71 let f x n x2 0 x 1 and xi n 1 n the appropriate

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Unformatted text preview: 9 4 0 9 9 9 4 9 2 3 2 2 9 2 9 4 9 2 x2 0 3x 32 S ection 4.4 The Fundamental Theorem of Calculus 193 3 2 3 25. 0 x2 4 dx 0 4 4x 8 23 3 x 2 dx 2 x2 x3 3 4x 4 dx 3 2 x3 3 8 3 2 0 9 12 8 3 8 27. 0 1 sin x dx x cos x 0 1 0 1 2 6 6 29. 6 sec2 x dx tan x 6 3 3 3 3 3 23 3 3 31. 3 4 sec tan d 4 sec 3 42 42 0 3 33. 0 10,000 t 3 6 dt 10,000 3 t2 2 3 1 6t 0 $135,000 35. A 0 x x2 dx x2 2 12 3 5 x3 3 1 0 1 6 37. A 0 3 x x dx 0 3x1 2 x3 2 dx 2x3 2 25 x 5 3 2 0 xx 10 5 3 2x 0 2 2 39. A 0 cos x dx sin x 0 1 41. Since y ≥ 0 on 0, 2 , 2 2 A 0 3x2 1 dx x3 x 0 8 2 10. 43. Since y ≥ 0 on 0, 2 , 2 2 45. 0 x 2 x dx fc 2 c 0 2c 1 1 c 2 x2 2 6 3 3 6 4x3 3 82 3 42 3 42 3 42 3 6 22 2 0 82 3 A 0 x3 x dx x4 4 x2 2 2 4 0 2 6. c 2c c 1 1 c c ± 42 3 6 42 3 2 1± 0.4380 or c 1.7908 194 Chapter 4 Integration 4 4 47. 4 2 sec2 x dx fc 2 tan x 4 21 2 1 4 4 4 2 sec2 c sec2 c sec c c 4 8 4 2 2 ± ± arcsec ± arccos 2 13 x 3 ± 0.4817 2 2 49. 1 2 2 2 4 2 x2 dx 8 3 1 4x 4 1 4 8 8 3 8 8 3 8 3 (− 2 3 3 ,8 3 ( 5 ( 2 3 3 ,8 3 ( Average value 4 x2 −3 3 −1 8 when x2 3 4 8 or x 3 ± 23 3 ± 1.155. 51. 1 0 0 sin x dx 2 2 1 cos x 0 2 2 2 (0.690, π ( 2 (2.451, π ( 53. If f is continuous on a, b and F x b f x on a, b , Average value sin x x 2 then a f x dx Fb Fa. − 2 −1 3 2 0.690, 2.451 6 2 6 55. 0 6 f x dx area of region A 6 6 1.5 57. 0 f x dx 0 f x dx 2 f x dx 1.5 5.0 6.5 59. 0 2 f x dx 0 2 dx 0 f x dx 15.5 (b) 1 3 3 12 61. (a) F x F0 Fx k sec2 x k 500 3.5 0 500 sec2 x dx 0 1500 1500 3 tan x 0 500 sec2 x 3 0 826.99 newtons 827 newtons 1 5 0 5 63. 0.1729t 0 0.1552t2 0.0374t3 dt 1 0.08645t2 5 5 0.05073t3 0.00935t4 0 0.5318 liter S ection 4.4 The Fundamental Theorem of Calculus 195 65. (a) 0 1 (b) 24 10 −1 0 0 24 The area above the x-axis equals the area below the x-axis. Thus, the average value is zero. 67. (a) v (b) 8.61 90 The average value of S appears to be g. 10 4t3 0.0782t2 0.208t 0.0952 − 10 − 10 70 60 (c) 0 v t dt 8.61 4 t2 2 8 25 2 8 10 4t 4 0.0782t3 3 x2 2 0.208t 2 2 60 0.0952t 0 2476 meters x x 69. F x 0 t 4 2 25 2 64 2 5 dt 52 55 58 5t 0 5x 71. F x 10 2 dv 1v 10 x 10 5 8 35 4 x x 10v 1 2 dv 10 v x 1 F2 F5 F8 10 1 1 x F2 F5 F8 10 10 10 1 2 4 5 7 8 x x x 73. F x 1 cos d sin 2 sin 5 sin 8 x 3 8 sin 1 sin x sin 1 75. (a) 0 t d 12 x dx 2 2 dt 2x t2 2 x x 2t 0 12 x 2 2x F2 F5 F8 sin 1 sin 1 sin 1 34 t 4 x 3 8 0.0678 1.8004 0.1479 34 x 4 x1 3 3 (b) 2 77. (a) (b) t dt 3 3 16 34 x 4 x 3 x 12 79. (a) x4 sec2 t dt d tan x dx 1 tan t x4 tan x 1 d 34 x dx 4 x 12 x x (b) sec2 x x 81. F x 2 t2 x2 2x 2t dt 83. F x 1 t4 x4 1 1 dt 85. F x 0 t cos t dt x cos x Fx Fx Fx 196 Chapter 4 Integration x 2 87. F x x 4t x 1 dt 2 Alternate solution: x 2 2t 2 t x Fx x 0 4t 4t x x 1 dt x 2 2x 8x Fx 8 2 10 2 x 2 2x2 x 1 dt 0 x 4t 2 1 dt 4t 1 dt 8 4t 0 1 dt 0 Fx sin x 4x x3 1 4x 2 1 89. F x 0 t dt sin x 1 2 cos 23 t 3 x 2 sin x 0 2 sin x 3 32 91. F x 0 sin t 2 dt sin x3 2 Fx cos x sin x Fx 3x2 3x2 sin x 6 Alternate solution sin x Fx 0 t dt sin x x Fx 93. g x 0 d sin x dx sin x cos x x f t dt 0, g 1 1 ,g 2 2 1, g 3 1 ,g 4 2 0 95. (a) C x 5000 25 5000 25 5000 25 3 0 t1 4 dt 45 t 5 4 x 4 0 g0 y 2 1 3 12 5 x 5 1000 125 $137,000 54 54 12x5 4 f (b) g x 1 2 3 4 C1 C5 1000 125 1000 125 1000 125 12 1 12 5 $214,721 $338,394 −1 −2 C 10 12 10 g has a relative maximum at x 97. True 2. 1 0 1 99. False; 1 x 2 dx 1 x 2 dx 0 x 2 dx 2 Each of these integrals is infinite. f x has a nonremovable discontinuity at x 1x x 0. 101. f x 0 1 t 2 x 1 dt 1 1 2 0t dt By the Second Fundamental Theorem of Calculus, we have fx 1 1x 1 Since f x 2 1 x2 1 x2 1 x2 1 1 x2 0. 1 1 0, f x must be constant. S ection 4.5 103. x t xt t3 3t 2...
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