin Check: d tan d tan d 13 t dt 3 cos t C t2 5e x 2 3x2 Check: 38. C sec2 d1 d3 1 3 d 3 d3 x dx 3 sin t tan tan C 2 C x3 2e x dx sec y tan y 2e x 2e x sec y dy 2e x sec2 y dy sec y tan y sec y d sec y dy sec2 C 3x2 C sin tan y C tan2 y Check: sec2 y dy 1 dy d tan y dy tan y sec2 y C 40. C tan2 y 1 tan y cos x dx cos2 x 4x dx x2 4x ln 4 Check: d2 x dx 4x ln 4 C x 5 dx x x2 2 Check: d x2 dx 2 5 ln x 43. 45. f x 5 ln x cos x 42. C 4x 2x C 5 x Check: 46. f x d dx csc x C 6 C sin x 3x ln 3 d sin x dx 3 C 0 2 3x ln 3 C sec2 x dx 4 ln x d 4 ln x dx tan x −4 1 sin x C 3x cos x tan x C C 4 x sec2 x ln x C=3 4 2 C=0 C = −2 4 −2 C cos x cos2 x x x −2 csc x 6 C=3 C=0 2 cos x dx sin x y 4 x 2C 1 sin x csc x cot x 47. f x x 2 3 3 x dx y 3 2 4 x 44. y 2 cos x Check: C x sec y csc x cot x dx 1 2x sec2 y C cos x dx sin2 x 1 Check: 41. sec y tan y sec y tan y Check: 39. 6 8 397 C 36. C sec2 C cos t Check: 2 sin x cos cos 13 t 3 34. 5e x C sin t dt Check: 3 cos x csc t cot t 5e x d 2 sin x t2 32. C C 2 cos x 2 cos x sec2 37. csc t C csc t 3 sin x C t 5e x dx 2 sin x 35. 3 sin x csc t cot t dt Check: 2 cos x Antiderivatives and Indefinite Integration 6 −2 −4 C = −2 8 cos x sin x 398 Chapter 5 Integration 1x e 2 49. f x y 48. f x fx 2 50. f x 2x C 6 C=3 2x C 2 y 4 2 f ( x) = x + 2 y 2 5 C=0 −2 4 2 3 f′ x2 8 4 x C = −2 x2 2 fx f )x) 2 −4 x f )x) f ( x) = 2 6 2x −4 f′ 4 x 3 2 1 2 3 −4 x −2 2 4 Answers will vary. 51. f x x2 1 1 x2 52. f x 53. dy dx 2x 1, 1, 1 3 fx x 3 x x3 3 f )x) C 1 x fx C 1 y x3 3 f )x) x 3 f′ 4 3 −4 C⇒C 1 x x x 1 x −2 4 f ( x) = −2 x 2 3 1 f 2 2 x2 1 dx f ( x) = − 1 + 1 x 2 3 1 y y x 2x y −1 x −4 2 Answers will vary. 54. dy dx 2x y 1 2x 2 3 y 2 x x2 2x 2x C⇒C 23 cos x, 0, 4 C y cos x dx 1 4 sin 0 C⇒C sin x 1 dy dx 1 x 2 x2 4 x C 4 C x 2 56. y 4 6 2 2 4 4 2 C y x2 4 x −3 5 −3 −4 8 −2 3 dx x 3 C y 5 3 , x > 0, x 3 ln e y sin x 1, 4, 2 y (b) 57. (a) Answers will vary. dy dx 4 55. 2, 3, 2 1 dx 2 dy dx y 2x 3 ln x e, 3 3 ln x C⇒C C 0 C 1 S ection 5.1 y (b) 5 dy dx x2 1, y 58. (a) x3 3 x x −4 4 1 3 3 −5 C 1 (b) y 4 x3 3 −5 C C 7 3 x dy dx cos x, y 59. (a) −4 7 3 y (−1, 3) C 1 399 5 1, 3 3 1 3 3 Antiderivatives and Indefinite Integration cos x dx 0, 4 7 6 4 y x −4 3 sin x C C⇒C sin 0 sin x 4 −6 6 −1 4 −2 60. (a) (b) y dy dx 1 , x > 0, x2 (1, 3) 1 dx x2 y 61. (a) 7 (b) 9 −3 1 1 C⇒C y 1 1 x dy dx 2 x, y 3 y 2, 2 0 C 1 x C −1 8 −1 2 (c) C C − 15 C⇒ C 4 12 15 6 −8 6 x dy dx 2 x, 2x1 6 0 2 2 y (b) 1 2 x2 2x dx 2 20 1 x dx 2 −9 62. (a) 2 x 3 x 5 1, 3 (c) 4, 12 2 12 4 4 3 32 y 43 x 3 2 43 x 3 dx C 4 3 2 4 8 3 20 C C 32 3 C⇒C 4 3 0 6 0 400 Chapter 5 Integration 63. f x 4x, f 0 fx 4x dx f0 6 fx 2x 65. h t 8t3 5, h 1 h1 4 5 dt 2t 4 4 ht 2 C 1 3 2x3 1 6s 8s 3, f 2 fs C⇒C 5 5t 2t4 5t 2x3 g0 6 6x 2 dx 66. f s C⇒C 2 8t3 1 gx C 6 ht 6x 2, g 0 gx 2x 2 20 2 64. g x 6 11 11 20 3 fs C⇒C 3s 2 2 32 3s 2 4 22 2s 4 1 3 8s 3 ds 6s f2 C 2s 4 C 12 32 C C⇒C 23 x2 67. f x 2 68. f x f2 5 f0 6 f2 10 f0 3 fx 2 dx 2x f2 4 fx 2x f0 1 fx 2x 1 dx x2 1 x f2 6 C2 10 ⇒ C2 fx x2 x C2 0 fx 5 ⇒ C1 C1 13 x 3 4 6 6 f0 0 fx 14 x 12 6 dx 14 x 12 70. f x 32 C1 6 ⇒ C1 C1 13 x 3 fx 4 13 x 3 x 2 dx fx C1 3 ⇒ C2 x f4 2 f0 0 f0 6x 3 3 1 f0 C2 C2 sin x 69. f x 0 6x 6 32 fx x f4 2 2 fx 2x 0 2 ⇒ C1 0 4x1 x C1 3 sin x dx 1 cos x fx 12 3 dx C2 2 fx f0 C1 3 x fx 2 fx 12 2x C1 2 fx f0 dx 3x 4x1 0 ⇒ C2 4x 2 0 3x 3x C2 f0 fx cos x 1 ⇒ C1 C1 0 sin x 2 2 cos x 0 C1 2 dx C2 2x sin x 6 ⇒ C2 6 6 2x C2 23 S ection 5.1 e x dx fx f0 2 x2 72. f x ex 71. f x Antiderivatives and Indefinite Integration ex ex fx ex 1 dx f0 5 e0 fx x 1 x C2 ⇒ C2 0 ex f1 C2 1.5t 4 4 f1 3 0.75t 2 h0 0 0 C 12 ⇒ C ht 0.75t2 5t 0.75 6 74. C 12 6x C 500 ⇒ C 75. f 0 P0 69 cm 4. Graph of f is given. (a) f 4 0 2 k 3 2 150 t3 3 100 7 (g) C 500 2 500 32 500 100t3 150 2 500 2352 bacteria 3. y 6 (b) No. The slopes of the tangent lines are greater than 2 on 0, 2 . Therefore, f must increase more than 4 units on 0, 4 . 4 2 x −2 (c) No, f 5 < f 4 because f is decreasing on 4, 5 . (d) f is an maximum at x 3.5 because f 3.5 the first derivative test. 2 600 ⇒ k 500 (f) f is a minimum at x 1.0 C2 3 23 kt 3 P7 12 6x 3 kt 1 2 dt Pt Pt 56 2 ln x k t, 0 ≤ t ≤ 10 P1 2 C1 6 C2 ⇒ C2 6 2 ln x dP dt 2 x dx 6 dx 12 (b) h 6 5t 2 C1 ⇒ C1 2 x fx 5 dt 2x 2 fx 4 73. (a) h t 2 dx x2 fx C1 ⇒ C1 e0 2 C1 0 and 2 6 4 8 −6 (e) f is concave upward when f is increasing on ,1 and 5, . f is concave downward on 1, 5 . Points of inflection at x 1, 5. 76. Since f is negative on , 0 , f is decreasing on , 0 . Since f is positive on 0, , f is increasing on 0, . f has a relative minimum at 0, 0 . Since f is positive on , , f is increasing on , . 60 32t C1 C1 32t 60 dt 16t 2 60t C2 2 f ′′ 1 −3 32 dt st 3 f′ vt v0 y 32 ft sec2 77. a t −2 1 −2 f s0 x −3 2 3 st 6 C2 16t2 60t 6, Position function The ball reaches its maximum height when vt 32t 60 32t t s 15 8 16 15 2 8 0 60 15 8 60 seconds 15 8 6 62.25 feet 401 402 Chapter 5 78. f t Integration 32 ft sec2 at 79. From Exercise 78, we have: v0 f0 s0 ft vt f0 0 32 dt 32t ft 32t 0 80. v0 s C2 s0 ⇒ C2 v0t v02 64 v0t C2 8 2t2 t 8 v0 8t 64 v0 1± 65 1 65 vt 1 st 65 f0 32t 32 4 82. From Exercise 81, f t 0 4.9t2 1600 ⇒t 4.9 83. From Exercise 81, f t 9.8t 10 t 10 9.8 10 9.8 7.1 m 9.8t C1 C1 ⇒ v t 9.8t 9.8t v0 10 s0 1 8 65 4 8 64.498 ft sec 4.9t2 1600. (Using the canyon floor as position 0.) 1600 1600 9.8t 9.8 dt 2.266 seconds. 8 65 f 9.8 4.9t 2 v0t C2 4.9t 2 v0t s0 v0 dt v0 4 4 (b) vt 550 187.617 ft sec ft Choosing the positive value, t2 v0 32 0 0 t 4.9t2 v0 35,200 vt 16t2 st ft 2 550 81. a t (a) v v0 32 time to reach s0 64 ft t 16 v02 32 s0 8 ft sec s0 v0 32 v0 32 0 when t v02 16t2 v0 dt 16t 2 ft C1 v0 0 v0t st 32t v0 maximum height. v0 st f0 32t v0 ⇒ C1 C1 ft 16t2 st f0 326.53 4.9t2 10t 18.1 sec 2. 0 (Maximum height when v 0.) C2 ⇒ f t S ection 5.1 4.9t2 84. From Exercise 81, f t ft 2 200 4.9t 2. If v0t v0t Antiderivatives and Indefinite Integration 2, then vt 9.8t v0 0 for this t-value. Hence, t v0 9.8 4.9 v0 9.8 and we solve v0 9.8 2 v0 4.9 v02 9.8 2 2 v02 9.8 200 198 9.8 v02 9.8 2 198 4.9 v02 4.9 v02 9.8 2 198 3880.8 ⇒ v0 v02 85. a 62.3 m sec. 1.6 vt 1.6 dt 1.6t v0 1.6t, since the stone was dropped, v0 0.8t2 1.6t dt 0⇒ s 20 s0 2 s0 0 s0 st 0. 320 0.8 20 Thus, the height of the cliff is 320 meters. vt 1.6t v 20 86. v dv 32 m sec GM GM y 12 v 2 1 dy y2 (a) v t C R, v 12 v 20 GM R C C 12 v 20 at v0. GM R GM y v2 2GM y v2 v02 6t2 0≤t≤5 2, 3t2 12t t2 4t 3 vt 6t 12 9 3t 6t 1t 3 2 (b) v t > 0 when 0 < t < 1 or 3 < t < 5. (c) a t v2 12 v 20 GM R 2GM R v02 2GM 9t xt 3 When y 12 v 2 t3 87. x t 1 y 1 R 6t 31 0 when t 2 1 3 2. 403 404 Chapter 5 88. x t Integration 0≤t≤5 3 2, t 1t t3 7t2 15t (a) v t xt 3t2 14t at vt 6t 1 14 v 6t 7 3 90. (a) a t 15 3t 0 when t 7 3 5 7 3 x1 4 3 ft v t dt sin t dt 3 ft cos 0 cos t 0 sin t C1 cos t C2 1 sin t since v0 C2 C2 ⇒ C2 4 sin t for t k ,k 0, 1, 2, . . . 91. (a) v 0 25 km hr 25 1000 3600 250 m sec 36 v 13 80 km hr 80 1000 3600 800 m sec 36 at a constant acceleration vt at v0 250 ⇒ vt 36 C s 13 550 468 t2 2 250 t 36 250 36 13a a st 13a 550 36 (b) 800 36 at a 275 13 234 2 2 250 36 275 234 s0 250 13 36 1.175 m sec2 0 189.58 m 4 0 4 xt 2 3 t v t dt 2t1 2 at 2 cos t dt v 13 xt cos t a t dt (b) v t 3 7 . 3 3 vt f0 5t 5 and 3 < t < 5. 3 14 3 t 9 (b) v t > 0 when 0 < t < (c) a t 89. v t vt 1 2, t>0 2t1 2 C C⇒C 21 2 2, position function 1 t 2 32 1 , acceleration 2t 3 2 S ection 5.1 v0 92. 45 mph 66 44 ft sec 44 22 16.5 22 ft sec a vt at st a2 t 2 vt 66 66t Let s 0 33 2 132 when a at 16.5t st 8.25t 2 16.5. 0 132 73.33 feet 117.33 feet 16.5 vt 66 a ph = 66 (c) 2.667 117.33 ft 30 m a 66 2a 2 44 16.5 44 16.5 66 ft/s ec 66 a s ph = 0 when t 22 t 66 . a 1.333 73.33 ft 66 0. 0 after car moves 132 ft. 66 22 16.5 16.5t (b) 45 m at s t s at 93. Truck: v t It takes 1.333 seconds to reduce the speed from 45 mph to 30 mph, 1.333 seconds to reduce the speed from 30 mph to 15 mph, and 1.333 seconds to reduce the speed from 15 mph to 0 mph. Each time, less distance is needed to reach the next speed reduction. 66 66 t (a) s 10 30t Let s 0 Automobile: a t 0. 3 10 (b) v 10 30 st 2 6 10 300 ft 60 ft sec 41 mph 6 vt 6t Let v 0 0. st 3t2 0. Let s 0 At the point where the automobile overtakes the truck: 30t 3t2 0 3t2 30t 0 94. 3t t 10 when t 1 mi hr 5280 ft mi 3600 sec hr (a) 405 44 ft/s ec 15 mp 0m h= ph 22 f t/se c 15 mph 16.5t (a) 66 ft sec 30 mph Antiderivatives and Indefinite Integration 10 sec. 22 ft sec 15 t 0 5 10 15 20 25 v1 ft sec 0 3.67 10.27 23.47 42.53 66 95.33 v2 ft sec 0 30.8 55.73 74.8 88 93.87 (b) v1 t 30 95.33 v2 t (c) s1 t v1 t dt s2 t 0.1068 3 t 3 v2 t dt 0.1208t3 3 0.0416 2 t 2 6.7991t2 2 0.3679t 0.0707t In both cases, the constant of integration is 0 because s1 0 s1 30 953.5 feet s2 30 1970.3 feet The second car was going faster than the first until the end. s2 0 0. 0.1068t2 0.1208t2 0.0416t 6.7991t 0.3679 0.0707 406 Chapter 5 Integration 95. True 96. True 99. False. For example, x x dx 97. True x dx because x dx x3 3 x2 2 C 100. False. f has an infinite number of antiderivatives, each differing by a constant. 98. True x2 2 C1 101. f x C2 . 2x fx x2 f2 0⇒4 C x3 fx 4x 3 0⇒ f2 8 3 8 0≤x<2 2<x<3 3<x≤4 2, 0, 102. f x x 2x C3, fx 1 ⇒ C1 f0 x3 3 2⇒ f continuous at x x 1 2 1 5 4 C3 C2 ⇒ C2 4 5 1 104. C1 3 ⇒ C1 f is continuous: Values must agree at x fx 3 0≤x<2 2≤x<3 3≤x≤4 1, 5, 3⇒1 f1 4 2 d sx dx 2 cx 6 C2 ⇒ C2 2s x s x 2c x c x 2c x s x 0 Thus, s x 2 c x 2 k for some constant k. Since, s0 0 and c 0 1, k 1. 2 2: 2 Therefore, s x [Note that s x properties.] 0≤x<2 x 2, 3x 2 2, 2 ≤ x ≤ 5 2 The left and right hand derivatives at x Hence f is not differentiable at x 2. 2 2s x c x x C1, 0≤x<2 3x2 C2, 2 ≤ x ≤ 5 2 fx 16 3 4x 1 1, 0 ≤ x < 2 3x, 2 ≤ x ≤ 5 103. f x 16 3 2 3⇒6 x 2x 1, 0 ⇒ C1 C1 1 f continuous at x fx 4 y 0≤x<2 2<x<3 3<x≤4 C1, C2 , 0⇒C C1 Answer: f x 1, C 2 do not agree. 2 cx 2 sin x and c x 1. cos x satisfy these S ection 5.2 105. d ln Cx dx d ln C dx ln x 107. f x y f xf y y f xg y 106. d ln x dx 1 x C 407 1 x 0 gxg y gx 1 x 1 x 0 Area gxf y f0 0 Note: f x cos x and g x sin x satisfy these conditions. fx y f xf y g x g y Differentiate with respect to y. gx y f xg y g x f y Differentiate with respect to y. 0, f x f xf 0 gxg 0 gxg 0 gx Letting y f xg 0 gxf 0 f xg 0. Hence, 2 f x f x 2f x g x g 0 2g x g x 2g x f x g 0 . Adding, 2 f x f x Integrating, f x Clearly C 2g x g x 2 2 gx C. 0, for if C Now, C fx 0, then f x 2 y 0. gx 2 y 2 2 gx fxfy f x 2f y fx gxg y 2 2 ⇒fx gx 2 g x 2g y gx 2 fy 2 2 0, which contradicts that f, g are nonconstant. f xg y 2 gxf y f x 2g y 2 g x 2f y 2 2 gy C 2. Thus, C 1 and we have f x Section 5.2 5 2i i 2 1. 2 gx 3 Area 5 1. 2 1 5 2 1 i i 1 1 i 21 4 5 5 35 1 6 4 2. kk k 2 31 42 53 64 50 3 5 k j 1 3j 1 3 i 4. 1 3. 1 4 1 1 5 0 k2 1 2 1 1 1 5 1 10 1 17 158 85 4 47 60 5. c k c c c c 4c 1 4 6. i 9 7. i 11. i 1 3 0 8 1 15 1 3i 1 2n ni 1 8. i 3 2i n 20 15. 1 2i n 20 2i i 2 27 4 64 9 125 238 1 2 i i 1 2 12. 20 21 2 1 8 5 1 2n 1 ni 1 5 9. i j 2i n 2 1 13. 1 4 j 8 10. 3 j 3n 21 ni 1 2 3i n 14. 15 420 16. 1 3 2 i i 1 3 15 1 1n ni 15 2i i 1 2 j 4 2 1 i n 1 0 15 16 2 45 2 195 408 Chapter 5 Integration 20 i 10 19 i 17. 2 1 1 i i 0 19 20 39 6 15 19. i 15 2 1 i3 1 i 1 i i i 15 16 31 6 2480 1 1 i 1 1 10 i i2 20. i 1 2 14,400 i 10 15 2 152 16 4 1 10 11 21 6 i2 2 10 i2 1 2470 15 ii 10 i2 18. i2 i 15 16 2 375 10 i3 1 1 10 i 1 i 102 11 4 1 2 10 11 2 3080 120 12,040 i2 i 2930 (TI-82) 3, x, 1, 20, 1 20 20 20 3 1 2 20 6 1 20 21 41 6 22. sum seq x 2x, x, 1, 15, 1 15 1 15 2i 2 15 i3 3 20 i 60 3 > 2 > 21. sum seq x 2 1 2930 15 4 1 14,160 (TI-82) 2 2 15 16 4 3 4 9 2 51 33 2 16.5 24. S 5 5 4 21 16 s 1 3 4 9 2 25 2 12.5 s 4 4 2 01 10 25. S 3 3 51 26. S 4 2 4 3 1 2 4 3 3 2 s 2 2 31 0 0 s4 4e 0.5 s5 30. S 5 1 5 e 1 1 e 1.5 e 11 655 1 5 1 24 5 1 1 5 21 5 2 21 1 5 2 1 2 8 3 2 3 29 5 9.13 5.8 0.768 8 1 4 5 137 15 0.518 2.666 e 2 1 4 5 4.395 2 2 11 855 11 855 1 5 1 4 31 44 1.5 1 e 11 755 11 755 1 1 5 s5 0.5 e 1 11 24 11 655 1 31 44 11 44 4e 1 11 24 1 4 28. S 4 29. S 5 s 7 11 44 27. S 4 s4 11 1 5 1 6 1 7 1 8 1 9 11 25 1 6 1 7 1 8 1 9 1 10 11 955 1 5 2 5 1 16 5 1 11 955 9 5 2 5 2 1 5 2 1 5 3 5 1 2 1 5 0.746 0.646 4 5 1 2 1 5 0.859 1 3 5 2 1 5 1 4 5 2 1 5 0 1 2 16 23. S 1 15 15 2 0.659 14,160 S ection 5.2 31. lim 81 n2 n 1 n4 4 32. lim 64 n n n3 33. lim 18 n n 1 n2 2 n→ n→ n→ n 2i 35. i 1 n2 1 S 10 2 1 2n 6 1n 2i n2 i 1 12 10 64 lim 6 n→ 1 n2 18 lim 2 n→ n2 2n3 n 81 1 4 n 3n2 n3 18 1 2 n2 64 2 6 1 1 nn 1 2 n2 2 n 1 1 4n n 1 n2 2 n 64 3 34. lim 9 n→ n 2 Sn n 1.002 S 10,000 n 4j 36. j 1 1 n2 S 10 1.0002 1n 4j n2 j 1 23 10 2n 3 Sn n 2.3 S 100 2.03 S 1000 2.003 S 10,000 n 37. k 6k k 1 n3 1 2.0003 6n 2 k n3 k 1 6 2n2 n2 S 10 S 100 S 1000 S 10,000 n 4i2 i 1 n4 1 6 nn n3 k 3n 1 6 3n 1 2n 6 3 1 nn 1 2n2 n2 2 1.999998 1.99999998 i2 4 n2 n 1 n4 4 2 2n2 4 nn 1 2n 6 1 3n 6 1 2n2 n 1 3n3 3n3 S 1000 S 10,000 1.056 1.006566 1.00066567 1.000066657 6n2 3n 4n2 1 3n3 3n3 S 100 Sn 1.9998 4 n3 n3 S 10 1 2 1.98 4n3 i n4 i 1 409 81 4 1.02 S 1000 i 2n3 n4 1.2 S 100 38. n4 81 lim 4 n→ Area 2n2 3n 2 6n Sn 2 1 nn 1 n2 2 1 n2 n lim 2 n→ n2 1 1 2 1 2 410 Chapter 5 n 1 16i n2 1 2i n 39. lim n→ i n 40. lim n→ i n 41. lim n→ i Integration 16 n i n2 i 1 lim n→ 2 n 1 i n3 1 4n i n2i 1 lim n→ 1 2 1 i2 1 lim n→ n n→ 2i n 1 i 1 lim 2 n 2 n3 n→ n→ n 2 lim 1 2 21 2 4 3 n→ n 43. lim n→ i n 1 i 1 n 44. lim n→ 3 2i n 1 i 2 n 1 1 n 2 lim n→ 2 n 2 lim n→ i 2 1 1 n2 3n 1 3 1 1 1 4 3 4n n 2 n 1n i ni 1 1 1 1n n n4i 1 2i 14 n n4 2 lim 1 1 2n 6 1 2 3n2 2 lim n→ 1 n n 1 nn 1 n 2 6n2i 6n2 12ni 2 nn 13 n 2 lim 10 n→ n2 n 2n2 1 n→ 1 2 3 2 2 <n n n 2 21 8i 3 1 12n 2 3 n 3 2 lim 3 nn 1 2n 6 4 4 n2 6 n 2 n2 2 4 n (b) n→ 8 26 3 2 lim 45. (a) 1 n 2 1n 3 n n4 i 1 n→ 1 2 n 2 lim n→ n→ i2 4 nn 4n n i i i 23 n n3 1 n 8 lim 1 n 4n 1 lim n→ n n2 2 2i n i n2 4 1 2 12 6 lim n→ n2 8 1 n 2n 6 n 2n n n3i 1 lim 42. lim 2 3n2 n3 1 2n 6 lim n→ 1n n3 lim n→ 3 lim n→ 4 nn 1 n2 2 lim n→ 1n n3 i lim n→ 16 n n 1 n2 2 lim n→ x 1 n2 n 8 1 2 4 2 n2 20 y 3 2 0 n 2 n Endpoints: 2 0<1 1 x 1 2 2 <2 <...< n n n (d) f Mi 3 f xi on xi 1, n (c) Since y x is increasing, f mi f xi 1 on xi 1, xi . f xi i n sn f xi i 1 n f i 1 x 1 —CONTINUED— 2i 2 n 2 n n i i 1 1 2 n 2 n xi n Sn 1 x f i 1 1 2i 2 nn n i i 1 2 n 2 n S ection 5.2 Area 411 45. —CONTINUED— n (e) x 5 Sn 50 1.8 1.96 i 2 n 1 2 n 1.98 2.4 2.2 2.04 n→ i 1 lim 4n i n2i 1 lim (f) lim 100 1.6 sn 10 4 nn 1 n2 2 n→ n→ 2.02 2n lim n→ n i lim n→ i 1 2 n 2 n lim n→ (b) 3 x 3 2 <1 n 1<1 2 1 1<1 1 x 2 (c) Since y 2 <1 n 4 ... < <1 n 2 2 <. . .