5 30 ec t 70 30ekt 52 t 1 70 30ek k ln 30 3 t

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Unformatted text preview: 9270 cm3 ≈ 9.3 L 600 600 [0 + 4(7) + 2(16) + 4(24) + 2(25) + 4(16) + 0] = 9000 ft2 , (3)(6) 0 V = 75A ≈ 75(9000) = 675, 000 ft3 40. A = h dx ≈ b f (x) dx ≈ A1 + A2 + · · + An = 41. a = b−a 1 1 1 (y0 + y1 ) + (y1 + y2 ) + · · · + (yn−1 + yn ) n 2 2 2 b−a [y0 + 2y1 + 2y2 + · · · + 2yn−1 + yn ] 2n 42. right endpoint, trapezoidal, midpoint, left endpoint 43. (a) The maximum value of |f (x)| is approximately 3.844880. (b) n = 18 (c) 0.904741 44. (a) The maximum value of |f (x)| is approximately 1.467890. (b) n = 12 (c) 1.112062 Exercise Set 9.8 330 45. (a) The maximum value of |f (4) (x)| is approximately 42.551816. (b) n = 8 (c) 0.904524 46. (a) The maximum value of |f (4) (x)| is approximately 7.022710. (b) n = 8 (c) 1.111443 EXERCISE SET 9.8 1. (a) (b) (c) (d) (e) (f ) improper; infinite discontinuity at x = 3 continuous integrand, not improper improper; infinite discontinuity at x = 0 improper; infinite interval of integration improper; infinite interval of integration and infinite discontinuity at x = 1 continuous integrand, not improper 2. (a) improper if p > 0 (b) improper if 1 < p < 2 (c) integrand is continuous for all p, not improper 3. 4. 5. 6. 7. 8. 9. lim (−e−x ) = lim (−e− + 1) = 1 →+∞ lim →+∞ 1 ln(1 + x2 ) 2 lim ln →+∞ lim − →+∞ 0 √ lim 2 ln x lim − ln →+∞ = lim 1 2 ln2 x →+∞ −1 →+∞ 4 2 1 lim − e−x 2 →+∞ = lim = lim x−1 x+1 →+∞ →−∞ →+∞ 0 →+∞ = − ln 3 5 = ln 5 3 − 1 1 1 2 +2 =2 2 ln →+∞ 1 4(2x − 1)2 0 = lim →−∞ 2 10. 1 x tan−1 →−∞ 2 2 11. 1 3x e →−∞ 3 12. 1 lim − ln(3 − 2ex ) →−∞ 2 lim 3 −1 − ln +1 5 ) − ln 2] = +∞, divergent √ √ = lim (2 ln − 2 ln 2) = +∞, divergent 2 lim 2 2 1 −e− + 1 = 1/2 2 = lim e 1 [ln(1 + 2 = lim →−∞ 0 1 [−1 + 1/(2 − 1)2 ] = −1/4 4 1 1π − tan−1 = [π/4 − (−π/2)] = 3π/8 24 2 2 1 13 −e 33 = lim →−∞ 0 = 1 3 1 1 ln(3 − 2e ) = ln 3 →−∞ 2 2 = lim 331 Chapter 9 +∞ 13. x3 dx and −∞ −∞ +∞ diverge. +∞ 14. √ 0 ∞ so −∞ x 14 x →+∞ 4 x2 + 2 √ +∞ 15. (x2 0 +2 −∞ 0 0 −∞ −∞ 18. 19. 20. 21. 23. 24. 