05 - CHAPTER Integration Section 5.1 Section 5.2 Section 5.3 Section 5.4 Section 5.5 Section 5.6 Section 5.7 Section 5.8 Section 5.9 5 Antiderivatives

# 05 - CHAPTER Integration Section 5.1 Section 5.2 Section...

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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...
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