< 1 n 4 x is increasing, f mi f xi 1 on xi 1, xi . n sn f xi i x 1 1 n f1 i (d) f Mi i 1 f xi on [xi n f xi (e) x f1 1 x n 2 n i 2 n 2 n 1 i 1 1 i 1 2 n 2 n 1, xi n Sn i 2 n 1 i 1 i 5 10 50 3.6 3.8 3.96 4.4 4.2 4.04 2 n 1 2 n 3.98 Sn i 100 sn n 4.02 n n→ 1 i i 1 1 2 n 2 n lim 2 n lim (f) lim 2 lim 2 n n lim 2 n→ n→ n 1 lim n→ i 1 i 2 n 2 n n→ n→ 2 nn 1 n 2 n 2n 2 n 4 n n lim n→ 2 n 4 4 2 nn 1 n 2 2n 1 n lim n→ 2 2n 1 n 2 2 n 1 n Endpoints: 4 4 n 4 nn 1 n2 2 n→ y 1 4n i n2i 1 lim 46. (a) n n n→ lim 1 4 2 n 4 2n n 3 n 1 2 n <1 n 2 n 412 Chapter 5 47. y 2x Integration 3, 0, 1 n sn i n f i 1 2 i 1 n i i n 2n 3 0 1 n n n 1 n 2 n2i 3 1 Note: x 1 n 3 1n 2 1 n 2 2n2 1 y 1 x lim s n Area 48. y n→ 3x 1 2 5 Note: x 4, 2, 5 2 3 n n 2 3 y 12 n Sn 3i n f2 i 1 n 1 6 2, 0, 1 n Sn i i f n 1 n 1 n i 50. y 2 lim S n 3 3 n 27 1 2 6 2n i i 4 12 1 2 x 1 n 4 6 8 10 12 39 2 i n 1 18 6 y 1 n Note: x 1n2 i n3 i 1 Area 3 n 4 27 2 6 n→ x2 49. y 8 27 n 1 n n2 2 lim S n Area 10 3i n 32 i 3 n 3 2 1 n 2 1 7 3 n→ x2 1, 0, 3 nn 1 2n 6n3 Note: x 3 1 0 2 1 2 6 3 n 1 n2 2 x 1 3 n n 2 3 y 12 n Sn i 3i f n 1 n 3 n i 1 3i n 2 3 n 1 10 8 6 27 n 2 i n3 i 1 27 n n n3 Area 3n 1 ni 1 1 2n 6 lim S n n→ 9 2 2 4 2 3 n n 1 3 12 9 2 2n2 3n n2 1 x 3 1 2 3 S ection 5.2 51. y x 2, 1, 3 16 n sn i 1 n 2 ni 15 4 nn n2 lim s n 1 2n 6 1 14 12 10 8 6 4 2 2 n 1 4 n n 8 3 4 70 3 23 i f n 1 1 n i n 1 Area 2 1 nn 1 lim s n 1 3 1 n→ 2 i n 1 i 1n2 i n3 i 1 1 3 1 3 1, 1 ; Find area of region over the interval 0, 1 . sn 2 1 x2, n x 1 4nn 1 n 2 1 2n 30 n→ 1 Note: x y 1 n 3 1 n 2 1 2n 6n3 1 1 2 6 1 3 n 1 n2 x −2 2 −1 2 3 4 3 Area 53. y 1 2 4i n n2 8 n 6n2 30 52. y i 2i n 1 −1 2 15n n Area 16 4i2 1 y 18 n 2 n 2i n f1 2 n Note: x Area x3, 1, 4 64 4 Note: x 1 y 3 n n 70 60 n sn 3i n f1 i 1 3 ni n 27i2 n2 81 n 4n2 lim s n 2 1 x3, 0, 1 1 3 n 50 40 30 20 9i n 2 10 81 n 6n2 1 2n 6 1 2n 1 27 2 513 4 0 1 1 n Tn i i f n 1 2 n2 i 1 Area n i 1 1 n4 i 1 n 1 4 i lim T n n→ n 1 n 1 n i 2 n nn i3 1 n2 1 2 4n 1 i n 1 4n2 1 4 3 4 3 2 3 5 6 27 n 1 2 n 128.25 y 2.0 Since y both increases and decreases on 0, 1 , T n is neither an upper nor lower sum. n 1 9nn 1 n 2 1 n 27 Note: x x −1 27 n n n2 81 4 189 3i n 1 27 n2 n 1 n3 4 n→ 2x i n3 63 189 54. y 64 27i3 1 3 63n n Area n 3 n 3 1 n 1.5 1.0 0.5 2 1nn 1 n4 4 2 x 0.5 1.0 2.0 413 414 Chapter 5 x2 55. y Integration x3, 1, 1 1 1 Note: x 2 n n y 2 Again, T n is neither an upper nor a lower sum. n Tn f i 1 n 10i n 2 16i 2 n2 1 4 n n 56. y x2 x3, sn f i 1 n 5 2n 1 5 2 2 Area 4i 2 n2 4 3 lim s n 3y, 0 ≤ y ≤ 2 57. f y n Sn f mi i i 12 n i n2 i 1 Area 12 n2 lim S n 1 4 2i f n 1 1 2i 3 n 1 2i n 2 1 n2 y 3 i n 3 4n2 i n3 i 1 1 n 2 1 1n3 i n4 i 1 −2 x −1 1 2 −1 y 4 2 n 1 2 6 n 6 n x 2 4 6 2 6 4 2 n 2 n y 5 4 2 n 1 2 12 1 4 2 n n i Note: y 2i n n2 n 16 n 3 i n4 i 1 1 4n2 0 6n 6 n 32 n 2 i n3 i 1 16 n4 2 n 1 1 2n 2 n 1 x 1 1 7 12 2 n n→ g2 2 n Sn i 1 4 2 1 y, 2 ≤ y ≤ 4 2 n 4 3 1 1 1 n 5n i n2 i 1 2 2 3n2 lim 6 n→ 1 n nn y 1 1 n i2 n Note: y n 58. g y 2 n 5 2 2 n→ 0 1 i3 n3 2 n 2 n 20 n i n2 i 1 41 3 2 3 4 1 i 2i n 8i3 n3 1 2n 6 1 n2 n 1 n 5i n 2 i 32 3 nn 3 n Note: x i n 1 32 n3 10 4n 1 ni 1 2 n 16 2 3 1, 0 n 1 1 n 4 n→ 8i3 n3 1 12i 2 n2 6i n 1 2 10 1 lim T n Area nn 20 n2 4 1 4i2 n2 1 n i i 2 2i n 1 4i n 1 i n 2 n 2i n 1 3 2 n i 2n 1 ni 1 i n 1 x 1 2 n n Area 1nn 1 n 2 lim S n n→ 2 2 1 3 n 1 n 2 3 4 5 S ection 5.2 y 2, 0 ≤ y ≤ 3 59. f y n Sn 3i n f i 1 27 n3 i i n f1 1 1 ni 9 2n2 4 2 9 1 9 27 2n 9 2n2 x 4 6 8 10 1 2 3 4 5 −4 2 Note: y 2 −2 1 1 n n y 5 3 i n 4 4i n 1 2i n 2 2 i n 1 1n 3 ni 1 1 x i2 i2 n2 1 1 3n n Area 2i n 2nn 1 n 2 n 3 61. g y 3n 2 y 6 1 n 1n 41 ni 1 n 3 n 27 2n y 2, 1 ≤ y ≤ 2 Sn 03 n n 27 n 2 i n3 i 1 9 2n2 n2 1 n→ n i 1 lim 9 n→ 4y 2 3i n 1 2n 6 lim S n Area 60. f y n 3 n nn 3 Note: y 1 n lim S n 3 1 nn n2 1 2n 6n2 n n→ 1 2n 6 1 1 1 3 1 y3, 1 ≤ y ≤ 3 4y2 n2 11 3 3 Note: y 1 n y 2 n 10 8 n g1 2i n 41 2i n 2n 41 ni 1 4i n 4i2 n2 2n 3 ni 1 10i n 4i2 n2 8i3 n3 6 10 8 3 Sn i 1 n i Area 1 lim S n n→ 2 n 6 2 1 2i n 2 3 2 n −4 12i2 n2 6i n 1 2 3n n 4 x −4 −2 −2 44 3 8i3 n3 10 n n 1 n 2 4 nn n2 1 2n 6 1 8 n2 n 1 n3 4 2 Area 415 416 Chapter 5 n 1 i n 1 i n Sn h1 i n 1 i n 1 ni n x 1 x 1 2 1 4 2 3 nn n2 1 2n n2 xi 2 1 n f ci i 1 i 9 16 xi 2 xi x 16 1 4 49 16 3 , c1 67. f x ,n 4 7 2 4ci 1 6 i 25 4 1 9 4 2 sin x, 0 ≤ x ≤ 66. f x 32 3 ,c 32 3 , c2 5 ,c 32 4 7 32 x tan tan ci 32 1 tan 10 49 4 5 ,c 16 4 7 16 14 Approximate area Exact value is 16 3. 16 f ci 3 32 tan i 5 32 tan 7 32 0.345 8 4 8 12 16 20 5.3838 5.3523 5.3439 5.3403 5.3384 ,n 4 . , c1 3 ,c 16 3 x 8 , c2 4 Area 16 x, 0, 4 n 1 2 n x i xi 2 xi Let ci 1 16 ci2 x 1 1 4 4 f ci i 5 ,c 24 4 3 . n Area 3 ,c 23 53 tan x, 0 ≤ x ≤ Let ci 1 ,c 22 f ci i 4 . n 69 8 65. f x 1 1, c1 Area 25 16 3 xi 2 xi x 1 2 3 1 3 7 4 4x, 0 ≤ x ≤ 4, n x2 64. f x 5 ,c 44 ci2 x 11 2 16 19 4 4 4 Area 10 3n 1 2n Let ci 3 ,c 43 8 3nn 1 n 2 1 . 1 ,c 42 1 ,c 21 1 2n 6 1 3 2 1 3, 0 ≤ x ≤ 2, n xi 6 3i n 1n 2 2 4 2 n2 n24 x2 Let ci 1 1 n 3i2 n→ 63. f x 2 3 lim S n Area 3 1 n2 n 1 n3 4 1 2n n 2 4 n3 2 5 1 n i3 1 y 1 n 1, 1 ≤ y ≤ 2 Note: y y3 62. h y Integration 1 sin sin ci i 16 sin 1 3 16 sin 8 5 16 sin 7 16 1.006 S ection 5.2 8 68. f x x2 1 4 8 12 16 20 2.3397 2.3755 2.3824 2.3848 2.3860 4 8 12 16 20 2.2223 2.2387 2.2418 2.2430 2.2435 4 8 12 16 20 1.1041 1.1053 1.1055 1.1056 1.1056 4 8 12 16 20 4.0786 4.0554 4.0509 4.0493 4.0485 4 8 12 16 20 8.1711 Approximate area tan 8.3341 8.3646 8.3753 8.3802 x , 1, 3 8 n Approximate area 70. f x cos x, 0, 2 n Approximate area 71. f x ln x, 1, 5 n Approximate area xe x, 72. f x 0, 2 n Approximate area 73. 417 , 2, 6 n 69. f x Area 74. y y 4 4 3 3 2 2 1 1 1 (b) A 2 3 x 4 6 square units 75. We can use the line y x bounded by x a and x b. The sum of the areas of these inscribed rectangles is the lower sum. 1 (a) A 2 3 x 4 3 square units The sum of the areas of these circumscribed rectangles is the upper sum. y y x a b x a b We can see that the rectangles do not contain all of the area in the first graph and the rectangles in the second graph cover more than the area of the region. The exact value of the area lies between these two sums. 418 Chapter 5 Integration 76. See the definition of area, page 297. 77. (a) y (b) y 8 8 6 6 4 4 2 2 x 1 3 2 Lower sum: s4 0 (c) x 4 1 51 3 4 151 3 6 46 3 Upper sum: 4 S4 15.333 y 2 3 4 51 3 6 62 5 2111 15 326 15 21.733 (d) In each case, x 4 n. The lower sum uses left endpoints, i 1 4 n . The upper sum uses right endpoints, i 4 n . The Midpoint Rule uses midpoints, i 1 4 n . 2 8 6 4 2 x 1 3 2 Midpoint Rule: M4 22 3 (e) n 4 44 5 55 7 62 9 6112 315 19.403 4 8 20 100 200 sn 15.333 17.368 18.459 18.995 19.06 Sn 21.733 20.568 19.739 19.251 19.188 Mn 19.403 19.201 19.137 19.125 19.125 (f) s n increases because the lower sum approaches the exact value as n increases. S n decreases because the upper sum approaches the exact value as n increases. Because of the shape of the graph, the lower sum is always smaller than the exact value, whereas the upper sum is always larger. 78. f x sin x, 0, y 2 1.0 Let A1 area bounded by f x sin x, the x-axis, x 0 and x 2. Let A2 area of the rectangle bounded by y 1, y 0, x 0, and x 2. Thus, A2 21 1.570796. In this program, the computer is generating N2 pairs of random points in the rectangle whose area is represented by A2. It is keeping track of how many of these points, N1, lie in the region whose area is represented by A1. Since the points are randomly generated, we assume that A1 A2 N1 ⇒ A1 N2 ( π , 1) 2 f (x) = sin(x) 0.75 0.5 0.25 π π 4 N1 A. N2 2 2 The larger N2 is the better the approximation to A1. 79. True. 80. True. (Theorem 5.2 (2)) (Theorem 5.3) 81. Suppose there are n rows and n 1 columns in the figure. The stars on the left total 1 on the right. There are n n 1 stars in total, hence 21 1 2 2 ... ... n n nn 1 2 nn 1 1. 2 ... n, as do the stars x S ection 5.2 2 n 82. (a) 419 r h r (b) sin Area h θ r h r sin A 1 bh 2 n (c) An Let x 1 r r sin 2 r 2n 2 sin 2 n 2 12 r sin 2 n n. As n → 2 lim An x→0 4.09 5 10 sin 2 n 2n r2 , x → 0. sin x x r2 lim n→ 83. (a) y 12 r sin 2 x3 r2 1 0.016x2 r2 2.67x (b) 452.9 500 (c) Using the integration capability of a graphing utility, you obtain 76,897.5 ft2. A 0 350 0 For n even, 84. For n odd, n 1, 1 row, 1 block n 2, 1 row, 2 blocks n 3, 2 rows, 4 blocks n 4, 2 rows, 6 blocks n 5, 3 rows, 9 blocks n 6, 3 rows, 12 blocks n, n rows, 2 n n, 1 2 rows, n 1 2 blocks 2 n 85. (a) nn 1 i The formula is true for n 1: 2 11 Assume that the formula is true for n 1 2. 2i kk 1. n2 n 2 1 4 1 Assume that the formula is true for n 1 k 1 i 1 Then we have blocks The formula is true for n 1 because 12 1 1 2 4 13 1. 4 4 k: k i i3 (b) 1 2n 4 n 2i i n2 k k 2i 2i i 2k i3 1 i 1 kk k 1 1k 2k . 4 1 k i3 Then we have i 1 i3 i k 1 3 1 k2 k k k: 2 1 k 1 2 which shows that the formula is true for n 1 k2 k 1. 1 2 k 4 1 2 k2 4 k 4 3 2 1 k 1 k 4k 2 1 2 which shows that the formula is true for n k 1. 420 Chapter 5 Integration 86. Assume that the dartboard has corners at ± 1, ± 1 . y A point x, y in the square is closer to the center than the top edge if y2 ≤ 1 x2 y y2 ≤ 1 x2 1 1 2 y≤ (x, 1) 1 (x, y) −1 y2 2y (0, 0) x2 . 1 x −1 By symmetry, a point x, y in the square is closer to the center than the right edge if 1 1 2 x≤ y2 . In the first quadrant, the parabolas y 1 1 x2 and x 1 1 y 2 intersect at 2 2 2 There are 8 equal regions that make up the total region, as indicated in the figure. 1, 2 1. y ( 1 2 − 1, 2 − 1( (1, 1) −1 x 1 −1 2 1 Area of shaded region S 0 8S Area square Probability Section 5.3 1. f x 1 1 2 x2 22 3 2 22 3 x dx 5 6 42 3 5 6 5 3 Reimann Sums and Definite Integrals x, y 0, x 0, x 3i2 n2 3, ci y 3 3i2 n2 xi 3i 1 2 3 2i n2 n2 n n lim n→ f ci xi i 1 i 1 3 3n lim 2i2 n→ n3 i 1 lim n→ 3 3 nn 2 n3 n lim 3 3 n→ 33 2 3 0 y= 2 3i2 3 2i n2 n2 lim n→ 1 1 1 1 2n 6 1 2n 3n2 23 ... 3 n2 i 1 nn 1 2 1 n 1 2n2 3.464 x 3(2)2 . . . 3(n − 1)2 3 n2 n2 x S ection 5.3 2 3 x, y 2. f x 0, x 0, x xi i3 n3 xi 2 lim n→ i 3 n f ci 3 n→ 1 3i 2 3i n3 1 i 3 3i 2 n3 1 3i n3 1 n3 n lim i3 n3 1, ci i i 1 1n 2 lim 4 3i 3 n→ ni1 3i 2 i 2 lim n2 n 1 1 3 n4 4 2 lim 1 3n4 n4 6n3 4 2 lim 1 3n4 n4 4 n3 2 n2 4 4 6 , n n→ n→ n→ 3. y 6 on 4, 10 . 10 Note: x n i xi 1 6i n f4 i 2 3n2 1 2n3 3n2 2 1 nn 1 2 n2 n n 2 3 4 n→ 1 2n 3 2 1 4n2 → 0 as n → n 6 n 6 i 1 2n 6 2 lim n 6 n nn 3 n n f ci Reimann Sums and Definite Integrals 1 i 1 36 n 36 10 6 dx 4 4. y x on lim 36 2, 3 . xi f 1 i 1 x dx x3 on 1, 1 . n f ci xi 1 f i 1 2 x3 dx 1 61 lim n→ 2 n 1 0 5 n 25 n i n2 i 1 10 1 n 10 25 1 2 2 , n n 12 n i n2 i 1 1 i 1 1 2 5i n 2 5 2 25 2n → 0 as n → 5 2 2i n 1 → 0 as n → n 5 n 25 2n Note: x n i 5 2 lim n→ 5 , n 25 n n 1 n2 2 10 2 5i n 2 3 2 n n f ci 5. y 3 Note: x n i 36 n→ 2 n n i 24 n 2 i n3 i 1 1 n 2i n 1 42 1 3 2 n n 1 i 6i n 1 n2 2 n 1 16 n 3 i n4 i 1 3 n 1 n2 41 2 n 12i 2 n2 8i 3 n3 2 n 421 422 Chapter 5 Integration n n i f1 2i n xi 1 i 1 n i 1 6n 1 ni 1 n 12 lim 1 6 x2 7. y 12 1 8 1 on 1, 2 . 1 2 1 dx 2 on n xi 1 f i 3i n 1 1 3n 3 ni 1 3n 31 ni 1 2 dx lim 15 i 1 2 1 6n2 2 2 i n 1 1 1 n 1 n n 3ci 10 3i n i 1 2 6 1 3 n on the interval 15 1 10 3 3 2n 1 n 1 6n2 10 3 1 3 ; n n → 0 as n → 2 9i2 n2 27 2 27 n n n2 27 n 2 27 n 1 2n 6 1 2n n2 1 n 27 10 dx 27 n 2 1 2n 1 1 2n n2 1 15 3x 1 1, 5 . 1 n2 i2 n2 2 6i n xi 1 2i n 1 5 n 9. lim →0 i → 0 as n → 1 3 2n 18 n n 1 n 2 n→ 1 1 , n n 1 n 2 3x 2 1 1n2 i n3 i 1 27 n 1 n 15 1 3 n 1 3 3n n 1 1 1 2n n2 n 1, 2 . Note: x n f ci 2 10 3 lim n→ 1 3x 2 4n 2n i n2 i 1 2 8. y 1 2n n2 i n f1 i 1 2n 6 26 xi 1 x2 n 4 n f ci 4 nn n2 Note: x n i 4i2 n2 12 n 1 n 6 → 0 as n → 2 n n 3 n→ 2 4nn 1 n 2 6 2 , n 2 n 4i n 6 n n 3x 2 dx 1 n 2i n 31 f ci i 3 3x 2 on 1, 3 . Note: x 6. y 4 n 10. lim →0 i 6ci 4 ci 2 1 on the interval 0, 4 . xi 6x 4 0 x 2 dx S ection 5.3 3 n ci2 11. lim →0 i n x2 4 xi 1 12. lim 4 dx →0 i 0 5 1 4 1 2 14. x sin x dx 15. 2 x2 dx 16. 0 0 2 18. 2e 4 x dx 0 tan x dx 20. sin x dx 19. 0 0 2 2 y 3 dy 21. 1 3 dx x2 3 dx 0 2 dx x 17. 3 xi 5 3 dx x 1 1 3 ci2 423 on the interval 1, 3 . on the interval 0, 3 . 13. Reimann Sums and Definite Integrals 22. 2 2 dy y 0 0 23. Rectangle A bh 34 A bh 3 A 25. Triangle 24. Rectangle 24 a a 4 dx 12 A 4 dx 0 8a 4 A a x dx 8 0 y y 1 44 2 1 bh 2 A y 5 5 3 3 Rectangle 2 4 Triangle Rectangle 2 2 1 1 x 1 2 3 4 −a 5 x a x 2 1 42 2 1 bh 2 A 4 A 0 28. Triangle 27. Trapezoid b1 b2 h A 2 26. Triangle 5 9 2 2 1 bh 2 A 2 x dx 2 A 4 2x 4 5 dx 1 88 2 32 8 14 A 0 8 x dx 32 0 y y y 3 10 9 2 8 Triangle 6 1 6 Trapezoid 1 2 3 4 3 x 4 2 −1 x 1 2 x 3 2 y 29. Triangle A 1 21 2 1 bh 2 1 A Triangle 8 10 y 1 bh 2 1 2a a 2 a Triangle a 1 1 6 30. Triangle 1 A 4 x dx 1 x 1 1 A a a x dx a2 −a a x 424 Chapter 5 Integration 31. Semicircle 1 2 12 r 2 A Semicircle 2 3 9 2 x2 dx 9 3 2 2 Semicircle r r x 4 y 12 r 2 A 3 A 32. Semicircle y 4 4 r2 A −r x2 dx x r r −r 4 4 x3 dx In Exercises 33–40, 60, 2 2 6, dx 2 2 2 2 4 33. x dx x dx 4 4 2 4 x dx 2 3x 8 dx 6 1 2 2 dx 7 4 3 2 2 dx 22 10 6 60 60 f x dx 4 1 3 3 3 6 f x dx f x dx 6 0 x3 dx 2 6 0 (b) 10 68 4 x dx 26 f x dx 0 f x dx 2 2 3 f x dx 42 4 dx 2 5 5 x3 dx 2x 42. (a) 5 60 4 6 62 13 dx 2 16 3 4 2 6 f x dx 4 x3 dx 2 2 36 40. 30 4 4 dx 7 f x dx 15 2 4 4 x dx 2 f x dx 0 1 1 3 3 5 f x dx (c) dx 2 x3 38. 10 4 x3 dx 0 (b) 15 2 5 f x dx 0 82 2 1 60 2 41. (a) 15 dx 4 4 x dx 2 13 x 2 4 36. 2 8 dx 2 4 24 4 x 39. 46 2 4 37. 0 2 4 4x dx x3 dx 34. 6 2 4 35. 4 x dx f x dx (c) 0 0 3 5 5 6 5 3f x dx (d) 3 0 f x dx 3 10 6 6 fx g x dx 2 2 8 6 gx f x dx g x dx 2 2 10 12 f x dx 2g x dx 2 (d) g x dx 2 2 3f x dx 3 5 10 f x dx 3 0 f x dx 3 10 0 1 3f x dx (d) 30 1 1 4 6 2 3 1 2 6 5 1 3f x dx (c) 6 2 5 1 1 (c) 0 0 2 2 6 5 0 f x dx 0 f x dx 1 f x dx 1 1 (b) 6 5 1 f x dx 1 2 10 6 1 f x dx 44. (a) g x dx f x dx 3 0 6 f x dx 2 (b) 5 3 0 43. (a) 6 5f x dx (d) 30 f x dx 35 15 0 30 2 45. Lower estimate: 24 12 4 20 36 2 Upper estimate: 32 24 12 4 20 2 48 88 46. (a) 6 8 30 (2) 64, left endpoint estimate (b) 8 30 80 2 236, right endpoint estimate (c) 0 18 50 2 136, midpoint estimate If f is increasing, then (a) is below the actual value and (b) is above. 5 S ection 5.3 1 4 47. (a) Quarter circle below x-axis: 1 2 bh (b) Triangle: (c) Triangle 1 2 42 1 4 r2 1 2 Semicircle below x-axis: (f) Answer to (d) plus 2 10 1 2 20: 3 1 f x dx 1 2 3 2 1 2 2 1 2 2 2 20 5 23 2 2 1 2 f x dx 0 2 2 21 (e) Sum of absolute values of (b) and (c): 4 1 0 y 2 (3, 2) (4, 2) 4 (b) (11, 1) 3f (x) dx 32 f (x) dx 1 2 1 6 3 x −2 7 (c) 425 4 (d) Sum of parts (b) and (c): 4 48. (a) Reimann Sums and Definite Integrals 0 11 1 2 f (x) dx (d) 5 11 0 2 1 22 2 2 1 42 2 1 2 f (x) dx (e) 1 22 2 1 2 2 1 2 5 10 14 −1 2 (0, 4 8 12 1) −2 (8, 2) 3 2 1 2 4 2 10 f (x) dx (f) 2 4 2 4 5 5 fx 49. (a) 2 dx 5 f x dx 0 2 dx 0 3 4 0 fx 2 dx f x dx 2 4 Let u x f x dx 2. 5 f x dx 2 5 44 44 0 0 5 (c) 4, x < 4 x, x ≥ 4 8 5 (b) 50. f x 1 44 2 40 y f x dx 24 8 f even 0 (8, 8) 8 5 f x dx (d) 0 f odd 5 4 x 4 51. The left endpoint approximation will be greater than the actual area: > 52. The right endpoint approximation will be less than the actual area: < 8 1 53. f x x 4 is not integrable on the interval 3, 5 because f has a discontinuity at x 4. 54. f x x x is integrable on 1, 1 , but is not continuous on There is discontinuity at x 0. To see that 1 1 1, 1 . y 2 x dx x is integrable, sketch a graph of the region bounded by f x 1 ≤ x ≤ 1. You see that the integral equals 0. 1 x x and the x-axis for x −2 1 −2 2 426 Chapter 5 55. Integration y 56. y 4 4 3 3 2 2 1 1 1 2 (a) A 3 x x 4 1 4 1 2 4 3 (b) A 5 square units 3 4 1 square units 2 57. y y ln x dx 58. 1 1 3 Area 1 3 x 1 1 −1 x 1 2 2 x2 2e 3 1 22 2 dx 0 2 3 x3 59. x dx 0 4 8 12 16 20 Ln 3.6830 3.9956 4.0707 4.1016 4.1177 Mn 4.3082 4.2076 4.1838 4.1740 4.1690 Rn 3.6830 3.9956 4.0707 4.1016 4.1177 4 8 12 16 20 Ln 7.9224 7.0855 6.8062 6.6662 6.5822 Mn 6.2485 6.2470 7.2460 6.2457 6.2455 Rn 4.5474 5.3980 5.6812 5.8225 5.9072 4 8 12 16 20 Ln 1.2833 1.1865 1.1562 1.1414 1.1327 Mn 1.0898 1.0963 1.0976 1.0980 1.0982 Rn 0.9500 1.0199 1.0451 1.0581 1.0660 n 3 60. 