1 x−3 3 lim x2/3 +2 →0 4 = lim −1 + + →3 8 = lim →0+ lim − ln(cos x) 2) = +∞ 1 [−1/( →+∞ 2 0 ∞ −∞ (x2 + 3) + 1/3] = →9 lim sin−1 x →+∞ − tan−1 (e− ) + 0 = lim →−∞ − π π =, 4 4 π π + tan−1 (e− ) = , 4 4 )=6 lim − ln(cos ) = +∞, divergent →9− 0 = lim sin−1 = π/2 0 →1− √ = lim + (− 8 + →−3 √ lim − 1 − 2 sin x →π/6− = 0 lim − ln(1 − tan x) →π/4− 1 = +∞, divergent −3 √ = lim 2(− 9 − + 3) = 6 lim + − 9 − x2 →−3 2/3 = 0 dx = lim ln |x − 2| x−2 →2− 9− 2) √ =− 8 √ lim (− 1 − 2 sin + 1) = 1 →π/6− lim − ln(1 − tan ) = +∞, divergent →π/4− 0 = lim (ln | − 2| − ln 2) = −∞, divergent − →2 1 , 6 x dx = 1/6 + (−1/6) = 0 + 3)2 = lim 0 2 →π/2− 0 √ lim −2 9 − x − →1− 3 (4 − 2 = →π/2− 0 √ π ππ e−t dt = + = 1 + e−2t 4 4 2 →3 2 +2− −∞ = lim e−t dt = lim − tan−1 (e−t ) →+∞ 1 + e−2t lim − + 25. 2 x3 dx is divergent. dx is divergent. 1 22. +∞ = +∞ so →+∞ 0 e−t dt = lim − tan−1 (e−t ) →−∞ 1 + e−2t +∞ 4 = lim ( x dx = −1/6 so 2 + 3)2 (x similarly +∞ 0 1 x dx = lim − 2 2 + 3) →+∞ + 3) 2(x 0 16. →+∞ x x2 1 →+∞ 4 = lim x2 + 2 dx = lim x3 dx both converge; it diverges if either (or both) 0 x3 dx = lim 0 17. +∞ 0 x3 dx converges if Exercise Set 9.8 2 26. 0 8 27. 0 0 332 dx = lim −1/x x2 →0+ x−1/3 dx = lim →0− 8 so −1 2 0 3 2/3 x 2 8 3 = lim (4 − →0+ 2 = lim →0− −1 dx = lim 3(x − 2)1/3 →2− (x − 2)2/3 similarly 2 +∞ 29. Define 0 0 →0+ 3 ( 2 2/3 dx is divergent x2 −2 ) = 6, − 1) = −3/2 2/3 x−1/3 dx = 6 + (−3/2) = 9/2 4 1 2 = lim (−1/2 + 1/ ) = +∞ so 3 x−1/3 dx = lim x2/3 →0+ 2 −1 28. 2 +∞ 1 →2 4 dx = lim 3(x − 2)1/3 →2+ (x − 2)2/3 1 dx = x2 a +∞ 1 dx + x2 0 a √ 3 = 3 2 so 4 0 +∞ = lim (1/ − 1) = +∞ so →0+ a dx √ = x x2 − 1 1 √ dx 3 =6 2 2/3 (x − 2) 1 dx where a > 0; take a = 1 for convenience, x2 1 1 dx = lim (−1/x) x2 →0+ 30. Define 0 √ 3 = lim 3[( − 2)1/3 − (−2)1/3 ] = 3 2, − 0 +∞ dx √ + x x2 − 1 a 1 dx is divergent. x2 dx √ where a > 1, x x2 − 1 take a = 2 for convenience to get 2 1 dx √ = lim sec−1 x →1+ x x2 − 1 +∞ 2 +∞ 31. +∞ √ x e− √ √ 0 +∞ 33. √ 0 +∞ 34. √ 0 x →1 + 0 1 0 e−x dx = − 1 − e−2x 0 −e−u = 2 lim 0 √ du √ = lim 2 u u →0+ √ 1 du = 1 − u2 39. (a) 2.726585 √ →0 √ 0 1 xe−3x dx = lim − (3x + 1)e−3x →+∞ 9 0 →+∞ 0 = lim 2(1 − + 1 1 − e− = lim tan−1 1 =2 2 = π 2 )=2 du = lim sin−1 u →1 1 − u2 = lim sin−1 = 0 →1 π 2 = 1/3 0 1 −x e (sin x − cos x) →+∞ 2 0 (b) 2.804364 (c) 0.219384 e−x cos x dx = lim lim →+∞ du 1 u = 2 lim tan−1 →+∞ 2 u2 + 4 2 0 dx √ = π/2. x x2 − 1 1 2 →+∞ +∞ +∞ = π/2 −...
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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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