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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 &gt; 0 (b) improper if 1 &lt; p &lt; 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 &gt; 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 &gt; 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 −...
View Full Document

## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online