5 x2 0 1 n 3 61. 1 dx 1 dx x n 2 S ection 5.3 Reimann Sums and Definite Integrals 427 4 e x dx 62. 0 n 4 8 12 16 20 Ln 31.1929 41.3106 45.1605 47.1772 48.4169 Mn 51.4284 53.0439 53.3508 53.4588 53.5089 Rn 84.7910 68.1097 63.0265 60.5768 59.1365 2 sin2 x dx 63. 0 4 8 12 16 20 Ln 0.5890 0.6872 0.7199 0.7363 0.7461 Mn 0.7854 0.7854 0.7854 0.7854 0.7854 Rn 0.9817 0.8836 0.8508 0.8345 0.8247 n 3 x sin x dx 64. 0 n 4 8 12 16 20 Ln 2.8186 2.9985 3.0434 3.0631 3.0740 Mn 3.1784 3.1277 3.1185 3.1152 3.1138 Rn 3.1361 3.1573 3.1493 3.1425 3.1375 66. False 65. True 67. True 1 1 x x dx 0 x dx 0 2 x dx 2 0 x2 71. f x x0 3x, 0, 8 0, x1 x1 1, x2 1, x2 c1 1, c2 4, x4 2, x3 2, c3 7, x4 3, x3 5, c4 8 1 8 4 f ci i x f1 x1 f2 x2 f5 x3 f8 x4 1 41 10 2 40 4 x dx 0 70. True. The limits of integration are the same. 69. False 68. True 1 88 1 272 428 Chapter 5 72. f x Integration sin x, 0, 2 x0 0, x1 x1 4 c1 4 , x2 , x2 12 , c2 3 f ci 6 3 xi , x3 , x4 2 , x4 3 , x3 2 ,c 34 , c3 2 3 2 4 i f x1 6 1 1 2 73. x b a n , ci ix 12 b i x3 2 3 f 3 2 x4 1 0.708 a n n x dx lim f ci →0 i a x 1 n lim n→ a i b lim n→ lim ab ab a b aa b aa 2 b a n b 2n a i n 1 b i a 2 1 nn n n b 1 2 a2n b a 1 2 2 a 2 b a 2 b2 b a2 2 a , ci b an n a n a i a b n a b n→ a n n lim x b i 1 n→ 74. a f 2 3 3 2 x2 3 3 2 4 a b f ix a b i a n n x 2 dx lim f ci →0 i a x 1 n lim n→ a i b n→ lim b a n b 2ai b n a2 i 1 a a2 b a ab 13 b 3 a3 a i2 b a ab a 2 n 1 n a 2 1 b 3 b 3 a 2 nn n b a 6 a 2 n 2a b a n n 1 n 2 na2 a2 b a n n a n 2 n lim n→ a 1 lim n→ b i 3 n 1 2n n2 1 2n 6 1 1 S ection 5.4 1, 0, 75. f x x is rational x is irrational The Fundamental Theorem of Calculus → 0, is not integrable on the interval 0, 1 . As f ci 1 or f ci 0 in each subinterval since there are an infinite number of both rational and irrational numbers in any interval, no matter how small. y 0, x 0 1, x 0<x≤1 76. f x 429 1 f (x) = x 2 The limit 1 n f ci lim →0 i xi 1 x 1 2 does not exist. This does not contradict Theorem 5.4 because f is not continuous on 0, 1 . 77. The function f is nonnegative between x 1 and x 1. 2 0 78. To find x dx, use a geometric approach. y y 2 3 f (x) = 1 − x 2 2 1 x −2 1 2 −1 x −1 1 Hence, Thus, b 2 x 2 dx 1 x dx a 1 and b 1 1. 1. x2, 0 ≤ x ≤ 1, and xi 79. Let f x n n f ci xi 1 i 1 lim 3 12 n→ n 1 22 i n 2 1n2 i. n3i 1 1 n 32 1 n. The appropriate Riemann Sum is ... n2 lim n→ lim n→ Section 5.4 4 1 2n2 n 2n 3n 6n2 5 x2 x2 1 n3 1n 6 1 1 lim n→ 1 3 1 2n 1 6n2 1 3 The Fundamental Theorem of Calculus 4 1. f x 12 0 is a maximum for a 0 3 −1 −2 i 2 2. f x 1 cos x dx dx is positive. −5 2 cos x 0 0 0 5 −2 3. f x x x2 −2 5 1 4. f x 2 x2 5 x 2 x x2 1 dx −5 0 x2 5 2 x dx is negative. −2 2 2 −5 2x dx 0 x2 1 0 7 7 1 1 5. −5 0 1 6. 3 dv 2 3v 37 2 32 15 430 Chapter 5 Integration 0 x 7. 0 x2 2 2 dx 1 2x 5 8. 3v 1 t2 5 32 v 2 4 dv 2 9. 1 2 1 3 2t 1 3 3x 2 10. 5x 1 1 1 2 dt 2t 4t2 0 27 1 t3 14 t 4 9t dt 2t2 9 2 1 4 3 2 2 1 2 1 1 dt 1 2 3 x2 13. 1 2 4 15. u 34 v 4 3 1 3 t 8 1 19. x x 3 0 12 3 3 4 3 3 3 4 3 3 1 3 1 2 2 2 x1 dx 1 3 2 x1 x 2 0 2 20. 2 t 2t1 t dt 0 2 t3 2 43 t 3 dt 0 0 t1 21. 3 t2 3 34 t 4 dt 1 1 22. 8 x x2 dx 23 x 1 2 3 4 3 35 t 5 0 3 0 1 3 2 3 2 3 4 2 4 8 2 1 x2 32 dx 4 0 2 2x 1 1 2 3 2 8 12 x 44 1 3 1 2 1 2 1 dx 1 2 4 38 1 2 1 4u1 2 4 0 4 2 3 3 4 2t 9 2 2 23 u 3 du 4 3 0 5 2 1 12 t 3 8 2 dx x 1 2u 34 t 4 2 dt 1 18. 2 2 1 v1 3 dv 17. 1 u1 3 16. 1 u 1 4 2 du u 1 2 x u2 2 1 du u2 u 1 3 x 1 dx 1 14. 1 10 3 1 43 t 3 1 4 4t 92 t 2 45 2 1 39 2 8 2 3 4x 0 12. 6 1 3 1 11. 20 2 5x 2 2 x3 4 dx 5 2 2 75 2 4v 1 t3 3 2 dt 1 2 0 1 8 22 1 23 x 3 1 0 25 t 5 2 3 4 2 3 1 18 tt 20 15 6t 11 32 2 2 2 0 3 5 2 0 22 20 15 12 27 20 1 x2 3 x5 3 dx 8 135 x 25 3 38 x 8 1 3 8 x5 3 24 80 1 15x 8 1 39 80 32 144 80 4569 80 16 2 15 S ection 5.4 3 32 2x 23. 3 dx 3 3 0 2x dx 2x 0 3 dx 32 3 x2 x2 0 5 32 4 3 x 3 x 4 dx 3 2 1 dx 7 2 4 2 0 2 x3 3 x2 2 x3 3 0 2x2 3x 13 2 8 4 dx 3 4x 2 4x x 9 12 4 x2 3 dx 1 2x2 4x 3 dx 2 3 4 3 9 0 cos x 3 2x2 18 9 4 3 0 3 x3 3 3x 0 4 3 sin x dx 8 3 8 1 1 3 x3 3 3x 1 1 3 2 split up the integral at the zeros x 1, 3 4 3 64 3 3 32 12 9 18 9 4 1 0 1 2 0 0 4 28. 1 0 sin2 cos2 4 d 2x cot x 0 4 e 6 1 x 1 x 1 2 x2 e2 ln x 5 dx 1 1 e2 0 3 2 1 2 2 1 1 1 dx x 5 x ln x 5 ln 5 1 tan d 0 ln 1 34. 4 ln 5 1 2 3 4 sec 23 3 1 3 33. 3 3 e 1 dx x 2x 1 2 4 3 3 tan x 6 2 csc2 x dx 2 31. sec2 x dx 29. 4 0 0 6 6 4 d 2 30. 5 3 dx 4 dx 3 x2 x3 3 32. x 28 0 1 4x 9 2 4 1 3 dx 0 27. 23 3 x2 9 4 2 4 4x 8 9 2 3 x 2 dx 4 4x x2 8 3 2 5 25 2 35 0 26. 9 2 x dx x2 2 7x x 4 0 9 4 9 4 x2 2 4 dx 9 5 x x2 0 4 4 3 9 4 5 4 dx 2 25. 9 2 3x 431 3 2 split up the integral at the zero x 32 3x 24. The Fundamental Theorem of Calculus 4 sec 42 3 42 2 2t cos t dt t2 sin t 2 2 2 4 2 1 4 1 2 432 432 Chapter 5 Integration 2 2x 35. 2x ln 2 6 dx 0 3 36. t2 2 5t dt t 0 2 4 ln 2 6x 0 5t ln 5 3 9 2 0 1 1 ln 2 12 125 ln 5 sin d e cos 2e 38. e sin x e cos 1 e 1 cos 1 1 e e x3 3 0 3 ln 2e x x dx 0 2 1 1 dx x2 x 44. A 2 1 x 2 0 1 2 1 x3 2 2x3 dx 3 2 0 x3 2 2 2 0 8 4 2 Area 10. 2c 2c c 1 1 c 2 1 c c x1 1 3 dx x 4 2 3 x2 2 4 2 A 6. 0 34 x 4 2 4 ln e 4 ln 1 8 3 8 0 4x3 3 6 42 3 6 0 42 3 3 82 3 2 82 3 3 22 42 3 ± 1± 6 3 52. 1 ex e2 9 dx x3 9 2x2 3 1 0.4380 or c 2 1.7908 9 2 2 c3 42 3 e2 4 9 2 c 6 1 2 1 9 c3 42 3 e0 1 0 fc 3 1 x3 3 2 e x dx 50. 4 0 x2 2 32 x 2 x2 dx 3x 0 1 c 1 0 2 e 4 ln x 0 sin x 48. Since y ≥ 0 on 0, 3 , x dx 0 8 5 1 2 cos x dx 0 x4 0 fc 2 0 8 x 2 2 x dx 1 12 3 5 2x 0 47. Since y ≥ 0 on 0, 2 , 2 15 x 5 2 0 4 dx x ln 2 46. Since y ≥ 0 on 0, 8 , 1 dx x3 x 3 xx 10 5 2 A x 4 dx 1 43. A 2 0 c 25 x 5 2 1 2 cos x 2 3x2 sin e 1 45. Since y ≥ 0 on 0, 2 , A sin 2e 1 1 x2 2 sin x dx 0 x ln e 40. A 0 42. A sin e 1 6 3 3 41. A sin 2e 1 3x1 x2 2 x2 dx x ln x e 1 39. A 51. 72.546 2e 1 dx x cos x 1 124 ln 5 1 1 e 9 2 1 ln 5 0 12 1 e 37. 49. 3 ln 2 0 3 9 2 1.6510 4 3 0 9 . 2 3 16 4 20. S ection 5.4 4 2 tan x 21 2 1 4 cos x dx 54. 4 4 fc 4 fc 4 2 5 56. f x 1 5x ln x 2x dx 0 ln 4 5 0 15 ln 4 0 30 2c 3 30 15 7 ln 2 ln 4 ln 4 3 ln 4 2c 2.1640 1 4x 4 x2 dx 4 2 Average value 13 x 3 2 2 1 4 8 3 8 8 8 3 7 3 ln 2 log2 8 3 7 3 ln 2 (− 2 3 3 ,8 3 ( 1.7512 5 −3 4 58. x 3 1 3 1 1 ( 2 3 3 ,8 3 ( 8 3 8 when x2 3 2 1 ln 2 0 7 ln 2 ln 4 2 2 0 7 ln 2 c 2 3 8 ln 2 30 ln 4 10 c 1 2x ln 2 10x 3 2c 15 3 c 57. 0, 3 1 fc 3 3 c 15 10 4 15 1 3 c 5 2x, 10 3 20 fc 4 ± 0.4817 2 1, 4 1 dx x 1 ± 0.5971 2 ± arccos 1 , x 33 2 c ± arcsec c 3 3 cos c ± sec c 5 3 4 sec2 c 3 3 8 2 sec2 c 55. f x sin x 3 4 4 433 3 3 4 2 sec2 x dx 53. The Fundamental Theorem of Calculus 4 x2 1 dx x2 8 or x 3 4 23 ± 3 ± 1.155. 3 2 1 x 2 dx 1 23 1 3 16 3 2x 1 x 3 1 3 −1 30 7 ln 2 434 434 Chapter 5 Integration 2e x, 59. f x 1 1 Average value 1 , 2x 60. f x 1, 1 1 2e x dx 1 1, 4 1 e x dx 1 1 4 1 Average value 4 1 1 1 ex e 1 e 2e x e 1 e x ln 1 2x 6 ln 4 x 3 ln 4 0.2310 1 e e 1 ln 4 6 1 1 ln 4 6 2x 1 e 2 ex 4 1 ln x 6 2.3504 1 1 dx 2x 1 e 2 0.1614 5 2.1640 1 (0.16, 2.35) −1 1 −1 4 0 0 61. 1 1 sin x dx 0 2 cos x 62. 0 0 2 Average value 2 2 (0.690, π ( 2 sin x x − 2 1 2 0 sin x 0 2 1.5 2 (0.881, π ( 2 (2.451, π ( 3 2 2 cos x −1 0 x 8 2.71 − 0.5 0.881 5 63. The distance traveled is 0 v t dt. The area under the curve from 0 ≤ t ≤ 8 is approximately 18 squares 30 540 ft. 64. The distance traveled is 0 v t dt. The area under the curve from 0 ≤ t ≤ 5 is approximately 29 squares 5 145 ft. b 65. If f is continuous on a, b and F x f x on a, b , then f x dx Fb Fa. a 7 1 7 f x dx 66. (a) (b) Average value Sum of the areas A2 1 3 2 A3 1 1 1 2 A4 2 (c) A 1 2 2 1 31 8 62 1 20 20 6 Average value 8 f x dx 7 1 A1 10 3 y 7 y 6 4 5 A1 4 3 A2 A3 3 A4 2 2 1 1 x 1 x 1 2 3 4 5 6 7 2 0 Average value 2 0.690, 2.451 2 2 cos x dx 2 3 4 5 6 7 8 6 4 3 S ection 5.4 6 2 f x dx 67. The Fundamental Theorem of Calculus area of region A 68. 1.5 6 f x dx area of region B 0 1.5 3.5 6 2 f x dx 69. 6 f x dx 0 0 6 2 f x dx 5.0 6.5 0 3.5 1 6 72. Average value 3 1 3 (b) Fx 0 1500 500 sec2 x dx 75. 2 R 0 1500 k2 Rr R 2 r dr 0 3R r 3 0 76. (a) 0.1522t2 0.1729t 0 1 0.08645t2 5 0.0374t3 dt 0.5833 3 0 827 newtons 0.00935t4 0.5318 liter 0 10 24 0 −1 24 0 The average value of S appears to be g. The area above the x-axis equals the area below the x-axis. Thus, the average value is zero. 8.61 10 4t3 0.0782t2 0.208t 0.0952 60 (b) 0 5 0.05073t3 (b) 1 77. (a) v 1 3.5 6 f x dx 2 0 0 6 2kR 3 5 1 5 kR 3.0 0 500 R 1.5 3 0 500 sec2 x 1 2 tan x 826.99 newtons 74. f x dx 15.5 k sec2 x k 5.0 0 0 12 F0 2 6 f x dx f x dx 0 2 2f x dx 70. 0 2 dx 0 73. (a) F x 1.5 2 6 71. 2 f x dx 2 f x dx 2 0 435 v t dt (c) 90 0.0782t3 3 4 0 − 10 10 4t 4 8.61 0.208t 2 2 60 0.0952t 2476 meters 0 70 − 10 78. (a) histogram (b) 6 N 18 16 14 12 10 8 6 4 2 9 12 15 14 11 7 2 60 83 60 4980 customers t 123456789 —CONTINUED— 7 436 436 Chapter 5 Integration 78. —CONTINUED— 0.084175t3 (c) Using a graphing utility, you obtain N t (d) 0.63492t2 0.79052 4.10317. 16 −2 10 −2 9 (e) N t dt 85.162 0 The estimated number of customers is 85.162 60 5110. (f) Between 3 P.M. and 7 P.M., the number of customers is approximately 7 N t dt 60 50.28 60 3017. 3 Hence, 3017 240 12.6 per minute. x 79. F x t t2 2 5 dt 0 F2 4 2 F5 25 2 55 64 2 58 x2 2 0 5x 25 2 F8 x 5t 8 52 8 x t3 80. F x 2t t4 4 2 dt 2 x4 4 x4 4 x t2 2t 2 x2 x2 2x 2x 4 4 4 t3 2t 4 2 F2 4 4 4 4 0 Note: F 2 2 dt 0 2 F5 625 4 F8 84 4 25 10 64 16 10 dv 2 1v 4 167.25 x x 81. F x 4 10v 1068 2 dv 1 10 x 10 10 1 1 x 10 v x x 82. F x 1 2 F2 1 4 x 2 dt t3 2t 3 dt 2 1 4 0 F2 10 1 2 5 F5 1 25 1 4 21 100 F5 10 4 5 8 F8 1 64 1 4 15 64 F8 10 7 8 35 4 0.21 1 t2 x 2 1 x2 1 4 S ection 5.4 x 83. F x The Fundamental Theorem of Calculus 437 x cos d sin sin x sin 1 1 1 F2 sin 2 sin 1 F5 sin 5 sin 1 F8 sin 8 sin 1 0.0678 1.8004 0.1479 x 84. F x x sin d cos cos x cos 0 1 cos x 0 0 F2 1 cos 2 1.4161 F5 1 cos 5 0.7163 F8 1 cos 8 1.1455 x x 85. g x 86. g x f t dt f t dt 0 0 0 0 (a) g 0 f t dt (a) g 0 0 f t dt 0 0 0 2 2 g2 f t dt 4 2 1 f t dt 24 4 1 2 44 8 f t dt 8 2 4 f t dt 2 6 4 0 4 0 4 g4 f t dt 7 2 g4 9 0 6 0 6 g6 f t dt 9 1 g6 8 2 0 8 0 8 g8 1 2 f t dt g2 7 f t dt 8 3 g8 5 0 0 (b) g increasing on 0, 4 and decreasing on 4, 8 (b) g decreasing on 0, 4 and increasing on 4, 8 (c) g is a maximum of 9 at x (c) g is a minimum of 4. (d) y (d) 8 at x 4. y 10 4 2 8 6 −2 4 x −2 2 4 8 10 −4 2 −6 x 2 4 6 x t 87. (a) 0 (b) d 12 x dx 2 2t x 2 2x x t3 1 dt 0 (b) 89. (a) 12 x 2 34 t 4 t dt 8 (b) d 34 x dx 4 3 12 x3 x x x2 x 34 x 4 3 8 x1 3 2x 14 t 4 t dt 0 d 14 x dx 4 x 3 12 x 2 0 x t t2 88. (a) x t2 2 2 dt −8 8 3 3 x 16 12 t 2 x 0 14 x 4 12 x 2 x2 2 x 4 2 1 34 x 4 x 3 12 t dt 90. (a) 4 (b) d 23 x dx 3 2 23 t 3 16 3 x 23 x 3 2 4 x1 2 16 3 2 x 23 x 3 2 8 438 438 Chapter 5 Integration x tan t tan x (b) d tan x dx e t dt dx e dx e 1 94. (a) F x 1 ex (b) x t2 96. F x 2t dt x2 2x x 4 98. F x Fx 4 x x Fx x 1 t4 1 dt x4 Fx x sec3 t dt 100. F x t cos t dt 0 2 2x 2 8x Fx x 2x2 2 x x t3 dt x 0 x t3 dt x Alternate Solution: x 1 dt x Fx 2 1 dt 4t x 1 dt 0 x x 4t 1 dt 0 1 4x sin x 103. F x t dt 0 sin x 2 1 dt 0 4x 12 2 23 t 3 cos x 2 1 sin x 0 8 2 sin x 3 cos x sin x Alternate Solution: sin x Fx t dt 0 sin x t3 dt 0 0 4t x t3 dt x 4t t3 dt 0 2 4t 0 x 0 Fx 8 Fx x t4 4 t 3 dt Alternate Solution: 2 10 x sec3 x Fx x cos x x x d sin x dx sin x cos x 32 1 1 102. F x t 1 dt 1 x 4t x Fx 1 x 97. F x 2 2t2 Fx ln x 1 dt x Fx ln t 0 101. F x Fx x 1 dt t x 99. F x t dt 1 Fx d ln x dx sec x tan x x2 2 2 x t2 t2 1 2 Fx 2 x ex x 95. F x d sec x dx 1 1 e (b) et 1 sec x 3 x x sec t 3 sec2 x 1 93. (a) F x (b) sec t tan t dt 92. (a) 1 4 4 x x x sec2 t dt 91. (a) 0 x 3 1 x3 0 S ection 5.4 x2 104. F x t 3 dt t 2 x2 2 2 1 2t 2 2 x2 Alternate Solution: F x 3 x2 1 2x4 2 2x 1 ⇒F x 8 The Fundamental Theorem of Calculus 5 2x 5 2x x2 x3 105. F x sin t2 106. F x dt sin x3 Fx 2 3x2 sin x 2 Fx 3x2 sin x 6 x 107. g x 2 sin 2 g0 y 2 1 ,g 2 2 0, g 1 1 ,g 4 2 1, g 3 f 1 0 g x 1 g has a relative maximum at x 2. 2 3 4 −1 −2 108. (a) gt lim g t 4 t2 4 4 t→ Horizontal asymptote: y x (b) A x x 4 t 4t 4x2 lim A x 4 x 8 4x 1 4x 1 8x x 4 2 x 4 x lim 4x x→ 4 4 dt t2 4 1 x→ 0 8 8 The graph of A x does not have a horizontal asymptote. t3 6t 2 9t 2 3t 2 12t 9 3 t2 4t 3 3t xt 2x f t dt 0 109. x t d 0 0 3t 1 5 5 Total distance x t dt 0 3t 3t 1 3 t2 3 4t 0 4 1 dt 0 3 dt 1 4 28 units 20 5 t2 3 4t 3 dt t2 3 3 4t 3 dt 2x sin x 4 439 440 440 Chapter 5 Integration 4 110. x t t 1t t3 7t 2 15t 9 111. Total distance x t dt 1 3t2 xt 2 3 14t 15 4 v t dt Using a graphing utility, 1 4 1 5 Total distance x t dt 27.37 units. t 1 dt 0 4 2t1 2 1 22 2 2 112. P 2 2 sin d 2 cos 0 1 2 1 2 units 63.7% 0 0 114. True 113. True 115. The function f x x 1 2 is not continuous on 0 x 1 116. Let F t be an antiderivative of f t . Then, 2 dx 2 dx x 1 vx vx 1 2 dx x 1, 1 . f t dt Ft Each of these integrals is infinite. f x has a nonremovable discontinuity at x x 0. 2 d dx Fvx vx d Fvx dx f t dt ux F vx v x f vx v x 1x 117. f x x 1 t2 0 dt 1 1 t2 0 1 dt By the Second Fundamental Theorem of Calculus, we have 1 fx 1x 2 1 1 x2 1 1 x2 1 Since f x 1 x2 x2 1 0. 1 0, f x must be constant. x 118. G x s s 0 f t dt ds 0 0 (a) G 0 0 s s s f t dt ds (b) Let F s 0 s x f t dt. 0 0 x x (c) G x fx Gx f t dt F s ds 0 0 0 (d) G 0 0 f0 f t dt x 0 Gx Fx x f t dt 0 0 0 G0 Fux ux ux 0 0 f t dt 0 0 Fux F ux u x f ux u x. Section 5.5 Section 5.5 Integration by Substitution f g x g x dx 1. 2. 5x2 1 x2 x3 u 5x2 10x dx du gx 1 g x dx 10x dx 1 dx x 3. 2 2 x 1 Integration by Substitution x3 1 3x2 dx dx x2 1 2x dx 4. sec 2x tan 2x dx 2x 2 dx 5. tan2 x sec2 x dx tan x sec2 x dx 6. cos x dx sin3 x sin x cos x dx 7. 1 Check: 2x 4 1 2 dx 2x d 1 2x dx 5 x2 5 x2 8. C 5 9 5 C 21 Check: 4 2x 3 2x dx d x2 9 dx 4 9 4 C 4 4 C 4 x2 x2 9. 9 Check: 10. 1 x2 12 d2 9 dx 3 2x2 13 Check: 11. d3 1 dx 4 x3 x4 3 2 dx Check: 12. x2 x3 Check: 13. x2 x3 Check: 9 2x dx x2 5 4 dx 43 1 4 x4 d x3 5 dx 15 1 4 dx d x3 1 dx 15 2 9 3 x2 2 3 3 9 2 2x2 43 C 3 4 3 3 x3 5 5 5 C 4 3x2 dx 1 4 5 x3 2x2 32 C 2x x2 9 2x 13 1 3 C 2x2 x4 4x 3 3 5 13 5 4x C 12 2 4x3 x4 1 x3 5 3 5 5 4 3x2 15 3x2 dx 12 1 x4 3 4 3 4x3 dx 5 x3 C x3 2 x2 C 4 1 3 3 x4 3 12 C 1 3 C 3 1 4 2x2 1 3 32 C 32 4x dx d x4 3 dx 12 x2 32 2 x3 5 x3 C x3 1 x3 1 3 5 1 4 3x2 15 3 C 15 5 4x2 5 x2 x3 C 1 4 x3 1 15 5 C 9 3 3 2x 4 9 2x 441 442 14. Chapter 5 x 4x2 t t2 x2 5x 1 18. u2 u3 Check: 19. 20. 21. Check: 22. 9 d dx x2 dx x3 2 Check: x2 1 3 x3 9 C x2 21 1 3 2 x4 1 3 11 x3 x3 2 x2 2 u3 5 12 5 x2 32 C u2 x2 41 C 2 x x2 3 1 C 2 1 31 x2 1 x3 3x2 x3 1 C x4 41 2 1 C x2 9 x3 2 C x3 1 39 x3 43 5x 3 1 1 C C 3x2 13 9 1 19 3 15 1 8 2 1 1 2 t3 5x 1 u3 2x x3 1 C C C x3 1 x4 11 3 32 43 32 x4 5 12 2x x2 2 4x3 3x2 dx 19 13 x2 43 3u2 3 1 1 4 3x2 dx 2 12 5 1 11 2 2x dx 4x3 dx C x3 d 1 dx 3 9 x3 2 5 14 t 6 t4 1 u3 5 3 32 33 u 2 1 1 4 x3 1 1 3 2 C 1 31 1 4 x4 x2 3u2 du 3 C C 15 8 C 1 2t 2 4t3 C 32 2 3 32 5 2 4 1 3 t2 t2 3 2 3 C 2x dx 13 2 9 C d 1 dx 4 1 x 4 x2 dx 1 x3 2 12 12 4x2 3 24 C 32 2t 5 3 x 4x2 1 t4 5 4 32 34 t 2 C 32 1 1 4 12 2 4t3 dt x2 5 2 8x 1 t2 2 2 32 1 6 43 1 1 d dx 4 1 x2 12 1 u3 1 2 dx x3 dx 1 x4 2 Check: 1 3 d 2 u3 5 du 9 x2 3 Check: C x2 5 du x 1 32 15 1 8 d dx Check: 5 5 2 2 3 t4 5 3 24 3 2 t2 C 1 4 1 4x2 3 8 3 8x dx 2t dt 32 dx 13 12 2 2 3 4x2 C 3 d14 t dt 6 3 3 t2 2 5 dt Check: 17. 