ODDREV04 - Review Exercises for Chapter 4 209 Review...

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Unformatted text preview: Review Exercises for Chapter 4 209 Review Exercises for Chapter 4 1. y 3. f f 2x2 x 1 dx 23 x 3 12 x 2 x C x 5. x3 x2 1 dx x 1 dx x2 12 x 2 1 x C 7. 4x 3 sin x dx 2x2 3 cos x C 9. f x fx When x y C y 2x, 2x dx 1: 1 2 2 x2 1, 1 x2 C 11. at vt v0 a a dt 0 at at dt 0 a2 t 2 a 30 2 2 at C1 0. C1 0 when C1 C 1 vt st s0 st s 30 a v 30 a2 t 2 C2 0. C2 0 when C2 3600 or 8 ft sec2. 240 ft sec 2 3600 30 2 8 30 10 13. a t vt st (a) v t (b) s 3 (c) v t (d) s 3 2 32 32t 16t2 32t 144 32t 16 9 4 96 96t 96 288 96 0 when t 144 ft 96 when t 2 96 3 2 3 sec. 2 3 sec. 15. (a) i n 1 2i i3 i 1 10 1 (b) (c) i 1 4i 2 108 ft 210 Chapter 4 Integration 17. y 10 x2 1 S4 ,x 1 ,n 2 4 10 1 22 10 1 1 2 Sn 1 10 21 13.0385 1 10 3 22 1 sn s4 10 1 2 1 22 9.0385 10 1 1 1 10 3 22 10 1 22 1 9.0385 < Area of Region < 13.0385 4 , right endpoints n f ci i n n→ 1 y 8 6 19. y 6 x, x n Area n→ lim x 4i 4 nn 4nn 1 n 2 4 2 −2 −2 x 2 4 6 8 lim i 1 6 4 6n n 24 8 n→ lim lim n n 1 n→ 24 8 16 21. y 5 x2, x n 3 n f ci x 2 12i n 3i n 2 y 6 4 3 2 1 −4 −3 −1 −2 x 1 2 3 4 Area n→ lim i n 1 n→ lim i 1 5 3n 1 ni 1 3 n n 3 18 3 n n→ lim lim lim 9i2 n2 9 nn n2 1 2n n2 1 2n 6 1 1 n→ 12 n n 1 n 2 18 9 n n 12 3 n 2 4 9i2 n2 1 9n 2 n→ 3 23. x 5y y2, 2 ≤ y ≤ 5, y n y 6 Area n→ lim i 1 n 52 3 ni 10 1 3i n 15i n 3i n 3i n 12 i n 2 3 n 9i2 n2 4 3 2 1 x 1 2 3 4 5 6 n→ lim 3n lim 6 n→ ni 1 n→ lim 18 3 6n n 9 2 9 3nn 1 n 2 27 2 9 nn n2 1 2n 6 1 R eview Exercises for Chapter 4 211 n 6 25. lim → i 1 2ci 3 xi 4 2x 3 dx 27. 12 9 6 3 −3 −3 y x 3 6 9 5 5 5 5 0 x 5 dx 0 5 5 x dx 0 x dx 25 2 (triangle) 6 6 6 29. (a) 2 6 fx fx 2 6 g x dx 2 6 f x dx 2 6 g x dx g x dx 2 6 10 10 3 3 2 10 13 7 33 11 (b) (c) 2 6 g x dx 2 f x dx 6 2f x 5f x dx 2 3g x dx 6 2 2 f x dx 5 10 3 2 g x dx (d) 5 2 f x dx 50 8 31. 1 3 x 1 dx 34 x 4 8 3 x 1 3 16 4 16 2 9 8 3 4 1 73 (c) 4 1 1 4 33. 0 2 x dx 2x x2 2 4 8 0 16 35. 1 4t 3 2t dt t4 t2 1 0 9 9 37. 4 x x dx 4 x3 2 dx 25 x 5 3 4 2 4 2 5 2 2 9 5 4 5 2 243 5 2 2 32 422 5 3 4 39. 0 sin d cos 0 1 1 22 2 4 3 3 41. 1 2x y 1 dx x 2 x 1 6 43. 3 x2 9 dx x3 3 64 3 64 3 4 9x 3 36 54 3 10 3 9 27 6 4 2 8 y x 2 2 4 6 6 4 2 x 2 4 6 8 212 Chapter 4 Integration 1 45. 