Ch 5 solutions

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Unformatted text preview: =− + C. 3 2 4 + 2)2 u 40 u 40(5x 29. u = 5x4 + 2, du = 20x3 dx, 1 20 31. u = sin x, du = cos x dx; eu du = eu + C = esin x + C . 33. u = −2x3 , du = −6x2 , − 35. u = ex , 1 6 3 1 1 eu du = − eu + C = − e−2x + C . 6 6 1 du = tan−1 (ex ) + C . 1 + u2 37. u = 5/x, du = −(5/x2 )dx; − 1 5 39. u = cos 3t, du = −3 sin 3t dt, − 41. u = x2 , du = 2x dx; 1 2 sin u du = 1 3 u4 du = − √ 1 4 1 1 u1/2 du = − u3/2 + C = − (2 − sin 4θ)3/2 + C . 6 6 1 du = sin−1 (tan x) + C . 1 − u2 47. u = sec 2x, du = 2 sec 2x tan 2x dx; 49. 15 1 u + C = − cos5 3t + C . 15 15 1 1 tan u + C = tan x2 + C . 2 2 sec2 u du = 43. u = 2 − sin 4θ, du = −4 cos 4θ dθ; − 45. u = tan x, 1 1 cos u + C = cos(5/x) + C . 5 5 e−x dx; u = −x, du = −dx; − 1 2 u2 du = 13 1 u + C = sec3 2x + C . 6 6 eu du = −eu + C = −e−x + C . Exercise Set 5.3 133 √ 1 51. u = 2 x, du = √ dx; , x 53. u = 2y + 1, du = 2dy ; 55. sin2 2θ sin 2θ dθ = √ 1 du = −e−u + C = −e−2 x + C . u e 1 1 1 1 1√ 1 u + C = (2y + 1)3/2 − (u − 1) √ du = u3/2 − 4 6 2 6 2 u (1 − cos2 2θ) sin 2θ dθ; u = cos 2θ, du = −2 sin 2θ dθ, − 1 1 − cos 2θ + cos3 2θ + C . 2 6 57. 1+ 1 t 1 2 2y + 1 + C . 1 1 (1 − u2 )du = − u + u3 + C = 2 6 dt = t + ln |t| + C . 59. ln(ex ) + ln(e−x ) = ln(ex e−x ) = ln 1 = 0, so [ln(ex ) + ln(e−x )]dx = C . √ √ √ √ (c) (1/ π ) sec−1 |x/ π | + C . 61. (a) sin−1 (x/3) + C . (b) (1/ 5) tan−1 (x/ 5) + C . ember 10, 2008 16:01 ”ISM ET chapter 5” Sheet number 14 Page number 238 black 63. u = a + bx, du = b dx, (a + bx)n dx = 1 b un du = (a + bx)n+1 + C. b(n + 1) 238 65. 67. 69. 70. 69. 71. 72. 71. 73. 73. Chapter 5 1 1 1 u = sin(a + bx), du = b cos(1 + bx)dx, a un du = un+1 + C = sinn+1 (a + bx) + C . 1 b x2 + 15x −b1) + C = 25 x3 − 5x2 n + 1) 1 + C ; (n + 1) b( + x − 3 3 (b) (5x − 1) + C2 = (125x − 75 2 2 15 15 3 15 1 1 the answers differ by a constant. (a) With u = sin x, du = cos x dx; u du = u2 + C1 = sin2 x + C1 ; 2 2 ￿ √ 2 2 3/2 1 y= 5x + 1 dx = (5x + 1) + C ; −2 =1 (3) = y 64 + C , with u = cos x, du = − sin x dx; − u du = − u2 + C15 − cos2 x + C2 . 15 2= 2 2 2 158 2 158 so C = −2 − 64 = − , and y = (5x + 1)3/2 − 15 15 15 15 (b) ￿ Because they differ by a constant: 1 1 1 y1 = 2 + sin 3x) dx = 2x − cos 3x + C and2 (2 sin x + C1 − − cos2 x + C2 = (sin x + cos2 x) + C1 − C2 = 1/2 + C1 − C2 . 3 2 ￿ π ￿ 2π 1 2 2π + 1 2 1 2π + 1 0=y = + + C, C = − , y = 2x − cos 3x − 3 3 3 3 3 3 √ 2 2 2 158 3/2 (5x + 1) + C ; −2 = y (3) = 64 + C , so C = −2 − 64 = − , and y = y = ￿ 5x + 1 dx = 1 2t 15 1 1 2t15 13 15 15 y = − e2t dt =158 e + C, 6 = y (0) = − + C, y = − e + − 2 2. 