1 2 t2 d dt t 3 t4 4x2 d 4x2 3 dx 24 2 dt Check: 16. 1 8 3 2 dx Check: 15. Integration C C x2 S ection 5.5 x 23. x2 1 Check: Check: Check: 1 3x Check: Check: 1 2x dx Check: x2 t d dx 3x x Check: 30. d dx t C x4 x4 1 x 2 x 7 12 d 23 t dt 3 t2 t Check: d 14 t dt 4 2x3 2 2 dt t 2 45 t 5 2 C 2 1t 4 3 4 1 t2 1 3x x2 C 3 1 1 t2 1 t x3 3 1 9x C 14x1 2 C 3x4 1 9x C 2 2x C 1 2x C x 3x1 2t3 2 2 7x 14x1 2 t3 t2 x4 1 C x3 1 x4 C 2 1 x1 2 212 x3 t1 12 12 C 1 2x dx dt 2 C x2 1 1x1 91 1 2x 1 2 2 12 dx dx C 2 1 x 9 x2 1 C x 4x3 1 t 12 2x 1 x3 3 dx x2 1 2x 2 C 12 1 dt t2 2 C 1 22 2x 2x 1 x4 1 41 4 1 x 9 x2 1 x 9 3 1 t 12 1 1 x4 4 12 1 1 2 1 2 C dx x2 4x3 dx 4 1t 4 1 2 12 1 1 x2 2 12 2x dx 1 1 2 1 1 2 12 C 1 dt t2 Check: 31. 12 2 d 25 x dx 5 2t2 x2 1 1 d 13 x dx 3 1 dx 2x 27. 1 4 d dt x2 26. 3 1 x2 1 d dx 1 t 1 25. 29. d dx x3 dx 1 x4 24. 28. 1 2 dx Integration by Substitution 2 C 2 dt t1 C 2t dt C 23 t 3 t3 2 12 25 x 5 dx x2 45 t 5 2 2t3 14 t 4 t2 2t t2 t 3x x 2 C 2 t 2x3 2 7 C t 2 2 2t2 t 23 t 15 2 5 6t C C 2 5 x x2 5x 35 C 443 444 Chapter 5 32. t3 3 1 dt 4t2 9 y y3 2 y8 Check: 35. y 2 2 23 y 3 2 dy 2 2 9 d 6y3 dy 25 y 5 25 y 5 2 4y2 2 9y1 2x 27 y 7 27 y 7 2 2 1 x2 2 2x 3 2 12 10 3 2x dx 1 x3 12 C 4 8x 1 8x 1 1 x2 2 x2 C dy dx x4 x2 dx 8x 1 12 8x dx 12 2x 12 C 1 C x2, 2, 2 x4 3x2 dx C x2 1 2 1 2 x 2 2 4 3 x2 1 2 32 4 x2 12 1 4 3 C 2x dx x2 32 −1 1 4 3 1 4 3 2, 2 : 2 y 22 x2 32 2 −1 C⇒C 32 2 −2 2 12 x3 1 x C 3 y3 2y8 x2 2 dx (b) C 2 10 1 x3 3 12 1 3 2 10x2 dx 1 x3 20 3 2x y y3 2 y5 38. y y −2 y 16 y y 39. (a) 9 2 4 y2 14 7 C C 1 1 2x C y3 C dx 3 y 2 C 2 15 36. y x2 3 C 2 12 1 2x 1 4t 23 y 5 C C 4y2 2 d 2 dy C x2 12 x2 2 x2 dy y5 2 x2 16 16 2 x 1 2 2 8y 14 t 12 C 1 4t2 C 2 4 16 x2 y3 1t1 41 4x dx 16 x2 x2 2 2x2 2 y3 1 t4 34 dt 13 t 3 y d 4 y2 14 dy 7 4 x dx 37. y 15 dy 2 C 9y1 2 4x 4 1 t 4 1 4t y dy d 23 y dy 5 Check: 34. 13 t 3 d 14 t dt 12 Check: 33. Integration 2 2 C 8 dx S ection 5.5 (b) y dy dx x 2 x3 y 40. (a) x2 x3 Integration by Substitution 445 1 2, 1, 0 2 1 2 dx 1 3 x3 2 1 x3 1 x −2 1 x3 1 3 3 2 −2 0 41. (a) 13 x 9 3 C 13 x 9 1 3 C C y 1 3 42. (a) y y 4 3 x −4 4 x −3 3 −4 −3 dy dx x cos x2, 0, 1 y (b) 3x2 dx , u x cos x2 dx 1 cos x2 2x dx 2 2 sec 2 x tan 2 x , 0, 1 2 sec 2x tan 2x dx, u 2x C sec 2x 1 sin 0 2 1 sin x2 2 y dy dx y 1 sin x2 2 0, 1 : 1 (b) C⇒C 1 1 sec 0 y −6 C⇒C sec 2x 5 1 C 0 6 −3 43. (a) 44. (a) y y 4 5 (0, 1) x −4 4 x −2 5 −4 −2 (b) dy dx 2e x 2, 2e x2 4e y x2 dx 0, 1 : 1 4e x2 1 dx 2 xe xe C 4e0 y 4e dy dx y (b) 0, 1 C x2 4 C⇒C 0.2x2, 0.2x2 C 6 −4 0, 8 −2 3 2 2.5e0 y 5 3 : 2 1 0.4 0.2x2 dx 1 e 0.4 5 3 2 0, 2.5e e 0.2x2 2.5e 0.2x2 C 0.2x2 2.5 1 0.4x dx C C⇒C 1 446 Chapter 5 45. (a) Integration y 46. (a) y 5 6 4 x −5 2 5 x 10 −2 −5 dy dx (b) ex 3, 1 2 0, ex 3 dx y (b) 3ex 3 dy dx esin x cos x, esin x cos x dx y C esin x esin C 1 esin x 1 ,2 : 2 11 0, : 22 3e0 3 3e x y 5 2 C⇒ C 5 2 3 ,2 y C C⇒C 1 4 −6 6 −4 sin x dx 47. 49. 51. 1 2 sin 2x dx 1 2 cos 53. Let u 1 57. x cos 5x, du 1 ex 2x 1 cos 2x 2 1 1 d 2 sin 1 2 1 2 54. ex 52. C e 2 2x 2x x3 cos 2x sin 2x cos 2x dx 1 2 58. sec 2 x dx 59. tan4 x sec2 x dx 3x2 dx sec 2 C 15 tan x 5 2x 1 sin 2x 2 2 x sin x2 dx x3 e 1 6 cos x4 1 sin 6x 6 cos 6x 6 dx 1 2 C C 1 cos x2 2 sin x2 2x dx x2e 55. C x3 dx 1 2 x tan 2 C 2 sin 4x dx x 1 3 e x3 C x3 3x2 dx C C 12 sin 2x 4 C 1 cos 2x 2 2 2 sin 2x dx 2 sin 2x cos 2x dx tan5 x 5 1 x2 e 2 2 dx sin 2x 2 cos 2x dx 1 2 sin 2x cos 2x dx x tan 2 1 cos 6x dx sin x4 4x3 dx C dx sin 2x cos 2x dx 50. C 4x3 sin x4 dx 1 e 3 5 dx. e5x 2 48. C sin 2x 2 dx d e5x 5 dx 56. cos x 2 1 cos 4x 8 60. OR 1 cos2 2x 4 C1 1 dx C sec 2 C1 OR C2 x C tan x sec2 x dx tan x 3 32 2 C 2 tan x 3 32 C S ection 5.5 61. csc2 x dx cot3 x 3 cot x 62. sin x dx cos3 x 63. cot2 x dx 65. ex ex 3 csc2 x C 12 tan x 2 cos x 2 sin x dx 1 dx ex 1 2 dx 1 2 cot2 x C cos x 2 C C 1 sec2 x 2 1 2 cos2 x x C 64. csc2 1 sec2 x 2 C 1 sec2 x 2 C 66. cot x 1 x dx 2 ex 1 C C 3 x 2 2 csc2 3 1 C1 1 dx 2 1 3 3e x dx 3e x x 2 2 cot 3e x 1 1 3e x dx 2 6 67. Let u ex, du 1 ex 1 ex dx. ex dx ex 5 71. e sin x 5e dx 2x 1 73. e x sec2 e x x esin 3 x 2 dx 2 x 3x 2 x5 x2 dx x 2 70. e2x 2ex ex 1 x dx 72. etan 2x sec2 2x dx x e x dx 74. ln e2x 1 dx dx. 2 ex 2 x e 2x e x e x dx C 1 tan 2x e 2 sec2 2x dx 2 C 1 dx x C x 76. C 2 4 78. 3 x 4 ln 4 dx C C x 73 x 2 dx 1 2 x 73 23 x 2 15x 2 ln 5 1 5 2 ln 5 x2 C C dx C ex 2x x2 3x 2 ln 3 2x dx x 1 tan 2x e 2 C 2 x e ex x x2 2 e x e ex dx C 1 dx 2 1 5 2 ex ex dx 2x 3 ln 3 77. 2e x dx e x2 C cos sec2 e dx e x, du C e sin tan e 75. x 2ex ex ex dx 32 e e 1 cos x dx 12 ex 2x dx 5 e 2 e2x ex 68. Let u ex 1 2 1 3 69. 447 csc2 x dx 2 cot x 2 Integration by Substitution 1 73 2 ln 7 x 2 C 2 dx C 448 Chapter 5 79. f x Integration 1 4 2 f2 x2 1 4 3 fx 2 3 32 1 0 3 2 1 2 x2 dx x4 32 12 1 4 3 C C⇒C x2 x2 4 2x dx x2 32 C dy dx 0.4x 3, 0, y 80. 0.4x 3 dx 3 0.4x ln 0.4 2 2 f0 81. f x cos Since f 0 x dx 2 3 fx 2e dx C, C 3 ln 0.4 82. f x 3. Thus, e 1 dx 4 x4 1 C⇒C x4 33 : 22 1x e 2 f0 C1 e f0 1 fx 1x e 2 dx 1x e 2 e x C1 0 86. f x sin x x e dx 1 ⇒ C2 C2 e x 1x e 2 e x fx cos x C2 0.6x2 dx 19 6 19 6 0.2x 3 e2x dx 1 2 1 cos x 1 ⇒ C1 2 C1 1 2x e 2 1 0 fx cos x sin x f0 1 4 C2 fx x sin x 1. Thus C C⇒C 5 e 3 f0 1x e 2 fx x 0.2x3 5 3 fx C C, C 3 0.2x 3 e 5 e 3 9 sec x 0.2x 3 dx 1 0.6 9 3 ln 0.4 1. x2e fx 0, 85. f x sec sec x 1 2 3 ln 0.4 1 2 3 Since f 1 3 84. 1 ⇒C 2 sec x tan x dx fx 8 1 dx 3 3 C C C 8 8e C 3. x4 1 fx x 2 x4 8e f0 x 2 2 sin 0 2 sin 83. f x 2 sin 3 0.4x 3 3 0.4x ln 0.4 y 1 2 1 2x e 2 1 2x e 4 x 1 ⇒ C2 4 1 2x e 4 1 dx C2 0 1 2x e 2 1 C1 S ection 5.5 87. u x 2, x 2, dx 2 dx xx u u 2 u3 25 u 5 88. u du u du 2 2u1 43 u 3 2 2u3 2 3u 15 2 32 2 x 15 2 32 3 2 u 3 dx 3 3x 2 1 x, x x2 1 1 u, dx x dx 2 u 4 C u du u1 2 2u3 2 u5 23 u 3 2 45 u 5 2 27 u 7 2u3 2 35 105 x 32 35 2 1 105 2 x 32 15x2 x, x x 2 2 u, dx 1 2 x dx du 2 C 15u2 42u 2 1 105 90. u C 42 1 x 12x 15 1 8 C du 3 u 3u1 2u3 2u3 5 u du 2 u3 2 25 u 5 2 2 du C 2 5 2 2 5 x 2 2 5 x u 32 32 2 3u1 2 2u3 C 5 2 x 3 2 du 2 10 C C 1 2x 20 C du 1 u3 1 du 2 3 32 2 2x 1 2x 20 10 3 32 4x 1 2x 5 89. u u 2 u3 2 2u 20 C x C C x 2 C 449 1 du 2 , dx 125 u 45 C 3x u 1 4 du 10 2 3, x x 2x 2 2 x 15 2x Integration by Substitution 3 32 x 3 10 4 1 C C C 450 91. u Chapter 5 Integration 1 u 2 1, x 2x 1 du 2 1 , dx x2 1 dx 2x 1 12 u 1 u 8 1 8 12 2 1 u u2 11 du 2 2u 1 4 du u3 2 2u1 2 3u 12 125 u 85 2 43 u 3 2 6u1 2 u1 2 2 3u 60 10u 45 2x 1 3 2x 60 1 1 60 x 4, x 2x x 2x 1 3x2 4, du C C 2 1 12x2 1 15 92. Let u 2x du 10 2x 1 52 C dx. u 1 dx 4 2u 2u1 4 u 2 1 8x 2x C C 93. u x 1, x u 1, dx x du 12 7u 13 45 x 1 x 1 du u dx 1 du u u u du 1 u 43 u 3 2 21 u 3 2 14u1 2u 2 C 21 1 u 2 3 x 42x 4 21 2 3 x 4 2x 13 C C 1 x t 4, t u 4, dt 4 dt du u x2 95. Let u 1, du 1 t3 t 4 u1 3 du x x2 1 u4 37 u 7 3u4 7 3 3 4u1 3u4 3 3 C 7 3 t 7 4 43 3 t 7 4 43 C t t 4 3 7 C C 2x 1 1 2 1 x2 1 3 2x dx 1 1 1 4 0 1 1 1 C. 2x dx. 12 x 8 du 3 u 1 3 dx C 2x where C1 94. u C 2x x du du 2 2u x 1 1 12 u 2u1 u C u u 1 1 C C C1 S ection 5.5 x3 96. Let u 4 1 3 8 2 dx 2 x3 8 x3 2 1 dx x2 dx 0 99. Let u 2x 1, du 4 1 2x 1 dx 1 2x2, du 2 x 2x2 1 0 1 101. 2x e 3 1 x2 4 12 0 4 12 1 2 dx 2 2x2 1 12 1 1 e 2 2 dx 8 3 1 2 1 2 1 0 9 0 1 e 2 2x 0 3 2 2x2 1 e1 1 dx x e 8 3 1 2 2 1 3 e3 3 dx x2 x 1 13 e 3 3 1 1 e 3 x 1 e2 e 3 e3 1 x dx. x2 2 e x dx e 9 x 2 1 0 e e 1 1 e 1 dx. 2x x, du 1 2 x2 2 0 dx 2 1 x 2 1 4 x2, du x3 4 x2 dx 1 2 1 2x 9 2 dx 1 x 1 2 1 1 2 1 2x dx. 106. Let u 2 0 22 1 0 1 0 32 1 2 x1 1 4 27 9 2 x x 2 2 dx 1 x2 2 1 2 4x dx 0 2 9 32 0 0 1 3 2x 0 2x e x2 , du 2 105. Let u 1 4 2x 1 4 e1 e3 x dx x2 xe 2 43 x 9 1 2 1 4 3 2x dx 1 104. Let u 32 2 2 x3 1 3 32 3x2 dx 2 1 dx 1 103. 12 1 2 x 41,472 4x dx. dx 2 e1 3 2 dx. 1 2 1 2 dx 0 102. 8 2 2x dx. x4 100. Let u 8 1 1 2 2 0 x3 x2, du 4 3 2 1 3 2 1 98. Let u 8 34 1 x3 8 3 3 3x2 dx 3x2 dx. 1, du 2x2 x3 2 2 1 64 9 97. Let u 451 3x2 dx. 8, du 4 x2 x3 Integration by Substitution 2 4 0 x2 13 2x dx 3 4 8 2 x2 43 0 34 8 8 3 44 3 6 33 4 2 3.619 12 8 9 2 452 Chapter 5 107. u 2 Integration x, x When x 2 u, dx 1, u du 1. When x 2, u 1 2 0 x dx 1 1 u3 u du 1 108. Let u 2x When x 5 1 u 2 1. When x 1, u 5, u 9 1 dx 1 2u 1 1 4 2 2 x dx 3 cos 0 x 3 2 x 2 2 ln 2 x 2 25 u 5 du 23 2 3 2 3 2 0 3 2 2 9 u1 2 0 2 5 2 1 1 2 du 2 123 u 43 2 dx 1 1 2u1 2 1 16 3 2 1 7 2 ln 2 7 ln 4 18 52 72 3 2 3 2 2 0 0 3 112. 2 3 5 dx 27 2 25 dx 2x 2 0 0 2 113. dy dx 18x2 2x3 y 3 y 3 4 13 y 1 2, 0, 4 2x3 1 2x3 2 y 1 48 , 53 3x 48 C⇒C 3 1 3 C 3x 3 3x 5 8 3 y dy dx 2x 2x2 1 , 5, 4 y 1 2 y 1 2x2 1 2 12 2x2 1 12 4x dx , u 2x 2 1 12 C 4 49 C 7 y 2x2 1 3 2x2 C⇒C 1 3 C 3 5 5 5 8 3x 5 2 33 dx 2 C C 2 8 1 4 dx 2 3 3x 115. 3 5 1 3 48 2 1, 3 16 3x 2x3 C 3 1 2x3 u dy dx 3 1 2x3 114. 6x2 dx, 4 15 9 2 2 dx 111. 2 3 33 4 8 3 12 u 1 2 sin x 1 4 ln 2 23 u 3 2 9. 1 4 2 27 3 x2 2 cos x dx u1 1. 11 du u 2 3 2 sin x 2 3 2 2 1 2 dx, x 1, du x 2x 1 110. u 0 x 109. 0. 2 2 2 1 C 8 4 C⇒C 1 S ection 5.5 116. dy dx 4x 3x3 9x2 1 32 117. u , 0, 2 x 1, x When x y 3x3 4x 3 2 9x2 1 u 0, u dx 2x 2 2 2 1 0 118. u 2, x When x x3 x 2, u 37 u 7 4 4 6, u 2 dx 8 u 23 2 1 cos 2x 2 2 cos x 1 sin 2x 2 cos x 0 3 sec2 2 2x, du 2 x dx 2 3 sec2 2 3 4u4 2 Area csc 2x cot 2x dx 12 3 34 u 4 3 du 3 8 1 3 7 12 4u1 3 3 10 u 10 du x 2 1 dx 2 8 12 7 u 7 3 3 e b 3u4 3 0 4 0 2 0 2 tan x 2 2 3 2 3 1 2 1 2 4 4 1 csc 2x 2 csc 2x cot 2x 2 dx 12 12 1 2 b 5 ex e5 1 147.413 b x e 124. 0 0 u1 2 dx. 4 ex dx 3 0 sin 2x dx cos 2x dx 5 u7 u du 0 sin x 123. 3 dx e a x e a a 150 a 0 b 6 0 6 125. xe x2 4 dx 2e 2 6 x2 4 2e 32 2 e 126. 0 0 2x 0 1.554 1 e 2 4 4.5 −3 −2 4 0 2 1 e 2 2 dx 3 − 4.5 3 4 8. 8 0 122. Let u u du du 0. When x 2 sin x 2 3 1209 28 2, dx x2 3 x 121. Area 1 384 7 2 120. A u 1 8 6 119. A 1 dx C 1 u Area 8. 8 u4 1 2 3x3 x 7, u C C⇒C 2x2 y 1. When x 1 2 3x3 2x2 453 du 0 12 1 12 1, dx 7 Area 3x3 Integration by Substitution 2x 2x 0 4 4 1 2 4.491 4752 35 454 Chapter 5 4 x 2x 127. 0 1 Integration dx 2 10 3 3.333 7 x3 x 128. 2 dx 7.581 129. xx 0 −1 28.8 3 20 3 3 dx 15 5 0 0 5 3 x2 x 130. 8 0 2 0 −1 1 dx 67.505 2 131. 1 cos 0 6 d 7.377 132. sin 2x dx 5 50 2 0 −1 −1 0 2 xe 133. x2 2 dx −1 2 x2 2 e 2 134. 0 0 1 e 1 2 4 5 1 1.0 0 e 2x 0 0.632 1 e 2 2 2 1 e 2 2 dx 2x 4 2x 0 1 2 4 4.491 4 −1 2 −1 −2 4 0 2x 2x 1 2 1 2 dx 1 2 dx 135. 4x2 4x They differ by a constant: C2 136. sin x cos x dx cos x They differ by a constant: C2 137. f x x2 x2 C1 2x2 sin2 x 2 2 C2 C1 x2 dx 0 2 32 5 2x2 x 8 3 1 6 C1 C2 sin2 x 2 C2 1 2 C2 1 2 sin2 x cos x is even. 138. f x x4 2 x cos2 x 2 sin x dx 2 1 dx 43 x 3 C1 1 is even. 2 x2 x2 3 1 6 C1 sin2 x 2 1 C2 1 1 43 x 3 1 dx sin x 1 cos x dx sin x cos x dx cos2 x 2 1 2x 6 1 22 dx 2x 272 15 2 x5 5 x3 3 2 2 2 sin2 x cos x dx 0 sin2 x cos x dx 2 0 2 2 sin3 x 3 2 0 2 3 144 5 S ection 5.5 x x2 139. f x 1 3 is odd. 140. f x Integration by Substitution sin x cos x is odd. 4 2 x x2 1 3 dx sin x cos x dx 0 2 x3 3 x2 dx 141. 0 4 2 0 2 8 ; the function x2 is an even function. 3 0 0 2 x2 dx (a) 2 0 2 2 8 3 x2 dx 0 0 0 0 8 3 x2 dx 16 3 x2 dx 2 2 2 x2 dx (c) 2 x2 dx (b) 2 3x2 dx (d) x2 dx 3 2 8 0 4 142. (a) sin x dx 0 since sin x is symmetric to the origin. cos x dx 2 4 4 (b) 4 4 cos x dx 2 since cos x is symmetric to the y-axis. 2 sin x 0 0 4 2 2 (c) cos x dx 2 2 cos x dx 2 sin x 2 0 0 2 2 (d) 0 since sin sin x cos x dx x cos sin x cos x, and hence sin x cos x is symmetric to the origin. x 2 4 4 x3 143. 6x2 2x 4 144. 4 x3 3 dx 4 sin 3x cos 3x dx 4 6x2 2x dx 0 3 dx 4 sin 3x dx 4 6x2 2 x2, then du 5 146. f x x x2 2x dx and cos 3x dx x5 3x 0 2 1 2 x2 3 dx 2 sin 3x 3 cos 3x dx 5 x2 3 232 0 0 0 145. If u 2 2x3 3 dx 0 0 13 u du. 2 2x dx 2 1 2 is odd. Hence, x x2 1 2 dx 0. 2 147. dQ dt k 100 Qt Q 100 t t 3 C Q0 k 100 3 dV dt Vt k t 1 2 k t 3 2 100 50, Q 50 2,000,000 ⇒ k t 3. 6 1 k 1 k 2 2 C V1 3 t V0 k 100 3 When t k 100 3 0 Qt Thus, Q t 148. t 2 dt k 100 C 2 C dt k t 1 500,000 400,000 Solving this system yields k C 300,000. Thus, Vt $250,000. When t C 200,000 t1 300,000. 4, V 4 $340,000. 200,000 and 455 456 Chapter 5 149. R 3.121 (a) Integration 2.399 sin 0.524t 1.377 8 Relative minimum: 6.4, 0.7 or June Relative maximum: 0.4, 5.5 or January 0 12 0 12 (b) 37.47 inches R t dt 0 (c) 150. 12 1 3 9 a (a) (b) (c) 4.33 inches b 1 b 1 13 3 R t dt 74.50 a 1 74.50t 3 262.5 1 74.50t 3 262.5 1 74.50t 12 151. (a) t dt 6 43.75 sin cos t 6 3 t 6 6 cos 262.5 cos a 1 223.5 3 0 12 0 262.5 74.50t 262.5 1 447 3 3 t 6 1 b t 6 b a 102.352 thousand units 262.5 1 894 12 cos 102.352 thousand units 223.5 262.5 262.5 74.5 thousand units 70 0 24 0 Maximum flow: R 61.713 at t 9.36. 18.861, 61.178 is a relative maximum. 24 (b) Volume R t dt 1272 thousand gallons 0 152. b 1 b (a) (b) a 2 sin 60 t cos 120 t a 1 1 60 1 cos 60 t 30 0 1 1 240 0 1 cos 60 t 30 dt 1 b a 1 cos 60 t 30 1 sin 120 t 120 1 60 1 30 60 0 1 240 1 sin 120 t 120 240 0 2 (c) 1 1 30 0 1 cos 60 t 30 1 sin 120 t 120 1 30 30 0 b 1 120 5 1 30 sin 120 t a 1 30 0 1 1 120 30 2 22 4 1 30 1.382 amps 1 30 1.273 amps 0 amps S ection 5.5 153. u 1 x, x When x 1 a, u b Pa, b a 15 4 u, dx 1 1 1 2 u1 1 x 2 b 3b x, x When x Pa, b a 1 0.586 a b 1 x 2 b 1 a b 32 1 1 5 a 3x 2 a 35.3% b 32 3x 2 0 58.6% x 32 dx b, u 1 1155 32 1 b. b u 3u3 2 du 1 1 a 2 3u7 2 3u5 2 u3 2 1155 2 11 2 u 32 11 du 1 385u2 495u b 1 29 u 3 a 67 u 7 2 2 25 u 5 2 1 385u2 495u b 1 u5 2 105u3 16 231 a 231 0.75 105u3 385u2 495u 231 0.025 2.5% 1 0 0.0139 t 48 2 385u2 495u 231 0.736 73.6% 0.5 x dt x et dt ≥ 156. 48 1 dt 0 Graphing utility: 0.4772 0 x 47.72% et 0 ex 157. (a) 1 15 2u3 2 3u 4 15 du u5 2 105u3 16 e 0.353 a 16 60 155. 0.0665 x 2 a. When x u9 1 b 1 b u5 2 (b) P0.5, 1 1 2 1 1155 2u5 2 105u3 32 1155 (a) P0, 0.25 2 1 1155 3 x1 32 1155 32 23 u 3 2 0.75 u, dx 1 u du 0.5 2 1 a, u b 15 2 5 u 45 du 3x b 1 u a 32 P0, b 1 b. 1 a (b) 154. u 2 1 b 1 b (a) P0.5, 0.75 32 b, u 1 15 4 x dx u3 457 du a. When x 15 x1 4 Integration by Substitution x ≥ t 0 1 ≥ x ⇒ ex ≥ 1 x for x ≥ 0 (e) The graph of h is that of g shifted two units downward. 4 g 4 9.4 0 f 9.4 0 −4 (b) g is nonnegative because the graph of f is positive at the beginning, and generally has more positive sections than negative ones. (c) The points on g that correspond to the extrema of f are points of inflection of g. 2, do not correspond (d) No, some zeros of f, like x to an extrema of g. The graph of g continues to 2 because f remains above the increase after x x-axis. −4 t gt f x dx 0 2 t f x dx 0 f x dx 2 2 ht 458 Chapter 5 sin x, 0 ≤ x ≤ 1. 158. Let f x n lim n→ i 1 x, du 0⇒u x dx, x x 5 dx 0 1 u 2u5 du 1 1 u5 1 lim f ci x →0 i x 1 1 0 sin x dx (b) Let u 1 x 0 0⇒u 1 cos x 0 1 1 x 2 dx 0 1 1 u 2 du x5 1 n sin i n n 1 u. 0 0 1, 2, . . . , n 1 1⇒u 1, x 1 x2 1 i ,i n ci 159. (a) Let u 1 and use righthand endpoints. n x Let Integration x, du dx, x u. 0 0 xa 1 2 1 1⇒u 1, x 1 1 x b dx 1 0 u aub du 1 1 ub 1 u a du xb 1 x a dx 0 1 0 160. (a) sin x cos Let u x and cos x 2 2 x, du dx, x 2 x 2 u. 2 2 sin2 x dx 0 cos2 0 0 (b) Let u sin 2 2 cos2 u x dx 2 cos2 x dx 0 2 0 x as in part (a). 