# H v r2 h2 r y 2 dy r2 h3 0 h y 37 the two curves

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Unformatted text preview: 0 6 16. √ =8−4 2 0 π /2 2 sin u 3 cos u du = 4 = −506/15 0 10. 2 5/2 8 3/2 u −u 5 3 0 8 2 (4 − x)5/2 − (4 − x)3/2 5 3 u = x/2, 8 = 1192/15, 1 = 1192/15 (u − 4)u1/2 du = π /4 9. 9 0 9 or 2 5/2 2 3/2 u −u 5 3 8 2 2 (1 + x)5/2 − (1 + x)3/2 5 3 or (u3/2 − u1/2 )du = 1 1 1 += 12 8 24 1 = 0 1 dx = − e−2x 8 1 4 4 = 0 1 − e−8 8 243 22. 23. 25. Chapter 7 2 (3x + 1)1/2 3 23 (x + 9)1/2 3 1 2 28. 1 32. 2 sec πxdx = tan πx π −1/4 = 2/3 34. 35. √ 2√ ( 10 − 2 2) 3 = −1 28 1 2 28 u= √ 12 = 1/2 2 (tan x)3/2 3 30. sin u du = −2 cos u π /3 sec2 u du = π/4 √ π /4 = 2/3 31. 0 5 sin(x2 ) 2 π =0 0 = −4 −1 1 3 u2 du = 0 1 tan u 3 13 u 9 π /3 √ = ( 3 − 1)/3 π/4 −1 = −1/9 0 1 1 (4 − u), dy = − du 3 3 4 1 16 − 8u + u2 du = (16u−1/2 − 8u1/2 + u3/2 )du 27 1 u1/2 u = 4 − 3y , y = 1 27 1 4 u = 5 + x, e ln(x + e) 2 −2 (a) 0 u−5 √ du = u 4 16 2 1 32u1/2 − u3/2 + u5/2 27 3 5 9 (u1/2 − 5u−1/2 )du = 4 = 106/405 1 2 3/2 u − 10u1/2 3 √ = ln(2e) − ln e = ln 2 4 − u2 du = u = 3x + 1, 38. f (u)du = 5/3 0 (b) 1 4 f (u)du = −1/2 4 1 2 = 8/3 4 = (e−1 − e−2 )/2 1 4 1 3 u = x2 , 1/2 u = 1 − x, 1 2 − e−x 2 9 1 [π (2)2 ] = 2π 2 0 (c) 42. √ √ √ √ 28 − 12 = 2( 7 − 3) π 4 41. = 1/10 −1 2π 2π 9 40. 0 13 (t + 1)20 10 1 −3π/4 u = sin 3θ, = 38/15 1 2 =0 1 3 2 12 π /4 x, 2 u = 3θ, = 4 π 2 (5x − 1)3/2 15 26. u−1/2 du = u1/2 = 37. −1/4 0 1 1 dx = − (x − 3)2 x−3 1 sin2 x 2 − 36. = 24. π 33. 1/4 2 1 27. u = x2 + 4x + 7, 29. 1/4 1 fave = 1/4 − (−1/4) u = 3x, 1 3 9 f (u)du = 5/3 0 f (u)du = −1/2 0 0 xm (1 − x)n dx = − 1 1 (1 − u)m un du = 0 1 un (1 − u)m du = 0 xn (1 − x)m dx Exercise Set 7.8 43. 244 sin x = cos(π/2 − x), π /2 π /2 sinn x dx = 0 0 π /2 π /2 cosn u du = cosn x dx 0 0 u = 1 − x, − (by replacing u by x) 0 1 (1 − u)un du = 1 45. (u = π/2 − x) cosn u du π/2 = 44. 0 cosn (π/2 − x)dx = − 1 (1 − u)un du = 0 (un − un+1 )du = 0 1 1 1 − = n+1 n+2 (n + 1)(n + 2) e1.528t dt = 524.959e1.528t + C ; y (0) = 750 = 524.959 + C , C = 225.041, y (t) = (802.137) y (t) = 524.959e1.528t + 225.041, y (12) = 48, 233, 525, 650 46. 47. 275000 Vave = 10 − 0 s(t) = 10 10 −0.17t e −0.17t dt = −161764.7059e 0 = \$132, 212.96 0 (25 + 10e−0.05t )dt = 25t − 200e−0.05t + C (a) √ s(10) − s(0) = 250 − 200(e−0.5 − 1) = 450 − 200/ e ≈ 328.69 ft (b) yes; without it the distance would have been 250 ft k 1 2x e 2 e2x dx = 3, 48. 0 49. 