5 5lny 1 2x603210678ylny 1lim 1 magnitudeeof error is

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Unformatted text preview: 0 √ 3 0 √ 1 1 u √ du = √ sin−1 2 2 23 4−u √ 1 1 du = tan−1 u 1 + u2 3 4 (b) u = 3x, 1 3 3 = 0 0 3 1 =√ 23 4 1 (16u−1/2 − 8u1/2 + u3/2 )du = 4 1 3 f (u)du = 5/3. 1 9 f (u)du = 5/3. 0 0 (c) u = x2 , 1/2 4 4 f (u)du = −1/2 0 π 3 π = √. 63 1π π =. 33 9 1 1 sin5 x − sin7 x 5 7 2 sin x(1 − sin x) cos x dx = 53. (a) u = 3x + 1, 4 16 − 8u + u2 1 du = 1/2 27 u 1 ln 3 (ln(3e) − ln e) = . 2 2 = 0 √ π /6 51. (b) √ √ 12 = 2( 7 − 3). π 1 3x2 , √ 23 1 49. u = 3x, 3 √ = 1. 1 ln(2x + e) 2 47. u = 1 . 21 = 2. 1 1 1 (4 − u), dy = − du, − 3 3 27 4 1 16 2 32u1/2 − u3/2 + u5/2 = 106/405. 27 3 5 1 √ = π/18. 0 1 43. u = 4 − 3y , y = 45. 1/2 9 28 28 1 35. u = x + 4x + 7, 2 π /4 0 1 1 du = sin−1 u 2 3 1−u √ 2√ ( 10 − 2 2). 3 = 2 37. 2 sin2 x √ 0 √ 1 √ du = u u 1 23 (x + 9)1/2 3 1/2 x 11 1 dx = − 2 + 1)2 2+1 (5x 2 10 5x 0 1 2 31. u = 2x − 1, 33. 1 1 dx = 2 3 1 − 9x f (u)du = −1/2. π /6 = 0 1 1 23 − = . 160 896 4480 Exercise Set 5.9 151 π /2 55. sin x = cos(π/2−x), cosn (π/2−x)dx = − 0 0 π /2 0 π /2 sinn x dx = π/2 cosn u du (with u = π/2−x) = cosn u du = 0 π /2 cosn x dx, by replacing u by x. 0 4 5(e−0.2t − e−t ) dt = 5 57. Method 1: 0 4 Method 2: 0 4(e−0.2t − e−3t ) dt = 4 4 59. Method 1: 0 4 Method 2: availability. 1 −0.2t e + 5e−t −0.2 0 1 −0.2t 4 −3t e +e −0.2 3 5.78(e−0.4t − e−1.3t ) dt = 5.78 4.15(e−0.4t − e−3t ) dt = 4.15 4 0 ≈ 8.85835, 4 0 ≈ 9.6801, so Method 2 provides the greater availability. 4 1 −0.4t 1 −1.3t e e + 5.78 −0.4 1.3 1 −0.4t 4.15 −3t e + e −0.4 3 0 ≈ 7.11097, 4 ≈ 6.897, so Method 1 provides the greater 0 e1.528t dt = 524.959e1.528t + C ; y (0) = 750 = 524.959 + C , C = 225.041, y (t) = 524.959e1.528t + 61. y (t) = (802.137) 225.041, y (12) ≈ 48,233,500,000. 63. (a) (b) 1 [0.74 + 0.65 + 0.56 + 0.45 + 0.35 + 0.25 + 0.16] = 0.4514285714. 7 7 1 7 [0.5 + 0.5 sin(0.213x + 2.481) dx = 0.4614. 0 k e2x dx = 3, 65. 0 1 2x e 2 k 1 1 = 3, (e2k − 1) = 3, e2k = 7, k = ln 7. 2 2 0 1 67. (a) sin πxdx = 2/π . 0 −a a 69. (a) Let u = −x, then −a a f (x)dx = − a f (x)dx = − the latter integral, a f (x)dx, 2 −a 0 about the origin, so a a f (−u)du = −a a f (x)dx is the negative of a 0 −a a f (−u)du = −a a f (u)du = 0 0 f (x)dx = 0. The graph of f is symmetric f (x)dx thus 0 f (x)dx = −a 0 f (x)dx, let u = −x in a f (x)dx, so f (x)dx = −a 0 a a f (x) dx + −a a f (x)dx + f (u)du, so, replacing u by x in −a −a a 0 0 f (x)dx = a f (x)dx = 0, −a a −a (b) −a a f (−u)du = − f (x)dx = 0. 