2 2 0 sinn x dx cosn 0 0 cosn u x dx 2 du 2 2 2 cosn u du cosn x dx 0 0 161. False 162. False 1 2 dx 2x 2 cos2 u du du 1 2 2x 1 2x 6 1 22 dx 1 3 C x x2 1 dx 1 2 x2 1 2x dx 163. True 10 10 ax3 10 bx2 cx 10 ax3 d dx 10 10 bx2 cx dx d dx 10 Odd 0 bx2 2 0 Even d dx 12 x 4 1 2 C S ection 5.5 Integration by Substitution 164. True b b sin x dx b cos x cos b cos a cos b 2 cos a a a 2 sin x dx a 165. True 4 sin x cos x dx 2 sin 2x dx cos 2x C 166. False 1 2 sin2 2x cos 2x dx 167. Let u cx, du cb f cx dx c a 168. (a) d sin u du cb du c fu ca Thus, 1 sin 2x 2 3 2 cos 2x dx 3 C 13 sin 2x 6 C c dx. b c 2 sin 2x u cos u cb f u du f x dx ca C ca cos u cos u u sin u du sin u u cos u x, u2 x, 2 u du u sin u u sin u dx. (b) Let u C. 2 sin x dx sin u 2u du 0 0 2 u sin u du 0 2 sin u (part (a)) u cos u 0 2 cos 2 169. Because f is odd, f a f x . Then x 0 a f x dx f x dx a a Let x 0 f x dx f u du fx a. a 0 a a, u f x dx f u du x 0. When x 0 a f x dx 0 0 h, then du b 0, u a 0 170. Let u f x dx. 0 du in the first integral. When x a a a f x dx 0 u, dx a a f x dx dx. When x b h h dx a, u b f u du a h a h f x dx. a h h. When x b, u b h. Thus, 459 460 Chapter 5 171. Let f x Integration a0 1 ... a2 x2 a1x an x n. 2 172. 2 f x dx 2 1 0 1 f x dx a0 x a1 0 x2 2 a2 x3 3 ... xn 1 n1 an 1 1 2 0 f x x dx 2 2 2 0 a1 2 a0 a2 3 an ... n 0 1 1 (Given) f x x2 dx 2 0 By the Mean Value Theorem for Integrals, there exists c in 0, 1 such that Adding, 1 1 f x dx fc 1 2f 0 x x2f x dx 2 xf x 0 x 2 dx 0. 0 0 1 0 f c. fx 0 Thus, the equation has at least one real zero. Section 5.6 Since x functions. 2 ≥ 0, f 0. Hence, there are no such Numerical Integration 2 x2 dx 1. Exact: 0 2 x2 dx Trapezoidal: 0 2 x2 dx Simpson’s: 0 1 1 Trapezoidal: 0 1 Simpson’s: 0 0 2 x3 dx Trapezoidal: 0 2 x3 dx 0 2 4. Exact: 1 2 Trapezoidal: 1 2 Simpson’s: 1 2 1 0 6 1 2 2 4 x2 2 x3 dx 2 dx x2 2.6667 1 2 1 dx 2 3. Exact: 8 3 0 2 x2 2 0 2 1 0 4 x2 2 2. Exact: Simpson’s: 13 x 3 1 1 1 8 1 1 12 7 6 1.1667 2 1 dx 4 3 2 2 2 x 1 dx 2 21 x3 6 2 3 2 2 21 x4 4 0 14 2 2 1 2 11 4 2 2 14 2 4 2 2 8 3 12 2 2 2.6667 2 12 2 2 1 2.7500 2 1 2 4 1 34 2 34 2 2 4.000 0 1 0 4 1 2 3 2 1 0 6 1 2 3 4 2 x 2 dx x2 1 2 8 2 dx x2 1 2 12 3 2 3 2 3 21 3 4 3 2 3 21 2 3 17 4 4.2500 2 3 24 6 4.0000 2 1 2 1 1 32 25 2 4 32 25 8 9 2 2 32 49 2 8 9 4 32 49 1 2 1 2 1.0180 1.0008 2 1 2 1 12 2 12 2 1 1 75 64 1.1719 7 6 1.1667 S ection 5.6 2 14 x 4 x3 dx 5. Exact: 0 2 0 2 0 8 3 6. Exact: 4.0000 0 0 8 3 2 4 2 3 1 4 4 34 x 4 x dx 3 1 4 2 1 0 12 x3 dx Simpson’s: 2 3 2 4 3 3 4 2 21 3 4 4 3 3 3 21 3 5 4 2 3 5 4 4 3 6 4 2 3 6 4 2 3 7 4 2 8 3 7 4 4 4.0625 8 4.0000 8 3 12.0000 0 3 x dx 1 0 2 2 23 2 23 3 23 4 23 5 23 6 23 7 2 11.7296 3 x dx 1 0 3 4 23 2 43 3 23 4 43 5 23 6 43 7 2 11.8632 x dx Trapezoidal: 23 x 3 0 8 Simpson’s: 461 2 1 0 8 x3 dx Trapezoidal: Numerical Integration 0 9 7. Exact: 4 9 Trapezoidal: 9 2 5 2 16 x dx 4 16 3 18 4 2 37 8 4 38 3 37 8 12.6667 21 4 2 47 8 2 26 4 2 57 8 2 31 4 2 67 8 2 3 12.6640 9 Simpson’s: 5 2 24 x dx 4 3 8. Exact: x2 dx 4 x3 3 4x 1 3 3 x2 dx 1 3 4 24 4 x2 dx 1 3 6 44 1 3 Simpson’s: 1 2 1 9. Exact: 1 dx 1 2 dx 1 1 x 2 1 Simpson’s: x 1 1 2 dx 2 x x2 10. Exact: 1 dx 0 2 x x2 Trapezoidal: x x2 1 11 84 2 11 12 4 4 64 81 1 1 2 1 6 2 1 2 8 25 32 121 8 25 2 12 x 3 1 32 0 1 64 121 13 5 3 12.6667 0.6667 1 1 2 2 1 74 1 9 2 1 0.1676 1 32 1 9 2 0.7500 3 2 5 32 2 2 5 67 8 4 0.1667 1 9 1 54 5 2 25 4 44 1 54 24 20 0 31 0.6667 2 1 3 32 81 2 3 57 8 4 1 2 1 4 74 1 1 9 2 0.1667 1 3.393 1 0 4 2 1 2 12 2 1 21 12 1 2 3 2 32 2 1 2 22 1 3.457 1 dx 1 0 6 4 1 2 12 2 1 21 12 1 4 3 2 32 2 1 2 22 1 3.392 2 0 9 4 26 1 dx 0 Simpson’s: x 11 12 4 2 Trapezoidal: 3 2 2 1 11 84 1 x 2 47 8 4 11 3 3 1 4 Trapezoidal: 21 462 Chapter 5 Integration 2 1 x3 dx 1 1 4 21 18 22 21 27 8 3 3.283 1 11. Trapezoidal: x3 dx 1 1 6 41 18 22 41 27 8 3 3.240 dx 1 1 4 2 dx 1 1 6 4 0 2 Simpson’s: 0 Graphing utility: 3.241 2 1 12. Trapezoidal: x3 1 0 2 1 Simpson’s: x3 1 0 1 12 12 1 3 1 1 1 2 3 1 1 2 32 3 1 3 1.397 32 1 3 1 3 1.405 1 4 13 1 1 2 13 1 Graphing utility: 1.402 1 13. 1 x1 x dx x1 0 x dx 0 1 x1 x dx 1 0 8 x1 Trapezoidal: x dx 1 0 12 0 1 Simpson’s: 0 1 1 4 2 1 4 1 1 4 4 1 1 2 2 1 4 1 2 1 1 2 2 3 1 4 2 1 2 3 4 3 1 4 4 0.342 3 4 0.372 Graphing utility: 0.393 x sin x dx 14. Trapezoidal: 2 x sin x dx Simpson’s: 2 16 2 24 1 2 5 5 sin 8 8 2 4 5 5 sin 8 8 3 3 sin 4 4 2 3 3 sin 4 4 2 7 7 sin 8 8 2 4 7 7 sin 8 8 0 0 1.430 1.458 Graphing utility: 1.458 2 cos x2 dx 15. Trapezoidal: 0 2 8 cos 0 2 cos cos 0 4 cos tan 0 2 tan tan 0 4 tan 2 2 2 cos 4 2 2 2 cos 2 3 2 2 2 cos 4 2 0.957 2 cos x2 dx Simpson’s: 0 2 12 2 2 2 cos 4 2 2 4 cos 2 3 2 2 2 cos 4 2 0.978 Graphing utility: 0.977 4 tan x2 dx 16. Trapezoidal: 0 4 8 4 2 2 tan 4 4 2 2 tan 2 3 4 2 2 tan 4 4 0.271 4 tan x2 dx Simpson’s: 0 4 12 0.257 Graphing utility: 0.256 4 4 2 2 tan 4 2 2 4 tan 3 4 4 2 2 tan 4 S ection 5.6 1.1 1 sin 1 80 sin x2 dx 17. Trapezoidal: 1 1.1 1 sin 1 120 sin x2 dx Simpson’s: 2 sin 1.025 1 2 2 sin 1.05 2 4 sin 1.025 2 2 sin 1.05 2 sin 1.075 2 2 4 sin 1.075 Numerical Integration sin 1.1 2 2 sin 1.1 463 0.089 2 0.089 Graphing utility: 0.089 2 1 cos2 x dx 1 18. Trapezoidal: cos2 x dx 2 21 cos2 8 21 cos2 4 21 cos2 3 8 1 1.910 2 16 0 41 cos2 8 21 cos2 4 41 cos2 3 8 1 1.910 2 Simpson’s: 24 0 Graphing utility: 1.910 2 x ln x 19. 1 dx 0 Trapezoidal: 1.684 Simpson’s: 1.649 Graphing utility: 1.648 3 ln x dx 1 0 4 2 ln 1.5 2 ln 2 2 ln 2.5 ln 3 5.1284 4 1.282 ln x dx 20. Trapezoidal: 1 0 6 4 ln 1.5 2 ln 2 4 ln 2.5 ln 3 7.7719 6 1.295 1 3 Simpson’s: 1 Graphing utility: 1.296 4 21. Trapezoidal: x tan x dx 32 0 0 2 0 4 tan 16 2 2 2 tan 16 16 2 3 3 tan 16 16 4 2 16 2 2 tan 16 16 4 3 3 tan 16 16 4 4 Simpson’s: x tan x dx 48 0 tan 16 16 Graphing utility: 0.186 0 sin x dx x 8 0 sin x dx x 12 22. Trapezoidal: Simpson’s: 1 2 sin 2 sin 4 2 4 sin 1 2 sin 3 4 34 2 4 2 sin 4 4 0 4 sin 3 4 34 2 2 1.836 0 1.852 Graphing utility: 1.852 4 x e x dx 1 0 2 2e1 2 2 e2 2 3 e3 2e4 102.555 x ex dx 23. Trapezoidal: 1 0 3 4e1 2 2 e2 4 3 e3 2e 4 93.375 0 4 Simpson’s: 0 Graphing utility: 92.744 2 24. Trapezoidal: xe x dx 0 2 Simpson’s: 2xe 0 Graphing utility: 0.594 x dx 1 0 4 1 0 6 e 12 2e 12 2e 1 2e 3e 1 32 6e 32 2e 2.2824 4 2 2e 2 3.5583 6 0.5706 0.5930 0.194 0.186 464 Chapter 5 25. Integration 26. Trapezoidal: Linear polynomials y Simpson’s: Quadratic polynomials y = f ( x) x a b The Trapezoidal Rule overestimates the area if the graph of the integrand is concave up. 27. fx x3 fx 3x2 fx 6x fx 6 4 f x 0 3 (a) Trapezoidal: Error ≤ 20 12 42 (b) Simpson’s: Error ≤ 2 05 0 180 44 28. f x 2x fx 0.5 since f x is maximum in 0, 2 when x 0 since f 4 x 2. 0. 3 2 fx 12 0 The error is 0 for both rules. fx 29. cos x fx sin x fx cos x f x 4 f sin x x cos x (a) Trapezoidal: Error ≤ (b) Simpson’s: Error ≤ fx 30. fx 0 3 3 1 2 12 4 05 1 180 44 192 0.1615 because f x is at most 1 on 0, 5 46,080 0.006641 because f 4 x is at most 1 on 0, sin x cos x fx 2 sin x f 3 cos x f x 4 x 4 sin x (a) Trapezoidal: Error ≤ 10 12 42 (b) Simpson’s: Error ≤ 1 05 180 44 . 3 2 2 192 ≤ 0.0514 since f x 4 4 46,080 0.0021 since f 4 x 2 ≤ on 0, 1 . 4 on 0, 1 . . S ection 5.6 fx x fx 1 x 2 31. x 4 f fx fx 52 2 fx 72 2 1 2 4x Trapezoidal: Error ≤ n ≥ 76.8. (b) Maximum of f 4 x 52 15 x 8 105 x 16 72 92 Let n 32 15 2 5 10 180 256 Let n 3 (b) Maximum of f 4 3 ≤ 0.00001 4 Let n 224. 105 105 is on 1, 3 . 16x9 2 16 x 25 105 ≤ 0.00001 180n4 16 7 6 n4 ≥ 10 5 n ≥ 18.5. 25 10 96 3 on 1, 3 . 4 10 5 n ≥ 223.6. is 72 1 2 n2 ≥ is 2 31 12n2 Trapezoidal: Error ≤ 77. 15 2 3 4x5 Let n 20 (even). 8 (even). fx sin x, 0 ≤ x ≤ fx 34. cos x 2 sin x 2 cos x fx 3 4 cos x sin x x sin x f x 4 3 x 4 32 25 10 24 cos x , 0 ≤ x ≤ 1 fx f 1 x 2 25 15 2 ≤ 0.00001 180n4 256 n ≥ 6.2. fx 1≤x≤3 1 2, (a) Maximum of f x 0.0884. 2 ≤ 0.00001 16 16 x x x Simpson’s: Error ≤ n4 ≥ f 2 16 is 0.0829. Simpson’s: Error ≤ fx 3 20 12n2 32 82 5 10 12 16 n2 ≥ 4 f (a) Maximum of f x 33. fx 32 2 15 x 16 15 2 256 32. 12 2 3 x 8 x 0≤x≤2 1 2, 1 x 4 fx f 2 Numerical Integration cos x 2 (a) Maximum of f x Trapezoidal: Error ≤ n2 ≥ 10 12n2 cos x is (b) Maximum of f n4 ≥ 4 x sin x 2 ≤ 0.00001 (a) Trapezoidal: Error ≤ 2 12 10 5 1 180n4 n2 ≥ Let n 4cos x Simpson’s: Error ≤ f All derivatives are bounded by 1. 3 n ≥ 286.8. 4 2. x 4 287. x is ≤ 0.00001 4 180 n ≥ 15.3. 10 5 Let n 4. 23 1 ≤ 0.00001 12n2 3 96 n ≥ 179.7. Let n (b) Simpson’s: Error ≤ n4 ≥ 180. 25 1 ≤ 0.00001 180n4 5 5760 n ≥ 8.5. 16. 10 5 10 5 Let n 10 (even). 465 466 Chapter 5 35. f x Integration 1 x 1 (a) f x 41 x 32 in 0, 2 f x is maximum when x 81 12n2 4 Trapezoidal: Error ≤ (b) f 4 f x 4 16 1 15 x 72 36. f x x 1 0 and f 9x 1 43 f 4 15 . 16 0 < 0.00001, n4 > 16,666.67, n > 11.36; let n 12. in 0, 2 81 x 56 1 10 3 < 0.00001, n2 > 14,814.81, n > 121.72; let n 122. in 0, 2 x is maximum when x Simpson’s: Error ≤ 2 . 9 0 and f 0 82 12n4 9 Trapezoidal: Error ≤ x 4 23 f x is maximum when x 4 130. in 0, 2 32 15 180n4 16 2 (a) f x (b) f < 0.00001, n2 > 16,666.67, n > 129.10; let n x is maximum when x Simpson’s: Error ≤ 1 . 4 0 and f 0 0 and f 32 56 180n4 81 4 56 . 81 0 < 0.00001, n4 > 12,290.81, n > 10.53; let n 12. (In Simpson’s Rule n must be even.) tan x2 37. f x 2 sec2 x2 1 (a) f x 4x2 tan x2 in 0, 1 f x is maximum when x Trapezoidal: Error ≤ (b) f 4 f x 4 1 and f 1 10 12n2 8 sec2 x2 12x2 3 49.5305 < 0.00001, n2 > 412,754.17, n > 642.46; let n 32x4 tan x2 3 x is maximum when x Simpson’s: Error ≤ 49.5305. 1 and f 4 36x2 tan2 x2 1 643. 48x4 tan3 x2 in 0, 1 9184.4734. 5 10 9184.4734 < 0.00001, n4 > 5,102,485.22, n > 47.53; let n 180n4 48. sin x2 38. f x (a) f x 2 2x2 sin x2 cos x2 in 0, 1 f x is maximum when x Trapezoidal: Error ≤ (b) f 4 f x 4 16x4 1 and f 1 10 12n2 12 sin x2 3 2.2853 < 0.00001, n2 > 19,044.17, n > 138.00; let n 139. 48x2 cos x2 in 0, 1 x is maximum when x Simpson’s: Error ≤ 2.2853. 0.852 and f 4 0.852 28.4285. 5 10 28.4285 < 0.00001, n4 > 15,793.61, n > 11.21; let n 180n4 12. S ection 5.6 39. (a) b a 4 4 0 4, 0 4 27 29 (b) 0 49 2 a 27 0 f x dx (a) 47 29 8 0 8 0 16 2 1.5 2 10 47 1 88 2 0 8 f x dx (b) 77 3 0 25.67 8 0 24 4 1.5 4 10 41. The program will vary depending upon the computer or programmable calculator that you use. n 2 3x2 on 0, 4 Ln Mn Rn Tn Sn 4 12.7771 15.3965 18.4340 15.6055 15.4845 8 14.0868 15.4480 16.9152 15.5010 15.4662 10 14.3569 15.4544 16.6197 15.4883 15.4658 12 14.5386 15.4578 16.4242 15.4814 15.4657 16 14.7674 15.4613 16.1816 15.4745 15.4657 20 14.9056 15.4628 16.0370 15.4713 15.4657 Mn Rn Tn Sn 43. f x n 1 x2 on 0, 1 Ln 4 0.8739 0.7960 0.6239 0.7489 0.7709 8 0.8350 0.7892 0.7100 0.7725 0.7803 10 0.8261 0.7881 0.7261 0.7761 0.7818 12 0.8200 0.7875 0.7367 0.7783 0.7826 16 0.8121 0.7867 0.7496 0.7808 0.7836 20 0.8071 0.7864 0.7571 0.7821 0.7841 Mn Rn Tn Sn 44. f x n sin x on 0, 4 Ln 4 2.8163 3.5456 3.7256 3.2709 3.3996 8 3.1809 3.5053 3.6356 3.4083 3.4541 10 3.2478 3.4990 3.6115 3.4296 3.4624 12 3.2909 3.4952 3.5940 3.4425 3.4674 16 3.3431 3.4910 3.5704 3.4568 3.4730 20 3.3734 3.4888 3.5552 3.4643 3.4759 23 29 2 5.5 26 29 0 44 1 134 3 42. f x 467 8 24.5 4 3 12 f x dx 8, b 0 1 49 2 4 40. n 8 4 3 8 f x dx n Numerical Integration 134 3 23 29 4 5.5 46 29 0 468 Chapter 5 Integration sin x on 1, 2 x 45. f x n Ln Mn Rn Tn Sn 4 0.7070 0.6597 0.6103 0.6586 0.6593 8 0.6833 0.6594 0.6350 0.6592 0.6593 10 0.6786 0.6594 0.6399 0.6592 0.6593 12 0.6754 0.6594 0.6431 0.6593 0.6593 16 0.6714 0.6594 0.6472 0.6593 0.6593 20 0.6690 0.6593 0.6496 0.6593 0.6593 Mn Rn Tn Sn 46. f x x2 2 6e n on 0, 2 Ln 4 8.4410 7.1945 5.8470 7.1440 7.1770 8 7.8178 7.1820 6.5208 7.1693 7.1777 10 7.6911 7.1804 6.6535 7.1723 7.1777 12 7.6063 7.1796 6.7416 7.1740 7.1777 16 7.4999 7.1788 6.8514 7.1756 7.1777 20 7.4358 7.1784 6.9170 7.1764 7.1777 47. f x x ln x n 1 on 0, 3 Ln Mn Rn Tn Sn 4 2.5311 3.3953 4.3320 3.4316 3.4140 8 2.9632 3.4026 3.8637 3.4135 3.4074 10 3.0508 3.4037 3.7711 3.4109 3.4068 12 3.1094 3.4044 3.7097 3.4095 3.4065 16 3.1829 3.4050 3.6331 3.4050 3.4062 20 3.2273 3.4054 3.5874 3.4073 3.4061 5 0 3 12 5 12 125 5 12 3 400 15 12 125 15 12 3 400 5 48. W 100x 125 x 3 dx 0 Simpson’s Rule: n 12 5 100x 125 0 x3 dx 200 ... 10 12 125 0 10 12 3 10,233.58 ft lb S ection 5.6 Numerical Integration 49. (a) Trapezoidal: 2 2 4.32 28 2 4.36 2 4.58 2 5.79 2 6.14 2 7.25 2 7.64 2 8.08 8.14 12.518 2 4.32 38 f x dx 0 4 4.36 2 4.58 4 5.79 2 6.14 4 7.25 2 7.64 4 8.08 8.14 12.592 Simpson’s: 2 f x dx 0 (b) Using a graphing utility, 1.3727x3 y 4.0092x2 0.6202x 4.2844. 2 Integrating, y dx 12.53. 0 12 6 50. x2 1 0 dx, Simpson’s Rule, n 1 0 2 36 6 4 6.0209 1 113.098 36 4 1 1 0 2 6.0851 4 6.1968 2 6.3640 4 6.6002 6.9282 3.1416 51. Simpson’s Rule: n 1 6 x2 6 4 1 36 dx 4 16 1 2 1 2 26 2 4 36 1 2 2 46 1 2 4 56 1 1 2 2 3.14159 52. Area 1000 125 2 10 53. Area 120 75 2 12 2 125 2 81 2 120 2 84 2 112 2 76 2 90 2 67 2 90 2 68 2 95 2 69 2 88 2 72 2 75 2 68 2 35 2 56 89,250 sq m 2 42 7435 sq m t 54. Let f x Ax3 Bx2 D. Then f Cx 4 x 0. sin x dx 55. 2, n 10 0 Simpson’s: Error ≤ b a5 0 180n4 0 By trial and error, we obtain t Therefore, Simpson’s Rule is exact when approximating the integral of a cubic polynomial. 1 x3 dx Example: 0 1 0 6 4 1 2 3 1 1 4 This is the exact value of the integral. 56. The quadratic polynomial px x x1 x2 x x2 x1 x3 y x3 1 passes through the three points. x x2 x1 x x1 x2 x3 y x3 2 x x3 x1 x x1 x3 x2 y x2 3 2.477. 2 23 0 469 470 Chapter 5 Integration Section 5.7 1. 5 dx x The Natural Logarithmic Function: Integration 1 dx x 5 5 ln x C 10 dx x 2. 10 1 dx x 10 ln x 3. u C x 1, du 1 x 4. u x 5, du 1 x 5 dx 5. u dx ln x 5 3 2x, du 1 C 3 2x dx 2 dx 1 ln 3 2 x2 7. u 1, du x x2 1 2x dx 8. u 1 ln x2 2 1 ln x2 10. u 1 x2, du 9 x x2 9 xx x3 1 2 2 4 x2 3x x 2 1 dx x3 x 3x2 3 5 12 x2 x2 x 17. x4 x x2 4 2 x3 3 4 x 2 C x2 2 3x2 4 ln x C C x3 4 dx x x dx 4 x 1 6 ln x 5 x 2x u 2x 3x2 3 x2 3 dx 1 3 x2 2x 3 x3 3x2 3 dx 9x 2x 3 dx 9x dx 1 3 1 ln x2 2 3x2 2 9x C 4 14. 2x2 x 7x 2 3 dx 2x x2 C 16. x3 6x x 20 5 x3 3 dx 2 18. C x3 3x2 x2 4x 3 9 19 11 x 11x x2 dx C x x2 x3 3x2 C dx 3 5 ln x 2 x2 x3 9x, du 1 ln x3 3 6 4 x2 dx x3 2x dx 3x2 x2 x3 3 x2 3x2 dx C 4x dx 1 1 3 dx 3 3x 2 dx 1 ln 3x 3 9. 1 1 3 3 x3 C C 1 ln 3 3 1 3x2 6x dx, 3 3x 3x2 4 dx 2x 11. u 9 x2 2 15. dx C 1 ln x3 3 13. x3 2x dx dx 3x2 3 2 1 C 9 12. x2 1 1 2x dx 2 x2 1 dx 3x ln x 2 dx 3x2 dx x3, du 3 1 6. 1 1 2 3 2x dx 1 dx 19 ln x 5x 5x2 2 dx 19 19x 3 3x 2 2 C 115 dx x5 115 ln x x x2 2 dx x x2 1 ln x2 2 3 5 C dx 3 C S ection 5.7 1 dx x ln x, du 19. u ln x 2 dx x 21. u x 1 x 1 20. 1 ln x 3 1, du 3 x 12 1 x 1 2 12 1 2x 2 dx x 1 2 ln x 1 1 2x 24. dx 1 2 x u dx x x 1 1 ln u 1 3 dx 3 3, du x x x 3 dx 2 2 x 26. u dx 1 2 u2 C1 3x 3 du dx 6u 9 ln u du 2 u C1 9 du u C1 18 ln u 2 6 C1 12 3 18 ln 27. x x 18 ln 3 C x 3 C1 1 dx 1 x 1 1 1 3 1 1 2x 12 u u3 dx 1 1 dx dx C 2 2 u 3 1 du 3x C 1 du u ln u 3x 3x 3 1 du 2 u 3 C du 9 6x where C C x 2 1 3 6u u 3 3 dx 1 2 3 2 12u x x 3 u u2 2 dx 3 3 dx ⇒ dx 2 3x 3x, du 1 1 1. 1 dx ⇒ 2 u 2x u2 x1 12 dx 13 ln x 2x 2x C1 u 1 3x2 3 2x 1 x 13 1 dx 1 du u ln 1 x2 2 dx 13 2 3 27. u 1 x1 1 x x C1 ln 1 where C 2 xx x dx 1 du 1 2x 2x x1 C 1 du u u 1 2 1 2 1 2 1 dx ⇒ u 2x 2x, du 1 2 dx 1 1 C 3 ln 1 1 dx 12 x 1 dx x C 2 1 2 3 C 1 2x x 25. u 1 dx 471 1 dx 3x2 3 x1 3, du 1 x2 2x 2x 11 3 ln x 1 ln ln x 3 22. u 2x 23. 