2 2 2 (5x + 1)3/2 − 15 15 ￿￿ ￿ ￿ ￿ 1 1 3 π 5 1π y= dt = tan−1 t + C, 1 = y − 1 = − 13 + C, 1 15 2t 9t2 2t 2t 15 4 y = − 25 + dt = − e ￿+ C, 6 =5 (0) = − 30+ C, y = 3 e + . e y − ￿ 2 2 2 2 π 1 3 π C= ,y = tan−1 t+ 60 15 5 60 √ 1 1 2 (a) u = x + 1, du = 2x dx; x ￿ √ du = u + C = ￿ 2 + 1 + C . √ 21 u 1 √ du = u + C = x2 + 1 + C (a) u = x2 + 1, du = 2x dx; 2 u 5 (b) –5 5 0 (b) 1 74. (a) u = x + 1, du = 2x dx; 2 2 ￿ 1 1 1 du = ln u + C = ln(x2 + 1) + C u 2 2 134 Chapter 5 √ 75. f (x) = m = (3x + 1)1/2 dx = 3x + 1, f (x) = 2 7 (3x + 1)3/2 + . 9 9 7 2 2 (3x + 1)3/2 + C , f (0) = 1 = + C , C = , so f (x) = 9 9 9 (ln 2) 2t/20 dt = 20 · 2t/20 + C ; 20 = y (0) = 20 + C , so C = 0 and y (t) = 20 · 2t/20 . This implies that 77. y (t) = y (120) = 20 · 2120/20 = 1280 cells. du 1 1 u √ = θ = sec−1 + C . a a a u u2 − a2 79. If u > 0 then u = a sec θ, du = a sec θ tan θ dθ; Exercise Set 5.4 1. (a) 1 + 8 + 27 = 36. (d) (b) 5 + 8 + 11 + 14 + 17 = 55. (c) 20 + 12 + 6 + 2 + 0 + 0 = 40 (e) 1 − 2 + 4 − 8 + 16 = 11. 1 + 1 + 1 + 1 + 1 + 1 = 6. (f ) 0 + 0 + 0 + 0 + 0 + 0 = 0. 10 k 3. k=1 10 5. 2k k=1 6 (−1)k+1 (2k − 1) 7. k=1 50 50 2k 9. (a) (b) k=1 k=1 11. 1 (100)(100 + 1) = 5050. 2 13. (2k − 1) 1 (20)(21)(41) = 2870. 6 30 15. k=1 n 17. k=1 19. n−1 k=1 30 k (k 2 − 4) = 3k 3 = n n 30 (k 3 − 4k ) = 30 k3 − 4 k= k=1 k=1 k= 31 3 · n(n + 1) = (n + 1). n2 2 n k=1 k3 1 =2 n2 n n−1 k=1 k3 = k=1 1 1 (30)2 (31)2 − 4 · (30)(31) = 214,365. 4 2 1 11 · (n − 1)2 n2 = (n − 1)2 . n2 4 4 21. True. 23. False; if [a, b] consists of positive reals, true; but false on, e.g. [−2, 1]. 25. (a) 2+ 3 n 4 3 6 , 2+ n n 4 3 9 , 2+ n n 4 3 3(n − 1) ,..., 2 + n n 4 3 3 , (2 + 3)4 . When [2, 5] is subdivided into n n n Exercise Set 5.4 135 3 3 3 3 , 2 + 2 · , 2 + 3 · , . . . , 2 + (n − 1) , 2 + 3 = 5, and the right endpoint n n n n approximation to the area under the curve y = x4 is given by the summands above. equal intervals, the endpoints are 2, 2 + (b) n−1 k=0 2+k· 4 3 n 3 gives the left endpoint approximation. n 27. Endpoints 2, 3, 4, 5, 6; ∆x = 1; 4 f (x∗ )∆x = 7 + 10 + 13 + 16 = 46. k (a) Left endpoints: k=1 4 f (x∗ )∆x = 8.5 + 11.5 + 14.5 + 17.5 = 52. k (b) Midpoints: k=1 4 f (x∗ )∆x = 10 + 13 + 16 + 19 = 58. k (c) Right endpoints: k=1 29. Endpoints: 0, π/4, π/2, 3π/4, π ; ∆x = π/4. 4 f (x∗ )∆x = 1 + k (a) Left endpoints: √ k=1 2/2 + 0 − √ 2/2 (π/4) = π/4. 4 f (x∗ )∆x = [cos(π/8) + cos(3π/8) + cos(5π/8) + cos(7π/8)] (π/4) = k (b) Midpoints: k=1...
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This document was uploaded on 02/02/2014.

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