Chapter 7 a 6 displacement s6 s0 6 2t 4dt

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Unformatted text preview: c2 u − 1)du = 1 1 (tan u − u) + C = (tan 3θ − 3θ) + C 3 3 (1 − cos2 2θ) sin 2θ dθ; u = cos 2θ, du = −2 sin 2θ dθ, 1 1 1 1 (1 − u2 )du = − u + u3 + C = − cos 2θ + cos3 2θ + C 2 6 2 6 Exercise Set 7.3 45. 1+ 216 1 t dt = t + ln |t| + C 2 e2 ln x dx = 46. e2 ln x = eln x = x2 , x > 0, so 47. ln(ex ) + ln(e−x ) = ln(ex e−x ) = ln 1 = 0 so 48. 49. cos x dx; u = sin x, du = cos xdx; sin x (a) with u = sin x, du = cos x dx; [ln(ex ) + ln(e−x )]dx = C 12 1 u + C1 = sin2 x + C1 ; 2 2 12 1 u du = − u + C2 = − cos2 x + C2 2 2 u du = because they differ by a constant: 1 1 sin2 x + C1 − − cos2 x + C2 2 2 (a) (b) First method: (25x2 − 10x + 1)dx = second method: 50. 1 5 u2 du = 13 x +C 3 1 du = ln |u| + C = ln | sin x| + C u with u = cos x, du = − sin x dx; − (b) x2 dx = = 1 (sin2 x + cos2 x) + C1 − C2 = 1/2 + C1 − C2 2 25 3 x − 5x2 + x + C1 ; 3 13 1 u + C2 = (5x − 1)3 + C2 15 15 1 1 25 3 1 (5x − 1)3 + C2 = (125x3 − 75x2 + 15x − 1) + C2 = x − 5x2 + x − + C2 ; 15 15 3 15 the answers differ by a constant. √ 2 (3x + 1)3/2 + C , 9 29 29 2 16 + C = 5, C = so y (x) = (3x + 1)3/2 + y (1) = 9 9 9 9 3x + 1dx = 51. y (x) = 52. y (x) = 53. f (x) = m = 5 cos 2x + C, 2 1 5 1 5 y (0) = + C = 3, C = so y (x) = 6x + cos 2x + 2 2 2 2 f (x) = (6 − 5 sin 2x)dx = 6x + √ 3x + 1, f (x) = 2 7 (3x + 1)3/2 + 9 9 54. y 4 -4 4 x (3x + 1)1/2 dx = 7 2 2 (3x + 1)3/2 + C ; f (0) = 1 = + C , C = , so 9 9 9 217 55. 56. Chapter 7 8 8 256 (4 + 0.15t)5/2 + C ; p(0) = 100, 000 = 45/2 + C = + C, 3 3 3 8 8 256 ≈ 99, 915, p(t) = (4 + 0.15t)5/2 + 99, 915, p(5) = (4.75)5/2 + 99, 915 ≈ 100, 416 C = 100, 000 − 3 3 3 (4 + 0.15t)3/2 dt = p(t) = √ 2 dr = −k t, r = −k t1/2 dt = − kt3/2 + C ; r(0) = 10, 000 = C so C = 10, 000 and dt 3 2 250 2 3/2 k , so r = − kt + 10, 000; r(25) = 9, 000 = − k (25)3/2 + 10, 000 = 10, 000 − 3 3 3 3/2 k = 12, r = −8t + 10, 000, and r(60) = 6281.94 m EXERCISE SET 7.4 1. (a) 1 + 8 + 27 = 36 (c) 20 + 12 + 6 + 2 + 0 + 0 = 40 (e) 1 − 2 + 4 − 8 + 16 = 11 (b) 5 + 8 + 11 + 14 + 17 = 55 (d) 1 + 1 + 1 + 1 + 1 + 1 = 6 (f ) 0 + 0 + 0 + 0 + 0 + 0 = 0 2. (a) 1 + 0 − 3 + 0 = −2 (c) e2 + e2 + · · · + e2 = 14e2 (14 terms) (b) 1 − 1 + 1 − 1 + 1 − 1 = 0 (d) 24 + 25 + 26 = 112 (e) ln 1 + ln 2 + ln 3 + ln 4 + ln 5 + ln 6 = ln(1 · 2 · 3 · 4 · 5 · 6) = ln 720 (f ) 1−1+1−1+1−1+1−1+1−1+1=1 10 20 k 3. 