1 dx x ln x3 C dx dx The Natural Logarithmic Function: Integration C ln 1 2 ln 1 3 3x C1 472 Chapter 5 x1 28. u 3 3 3 Integration 1 3x2 1, du x x 1 dx ⇒ dx 3 u dx 1 3 u 3 3 1 u3 3 cos sin u 31. d sin , du 1 du u 3u ln u 3 1 1 3 x1 3 3 C 1 2 3 x1 2 3 3x2 2 1 3 ln x1 1 3 1 C 3 3x1 3 x C1 30. C 1 5 sin 5 d 5 cos 5 tan 5 d 1 ln cos 5 5 cos t dt sin t 32. csc 2x 2 dx 1 ln csc 2x 2 33. 3 1 du cos d 1 2 csc 2x dx 3 3u2 2 x1 ln sin 2u 3u u2 3 ln x1 29. u2 u 3 1 2 du 3u u 1 2 du 3u ln 1 cot 2x sin t sec x dx 2 2 sec 34. u C sec x tan x dx sec x 1 ln sec x 1 C 36. sec t x 2 tan x 2 C csc2 t dt cot t, du csc2 t dt cot t 35. x1 dx 22 2 ln sec C C ln cot t tan t dt C ln sec t ln tan t sec t tan t cos t ln sec t sec t 37. e x tan e x dx tan e x ln cos e ln cos e x e x C x dx C 38. sec t sec t tan t dt ln cos t C tan t sec2 t dt tan t C sec t C sec t tan t dt C S ection 5.7 3 39. y 2 x The Natural Logarithmic Function: Integration 10 dx 2x 40. y 473 dx x2 9 ln x2 9 (1, 0) −10 1 3 x 2 3 ln x 10 dx C 0, 4 : 4 2 ln 0 9 C⇒C y ln x2 9 4 4 ln 9 −10 C 1, 0 : 0 3 ln 1 y 3 ln x C⇒C 2 ln 9 8 (0, 4) 0 2 −9 9 −4 41. s tan 2 d 1 tan 2 2 (0, 2) ln tan t 2d −3 1 ln cos 0 2 s 1 ln cos 2 2 f1 1 fx f1 1 fx 45. dy dx 2 ln x 1 x 2 (a) 44. f x C1 ⇒ C1 3 4 (π , 4) 2, x>1 0 2x 1 4 ln x 3 2 1 C C⇒C 4 4 x fx 2 x2 1 40 4 ln x C1 ⇒ C1 4 1 C1 x 2 7 7 , 0, 1 (b) y 1 y 3 y0 x 4 −3 0 f2 2 x (0, 1) −2 10 4x 2x 1 fx C1 2 2 4 x f2 3 4 1 x fx 3x 4 2 C⇒C 20 1 −2 fx 3x ln tan t 4 −8 3 2 ln x C⇒C 1 2 C 2 2 x fx C⇒C 2x 2, x > 0 2 x fx C ln 0 r −3 C 0, 2 : 2 2 x2 1 3 ,4 : 4 1 ln cos 2 2 43. f x sec2 t dt tan t 1 42. r 4 3 2 dx 1⇒1 Hence, y ln x ln x C C⇒C ln 2 2 2 1 ln 2 1 ln −3 ln 2 x 2 2 1. 6 −3 8 474 Chapter 5 dy dx ln x , 1, x (a) 46. Integration y 47. (a) 2 y (1, 4) 5 4 3 2 2 1 x 1 8 −1 x −2 4 −3 −1 −2 dy dx ln x dx x y (b) 2⇒ y1 ln x 2 Hence, y 1 , x x ln x 4 1 0 y 2 ln x 2 1 y (b) x ln x C ln 1 2 2 2 C⇒C 1, 4 C C⇒C 2 2 3 3 8 2. −1 8 −1 48. (a) y (b) 4 dy dx sec x, y ln 1 y π 2 −π 2 ln sec x 1 x 5 0, 1 ln sec x tan x 0 C − C⇒C 2 2 1 −3 tan x 1 −4 4 49. 5 3x 0 1 dx 4 5 ln 3x 3 5 ln 13 3 1 1 50. 0 1 1 1 x 2 dx ln x ln x, du 1 dx x 1 ln x 2 dx x 1 1 3 1 ln 3 4.275 1 51. u 2 ln 1 e ln 3 1 e ln x 7 3 52. u ln x, du e2 e 1 54. 0 1 dx x ln x x x 1 dx 1 2 1 dx x e2 e 0 1 1 dx 0 0 2 2 dx 1 x ln ln x ln 2 e 2 dx x1 2 55. 1 1 cos sin 1 x 0 e2 11 dx ln x x 1 x2 x 12 x 2 53. 2 ln x 1 1 2 ln 2 0.2 56. ln x 1 ln 3 0 sin 1 2 1 sin 2 sin 1 1.929 0.2 csc 2 0.1 dx 2 ln ln 0 1 2 d 1 x 1 x cot 2 2 csc2 2 d 2 csc 2 cot 2 cot2 2 d 0.1 0.2 0.2 2 csc2 2 0.1 2 csc 2 cot 2 1d cot 2 csc 2 0.0024 0.1 3 1 S ection 5.7 1 57. 1 dx x 21 x 2 1 1 58. x x dx x 2 ln 1 ln 1 1 x 2 x x x x 1 1 2x x dx 1 ln csc x x x 4 ln 1 sin x dx 4 ln 1 x x C where C C1 C1 C 5. 60. x2 x 1 dx ln x 1 x2 2 x C 2 ln csc x cot x cos x 4 4 ln 2 2 2 1 0.174 4 4 sin2 x cos2 x dx cos x 4 62. 2. C1 2 61. 475 C1 C where C 61 4x 59. x The Natural Logarithmic Function: Integration ln sec x tan x 2 sin x 4 ln 2 2 1 1 22 1.066 Note: In Exercises 63– 66, you can use the Second Fundamental Theorem of Calculus to integrate the function. x 63. F x 1 64. F x Fx 3x Fx 1 tan x x2 1 dt t 1 3 3x Fx tan t dt 0 1 x Fx 65. x 1 dt t 1 1 dt t 2x x2 2 x 66. F x 1 x Fx (by Second Fundamental Theorem of Calculus) Alternate Solution: 3x Fx 1 1 3 3x Fx 3x 1 dt t ln t ln 3x 1 1 x 3 67. y 68. 69. A y 1 2 2 1 x 1 2 3 4 −1 x −1 A −1 2 1 2 −2 1 1.25; Matches (d). A 3; Matches (a). 4 dx x 3 4 ln x 4 ln 3 1 476 Chapter 5 4 70. A 2 Integration 4 2 dx x ln x 2 2 4 4 11 dx ln x x 71. A tan x dx 0 0 4 2 2 ln 2 ln ln x 2 2 ln ln 4 ln ln 2 2 ln 2 ln 2 2 ln 3 4 4 0.3466 2 ln 2 4 73. A 1 3 ln 1 0 ln 2 2 ln 2 sin x dx 1 cos x 72. A ln cos x x2 4 x 4 dx 4 2 2 2 2 15 2 2 2 ln 1 4 ln x 8 1 2 4 ln 4 1 8 ln 2 13.045 square units 10 2 2 ln 3 4 x2 2 4 cos x ln 1 4 dx x x 1 22 ln 0 6 0 4 74. A 1 x 4 x 4 dx 2 4 dx x 1 1 2 sec 75. 0 x dx 6 4 x 12 4 ln x 4 ln 4 3 4 ln 4 1 12 ln sec ln sec 8.5452 12 6 ln 2 10 9 0 0 0 4 0 4 2x 76. x2 tan 0.3x dx 1 16 8 0 −2 5 10 ln cos 0.3x 3 10 ln cos 1.2 3 4 1 1 10 ln cos 0.3 3 11.7686 x dx 66 sec 0 1 4 2 12 x 6 tan tan 3 3 x 6 2 0 12 3 5.03041 ln 1 0 S ection 5.7 77. f x 12 ,b x a 4 24 Trapezoid: 1 f1 4 34 Simpson: 5 2f 2 f1 5 Calculator: 1 4, n 12 dx x 477 4 2f 3 4f 2 The Natural Logarithmic Function: Integration 2f 4 2f 3 1 12 2 f5 4f 4 12 8 6 2.4 20.2 2.4 19.4667 1 12 3 24 8 12 1 0 2 f5 3.2 4 3.6923 19.3133 Exact: 12 ln 5 78. f x 8x ,b a 4 4 f0 Trapezoid: 24 4 x2 Simpson: 4 34 4 0 4, n 2f 1 f0 Calculator: 0 2f 2 4f 1 8x x2 4 dx 4 2f 3 2f 2 f4 4f 3 f4 1.6 6.2462 6.4615 6.438 Exact: 4 ln 5 79. f x ln x, b 4 24 Trapezoid: Simpson: a 6 2 f2 4 34 f2 4, n 2f 3 4 2f 4 4f 3 2f 5 2f 4 1 0.6931 2 f6 4f 5 f6 2.1972 2.7726 3.2189 1.7918 5.3368 5.3632 6 ln x dx Calculator: 5.3643 2 80. f x sec x, b 3 23 f 24 Trapezoid: Simpson: a 2f 3 23 f 34 2 ,n 3 3 6 4f 3 6 4 2f 0 2f 2f 0 4f f 6 f 6 3 12 2 2.3094 2 2.3094 2 2.780 2.6595 3 3 sec x dx Calculator: 2.6339 3 82. Substitution: u Power Rule 81. Power Rule x2 ln cos x ln 87. ln sec x tan x tan x and C ln ln 85. sec x tan x sec x sec x tan x 1 sec x tan x C C tan x ln sec x C 1 cos x C ln sec x 84. Substitution: u Log Rule C ln tan x sec2 x sec x C x2 83. Substitution: u Log Rule 4 and 86. ln sin x C ln 4 and 1 csc x ln csc x tan2 x tan x C C C 478 88. Chapter 5 ln csc x Integration cot x C ln 89. Average value 4 1 4 csc x ln 2 cot x csc x csc x cot x 1 csc x C cot x 8 dx x2 2 cot x ln csc x ln C cot x csc2 x csc x cot2 x cot x C 90. Average value 4 2 4 2 dx 1 4 x 4 1 4 x C 2 2 2 2 2 1 2 4 1 x 2 ln x 1 4 2 91. Average value e 1 e 1 1 e 1 e 1 12,000 ln 1 P0 12,000 ln 1 C Pt 1 3 0.291 0.25t 0.25 dt 1 0.25t 10 ln 2 94. t C 0.25 0 C 50 40 40 300 250 T 10 ln T ln 2 1000 ln 2 3 ln 2 tan ln 1 x 6 2 0 0 3 300 10 ln 200 ln 2 100 250 ln 150 4.1504 units of time 1000 0.25t 90,000 dx 400 3x 0 1 dT 100 4 10 ln ln 2 3 0.25t 1.8863 x dx 6 sec 0 1 2 ln 4 16 x ln sec 2 6 3000 4 1000 12 ln 1.75 1 50 2 1 2 ln 2 2 1 3 2 1000 12 ln 1 95. 1 4 2e 1000 12,000 ln 1 P3 92. Average value 1 4 1 12 3000 dt 1 0.25t 93. P t ln x dx x ln x 1 2 1 2e 2 ln 4 1 dx 1 dx x2 1 x 2 ln 2 4 1 x2 4 4 4x 1 1 7715 50 3000 ln 400 96. 3x 40 $168.27 dS dt k t St k dt t S2 k ln 2 C 200 S4 k ln 4 C 300 k ln t C Solving this system yields k St 100 ln t ln 2 100 100 k ln t C since t > 1. 100 ln 2 and C ln t ln 2 1. 100. Thus, S ection 5.7 97. (a) 2x2 y2 8 y2 2x2 The Natural Logarithmic Function: Integration (b) y2 8 2x2 y1 ln x e C eln 1 4 . x 4 and graph y 2 Let k 8 2x2 y2 1 x dx e x y1 479 1 k x eC 2 x 2 , y2 x 10 8 10 − 10 − 10 10 10 − 10 − 10 2x2 (c) In part (a): 4x y2 2yy y 8 0 1: f1 x x k 2 yx 2 k 10 x1 0.1 10 x f1 1 8 6 10 x 1 f 4 0.5 f 0.1 ln x k→0 x 2 99. False 100. False 1 ln x 2 ln x1 ln x 6 8 10 1 x 1 dx x ln x C1 ln x 12 x (a) y x2 1 1 x intersects f x 2 1 x 2 ln C x2 x 1 2 1 x x2 0.5 1 x 10 A 0 x x2 1 1 ln x 2 2 —CONTINUED— 1 ln 2 2 : (b) f x 1 1 4 0 1 x2 x 2x 1 x2 2 1 1 x2 x2 2 1 Hence, for 0 < m < 1, the graphs of f and y mx enclose a finite region. 1 1 x2 f0 1 x y 5 102. False; the integrand has a nonremovable discontinuity at x 0. 101. True d ln x dx 2 4 ln Cx , C 103. f x y 2x 2y 4x 10 2 2 lim fk x 2y y 2x 2 y x1 0.5 0.1: f0.1 x 1 4 x2 y 1 0.5: f0.5 x 4x 2yy 2x y Using a graphing utility the graphs intersect at 2.214, 1.344 . The slopes are 3.295 and 0.304 1 3.295, respectively. 98. k 4 x y2 In part (b): 1 x dx 2 x2 4 1 0 480 Chapter 5 Integration 103. —CONTINUED— x (c) mx x2 1 y m 1 1 x2 m y = mx 1 x m , m 1 mm 0 1 ln 1 2 1 ln 1 2 1 1 [m 2 2x Fx x m 1 1 2 1 2 2x 0 m ln m 1 dt, t mm 11m m 2 m m 1 1 ln 2 m 1 mx2 2 x2 0<m<1 mx dx, x2 1 1−m m intersection point x A Fx 0.5 m x 104. f (x) = 2x x +1 mx 2 1 105. x>0 1 x 0 ⇒ F is constant on 0, d ln x dx 1 implies that x 1 dx x . ln x C. The second formula follows by the Chain Rule. Alternate Solution: 2x Fx ln t ln 2x ln x x ln 2 ln x ln x ln 2 Section 5.8 5 1. 3. 5. 7. 8. x2 9 7 16 x2 1 x 4x2 x3 Inverse Trigonometric Functions: Integration dx dx 2x x 1 dx x x4 x2 1 dx 1 x2 2 x 3 C 7 x arctan 4 4 dx 1 5 arcsin C 2 2x 2 x x2 1 dx 1 4. 1 dx 13 x 3 3 2. dx arcsec 2x x dx x C C 1 2x dx 2 x2 1 6. 12 x 2 9. 4x2 1 4 1 9x2 4 2 1 1 ln x2 2 2 4x2 C 1 2 dx dx 3 2 arcsin 2x 4 arctan 3x 3 1 x1 arctan 2 2 dx 1 1 x 2 1 4 3 dx 3 1 9x2 dx 1 x 1 3 dx C (Use long division.) arcsin x 1 C C C S ection 5.8 t 2, du 10. Let u t t 4 1 2 16 x2, du 12. Let u 1 x x4 1 t 1 x 3 t C 1 2 x2 1 22 2 1 1 2 dt e2x, du t2 1 1 2 3 dx, u x 3 2u du 2 u 1 u2 3 x 2 2 dx 15. 1 1 x 1 u1 C 1 dx, dx 2x x, du du 3 arctan u u2 1 1 2e2x dx 24 e2x 2 x u2 dx, u 2u du 4x 1 3 dx x2 x 19. 2x 2 5 1 e2x arctan 4 2 u2, dx x, x x x 20. 3 x dx 2 x 1 2 4 dx 3 9 x 3 2 x 3 x 2 1 1 ln x2 2 2 4 2x x x2 17. 3 dx 1 1 x2 1 dx dx x 41 8 x 9 8 arcsin dx x 3 2 3 3 12 4 x2 3 arcsin x 3x, du 3 dx. 1 9x2 1 0 dx 1 1 0 3x 1 arcsin 3x 3 2 1 4 2x, du 32 0 1 1 4x2 dx 0 1 arctan 2x 2 dx x2 1 3 2 1 C 8 arcsin x 3 arcsin x 2 1 C 1 6 0 18 0 24. 32 3 arctan x 16 2 dx. 1 2 1 3 dx 3 23. Let u x2 C 22. 16 1 3 1 3 dx 0 16 x2 6x 1 21. Let u C dx C 3 3 x1 arctan 2 2 5 x 1 2x dx 2 x2 1 1 2 arcsin u C 2 2x 1 2 C 2u du du 1 u2 2 1 ln x2 2 3 9 9 2 dx x2 1 2u du C 3 arctan x 18. C 2e2x dx. 2 arcsin 3 2 x1 1 arcsin t 2 2 2t dt 2 C 1 x2 arctan 3 3 16. 481 2t dt. e2x dx 4 e4x 2x dx 2 x dx 2 2 t4 1 13. Let u x2 1 arcsec 4 2 14. 1 t2 arctan 8 4 2t dt 22 2x dx. dx 4 2 4 t 2, du 11. Let u 2t dt. dt Inverse Trigonometric Functions: Integration 2x 2 dx 32 0 6 9 x2 dx 1 x arctan 3 3 3 3 36 dx C C 482 Chapter 5 25. Let u 1 Integration 1 arcsin x, du 2 0 1 arcsin x dx 1 x2 26. Let u dx. x2 1 arcsin2 x 2 1 2 1 2 2 0.308 32 0 1 arccos x, du 0 1 arccos x dx 1 x2 dx. x2 1 2 arccos x dx 1 x2 0 1 1 arccos2 x 2 27. Let u x2, du 1 0 x 28. Let u 0 1 2 dx x2 1 12 2x dx. 0 x2 1 12 x 2x dx 12 0 2 1 3 x dx 12 32 32 0 2x dx. 0 1 2 32 2 x2 1 x2, du 1 2 1 2x dx x2 31 0 1 ln 1 2 x2 ln 2 3 0.134 2 29. Let u cos x, du sin x dx. 30. 0 sin x dx cos2 x 21 2 1 0 2 dx 2x x2 2 0 2 1 x 1 4 0 4 2 2 arctan sin x sin x dx cos2 x arctan cos x 31. 2 cos x dx 1 sin2 x 1 2 dx 32. 2 2 x 2 dx 4x 13 2 arctan x 1 9 2 1 x2 arctan 3 3 2 0 dx 22 x 2 2 4 1 arctan 3 3 33. 2x 6x x2 13 dx x2 2x 6 dx 6x 13 ln x2 34. 2x x2 5 2x 2 1 x2 35. 4x 6x 2x dx x2 dx 13 2 2x 1 x 4 arcsin 2 x 2 2 1 6x x2 3 arctan dx 2 6 7 1 x 13 3 1 dx 2 x2 2x 6 dx 6x 13 6 dx ln x2 2x 2 2 x2 36. 7 arctan x 4x 1 x2 x 2 x2 4x 4x, du dx 2x x2 1 2 x2 4x 38. Let u x2 4x 12 C 2x 4 dx x2 2x, du x1 dx x2 2x 1 2 2x x2 x2 4x 2 x 4 4 dx. dx C 4 2 arcsin 37. Let u 2 3 2 dx C 2 1 x 4 C 2 1 x dx 2 x 2 2 4 dx dx C 2 2 dx. 2x 2x 12 C 2x 2 dx 0.925 S ection 5.8 3 2x 4x 39. 2 3 3 dx x2 2x 4x 2 1 x x2 1 2x dx x 1 x arcsec x 2 3 x2 4x 4 2x dx 2 1 x 4 2 4 23 dx x2 41. Let u C 1, du x2 4, du 9 x 8x2 43. Let u et et x 1 2 x x 2u2 u2 3 2, u2 x 3 x 2x2 x4 Let u 3 1 e t, 2u du du u2 3 4 2 du e t dt, and 6 u 3 1 u2 3 2 1 2 dx 2x 12 x2 dt. et 2 3 arctan u 6 arctan 3 3 C 2 du 2x 2 6 x, 2u du 3 2 u2 1 1 dx, 1 3 C 3 1 u2 3 du x 2 3 arctan 2 x u2. 1 2 3 dx xx 23 Let u 3x du 1 2 arctan u C 3 0 2u du u 1 u2 x 4 1 1, u2 x u2. 1, 2u du 2 2u du 2 4 u2 u 1 3 4 6 (b) (c) 1 x2 1 x 1 dx arcsin x C, u x 48. (a) 1 2 2 arcsin 4 2 u 2 arcsin 2 dx, du 4 u2 1 47. (a) 6 arcsin 1 x2 C, u 1 2 ex dx cannot be evaluated using the basic x2 1 dx cannot be evaluated using the basic x 1 x2 integration rules. 1 2 12 integration rules. dx x2 C dx. 2u2 6 6 du u2 3 du 2u du u2 3 3 x x, 1 dx C 46. u2 1 du 2 et C x, 2u du 1 x2 4 arcsin 2 5 dx 2 1 dx x1 1 3 2 2u2 2 dx 1 2x x2 2 3 arctan 2u 45. 25 3. Then u2 2u 44. Let u dx 2x dx. dx 4 3 dt 2 2x dx. 1 arctan x2 2 42. Let u 2 1.059 6 2 1 12 3 2 2 1 1 dx x2 x arcsin 1 3 1 4x 2 x2 2 4x 40. 3 4 dx x2 Inverse Trigonometric Functions: Integration 2 (b) xex dx (c) 1 1x e dx x2 1 x2 e 2 e1 C, u x C, u x2 1 x 483 484 Chapter 5 x 49. (a) Integration 2 x 3 1 dx (b) Let u u2 1 dx C, u u2 1. Then x x xx 32 1 x x 1 u2 dx 1 u 5 2u du. u4 2 2 x 15 C u2 1. Then x x 1 1 and dx 1 u 2u du 23 2 u 3u 15 (c) Let u x 1 and dx 2u du 32 1 2 u5 5 u3 3 1 3x 5 C 2 x 15 C 1 32 3x 2 C 2u du. u2 2 u2 du 1 du u3 3 2 u 22 uu 3 C 3 C 2 3 x 1x 2 C Note: In (b) and (c), substitution was necessary before the basic integration rules could be used. 1 dx cannot be evaluated using the basic 1 x4 integration rules. 50. (a) 51. Area 11 y 1 2 Matches (c). 3 2 x (b) x4 1 1 2x dx 21 x2 2 dx 1 arctan x 2 2 x3 (c) 1 x2 u −1 x −1 1 2 2 1 1 4x3 dx 4 1 x4 dx x4 C, 1 ln 1 4 x4 C, u x4 1 52. No. This integral does not correspond to any of the basic differentiation rules. 53. 1 y x2 4 , 1 y x2 4 y0 54. y 0, ) dx arcsin x 2 x2 4 8 y (b) y 5 dy dx y 3 x2 1 , 2, dx dx 3 1 x2 −5 5 C 7 8 C⇒C x 1 arctan 2 2 0, 0 : 0 3 arctan 0 3 arctan x C 7 8 3 2 3 arctan x y (0, 0) 1 x arctan 2 2 0, 0 x −5 , 1 2 arctan 2 2 x arcsin 2 55. (a) x2 1 y C C y 1 4 C C⇒C −8 8 0 −3 2 S ection 5.8 56. (a) y (b) y 4 2 x2 9 , 0, 2 2 y x2 9 Inverse Trigonometric Functions: Integration 5 2 x arctan 3 3 dx C −4 x 4 2 57. (a) C y −4 2 x arctan 3 3 (b) y y 4 3 1 x x2 4 2 , 2, 1 1 x x2 1 1 arcsec 1 2 C y x 1 arcsec 2 2 1, 1 4 −1 y 2 485 4 dx 4 x 1 arcsec 2 2 −4 C 4 x −1 1 4 −2 −3 −4 58. (a) (b) y y 2 25 x2 , −4 C x≥2 5, 5 4 y 4 2 arcsin 1 −4 dy dx x2 x −4 59. 2 25 10 x x2 y 1 , y3 0 2 arcsin dx x 5 2 arcsin C⇒C 0 60. dy dx 1 12 x2 , y4 12 −6 6 −8 dy dx 2 4 −6 −4 2y , y0 16 x2 2 62. dy dx 3 −3 1 y , y0 x2 4 7 3 −1 −1 6 0 5 −5 x 5 4 61. −5 C 486 Chapter 5 3 63. A 1 Integration 3 1 2x x2 dx 5 1 4 64. Area dx 2 x arctan dx 8 2 2 22 x 4 dx 0 2 2 1 2 arctan 1 1 arctan 0 2 0 2 4x x2 3 1 x1 arctan 2 2 1 arctan 1 2 0 1 12 x arctan 0 4 8 1 y 1 65. Area 4 0 arcsin arcsin 2 dx x2 2 1 2 1 x 2 0 1 2 arcsin 0 arcsec 2 x 2 3 2 ln 1 cos x dx sin2 x 1 4 3 68. A ex e2x ln 2 arctan 1 3 2 3 arctan x x x2 1 x C x x2 3 (b) A 1 x2 x arctan x x x2 ln 1 ln 1 2 1 ln 4 2 3 1 ln 3 2 d x arcsin x dx 1 ln 2 2 2 2x arctan x x x2 1 arctan x x2 x2 x1 arctan x x2 C. 1 3 9 21 1 ln 2 2 arctan 1 0.3835 4 x2 arcsin x 2 C 2x arcsin x 1 1 x2 2x arcsin x 1 x2 2 1 1 arcsin x 2 dx 0 arctan x 3 arctan 3 3 arcsin x (b) A 12 arctan x dx x2 ln x 70. (a) 1 ln 1 2 ln x 4 arctan 1 x2 2 x1 arctan x dx x2 3 3 x2 x1 1 x2 1 1 Thus, ex u 0 3 arctan 1 ln 1 2 dx, arctan ex 2 3 4 12 1 0 3 arctan sin x 3 arctan 1 2 2 arcsec 1 1 2 3 cos x dx 1 sin2 x 2 d ln x dx 2 2 2 69. (a) dx 2 3 3 4 1 arcsec x 6 67. Area 1 x x2 66. Area x arcsin x 2 2x 21 x2 arcsin x 0 2 2 1 x2 21 1 x2 2 2 0 4 2 0.4674 arcsin x 2 S ection 5.8 2 Area rectangle π 2 base height 1 Area rectangle arcsin x dx x 1 2 2 sin y dy 0 0 2 1 2 arcsin x dx 2 arcsin x dx. 0 cos y 0 0 1 Shaded area is given by 1 arcsin x dx 1 0 1 (b) arcsin x dx 0.5708 1 Hence, 0 arcsin x dx 2 0 1 1 0 dx x2 (b) Let n 4 1, 0.5708 . 1 4 1 0 2 1 1 72. (a) 487 (c) Divide the rectangle into two regions. y 71. (a) Inverse Trigonometric Functions: Integration 4 arctan x 4 arctan 1 4 arctan 0 4 40 4 0 6. 