0 x x3 dx y x2 2 x4 4 1 0 1 2 1 4 1 4 9 47. Area 1 y 10 4 x dx 4x1 2 12 9 83 1 1 16 1 8 6 x 1 −2 4 2 x −2 2 4 6 8 10 1 49. 1 9 2 5 x x 4 9 4 1 x dx 1 2x 5 9 4 2 3 5 2 2 Average value 5 2 y 1 x 5 2 25 4 1 25 2 , 45 x 2 4 6 8 10 51. F x x2 1 x3 53. F x x2 3x 2 55. x2 1 3 dx x6 3x4 3x2 1 dx x7 7 35 x 5 x3 x C 57. u x3 x2 x3 3, du 3 dx 3x2 dx x3 3 12 x2 dx 1 3 x3 3 12 3x2 dx 23 x 3 3 12 C 59. u 1 x1 3x2, du 3x2 4 dx 6x dx 1 6 1 3x2 4 6x dx 1 1 30 3x2 5 C 1 3x2 30 1 5 C 61. sin3 x cos x dx 14 sin x 4 C 63. sin 1 cos dx 1 cos 12 sin d 21 cos 12 C 21 cos C 65. tann x sec2 x dx tann 1 x n1 C, n 1 67. 1 sec x 2 sec x tan x dx 1 1 sec x 2 sec x tan x dx 1 1 3 9 4 sec x 3 C 2 69. 1 x x2 4 dx 1 2 2 x2 1 4 2x dx 1 x2 4 2 2 22 1 1 0 4 9 R eview Exercises for Chapter 4 213 3 71. 0 1 dx 1x 1 y, y 0, u 1 3 3 1 0 x 12 dx 21 x 12 0 4 2 2 73. u 1 u, dy 1. When y du 1, u 0 When y 2 0 0. 1 1 0 y 1 1 y dy 2 2 1 u 2u1 2 1 du u du 2 25 u 5 2 u3 2 43 u 3 0 2 1 28 15 75. 0 cos x dx 2 x, x a, u b 2 0 cos x1 dx 22 du 2 sin x 2 2 0 77. u 1 1 1 u, dx When x Pa, b a. When x x dx 2 b, u 1 1 b 1 1 b. u 2 a 15 x1 4 1 1 b 15 4 2 u du 23 u 3 1 2 1 a b a 15 4 (a) P0.50, 0.75 (b) P0, b 1 b 79. p C b 32 u3 a u1 32 du 2 15 2 5 u 45 0.75 15 2u3 2 3u 4 15 1 b a 5 1 1 x 2 32 b 3x 2 a 1 1 3b x 2 2 x 2 32 3x b 0.353 0.50 35.3% 3b 2 1 0.5 3x 1 2 0 1 b 2 32 0.586 58.6% 1.20 15,000 M 0.04t t t 1 p ds 10. 0.04t dt 11 (a) 2000 corresponds to t C 15,000 M 11 (b) 2005 corresponds to t C 24,300 M 1 1 8 1 13 1 1 12 1 13 2 1.25 4 1.25 2 1.5 2 1.5 15,000 1.20t M 15. 16 1.20 10 0.02t2 15 27,300 M 15,000 1.20t M 0.02t2 10 2 81. Trapezoidal Rule n Simpson’s Rule n 4: 1 2 1 1 1 x3 x3 dx 1 1 3 1 1 3 1 1 2 1.75 4 1.75 1 3 1 1 1 23 23 0.257 0.254 4: 1 1 dx 3 3 3 Graphing utility: 0.254 2 83. Trapezoidal Rule n Simpson’s Rule n 4: 0 x cos x dx 0.637 85. (a) R < I < T < L (b) S 4 40 f0 34 1 4 3 42 4f 1 21 2f 2 4 1 2 4f 3 1 4 f4 5.417 4 : 0.685 Graphing Utility: 0.704 214 Chapter 4 Integration Problem Solving for Chapter 4 1 1. (a) L 1 1 1 dt t 0 (b) L x L1 (c) L x 2.718 1 1 by the Second Fundamental Theorem of Calculus. x 1 x 1 1 1 dt for x t 2.718 (Note: The exact value of x is e, the base of the natural logarithm function.) 