2 Vrms = (a) k 0 1 1 = 3, (e2k − 1) = 3, e2k = 7, k = ln 7 2 2 1/f 1 1/f − 0 Vp2 sin2 (2πf t)dt = 0 1 fV 2 2p 1/f [1 − cos(4πf t)]dt 0 √ 1 1 1 f V 2 [t − sin(4πf t)] = Vp2 , so Vrms = Vp / 2 2p 4πf 2 0 √ √ Vp / 2 = 120, Vp = 120 2 ≈ 169.7 V 1/f = (b) 50. Let u = t − x, then du = −dx and t 0 f (t − x)g (x)dx = − 0 t f (u)g (t − u)du = t f (u)g (t − u)du; 0 the result follows by replacing u by x in the last integral. 51. 0 I=− (a) a a = f (a − u) du = f (a − u) + f (u) du − 0 (b) 3/2 52. a 0 a 0 f (a − u) + f (u) − f (u) du f (a − u) + f (u) f (u) du, I = a − I so 2I = a, I = a/2 f (a − u) + f (u) (c) π/4 1 1 1 1 1 1 , dx = − 2 du, I = du = −I so I = 0 which is impossible (−1/u2 )du = − 2 2+1 u u −1 1 + 1/u −1 u 1 because is positive on [−1, 1]. The substitution u = 1/x is not valid because u is not continuous 1 + x2 for all x in [−1, 1]. x= 1 53. sin πxdx = 2/π 0 245 55. Chapter 7 (a) Let u = −x then a −a −a f (x)dx = − a f (−u)du = −a a a f (−u)du = − −a f (u)du so, replacing u by x in the latter integral, a −a a f (x)dx = − a f (x)dx, 2 −a −a a f (x)dx = 0, −a f (x)dx = 0 0 The graph of f is symmetric about the origin so a −a 0 f (x)dx = −a a (b) −a 0 −a a f (x) + −a f (x)dx = 0 a f (x)dx + f (x)dx, let u = −x in 0 0 f (x)dx = − a f (−u)du = a a so a 0 −a f (x)dx to get a f (u)du = 0 a f (x)dx + 0 a f (−u)du = 0 f (x)dx = −a f (x)dx thus 0 0 0 f (x)dx = −a a f (x)dx is the negative of f (x)dx 0 a f (x)dx = 2 0 f (x)dx 0 The graph of 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. EXERCISE SET 7.9 1. (a) (b) y (c) y y 3 3 3 2 2 2 1 1 1 t 1 2. 2 3 0.5 t 1 1 y 3 2 1 2 3 3 2 1 t ac 3. (a) ln t (c) ln t 1 1/c = ln(ac) = ln a + ln c = 7 a/c 1 = ln(a/c) = 2 − 5 = −3 (b) ln t (d) ln t 1 a3 1 = ln(1/c) = −5 = ln a3 = 3 ln a = 6 e2 t Exercise Set 7.9 246 √ 4. (a) ln t (c) ln t a 1 2a = ln a1/2 = 2 (b) = ln 2 − 4 (d) ln t 2/a 1 ln t = ln 2 + 4 1 a 2 = 4 − ln 2 5. ln 5 ≈ 1.603210678; ln 5 = 1.609437912; magnitude of error is < 0.0063 6. ln 3 ≈ 1.098242635; ln 3 = 1.098612289; magnitude of error is < 0.0004 7. (a) x−1 , x > 0 (c) −x2 , −∞ < x < +∞ (e) x3 , x > 0 √ (g) x − 3 x, −∞ < x < +∞ 8. (a) f (ln 3) = e−2 ln 3 = eln(1/9) = 1/9 (b) f (ln 2) = eln 2 + 3e− ln 2 = 2 + 3eln(1/2) = 2 + 3/2 = 7/2 9. (a) 3π = eπ ln 3 (b) 2 10. (a) π −x = e−x ln π (b) x2x = e2x ln x 11. (a) 1+ lim x→+∞ 1 x (b) x2 , x = 0 (d) −x, −∞ &l...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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