1 x2 1 dx 1 18 4 1 4 1 36 1 2 19 1 4 14 1 1 2 49 4 25 36 1 1 2 3.1415918 (c) 3.1415927 1 2 73. F x x 2 2 t2 x 1 dt (a) F x represents the average value of f x over the interval x, x 1, 1 . greatest on x (b) F x 2 . Maximum at x 1, since the graph is 2 arctan t arctan x 2 1 arctan x 1 x Fx 1 6x 74. (a) 6x 1 x 1 x2 2 2 x2 1 x2 x2 x2 4x 5 1 x2 4x 5 x2 4x 1 x2 1 4x 0 when x 5 1. dx x2 x2 9 6x 9 9 9 dx x x 3 2 (c) 4 y2 1 6x (b) u x2 dx x, u2 x, 2u du 3 2 arcsin x 3 3 y1 C −1 7 dx −2 1 6u2 u4 2u du 2 6 du u2 u 6 2 arcsin 75. False, dx 3x 9x2 16 The antiderivatives differ by a constant, 3x 1 arcsec 12 4 C 2 arcsin Domain: 0, 6 C 76. False, C 77. True d dx x 6 dx 25 x2 dx 1 x arctan 5 5 78. False. Use substitution: u arccos x 2 C 12 1 x2 1 2 4 x2 9 C e2x, du 2e2x dx 2. 488 79. Chapter 5 Integration d u arcsin dx a 1 C 1 du Thus, a2 2 ua u a arcsin u2 u u a 2 2 80. 2 a u d1 u arctan dx a a 1 ua a1 ua C 2 1 u a2 a2 u2 a2 C. du Thus, a2 u u2 a2 u2 dx u a2 1 u arctan a a u2 C. 81. Assume u > 0. d1 u arcsec dx a a 1 a ua C ua ua 2 1 1 au u u u2 u u2 a2 a2 a2 . The case u < 0 is handled in a similar manner. Thus, du u u2 a2 1 82. (a) A 0 u u u2 1 (b) Trapezoidal Rule: n A u 1 arcsec a a dx C. dx x2 1 a2 8, b a 1 0 1 0.7847 (c) Because 1 x2 1 0 1 1 dx arctan x 4 0 , you can use the Trapezoidal Rule to approximate 4 0.785397 83. (a) v t 32t 4, and hence, . For example, using n 200, you obtain 3.141588. 500 (b) s t v t dt 550 32t 16t 2 s0 16 0 st 0 500t 16t 2 500 dt C 500 0 C 0⇒C 0 500t 20 0 When the object reaches its maximum height, v t 1 (c) 32 kv2 1 arctan 32k k v 32 arctan dt k v 32 t vt tan C t 500, C C 500 0 15.625 16 15.625 2 (d) When k v(t 32k t 0.001: 32,000 tan arctan 500 0.00003125 500 arctan 500 k 32 , and k 32 500 15.625 3906.25 ft Maximum height 32k t 32 tan C k 32 tan arctan 500 k 32k t . 0 7 0 vt —CONTINUED— 500 C1 32k t v 0, v 32t s 15.625 k v 32 When t we have vt 32t dv 0. 0 when t0 6.86 sec. 0.032 t S ection 5.9 Hyperbolic Functions 83. —CONTINUED— 6.86 (e) h 32,000 tan arctan 500 0.00003125 0.032 t dt 0 Simpson’s Rule: n 10; h 1088 feet (f) Air resistance lowers the maximum height. 84. Let f x x arctan x y 1 x2 1 1 x2 x2 2 5 1 fx x2 1 2x2 > 0 for x > 0. 1 x2 x arctan x 2 > 0 for x > 0. Thus, arctan x > 1 x2 Let g x x 1 y2 1 x x2 1 x2 1 y1 Since g 0 0 and g is increasing for x > 0, x x < arctan x < x. 1 x2 Section 5.9 e 2 10.018 sinh cosh 2 e e 2 2 eln 2 e2 e2 2 ln 2 e (b) coth ln 5 eln 5 (b) sech 1 7. tanh2 x 1 2 ln 2 2 3 ln 3 1 x ex ex 1 e2x cosh 2x 2 8 10 e 2x 0 0 (b) tanh 1 0 0 1 2 ln (b) coth 1 3 1 4 ln 2 2 13 12 2 2 ex 2 1 0.962 x2 0.648 1 ln 5 15 15 x 2 e e 4. (a) sinh 1.317 e e 2 (b) sech 1 4 3 1 2 ln 5 1 49 23 sech2 x e0 6. (a) csch e e eln 5 5 5 1 0.964 2 12 2 cosh ln 5 sinh ln 5 e0 2. (a) cosh 0 2 3. (a) csch ln 2 5. (a) cosh 6 arctan x > 0 for x > 0. Thus, x > arctan x. Therefore, 3 2 (b) tanh 4 Hyperbolic Functions e3 1. (a) sinh 3 2 > 0 for x > 0. x2 1 1 x2 . arctan x gx 8. 3 0 and f is increasing for x > 0, Since f 0 y3 4 e2x e e2x x 2 4 e 2x 2 ex e e ex e 2 2x x2 4 e2x e2x x2 cosh2 x 2 2 1 5 2 e e 0.347 2x 2x 0.481 1 489 490 Chapter 5 9. sinh x cosh y Integration ex cosh x sinh y x e ex y ex x x e x x e ex y ex y cosh x y 2 2 2 ex y2 ex ey sinh x cosh2 x 3 2 2 tanh x 14. 1 2 2 sech2 x cosh x 32 13 2 ex y e x sinh x 4 3 e2x x2 e 2 e ex x ex y2 2 ex x e 3x e 1x e 2 2x e e e x e3x x e2x e e x ex e x 2 y e 13 ⇒ cosh x 4 1 32 2 2 13 13 13 3 1 3 13 2 2 3 1 13 2 13 1 2 Putting these in order: 1 32 23 3 1 12 sinh x tanh x cosh x 1 33 3 2 2 3 1 2 23 3 3 3 3 3 csch x 3 cosh x 3 ⇒ sech x 4 sinh x 23 3 sech x 3 2 tanh x 1 ⇒ sech2 x 1 2 coth x 2 2x sinh 3x 2 y2 ey e 3x 2 y y y 3 13 13 coth x csch x y y x e y2 x cosh y 1 ⇒ cosh2 x sech x coth x e 3 2 tanh x csch x e 4 x y ex 3 2 cosh x 13. x e e x e 2 1 3x e 2 2 y ex sinh 2x 2 2 x y 2x e ex 12. 2 cosh y e 2 e2x 4 sinh2 x ey 2 x e ex y 2 sinh x 3 x e 2 y e y 2 4 sinh3 x 11. 3 sinh x e ex 2 1 2 ex 4 2 y e 2 1x e 4 10. 2 sinh x cosh x ey 1 S ection 5.9 15. y sech x y sech x 18. g x gx 16. y 1 1 tanh x coth 3x 19. y ln cosh x 1 sinh x cosh x ln tanh 21. h x 1 sinh 2x 4 hx 1 cosh 2x 2 25. x 2 y y 20. y 1 2 cosh 2x 2 x cosh x sinh x x sinh x 1 cosh t sinh2 t 1 cosh t cosh2 t x2 , cosh x cosh 1 x2 t coth t 1 csch2 t sech2 3x gx sech t 26. x cosh x, y ln y 2x 2x 2 sech 3x sech 3x tanh 3x 3 6 sech2 3x tanh 3x 2 0 coth2 t x 2x y cosh x ln x y y 1 cosh x x At 1, 1 , y 2 1, 1 sinh x ln x cosh 1 . Tangent line: y Note: cosh 1 sinh x 2, cosh x 2 cosh x At 0, 1 , y 0, 1 28. sinh x sinh x 21 1 1 2x 2x sin x sinh x cos x cosh x, fx sin x cosh x cos x sinh x 0 when x Relative maxima: ± , cosh Relative minimum: 0, 1 cosh 1 x 1.5431 0, 1 1 4≤x≤4 cos x sinh x (−π , cosh π ) 1 1x y Tangent line: y 1 29. f x 2 sin x cosh x e0 1 y0 0 cosh 1 x esinh x cosh x y 2. y Tangent line: y cosh x esinh x, y 1 y y cosh x csch x ht sinh2 1, 0 Tangent line: y 27. y coth x x sinh x 24. g x sinh 1 y1 y 22. h t arctan sinh t 1 1 cosh x sinh x 1 2 sinh x 2 cosh x 2 1 sinh x ft x 2 x 12 0 1 (π , cosh π ) sin x cosh x 0, ± . −2 (0, − 1) −2 491 ln sinh x fx 12 x sech2 tanh x 2 2 y tanh x 23. f t 17. f x 3 csch2 3x y 1 Hyperbolic Functions 2 1 cosh 1 1 492 Chapter 5 Integration 30. f x x sinh x 1 cosh x 1 fx x cosh x 1 sinh x 1 fx 0 for x gx sinh x 1 x cosh x −6 cosh 1 0, hx x sech x tanh x sech x 1 x tanh x −2 32. sech x x tanh x 6 (0, −1.543) 1.543 x sech x gx 1 0. By the First Derivative Test, 0, is a relative minimum. 31. 6 2 tanh x 2 sech2 x hx 1 1 0 1 2 sech2 x 0 x Using a graphing utility, x ± 1.1997. Using a graphing utility, x 0.8814. From the First Derivative Test, 0.8814, 0.5328 is a 0.8814, 0.5328 is a relative relative maximum and minimum. By the First Derivative Test, 1.1997, 0.6627 is a relative maximum and 1.1997, 0.6627 is a relative minimum. 2 1 (1.20, 0.66) (0.88, 0.53) −3 − 3 (− 0.88, − 0.53) −2 (−1.20, − 0.66) − 1 y 33. a sinh x 34. y a cosh x y a cosh x y a sinh x y a sinh x y a cosh x y a cosh x Therefore, y y Therefore, y 35. f x fx P2 x 0. f0 0 36. f x cosh x f0 cosh 0 1 sech2 f0 1 fx sinh x f0 sinh 0 0 f0 0 fx cosh x f0 cosh 0 1 P1 x f0 x, 2 sech2 x tanh x, f0 f0x f0 0 0 f0x x 1 2f 0x 0 2 x P2 x f 0x 1 3 P2 f −3 f P 2 3 P1 P1 −2 −2 10 2 0 15 cosh x , 15 15 ≤ x ≤ 15 (b) At x At x y (c) y 30 20 10 − 10 0 12 2x 1 2 37. (a) y 0. tanh x, fx P1 x y x 10 20 ± 15, y 0, y sinh 10 10 x . At x 15 15 cosh 1 15 cosh 0 15, y 33.146. 25. sinh 1 1.175. S ection 5.9 38. (a) y 18 x , 25 25 cosh 25 ≤ x ≤ 25 ± 25, y (b) At x At x 0, y Hyperbolic Functions 18 18 25 cosh 1 25 493 56.577. 43. 80 sinh (c) y −25 x . At x 25 sinh 1 25, y 1.175. 25 −10 39. Let u 1 sinh 1 2x, du 40. Let u 2 dx. 1 sinh 1 2 2x dx 2x cosh x cosh2 x 1 , du 1 sinh x 2x sinh x 1 dx x x 1 cosh 1 2 41. Let u cosh 2 dx 1 x, du dx dx. 2x 2 cosh x 1 dx 2x C 42. Let u 1 dx. 1 cosh3 x 3 1 C cosh x, du sinh x dx. sinh dx sinh2 x 1 sinh x dx cosh2 x sech x 43. Let u sinh x, du cosh x dx sinh x 44. Let u cosh x dx. ln sinh x 2x sech2 2x C 1, du 1 cosh x x2 , du 2 x csch2 46. Let u x dx. x2 dx 2 csch2 x2 x dx 2 coth x2 2 2 dx. 1 sech2 2x 2 1 dx sech x, du 1 2 dx 1 sech2 x sech x tanh x dx 1 sech3 x 3 1 , du x 47. Let u 1 dx. x2 48. Let u csch 1 x coth 1 x dx x2 csch 1 csch x x2, du 49. Let u x x4 1 dx 1 1 coth x x 2 cosh x dx sinh2 x 1 dx arcsin C 50. 2x x2 9 1 arctan x2 2 C 2 x1 4x2 dx C cosh x dx. arcsin 2x dx. 1 2 sinh x, du 1 dx x2 C sech x tanh x dx. sech3 x tanh x dx C C C 1 tanh 2x 2 45. Let u 2 sinh x 2 2 ln sinh x 3 ex 1 x C 6 1 1 2x e C 2x 1 2x 2 2 dx 4x2 C C 494 Chapter 5 Integration ln 2 51. ln 2 tanh x dx 0 0 1 sinh x dx, cosh x u 1 cosh2 x dx 52. cosh x 0 1 x 2 0 ln cosh ln 2 4 53. 0 1 25 ln 2 e 2x, du 24 5 4 4 1 1 dx 10 5 x 4 x x 1 ln 9 10 0 1 25 54. 0 1 dx 1 0 2x 2 1 sinh 1 cosh 1 2 dx x2 0 x 5 arcsin 4 4 5 arcsin 0 1 ln 3 5 x 2e 56. 24 4x2 1 12 2 1 sinh 2 2 1 2 2 dx. 2 0 2 1 1 dx 10 5 x 1 5 ln 10 5 55. Let u 5 4 ln 2 dx x2 0 eln 2 Note: cosh ln 2 ln cosh 0 1 1 sinh 2x 2 1 1 2 ln cosh x 5 4 cosh 2x dx 2 0 ln 2 ln 1 cosh x ln 2 2 dx ex x 2e x e 1 e 0 2 2x 1 2 ln 2 x 2e cosh x dx 1 0 2x e dx 0 24 arcsin 2x 4 0 1 e 2 x 57. y y sech 58. y 3x 3 9x2 y 60. y 1 cosh y 1 1 cos 2x , 0 < x < 1 cos2 2x cos 2x 1 y 63. y y tanh 1 1 2 sin 2x 1 2x sinh 2 sinh 2 sin 2x cos 2x sin 2x 2x 2 4x2 1 1 2x x 2 1 x2 59. y 2 1 2 2 4 2 cos 2x tan x 1 y x2 1 2 tan x sec2 x 1 sec x 2 sec 2x, 62. y 1 sinh 4. sin 2x 1 2 cos 2x sin2 2x 2x 1 1 ln 2 4 since sin 2x ≥ 0 for 0 < x < 61. y tanh 0 11 24 ln 2 3 8 ln 2 2x y 2 sec 2x 4x2 2 sinh 1 64. y 2x csch 4x2 y x 2 csch 1 x tanh 1 4x 1 1 x x x 1 1 2 x2 2 csch 1 x x 1 x2 1 x x2 1 x2 ln 1 tanh 1 x x tanh 1 x x2 1 x tanh 1 ln 1 2 1 x x2 S ection 5.9 Hyperbolic Functions 65. Answers will vary. 66. See the definitions and graphs in the textbook. 67. lim sinh x 68. lim tanh x 69. lim sech x 70. 1 x→ 71. lim x→0 sinh x x 0 x→ lim ex 1 1 x→0 ex dx e2x 1 ex 1 ex 1 dx 1 ex ln 2 csch e2x 1 ex 74. C x x4 9 for x → 0 , coth x → x2 x2 11 3 ln 26 3 C x2 x2 x x x3 1 1 77. 4x 78. x 79. 1 dx x2 1 2 1 3 2 x3 2, du 76. Let u 2 3 dx dx dx x2 1 4x 2x2 x3 1 22 8 dx 1 2x2 1 8 1 y 1, du 80 1 8x dx x 2 1 2x 3 4x 4x 2 sinh 3 2 1 x C 2 ln x C 2 ln x3 3 1 x C x 2 2 x3 2 1x4 ln 4 x 2 2 2 2 1 2 1 x3 C 2 1 ln 2 4 C x x 2 2 2 4 2 2x 3 3 1 dx 2 1 2x 2x 1 2x 1 1 2 1 1 ln 26 C 6 1 x dx 1 2 1 1 16x dx 1 4 81 4 4x 1 2 dx 3 3 C dx 1 x 2x 2x 3 2 1 ln 6 4 dx. 2 C dx 2 1 ln 23 dx 2 sinh 3 x dx 1 x ln 4 x dx 1 2 81. Let u 1 dx 2x 3 2 x 1 2 x 22 4 1 2 80. C x dx. 1 4x 2 x 1 x 2 C 1 dx. 2x x, du 1 x1 for x → 0 2x dx x2 2 1 29 dx 1 3 ln 12 3 75. Let u 0 72. lim coth x does not exist. coth x → 73. lim csch x x→ x e 2x x→0 x→ 1 4x 1 arcsin 4 9 C 3 x x 1 1 2 3 C 495 496 Chapter 5 82. Let u Integration 2x y x 1 , du 2 dx. 1 4x2 1 8x dx 1 2 2x 1 x3 83. y 5 4x x 2 1 x 3 4 21x dx x2 4 dx 20 4x 20 3 ln 6 3 x2 2 4x 10 1 ln 3 5 x x 4x 10 5 ln 3 x x 1 2 2 4x2 3 2x 8x 1 C 1 84. y dx 2x dx x2 x2 3 x ln 4 x ln 4x dx 1 4x ln 4x x2 4 4x 2x dx x2 x2 3x4 ln 4 x 3 2 2 x 2 2 1 22 4 C C x2 2 20 4x 5 1 x 32 x2 2 C 4 85. A 2 2x 1 ln 3 dx 2 sech 0 4 4 0 e ex ex C 2 x2 1 e2x 0 2 e4 e ln 0 8 arctan e2 2 0 2 x2 2 1 3 89. (a) 0 5 3 0 6 x2 3 0 4 4 e 1 ln 2 2 4 1.654 dx x4 6 ln 5 0 6 ln dx x2 1 dx x2 1 4 3 1 17 x2 6 ln x dx 5.237 90. (a) 1 0 3 sinh 1 2 1 21 6 ln 3 21 5 3.626 5 3 12 3 x2 ln x ln (b) dx 1 ln e 4 2 2x 2 88. A 2 5 ln x2 2 5 ln 4 2 2x e dx 5 2x 0 2x 5.207 5x dx x4 1 87. A 5 2 2 2x e e 2 e2x 2 e 4 8 arctan ex 2x e 1 ln e2x 2 dx 1 0 2 1 2 dx C e2x e2x tanh 2x dx 0 2 22 2 86. A 2 ex 2 0 2 2 x dx 2 4 2 x x 12 dx x2 21 3 x sinh 1 12 12 1x 1 ln 1 2 1.317 3 1.317 ln 12 (b) 12 ln 1 2 1 2 ln 1 2 ln 3 12 dx 1 3 2 ln 12 dx 1x 1 ln 1 2 x 3 1 ln 2 2 0 5 x2 tanh 1 x 12 ln 3 2 ln 3 2 ln 3 12 x 12 ln 3 2 dx S ection 5.9 3k dt 16 x2 1 12x 3kt 16 91. x 1 62 When x t 32 4 0: 1 x ln 22 x When x t 6 6 2 2 C 1x ln 4x 8 4 C When t 1: 3 16 10 1 ln 4 30k 16 1 ln 2 4 C 7 3 1 ln 2 4 1 7 ln 4 6 20: 2 7 ln 20 15 6 ln 2 7 ln 15 6 k 7 6 1 x ln 4 2x 2 ln x 2x t→ (b) s t v t dt s0 16 0 st 16t 2 2 When t k v, then du Let u 1 ln 2 32 k Since v 0 ln 32 32 0, C 32 32 k dv. s0 kv kv t s2 t kv kv e 2 e 2 32k t ke 2 32k t v 32k t 32k t 32 e 32 e ke 32 k 32 2 32k t e e 1 1 1 32k t 32k t 32k t 32 tanh k e e 32k t e e 32k t 32k t 32k t 32k t C 400 ⇒ 400 C 1 k ln cosh 32k t . 0.01: 400 100 ln cosh 0.32 t 16t2 0 when t 5 seconds 0 when t 8.3 seconds 400 When air resistance is not neglected, it takes approximately 3.3 more seconds to reach the ground. kv 32k t 2 2 k. C. s2 t 32k t kv 1.677 kg 32k t dt s1 t 2 32k t 32 k 32 s1 t 32 32 52 31 0, When k C 0. kv kv 8 8 32 k 1 ln cosh 32k 1 ln cosh k dt kv2 1 32 k dt dv 104 62 32k t 32 tanh k kv2 32 8 8 1 c ln cosh ct (which can be (e) Since tanh ct dt verified by differentiation), then 400 st 32 32 400 ⇒ C C 8 8 104 The velocity is bounded by C 400 dv kv2 16t2 32t dt dv dt (c) 32 tanh k (d) lim 32t x 2x x v 49 36 62x 92. (a) v t 497 dx dx 0 Hyperbolic Functions (f) As k increases, the time required for the object to reach the ground increases. 498 Chapter 5 93. y a sech dy dx 1 Integration x a a2 x2, a>0 xa a2 x 1 x2 1 a2 a2 x2 94. Equation of tangent line through P y a sech When x 1 x0 a a2 x a2 x2 a2 x a2 x2 x x2 a2 x2 x0, y0 : x2 y a2 x02 x x0 x02 a2 x x0 Q 0, a P y a sech 1 x0 a a2 a2 x02 Hence, Q is the point 0, a sech 95. Let u sinh u cosh u eu 1 tanh u e e2u 1 < x < 1, tanh u x, eu eu xe2u x 1 1 1 ln u b e xt dt 97. b a 96. Let y x. ex ex tanh x ex u tan y e Thus, y x e +e x x −x x e −e and y 2 sinh x. arctan sinh x . Therefore, arctan sinh x 1 1 e e x 2 x arcsin tanh x . x x 1 1 ln 2 1 e xt x arcsin tanh x . Then, sin y x x 2u L x e2u e2u 1 x x u xe 1 (a, 0) u e e xeu x02 2 x0 . a 1 x0 a . x02 Distance from P to Q: d a sech a2 1 x02 x , x 1<x<1 b 98. y ex cosh x b exb e x xb y x 2 exb e x 2 ex e e x 2 x sinh x 2 xb 2 sinh xb x 99. y sech y sech y tanh y y y sech 1 cosh y x sinh y y 1 1 sech y tanh y y 1 1 sech y y 100. x 1 sech2 y x1 x2 cosh 1 x x 1 1 sinh y 1 cosh2 y 1 1 x2 1 −x R eview Exercises for Chapter 5 y 101. sinh sinh y 102. y x x cosh y y 1 sinh2 y ex 1 x2 1 ex 2 ex y 1 cosh y 2 sech x 1 y 103. y 1 e 2 e x x 2 ex ex ex e e e x x sech x tanh x x 1 x c c cosh x e y Let P x1, y1 be a point on the catenary. P(x1, y1) x sinh c y (0, c) L x The slope at P is sinh x1 c . The equation of line L is y c When y 1 x sinh x1 c x ⇒x sinh x1 c 0, c c2 sinh2 0. x1 c c2 c cosh c sinh x1 c x1 . The length of L is c y1, the ordinate y1 of the point P. 104. There is no such common normal. To see this, assume there is a common normal. y cosh x ⇒ y sinh x. Normal line at a, cosh a is y cosh a 1 x sinh a y a. y = cosh x (a, cosh a) (c, sinh c) x Similarly, y 1 sinh a 1 x cosh c sinh c 1 ⇒ cosh c cosh c The slope between the points is y = sinh x c is normal at c, sinh c . Also, sinh a. sinh c c cosh a . Therefore, a a cosh a c sinh c cosh c sinh a. cosh c > 0 ⇒ a > 0 sinh x < cosh x for all x ⇒ sinh c < cosh c a cosh a sinh a < cosh a. Hence, c < a. But, c < 0, a contradiction. sinh c Review Exercises for Chapter 5 1. 2. y y f f′ f x x f 499 500 3. Chapter 5 2x2 x Integration 23 x 3 1 dx 12 x 2 x 4. C u 3x du 3 dx 2 dx 3x 3 x3 1 1 dx x2 12 x 2 1 x x3 2x2 x2 2 3 1 13 3x 3x x2 7. 4x 9. 5 11. dx x 2x2 3 sin x dx ex dx 5 dx x ex 5x 5 ln x 13. f x 2x, fx C 1, 1 1 C 2 2 10 dx x 2ex dx fx C 1 vt at 1 2ex st at dt 0 when C1 a2 t 2 0. C2 0 x st a2 t 2 s 30 a v 30 a 30 2 2 2 3600 30 2 8 30 0 when C2 3600 or 240 ft sec 1 1 44 ft sec a vt at 66 since v 0 a2 st 66t since s 0 t 2 Solving the system at a2 t 2 0. we obtain t 55 12 t s 8 ft sec2. C2 66 ft sec s(t C2 x C2 ⇒ C2 2 vt s0 2ex 1 dx 30 mph C1 at C1 C1 1 2ex fx at 0 C C 16. 45 mph v0 2 tan x C 2ex 2ex fx x2 a dt dx C 5 sin x et C Since the slope of the tangent line at 0, 1 is 3, f0 2 C1 3 ⇒ C1 1. a vt 1 x C fx at 2x 10 ln x f0 15. x 23 2ex, 0, 1 14. f x x2 t2 2 e t dt t 2 2 sec2 x dx 5 cos x 2 y 8. 12. 1: y C 6. 10. C 2x dx When x 3 cos x C x 12 x 2 5. dx 3 dx 72 5 66 66 ft sec. 0. 