1 1 1 dt t 0.999896 x1 (d) We first show that 1 1 dt t 1 dt. x1 t 1 dt. x1 1 1 To see this, let u x1 t and du x1 1 1 Then 1 1 dt t 1 x1 du x1 ux1 x1x2 1 1 1 1 du x1 u x2 1 1 1 x1 t dt. t x1 Now, L x1x2 1 dt t 1 1 du using u x1 u 1 du u 1 du x1 u 1 du u 1 x2 1 x2 x1 1 1 du u L x1 x L x2 . 3. S x 0 sin y 2 1 t2 dt 2 (b) 3 y (a) 2 x 1 −1 −2 3 1 x 1 2 3 2 5 6 72 23 The zeros of y extrema of S x . x2 2 x2 2 2 2 and x 0⇒ 1, 3, x2 2 2 5, and sin x2 correspond to the relative 2 (c) S x sin 0⇒ n ⇒ x2 2n ⇒ x 6 2n, n integer. 2.4495 Relative maximum at x Relative minimum at x (d) S x cos x2 2 x 1.4142 and x 8 2.8284 n 7. ⇒ x2 1 2n ⇒ x 1 2n, n integer Points of inflection at x P roblem Solving for Chapter 4 5. (a) 5 4 3 2 1 −1 −2 −3 −4 −5 y 215 (b) (6, 2) (0, 0) x 2 456789 x Fx 0 0 1 1 2 2 2 3 7 2 4 4 5 7 2 6 2 7 1 4 8 3 (8, 3) f (2, − 2) (c) f x 0≤x<2 x, x 4, 2 ≤ x < 6 1 x 1, 6 ≤ x ≤ 8 2 x 1, (d) F x fx 1, 1, 2 0<x<2 2<x<6 6<x<8 6 is not. Fx 0 f t dt x2 2 , x2 4x 1 4 x2 x 2 0≤x<2 4, 2 ≤ x < 6 5, 6 ≤ x ≤ 8 x 2 is a point of inflection, whereas x ( f is not continuous at x 6.) 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. 1 7. (a) 1 1 cos x dx cos x dx 1 cos 1 1 3 cos 2 sin 1 1 3 2 cos 1 3 1.6758 sin x 1 1.6829 Error. 1.6829 1 1.6758 1 1 13 2 0.0071 1 1 13 3 2 (c) Let p x 1 (b) 1 x2 11 dx ax3 bx2 a x4 4 cx bx3 3 b 3 x6 6 1 0 d. cx2 2 d 1 (Note: exact answer is 1.5708) 1 p x dx p 1 3 p dx 1 2b 3 d 2d 2b 3 2d 1 3 b 3 1 9. Consider F x b fx 2 b a ⇒F x 1 F x dx 2 b a 2f x f x . Thus, 11. Consider 0 x5 dx 1 . 6 f x f x dx a The corresponding Riemann Sum using right endpoints is Sn 1 n 1 n 5 1 Fx 2 1 Fb 2 1 fb 2 2 2 n 25 5 ... n5 ... n6 n n 5 Fa fa 2 n→ 15 1 n6 Thus, lim S n ... 15 25 n→ lim n5 1 . 6 216 Chapter 4 Integration b b 13. By Theorem 4.8, 0 < f x ≤ M ⇒ a f x dx ≤ a b b M dx f x dx. a Mb a. Similarly, m ≤ f x ⇒ m b b a a m dx ≤ Thus, m b Thus, 1 ≤ 0 a≤ a 1 f x dx ≤ M b x4 dx ≤ 2. a . On the interval 0, 1 , 1 ≤ 1 1 x4 ≤ 2 and b a 1. 1 Note: 0 1 x4 dx 1.0894 15. Since b fx ≤fx ≤ fx , b b b b f x dx ≤ a a f x dx ≤ a f x dx ⇒ a f x dx ≤ a f x dx. 17. 1 365 365 100,000 1 0 sin 2 t 60 365 dt 100,000 t 365 365 2 t 60 cos 2 365 365 100,000 lbs. 0 ...
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This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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