0 0 f (x)dx to get −a −a a f (x)dx + 0 f (x)dx = − a f (x)dx = 2 0 0 a f (−u)du = f (x)dx. The graph of 0 f (x) is symmetric about the y -axis so there is as much signed area to the left of the y -axis as there is to the right. a 0 f (a − u) du = a f (a − u) + f (u) so 2I = a, I = a/2. 71. (a) I = − (b) 3/2 (c) π/4 0 f (a − u) + f (u) − f (u) du = f (a − u) + f (u) a 0 a du − 0 f (u) du, I = a − I , f (a − u) + f (u) 152 (b)(b) = x −xπ−2, /2, = dx,dx, sin(uπ+2)/2)sin sin u, cos(uπ+2)/2)− sin sin u u u = / π du du = sin(u + / π = = u, cos(u + / π = = − u ￿ π￿ π ￿ π￿2 π/2 ￿ π￿2 π/2 π π/ 2 / π/ 2 / 8855 8 13 du sin8 x cos5 x dx = = sin8 u(8 usin5 u)5du = − − sin13 u13 u = 0= 0 Exercise 63(a). − (− 5 sin x cos x dx sin sin u) du = sin du by by Exercise 63(a). 0 0 −π /−π /2 −π /2 2 0 −π /−π /2 −π /2 2 EXERCISE SET 5.10 Exercise SetSET 5.10 EXERCISE 5.10 y (b)(b) y y 3 3 3 y (c)(c) y y 3 3 3 2 2 2 2 2 2 2 1 1 1 1 1 1 1 y 1. 1. (a) y y (a) 3 3 3 2 2 1 1 t t 1 1 1. (a) 2. 2. 3. (a) 3 3 2 2 y y y ac ln t 3 2 (c) ln t 1 1 1 1 12 2 23 3 t t t 3 (b) = ln(ac) = ln a + ln c = 7. a/c 1 = ln(a/c) = 2 − 5 = −3. t t t 0.5 0.5 1 1 0.5 1 (b) ln t (d) ln t (c) 1/c 1 a3 1 1 1 1 2 e2 e t e2 = ln(1/c) = −5. = ln a3 = 3 ln a = 6. tt 5. ln 5 midpoint rule approximation: 1.603210678; ln 5 ≈ 1.609437912; magnitude of error is < 0.0063. t 2 2 3 3 1 2 31 3 12 3 2 32 7. (a) x−1 , x > 0. (b) x2 , x = 0. (c) −x2 , −∞ < x < +∞. (d) −x, −∞ < x < +∞. ￿ac ￿ac ￿1/c 1/c ￿ ac 1/c 3. 3. (a) t ln t = ln(ac) ac)ln aln aln cln c7= 7 (a) ln (b)(b) t ln t = ln(1/c) /c)−5 −5 ln = ln( = = + + = = ln(1 = = √1 ex 3 3 1> 1 11 (e) x , x1 0. (f ) ln x + x, x > 0. (g) x − x, −∞ < x < +∞. (h) , x > 0. x ￿a/c a/c ￿a3 ￿a3 3 ￿ a/c a 3 (c)(c) t ln t = ln(a/ca/c)2= √ − 5−√−3 ln 3 (d)(d) t ln t = ln aln=33= 3 ln a6= 6 ln = ln( ) = −252 = = 2 ln 2 = 3 a ln a = π 1eπ ln 3 . 11 1 9. (a) 3 = 1 1 (b) 2 = e . √ ￿√a￿√a ￿2a ￿2a a 2a 1 1/ 4. 4. (a) t ln t = ln aln 2 1/2 = 1aln a9=29/2 (a) ln (b)(b) t ln t = ln 2ln 29+ 9 y 1/2 ln = 1/2 = 2 ln x = / a =+ 2x 1/2 11 11 1 1 12 1 1 = lim 1+ =￿a lim 1+ = e1/2 . 11. (a) y = ￿2x,￿ lim 2 /a 1+ 2/ax→+∞ 2/a a ￿a x→+∞ y →+∞ 2x 2x y (c)(c) t ln t = ln 2ln 29...
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This document was uploaded on 02/02/2014.

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