44 66t 264 24 5 and a 55 12. We now solve 66 0 and get t 72 5. Thus, 55 12 72 2 5 2 66 72 5 Stopping distance from 30 mph to rest is 475.2 264 211.2 ft. 475.2 ft. R eview Exercises for Chapter 5 9.8 m sec2 18. a t 17. a t 32 vt 32t st 16t2 vt 96t (a) v t 32t (b) s 3 144 (c) v t 32t 3 (d) s 2 3 sec. 0 when t 96 288 4.9t2 (a) v t v0 9.8t s0 40t, 9.8t 40 0 0 when t 40 144 ft 96 when t 2 96 9 16 4 9.8t st 96 3 96 2 (b) s 4.08 3 sec. 2 (c) v t 108 ft 501 40 9.8 4.08 sec. 81.63 m 9.8t (d) s 2.04 20 9.8 20 when t 40 2.04 sec. 61.2 m 10 19. (a) 2i i 1 1 n i3 (b) i 1 10 (c) 4i i 20. x1 (a) 2 1 2, x2 1 5i 5 (b) i 1, x3 5 1 2 5 xi 1 1 xi 1 1 2 1 5 3 1 5 1 3, x5 5, x4 1 3 7 16 5 7 1 7 37 210 2 2 5 xi2 2xi (c) i 22 2 1 1 2 25 5 2 23 2 3 27 7 2 56 1 5 (d) xi i 21. y xi 1 1 2 5 1 3 5 7 3 5 2 10 x2 1 ,x 1 ,n 2 Sn S4 1 10 21 sn s4 4 1 10 2 122 10 1 22 10 1 1 10 1 1 1 2 1 10 322 10 322 1 1 10 22 1 13.0385 9.0385 9.0385 < Area of region < 13.0385 22. y 2x, x 1, n 23. y 3 S3 1 21 22 23 1 20 21 22 7 7 < Area of region < 14 4 , right endpoints n x, x 14 s3 6 y 8 n Area 6 lim n→ f ci i n 4i 4 nn 6 i 1 lim 4 6n n lim 24 24 8 n→ n→ 4 1 lim n→ x 4nn 1 n 2 8 16 n 1 n 2 x −2 2 −2 4 6 8 502 Chapter 5 24. y x2 Integration 2 , n 3, x y right endpoints 12 10 n Area lim n→ f ci i 8 1 6 n 2i n lim n→ x i 1 2 3 lim x 4n 3 lim 8 3 25. y 5 1 2n 6 1 2n n2 1 1 3n 6 26 3 6 3 n x2, x y 6 n Area lim n→ f ci i 4 x 3 1 2 n lim n→ 5 i 2 3n 1 ni 1 lim 3 n n lim 3 n→ n→ 3 26. y 3i n 1 lim n→ 2 18 9 −4 −3 1 9 nn n2 1 2n n2 1 2n 6 12 y 20 f ci i lim n→ lim n→ i 1 10 1 2 4 1 2n i 2i n n 4n 1 ni 1 lim n→ 4 n n 4 6 3 2 n 15 2 3 n3 3i2 n2 3i n 1 x 1 8i3 n2 3nn 1 n 2 4 5 24i2 24i n 8 1 lim n→ 15 x 1 n 4 3 1 2 n lim 2 1 n n→ 1 −2 9n 2 n x −1 9i2 n2 12i n n 1 3 n 12 n n 1 n 2 18 13 x, x 4 Area 2 3 2 4 nn n n2 n→ 2 1 2 n 4i2 lim n→ n i 1 n2 n→ 4 2 n i3 n3 3 nn n2 1 2n 6 1 1 n2 n 1 n3 4 2 4 R eview Exercises for Chapter 5 27. x n Area lim 3i n 52 n→ i 1 3n lim 10 n→ ni 1 lim 18 9 2 28. (a) S b 4 m i 1 n 1 i 0 sn (c) Area b 4 1 2n 6 3 3 4 5 6 1 b n mbi n 1 n b 4 2b 4 m i 3b 4 b 4 b n 1 m b n m bi m n 0 b n b m n lim i mb2 n 2n 1 1 n→ 4b 4 m 3b 4 2 mb2 1 16 b 4 2n 3 2 1 2n 1 i 0 1 b mb 2 12 mb 2 y = mx mb2 n 2n 1 mb2 n 2n x=b 1 1 base height 2 3 n xi 2x 1 y 3mb2 8 mb2 n 1 n n2 2 i 5mb2 8 4 3 mb2 n n 1 n2 2 i i mb2 1 16 b 4 3 dx 30. → ci 2 3ci 9 lim 4 y 31. 2 6 2ci i m n n lim → 9 nn n2 b 4 mb2 n 2n n→ 1 1 b 4 b n bi f n lim 2 9i2 n2 2b 4 m bi n f 3 9i2 n2 27 2 m n (b) S n 4 x 9 b 4 m0 s 3 n i 12 n 3nn 1 n 2 b 4 2 3i n 4 3i n 3 6n n n→ y 6 2 15i n 3n lim 6 n→ ni 1 29. 3 n y2, 2 ≤ y ≤ 5, y 5y i xi 3x 9 1 32. x2 dx 1 y 6 12 5 9 6 3 Triangle 2 3 1 x x −3 3 6 − 4 −3 −2 −1 9 −3 1 2 3 4 −2 5 5 5 x 5 dx 5 5 0 5 x dx x dx 0 0 4 25 2 16 4 (triangle) 6 6 33. (a) fx g x dx 2 6 f x dx 2 6 fx g x dx 2 g x dx 10 3 13 g x dx 10 3 7 2 6 (b) 6 f x dx 2 6 2 6 (c) 2f x 3g x dx 2 2 6 f x dx 3 2 6 (d) 2 5 f x dx 2 g x dx 2 6 5f x dx 503 5 10 50 2 10 33 11 x2 dx 1 2 4 2 8 (semicircle) x 504 Chapter 5 Integration 6 34. (a) 3 f x dx 6 f x dx 0 f x dx 0 3 (b) 4 1 3 3 6 f x dx f x dx 6 1 1 3 4 f x dx (c) 0 4 6 6 10 f x dx (d) 10 f x dx 3 8 3 35. x 34 x 4 1 dx 1 3 36. 1 1 t2 1 3 16 4 x 1 6 x2 1 3 6 9 1 1 t3 3 2 dt 8 3 3 12x 2 2 12 dx x3 38. 10 10 2x2 73 , (c) 4 1 4 16 , (d) 3 6 1 2 37. x dx 4 x2 2 2x 0 16 2 8 0 1 16 1 4t 3 39. t4 2t dt t2 0 1 1 2 x5 5 2x3 3 5x 32 5 x4 3 4 8 14 3 2t 2 40. 1 3 16 3 10 5 dx 2 2 32 5 16 3 10 52 15 9 9 25 x 5 x3 2 dx x x dx 41. 4 4 2 42. 1 1 x2 3 2 x 3 4 2 2 2 x2 2 ex dx 0 4 1 2x2 2 1 5 2 243 5 1 2 1 8 6 2 e2 1 1 e2 46. 0 x x 1 1 4 6 3 dx x 3 ln x 1 2 6 1 1 4 tan t 4 2 2 1 dx 1 8 sec2 t dt 44. 1 3 2x 1 2 1 22 2 ex 3 422 5 32 4 2 2 0 1 5 1 x dx cos 0 47. x 3 x 9 1 sin d 45. 2 5 4 2 1 dx x3 4 43. 9 2 48. x x2 2 4 dx 0 y 3 ln 6 2 4x y 6 7 5 6 4 5 3 4 2 1 −2 −1 −1 2 x 1 2 3 4 5 6 1 −2 −1 x 1 2 3 4 5 10 0 1 2 R eview Exercises for Chapter 5 4 9 dx 3 2 4 x3 3 9x 64 3 x2 49. 36 x2 50. 3 x x3 3 2 dx 1 9 2 10 3 y 2 x2 2 8 3 10 3 27 2x 1 1 3 4 7 6 9 2 8 y 7 6 3 5 4 2 3 2 1 x 1 2 4 5 6 7 x −2 8 1 3 −1 −2 1 x2 2 x3 dx x 51. 0 x4 4 1 1 2 0 1 4 1 1 4 x1 52. x1 x dx 2 x3 2 dx 25 x 5 2 0 y 23 x 3 1 2 3 x 2 2 5 1 0 4 15 y 1 1 1 x 1 9 53. Area 1 4 x 4x1 2 12 dx 3 9 83 1 3 sec2 x dx 54. Area 16 1 tan x y 3 0 0 y 10 5 8 4 6 3 4 2 2 −2 x 2 −2 4 6 8 10 π 6 2 2 x 55. y x π 3 56. Area ex dx 1 Area 1 3 2 dx x 2 ln x 2 ln 3 2 ln 1 y 2 2 ln 9 1 e 1 1 y 10 5 8 4 6 3 4 2 1 x −1 1 2 3 4 5 x 1 2 0 0 3 −1 ex x 2 2 e 505 8.3891 1 2 2 506 57. Chapter 5 9 1 9 4 Integration 1 dx x 4 9 1 2x 5 2 3 5 2 5 2 1 58. 2 4 0 0 x3 1 3 2 0 2 x 2 , Average value 5 2 2 x4 8 x3 dx y 2 8 6 y x x x 4 5 2 25 4 2 ( 3 2 , 2) 2 x 1 1 25 2 , 45 2 x 2 59. F x x2 1 x2 1 3 dx 63. x3 x6 3x 4 x3 dx x3 3 x3 3, du 1 3x 2, du 6x x x2 71. 72. 3 6x sin cos cos x dx sin x x3 1 6 dx 1 x x 3x 2 62. F x x2 dx x3 3 1 3 csc2 x 2 C 5 2 3 1 23x2 1 dx 3x2 2x 5, du sin3 x cos x dx 1 64. 1 dx x x2 61. F x 2 2x x 1 x 2 dx C x3 3 1 23x2 32 23 x 3 dx C 3 12 C dx 23 x 9 3 6x dx 3x2 4 dx x2 68. u 1 2x2 3 1 3 3 dx x1 10 3x2 dx x2 x3 67. u x3 8 3x2 dx x2 69. 35 x 5 3, du x3 66. u 3x2 6 1 x2 60. F x x7 7 65. u 4 x2 14 sin x 4 d 1 sin x 1 1 30 2x 6 dx 6x 5 2 12 x 2 6x 5 1 3x2 30 1 C 5 70. C cos 12 6x dx 3x2 C 6 dx 1 2 4 12 cos x dx sin d 2 sin x 21 12 C cos 12 2 sin x 2 x2 x sin 3x2 dx C C 21 1 5 1 6x 1 6 cos C 5 C sin 3x2 6x dx C 1 cos 3x2 6 C R eview Exercises for Chapter 5 tann 1 x n1 tann x sec2 x dx 73. 75. 1 76. cot4 77. xe 79. x x 2 sec sec csc2 x tan d 1 5x 3x 2 dx 1 5x 2 2 1 2 x x2 1 1 x2 x3 82. 0 3 83. 0 6 84. 2x 3 85. u 1 8 2 3 22 1 x2 4 2 2 0 dx 21 1 0 x 12 4 2 1 6 dx 1 6 x2 12 8 C 1 dx x2 e1 x dt 1t 2 t 2 dt e1 x 1 2 ln 2 C 1t C 9 4 5 4 1 2 3 u, dy 8 1 27 3 12 3 1 du 1. When y 1, u 0. 0 y 1 1 y dy 2 1 u 2 2u1 1 u du 1 1, x When x u 1, dx u3 2 du 25 u 5 2 1, u 0. When x 0 2 1 28 15 0, u 1. 1 x2 x 2 43 u 3 2 du 0 1 dx 2 1 u 1 2 u du 0 1 u5 2 2 2u3 2 u1 2 0 0 1t 9 6 12 x 3 2x dx 1 cos 12 t2 1 16 12 4 1 0 87. 3 0 2 x x 3 12 x 0 86. u e1 x dx x2 1 0 4 1 1 x3 12 3x2 dx 1 2 78. C 0 0, u sec C 80. 3x 2 1 1 1 1 3 x dx 1 sec 2x 2 sec 2x tan 2x 2 dx C 4 2x dx x3 x tan 1 dx 1 1 y, y When y 1 sec 15 cot 5 d 1 e 6 2 x 3 1 dx 1x x 3 x2 x2 1 3 1 3 dx csc2 2 1 2 4 dx sec 1 2 sec 2x tan 2x dx 74. 1 6x dx 1 5x 2 ln 5 81. 1 4 cot 2 1 1 x dx 1 e 6 3x 2 dx C, n x dx 2 2 cos 0 x1 dx 22 2 sin x 2 du 2 27 u 7 45 u 5 2 2 23 u 3 1 2 0 32 105 4 2 0 88. sin 2x dx 4 507 0 since sin 2x is an odd function. C 508 Chapter 5 89. p 1.20 0.04t t 15,000 M C Integration 1 p ds t (a) 2000 corresponds to t 15,000 M C 1.20 10 11 2 1.75 sin 0 t 2 5.1 1.9 91. Trapezoidal Rule n 4: Increase is 7 10 15 2 2 1.75 1 7 1 2.2282 liters 0 0.6048 liters. 2 1 2 4: x3 1 1 Simpson’s Rule n 2 t 1.75 cos 27,300 M 0.02t2 24,300 M 0.02t2 2 dt 15. 16 15,000 1.20t M C 0.04t dt 15,000 1.20t M 90. (b) 2005 corresponds to t 10. 11 1 1 1 8 1 13 1 2 1.25 1 1 12 1 13 dx x3 1 1 dx 1 4 1.25 1 3 2 1.5 1 3 2 1.5 1 3 2 1.75 1 3 4 1.75 1 3 0.257 23 1 1 3 0.254 23 1 Graphing utility: 0.254 1 92. Trapezoidal Rule n x3 4: 0 1 Simpson’s Rule n 4: 0 2 x2 3 x3 2 2 3 x 1 0 8 21 432 3 1 42 21 232 3 1 22 23 432 3 3 42 1 2 0.172 1 0 12 41 432 3 1 42 21 232 3 1 22 43 432 3 3 42 1 2 0.166 dx dx Graphing utility: 0.166 2 93. Trapezoidal Rule n 4: x cos x dx 0.637 94. Trapezoidal Rule n 4: Simpson’s Rule n 3.820 0 4 : 0.685 Simpson’s Rule n Graphing utility: 0.704 4 : 3.820 Graphing utility: 3.820 1 95. Trapezoidal Rule n sin2 x dx 1 0 4: e x2 dx 1.463 96. (a) R < I < T < L 1 Simpson’s Rule n 4 40 f0 34 (b) S 4 Graphing utility: 1.494 97. u 7x 2, du 1 7x 99. 1 2 dx sin x dx cos x 7 dx x2 98. u 1 1 7 dx 7 7x 2 1 sin x dx cos x ln 1 cos x 1 ln 7x 7 2 x2 100. u C 21 1, du x C 4f 1 1 4 3 1.494 1 4 4f 3 1 2 1 4 1 ln x2 2 1 C 1 dx x ln x, du 1 2 ln x 1 dx x 1 ln x 4 2 C f4 5.417 2x dx 1 2x dx 2 x2 1 dx ln x dx x 42 2f 2 R eview Exercises for Chapter 5 4 101. x 4 1 dx x 1 4 1 dx x 1 1 x 102. ln 4 1 1 sec d ln sec ln x 1 1 1 dx x ln 2 104. 3 0 e2x e2x 0 x dx 4 2x e2x e2x e e e2x dx. Let u 2 e2x du 2x e 2x dx 1 ln e2x 2 e 2x dx ln cos x x4 111. Let u 16 x 16 x2 1 2 1 e2x 4 4 113. Let u arctan 1 2 arcsin x, du arcsin x dx 1 x2 x2 4 x , du 2 1 5 1 1 ln 2 2 C 1 3 1 53 1 arcsin x2 2 2x dx 2 x2 arctan 1 2 1 x2 16 dx 1 x arctan 4 4 C 2 4 1 x2 2 4 x2 x2 12 C 2x dx 4 arcsin dx. x 2 1 arcsin x 2 1 ln 16 2 dx 4 1 2 arctan 2x dx. 1 4 arctan x 2 dx 4 x2 114. Let u 2 1 1 2x dx 2 16 x2 x dx x2 112. 0 1 2 ln 5x 5x 3 C 1 x2 1 dx 25x2 2e2x dx 2 110. 1 x2, du dx 4 5 dx. 1 3 2x dx. dx 1 1 ln e2x 2 x C dx 1 arctan e2x 2 x2, du dx 5x, du e2x 108. Let u e4x 1 2x e e2x 1 21 109. Let u 106. . Then 2e2x dx. 2x 1 e2x 2x e 1 dx. e2x, du 107. Let u e2x 2x e e e 2 4 tan 0 105. 1 ln x 2 4 tan 0 e ln x dx x 3 3 103. e 3 ln x dx. C x2 dx 1 x arctan 4 2 2 C x 2 4 x2 C C 2 5 dx C 1 2 509 510 Chapter 5 dy 115. Integration k dt m A2 y2 y arcsin A Since y π 2 k t m 0 when t C y = arcsin x π 4 0, you have C k t m sin y 116. 0. Thus, x y A 0.25 0.5 0.75 1 2 y k t. m A sin Since the area of region A is 1 sin y dy , 0 1 arcsin x dx the shaded area is 2 0 117. y 2x y cosh x 1 2 2x sinh x2, du 119. Let u x x4 1 118. y x sinh x 2x 2 120. Let u 1 2 dx x y 2x dx. 1 x2 2 1 ln x2 2 1 1 1 1 4x2 x3, du tanh 1 2x 1 2x 4x2 1 1 1 dt t 1 3 sech x3 2 3x2 dx C L1 1 x (c) L x 1 1 2.718 1 0 1 by the Second Fundamental Theorem of Calculus. x (b) L x 1 dt t 1 dt for x t x1 1 Then x1 1 1 dt t L x1x2 2.718 0.999896 (Note: The exact value of x is e, the base of the natural logarithm function.) (d) We first show that 1 1 1 1 1 u L x1 du 1 x2 1 dt t 1 1 x1 u x2 1 du x1 u x1 1 1 1 1 dt. To see this, let u x1 t 1 x1 du x1 ux1 x1x2 1 1 1 dt t 1 x2 1 L x2 . 1 du x1 u du 1 du u 1 du u 1 1 1 x1 t using u t and du x1 dt. Now, t x1 1 dt. x1 tanh 1 2x 3x2 dx. Problem Solving for Chapter 5 1. (a) L 1 0.571. 2x 2 x2 sech x3 2 dx 2x dx x4 x tanh 1 1 tanh x3 3 C P roblem Solving for Chapter 5 511 x sin t2 dt 2. (a) F x 2 x 0 Fx 1.0 0.4945 0.8048 x 1.9 0.0265 2.0 0.0611 2.1 0 2.5 3.0 4.0 5.0 0.0867 0.3743 0.0312 0.0576 0.2769 x 1 (b) G x 1.5 sin t2 dt 2 2 x 1.9 1.95 1.99 2.01 2.05 2.1 Gx 0.6106 0.6873 0.7436 0.7697 0.8174 0.8671 lim G x 0.75 x→2 (c) F 2 lim x→2 lim x→2 Fx x F2 2 x 1 x 2 sin t2 dt 2 lim G x x→2 sin x2, F 2 Since F x Note: sin 4 x 3. S x sin 0 sin 4 lim G x . x→2 0.7568 t2 dt 2 y (a) y (b) 2 3 1 2 x 1 3 1 −1 x −2 1 The zeros of y (c) S x sin x2 2 0⇒ x2 2 ⇒ x2 n 1.4142 and x Relative maximum at x 2 Relative minimum at x 2 and x 2 (d) S x cos x 2 x Points of inflection at x 0⇒ 1, 2n ⇒ x 3, 8 sin 3 2 5 6 72 23 x2 correspond to the relative extrema of S x . 2 2n, n integer 6 2.4495 ⇒ x2 1 2.8284 2 x 2 2 2 5, and n 7 2n ⇒ x 1 2n, n integer 512 Chapter 5 Integration 4. Let d be the distance traversed and a be the uniform acceleration.We can assume that v 0 at at st 0. Then a vt 0 and s 0 12 at . 2 2d . a d when t st The highest speed is v 2d a a 2ad. The lowest speed is v 0. The mean speed is 1 2 2ad 0 ad . 2 The time necessary to traverse the distance d at the mean speed is 2d a d ad 2 t which is the same as the time calculated above. y 5. (a) 5 4 3 2 1 (c) f x (8, 3) (6, 2) f (0, 0) 0≤x<2 x, x 4, 2 ≤ x < 6 1 x 1, 6 ≤ x ≤ 8 2 x 2 −1 −2 −3 −4 −5 (b) 456789 (2, − 2) x Fx f t dt 0 x 0 Fx 1 0 2 1 2 3 x 5 7 2 4 7 1 4 2 3 6 is not. ( f is not continuous at x v 100 80 60 40 20 t 0.2 0.4 0.6 0.8 1.0 (b) v is increasing (positive acceleration) on 0, 0.4 and 0.7, 1.0 . v 0.4 0.4 (c) Average acceleration v0 0 60 0 0.4 150 mi hr2 (d) This integral is the total distance traveled in miles. 1 v t dt 0 1 0 10 2 20 2 60 2 40 2 40 65 385 10 38.5 miles (e) One approximation is a 0.8 v 0.9 0.9 v 0.8 0.8 50 40 0.1 100 mi hr2. 0≤x<2 4, 2 ≤ x < 6 5, 6 ≤ x ≤ 8 Fx f x . F is decreasing on 0, 4 and increasing on 4, 8 . Therefore, the minimum is 4 at x 4, and the maximum is 3 at x 8. 8 6<x<8 2 is a point of inflection, whereas x 6. (a) 6 0<x<2 2<x<6 1, 1, 2 fx 7 2 2 1, (d) F x 4 x2 2 , 2 4x 1 4 x2 x x2 (other answers possible) 6.) P roblem Solving for Chapter 5 1 7. (a) cos x dx 1 3 1 2 cos 1 3 1.6758 x cos x dx 1 1 x2 11 2 sin 1 ax3 cx 1 1 x x ft x t dt xf x bx3 3 b 3 d 2b 3 0 x d dx x b b f x f x dx a a dx 0 0 x 1 f v dv dt 0 0 Letting x x6 6 x5 dx 0 1 1 n 1 n Thus, lim S n 5 2 n lim n→ 1 n 2 fa 2 ... 25 n→ . 5 n5 n6 3 15 1 n6 ... 25 n5 . 1 . 6 9 —CONTINUED— 2 9 x2 dx y 0 2 27 2 n 10 3 8 7 6 5 4 3 2 1 0 36 −4 −2 −1 x 12 45 2 2 ... n3 2 3 x3 3 1 1 Thus, lim n n ... x2 dx 2 9x 1 n 1 n3 2 Sn Fa 3 9 0 2 . The corresponding 3 n n ... a 5 15 1 2 Riemann Sum using right endpoints is b n→ 12. (a) Area 23 x 3 x dx 10. Consider 2f x f x . Thus, The corresponding Riemann Sum using right endpoints is Sn 0. d 1 . 6 0 C. 0 0, we see that C 0 1 Fb 2 1 t t dt 2d 1 Fx 2 11. Consider f v dv. 0 ft x 1 F x dx 2 1 fb 2 f t dt. 0 x f v dv dt x 1 cx2 2 b 3 ⇒F x 2 fx xf x 0 1 9. Consider F x x f t dt Differentiating the other integral, 2d 1 3 p t f t dt 0 Thus, the two original integrals have equal derivatives, 2b 3 1 3 3 2 d. a x4 4 p x dx p d dx 1.5708) 2 bx2 x f t dt Thus, 1.6829 1 13 1 (Note: Exact answer is (c) Let p x 0 x 0.0071 1 13 1 t f t dt 0 0 1.6758 dx x x f t dt 0 1 Error: 1.6829 t dt x sin x 1 x ft x 8. 1 1 (b) cos 1 3 cos 513 ... n n. 2 . 3 514 Chapter 5 Integration 12. —CONTINUED— (b) Base 2 bh 3 9, area 6, height 2 69 3 b2 (c) Let the parabola be given by y 36 a2x2, a, b > 0. y ba Area b2 2 a2x2 dx 2 b2x 0 2b , height a Base 3 a2 b 3a b a 2 b2 b3 a 2 ba x3 a2 3 b2 0 1 b3 3a 4 b3 3a b2 − b 13. By Theorem 5.8, 0 < f x ≤ M ⇒ b f x dx ≤ M dx a Mb a. a b Similarly, m ≤ f x ⇒ m b x b a 4 b3 3a 2 2b 2 b 3a Archimedes’ Formula: Area b a b m dx ≤ a f x dx. a a b a≤ Thus, m b f x dx ≤ M b a . On the interval 0, 1 , 1 ≤ 1 x4 ≤ 2 and b a 1 Thus, 1 ≤ 1 x4 dx ≤ 1 2. Note: 3 14. (a) 1 i (b) 3i2 1.0894 0 1 3i2 3i i3 ⇒ 1 i 3i 3 1 n 1 3 i i3 3i2 3i 1 i3 n 3i2 i x4 dx 1 0 3i 1 i 1 i 3 1 i3 1 23 13 33 1 ... 23 1. n 1 3 n3 n Hence, n 1 3 3i2 i 3i 1 n (c) n 1 3 n 3i2 1 i 3i 1 i 1 n ⇒ 3i2 i n3 3n2 3n n 1 2 3n 1 2n3 3n n 2 3i2 1 6n2 3n2 6n 3n 1 n n 2n 2 2n3 nn n ⇒ i2 i 15. Since fx 1 2n 2 1 nn 1 2n 6 1 1 n ≤fx ≤ fx , b b f x dx ≤ a 3n2 2 b a b f x dx ⇒ f x dx ≤ a b f x dx ≤ a f x dx. a n 1 3 1 a 1. P roblem Solving for Chapter 5 22 16. (a) y fx arcsin x, sin y arcsin x dx (b) x Area C sin y dy 2 cos y 6 6 2 2 Area B 1 2 3 2 6 3 2 B 2 3 2 0.1346 y ey 3 1 2 y = ln x 0 0 12 1 12 2 8 ln 3 ey dy 2 2 0.1589 0.2618 12 3 8 2 ln 3 (c) Area A A 4 4 Area A 2 2 4 12 515 ln 3 3 Area B ln x dx 3 ln 3 A 3 ln 3 2 ln 27 2 ey = x A 1.2958 B 1 tan y (d) x x 1 2 3 3 Area A tan y dy 4 3 y ln cos y 4 1 ln 2 y = arctan x 2 ln 2 ln 2 3 Area C arctan x dx 1 12 17. Let u dx 1 43 x, 2u x 1 ln 2 2 3 u 4 2u 1, 4 1 1 dx xx x 1 tan x, du 4 Area 3 2 2u u 1 32 2 u u2 32 2 u 2 u2 0 2u 1 1 du u du sec2 x dx. 4 1 2 sin x 2 4 cos x dx 0 sec2 x dx tan2 x 4 1 du u2 4 1 u arctan 2 2 1 1 arctan 2 2 3 2 ln u 2 0.8109 3 0 du 2 ln 3 C B 1 18. Let u 2 du. Area A 0.6818 u2 1, x 1 ln 2 2 3 3 π 3 π 4 1 ln 2 2 2 ln 2 2 ln 3 2 1 0 516 Chapter 5 Integration 19. (a) (i) y ex y1 (ii) y 1 ex y2 1 x (iii) y 4 y 4 yy 3 −2 −1 2 −1 (b) nth term is xn n! in polynomial: y4 b 0 Let u b fb u u fb u u fb fb x x fb b b 0 b 0 1 x3 3! x2 2! x x3 3! ... dx. dx. fb A x x, du 0 x2 2! x fx fb fx du fu fu du dx fx Then, b fx fx fb 1 dx b b. 2A 0 x dx 0 fb fb x b 0 Thus, A b . 2 1 (b) b 1⇒ 0 sin 1 x3 6 y2 2 20. (a) Let A x2 2 x y −2 1 1 4 y1 (c) Conjecture: ex y3 x2 2 x ex sin x dx x sin x 1 2 x fx dx −2 2 −1 x4 4! ...
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This note was uploaded on 11/13/2010 for the course MATH MAT 231 taught by Professor Thurber during the Spring '08 term at Thomas Edison State.

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