4. k =1 49 3k k =1 k (k + 1) k =0 8 2k k =1 (−1)k+1 k =1 k =1 5 1 k (−1)k 11. k =1 3 1 k 12. (2k − 1) (b) 1 5 5 (−1)k+1 ak (a) k =0 5 ak xk (c) k =0 a5−k bk (d) k =0 1 (100)(100 + 1) = 5050 2 100 2 k− 16. k =1 17. n (−1)k+1 bk (b) k =1 15. kπ 7 50 2k 1 14. cos k =0 50 (a) (−1)k+1 (2k − 1) 9. k =1 5 10. 6 (2k − 1) 8. 2k 6. k =1 10 7. 13. 4 5. k= k =1 1 (100)(100 + 1) − (1 + 2) = 5050 − 3 = 5047 2 100 1 (20)(21)(41) = 2, 870 6 6 6 k3 − 2 19. 4 k =1 1=4 k =1 7 100 k+ k =1 6 k+ k =1 18. 1= k =1 1 122 (6) (7) − 2 (6)(7) + 6 = 1728 4 2 7 (100)(101) + 100 = 35, 450 2 Exercise Set 7.4 20 218 3 k2 − 20. k =1 k 2 = 2, 870 − 14 = 2, 856 k =1 30 30 k (k 2 − 4) = 21. 30 (k 3 − 4k ) = k =1 k =1 6 6 k− 22. k =1 k3 = k =1 k =1 n (4k − 3) = 4 k =1 n−1 k =1 n 25. k =1 n−1 26. k =1 n−1 27. k =1 n 28. k =1 k =1 n 3 3k = n n 1 3 = 4 · n(n + 1) − 3n = 2n2 − n 2 1 k2 = n n 3 31 · n(n + 1) = (n + 1) n2 2 k= k =1 n−1 k2 = k =1 1 k3 =2 n2 n 11 1 · (n − 1)2 n2 = (n − 1)2 n2 4 4 k3 = k =1 2k 5 − n n S − rS = 1 11 · (n − 1)(n)(2n − 1) = (n − 1)(2n − 1) n6 6 n−1 = n 5 n 1− k =1 n 30. k =1 1 1 (30)2 (31)2 − 4 · (30)(31) = 214, 365 4 2 1 1 (n − 1)[(n 1) + 1][2(n − 1) + 1] = n(n − 1)(2n − 1) 6 6 k2 = 24. n k− k =1 k= 1 1 (6)(7) − (6)2 (7)2 = −420 2 4 n 23. 30 k3 − 4 2 n n k= k =1 21 5 (n) − · n(n + 1) = 4 − n n n2 n ark − k =0 ark+1 k =0 = (a + ar + ar2 + · · · + arn ) − (ar + ar2 + ar3 + · · · + arn+1 ) = a − arn+1 = a(1 − rn+1 ) so (1 − r)S = a(1 − rn+1 ), hence S = a(1 − rn+1 )/(1 − r) 19 31. 19 3k+1 = (a) k =0 3(3k ) = k =0 25 25 2k+5 = (b) k =0 k =0 100 (−1) (c) k =0 25 2k = −1 2 k 32. (a) 33. 1 + 2 + 3 + ··· + n = n2 = 3 3(1 − 320 ) = (320 − 1) 1−3 2 25 (1 − 226 ) = 231 − 25 1−2 (−1)(1 − (−1/2)101 ) 2 = − (1 + 1/2101 ) 1 − (−1/2) 3 1.999023438, 1.999999046, 2.000000000; 2 n k =1 1 k =2 n2 n n k= k =1 (b) 2.831059456, 2.990486364, 2.999998301; 3 n+1 1 11 n+1 ; lim = · n(n + 1) = n2 2 2n n→+∞ 2n 2 219 34. Chapter 7 12 + 22 + 32 + · · · + n2 = n3 lim n→+∞ n 35. k =1 n−1 36. k =1 n k =1 1 k2 =3 n3 n n k2 = k =1 11 (n + 1)(2n + 1) · n(n + 1)(2n + 1) = ; n3 6 6n2 (n + 1)(2n + 1) 1 1 = lim (1 + 1/n)(2 + 1/n) = n→+∞ 6 6n2 3 5 5k =2 n2 n n k= k =1 2k 2 2 =3 n3 n 5 5(n + 1) 5(n + 1) 51 ; lim = · n(n + 1) = n→+∞ n2 2 2n 2n 2 n−1 k2 = k =1 (n − 1)(2n − 1) 21 · (n − 1)(n)(2n − 1) = ; n3 6 3n2 (n − 1)(2n − 1) 1 2 = lim (1 − 1/n)(2 − 1/n) = lim n→+∞ n→+∞ 3 3n2 3 5 37. 6 2j (a) j =0 7 2j −1 (b) j =1 j =2 5 38. 13 (k + 4)2k+8 (a) k =9 18...
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