# 2 3 3 2 sin2 sin2 1 1 2 2 cos xdx sin

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Unformatted text preview: , G(0) = 0, so F (0) − G(0) = 8/3 (c) F (x) = G (x) = 10x/(x2 + 5)2 F (0) = 0, G(0) = −1, so F (0) − G(0) = 1 (c) 49. (a) (b) 48. F (x) = (9x2 + 24x + 16)/6 = 3x2 /2 + 4x + 8/3 = G(x) + 8/3 F (x) = x2 (x2 + 5) − 5 5 = =1− 2 = G(x) + 1 2+5 2+5 x x x +5 (sec2 x − 1)dx = tan x − x + C (csc2 x − 1)dx = − cot x − x + C 50. 51. (a) 1 2 52. (a) F (x) = G (x) = f (x), where f (x) = (b) G(x) − F (x) = (c) no, because (−∞, 0) ∪ (0, +∞) is not an interval 53. (1 − cos x)dx = 1087 v= √ 2 273 2, 3, 1 (x − sin x) + C 2 (b) 1 2 (1 + cos x) dx = 1 (x + sin x) + C 2 1, x &gt; 0 −1, x &lt; 0 x&gt;0 so G(x) = F (x) plus a constant x&lt;0 1087 1/2 1087 1/2 ft/s T −1/2 dT = √ T + C , v (273) = 1087 = 1087 + C so C = 0, v = √ T 273 273 213 Chapter 7 EXERCISE SET 7.3 1. u23 du = u24 /24 + C = (x2 + 1)24 /24 + C (a) (b) − u3 du = −u4 /4 + C = −(cos4 x)/4 + C (c) 2 √ sin u du = −2 cos u + C = −2 cos x + C 3 8 1 3 3 1/2 3 u +C = 4x2 + 5 + C 4 4 1 1 u−1 du = ln u + C = ln(x3 − 4) + C 3 3 (d) (e) 2. (a) (b) (c) 1 4 1 4 1 π (d) (e) 3. u−1/2 du = 1 1 tan u + C = tan(4x + 1) + C 4 4 1 3/2 1 u1/2 du = u + C = (1 + 2y 2 )3/2 + C 6 6 2 3/2 2 u1/2 du = u +C = sin3/2 πθ + C 3π 3π 5 5 u4/5 du = u9/5 + C = (x2 + 7x + 3)9/5 + C 9 9 du = ln u + C = ln(1 + ex ) + C u sec2 u du = 1 1 u du = − u2 + C = − cot2 x + C 2 2 1 10 1 (b) u9 du = u +C = (1 + sin t)10 + C 10 10 1 (c) du = ln |u| + C = ln | ln x| + C u 1 1 1 (d) − eu du = − eu + C = − e−5x + C 5 5 5 1 1 1 1 (e) − du = − ln |u| + C = − ln |(1 + cos 3θ)| + C 3 u 3 3 (a) − 2 7/2 4 5/2 2 3/2 u − u + u +C 7 5 3 4 2 2 = (1 + x)7/2 − (1 + x)5/2 + (1 + x)3/2 + C 7 5 3 (a) (u − 1)2 u1/2 du = (b) csc2 u du = − cot u + C = − cot(sin x) + C (c) 4. eu du = eu + C = etan x + C (d) (e) 5. 1 2 u1/2 du = (u5/2 − 2u3/2 + u1/2 )du = 1 3/2 1 u + C = (1 + e2t )3/2 + C 3 3 1 du = ln |u| + C = ln x5 + 1 + C u u = 2x, du = 2dx; 1 2 eu du = 1u 1 e + C = e2x + C 2 2 Exercise Set 7.3 214 1 2 1 1 1 du = ln |u| + C = ln |2x| + C u 2 2 6. u = 2x, du = 2dx; 7. u = 2 − x2 , du = −2x dx; − 8. u = 3x − 1, du = 3dx; 1 3 1 2 u3 du = −u4 /8 + C = −(2 − x2 )4 /8 + C u5 du = 16 1 u +C = (3x − 1)6 + C 18 18 9. u = 8x, du = 8dx; 1 8 cos u du = 10. u = 3x, du = 3dx; 1 3 1 1 sin u du = − cos u + C = − cos 3x + C 3 3 11. u = 4x, du = 4dx; 1 4 sec u tan u du = 12. u = 5x, du = 5dx; 1 5 sec2 u du = 13. u = 7t2 + 12, du = 14t dt; 1 14 u = x3 + 1, du = 3x2 dx; 1 3 16. u = 1 − 3x, du = −3dx; − 18. u = 3x2 , du = 6x dx; 1 6 1 10 1 3 2 1/2 2 u +C = 3 3 cos u du = 4 − 5x2 + C x3 + 1 + C 1 −1 1 u + C = (1 − 3x)−1 + C 3 3 u−3 du = − 1 −2 1 u + C = − (4x2 + 1)−2 + C 16 16 1 1 sin u + C = sin(3x2 ) + C 6 6 eu du = eu + C = esin x + C u = sin x, du = cos x dx; 20. u = x4 , du = 4x3 dx; 21. u = −2x3 , du = −6x2 , − 22. u = ex − e−x , du = (ex + e−x )dx, 23. u = 5/x, du = −(5/x2 )dx; − 24. u= 25. u = x3 , du = 3x2 dx; 1 4 eu du = 1 6 1 x, du = √ dx; 2 2x 1 3 1 3/2 1 u +C = (7t2 + 12)3/2 + C 21 21 1 1 u−1/2 du = − u1/2 + C = − 5 5 u−2 du = 19. √ 1 1 tan u + C = tan 5x + C 5 5 u−1/2 du = 1 8 17. u = 4x2 + 1, du = 8x dx; 1 1 sec u + C = sec 4x + C 4 4 u1/2 du = 14. u = 4 − 5x2 , du = −10x dx; − 15. 1 1 sin u + C = sin 8x + C 8 8 1u 14 e + C = ex + C 4 4 1 1 3 eu du = − eu + C = − e−2x + C 6 6 1 5 1 du = ln |u| + C = ln ex − e−x + C u sin u du = 1 1 cos u + C = cos(5/x) + C 5 5 √ sec2 u du = 2 tan u + C = 2 tan x + C sec2 u du = 1 1 tan u + C = tan(x3 ) + C 3 3 215 26. Chapter 7 u = cos 2t, du = −2 sin 2t dt; − 1 2 27. e−x dx; u = −x, du = −dx; − 28. ex/2 dx; u = x/2, du = dx/2; 2 1 3 1 1 u3 du = − u4 + C = − cos4 2t + C 8 8 eu du = −eu + C = −e−x + C √ eu du = 2eu + C = 2ex/2 + C = 2 ex + C u5 du = 16 1 u +C = sin6 3t + C 18 18 29. u = sin 3t, du = 3 cos 3t dt; 30. u = 5 + cos 2θ, du = −2 sin 2θ dθ; − 1 2 u−3 du = 31. u = 2 − sin 4θ, du = −4 cos 4θ dθ; − 1 4 1 1 u1/2 du = − u3/2 + C = − (2 − sin 4θ)3/2 + C 6 6 1 5 32. u = tan 5x, du = 5 sec2 5x dx; 33. u = sec 2x, du = 2 sec 2x tan 2x dx; 34. u = sin θ, du = cos θ dθ; 35. u= 36. u= 1 2 14 1 u +C = tan4 5x + C 20 20 13 1 u + C = sec3 2x + C 6 6 u2 du = sin u du = − cos u + C = − cos(sin θ) + C √ 1 y , du = √ dy , 2 2y eu du = 2eu + C = 2e √ 1 y , du = √ dy , 2 2y 1 du = 2 eu √ 38. u = a + bx, du = b dx, dx = 1 b u3 du = u1/n du = 1 −2 1 u + C = (5 + cos 2θ)−2 + C 4 4 y +C √ e−u du = −2e−u + C = −2e− y +C 1 du b n n u(n+1)/n + C = (a + bx)(n+1)/n + C b(n + 1) b(n + 1) 39. u = sin(a + bx), du = b cos(a + bx)dx 1 1 1 un du = un+1 + C = sinn+1 (a + bx) + C b b(n + 1) b(n + 1) 41. u = x − 3, x = u + 3, dx = du (u + 3)u1/2 du = (u3/2 + 3u1/2 )du = 2 5/2 2 u + 2u3/2 + C = (x − 3)5/2 + 2(x − 3)3/2 + C 5 5 42. u = y + 1, y = u − 1, dy = du 2 2 u−1 du = (u1/2 − u−1/2 )du = u3/2 − 2u1/2 + C = (y + 1)3/2 − 2(y + 1)1/2 + C 1/2 3 3 u 43. u = 3θ, du = 3 dθ 1 1 tan2 u du = 3 3 sin2 2θ sin 2θ dθ = 44. − 1 2 (se...
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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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