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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

Ask a homework question - tutors are online