Business Calc Homework w answers_Part_69

Business Calc Homework w answers_Part_69 - 341 Section 8.3...

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Unformatted text preview: 341 Section 8.3 dx 44. 1 x4 1 dx 1 0 x4 1 1 on [1, x2 1 u lim 2 x b→ dx b 1 x 1 b lim b→ dv ln x x2 Area 1 1 ln x dx x2 dx x2 v 1 dx x 1 lim 1 1 x ). 1 b→ b→ ln x du b 1 dx x2 1 b lim ln x dx by parts. x2 exists on [0, 1]. x4 1 1 Integrate 1 0 on [1, ). ln x dx x2 1 x4 1 1 exists because x4 0 0, y Area dx x4 47. For x 1 dx 0 1 by symmetry about the y-axis x4 0 1 dx 0 dx 2 x4 ln x x dx x2 ln x x b→ Note that lim b→ C b ln x x lim 1 x 1 x ln b b b→ ln b b lim b→ 1 lim 1/b 1 1 b 1 1 0. Since this integral converges, the given integral converges. 48. For x 45. Integrate (1 dy y )(1 tan 2 1 y) by letting u 1 y2 dy y 2)(1 tan 1 (1 du 1 y) 1 u ln 1 (1 C ln 1 dy y 2)(1 tan u tan 1 b 1 lim y) b→ 0 (1 y C ln x ln x dx lim dx x 1 b→ 1 x ln x dx Integrate dx by letting u ln x so du . x x ln x 12 1 dx u du u C (ln x)2 C x 2 2 b 1 1 Area lim (ln x)2 lim (ln b)2 2 b→ b→ 2 1 Area 49. (a) The integral in Example 1 gives the area of region R. dy y 2)(1 tan 1 y) Area 1 lim ln 1 b→ tan 1 tan 1 lim (ln 1 b→ b 1 x2 y 0) The surface area of the solid is given by the following integral. 2 The integral converges. 2 1 e y dy y2 1 46. 0 lim y→ 0 e y dy y2 1 0 y 1 lim 2 y→ y ey 1 1 x 1 12 dx x2 1 ey 2y y→ lim ey y→ 2 lim 1 x 1 x4 1 on [1, x3 1 x4 dx x4 1 dx x3 2 Since 0 x4 1 x 2 e y dy y2 1 e y dy diverges since y2 1 e y2 dx x (b) Refer to Exploration 2 of Section 7.3. b y 0 ln 1 0 on [1, ). b y so dy du 0 tan 0, y ), the direct comparison test shows that the integral for the surface area diverges. The surface area is . Thus the given integral diverges. (c) Volume 1 12 dx x 1 1 dx x2 b lim b→ lim b→ lim b→ 1 1 dx x2 1 x 1 b b 1 1 (d) Gabriel’s horn has finite volume so it could only hold a finite amount of paint, but it has infinite surface area so it would require an infinite amount of paint to cover itself. 342 Section 8.3 1 50. (a) f (x) x 2/2 e 6, x 2 51. (a) For x x2 6x, so e 6x e 2 e x2 dx 6x e 6 dx 6 b lim , 0]. f is decreasing on [0, ). 1 (b) NINT e 2 1 NINT x 2/2 1, 1 x 2/2 2, 2 1 36 e 6 0.954 3, 3 b→ 0.683 e 2 1 NINT 0, 0.997 , x, , x, x 2/2 e , x, (b) x2 6 dx 4 x2 e 1 1 6 1 e 36 6 17 10 x2 dx 10 dx 17 e 6 x2 e dx 4 1 2 Thus, from part (a) we have shown that the error is (c) Part (b) suggests that as b increases, the integral approaches 1. We can make 1 large enough. Also, we can x2 NINT(e (c) . e dx x2 , x, 1, 6) 0.1394027926 1 (This agrees with Figure 8.16.) b f (x) dx and 10 f (x) dx as close to 1 as b we want by choosing b 17 bounded by 4 b make e 6 1 e 6b 6 lim 1 2 f has a local maximum at (0, f (0)) b 1 6x e 6 b→ f is increasing on ( dx 6 lim [ 3, 3] by [0, 0.5] 6x e b→ f (x) dx as small as we want b by choosing b large enough. This is because 0 f (x) 0 e f (x) e x/2 x/2 x/2 dx b for x 3 dx x2 e e x2 c lim c→ e x/2 dx b lim 2e x/2 lim [ 2e c/2 e e 3x 3 c dx 3. b dx 3x 3 3x for x lim e b→ 3x dx 3 b lim 2e 1 e 3x 3 lim 1 3b e 3 19 e 3 b c→ 2e since x 2 dx dx 0 dx x2 3 3 1) Thus, x/2 e dx 0 1. (Likewise, for x c→ As b→ , 2e x2 0 b e e e f (x) dx b (d) 0.000041 b→ b/2 b/2 b→ →0, so for large enough b, b/2 f (x) dx b 3 19 e 3 0.000042 is as small as we want. Likewise, for large enough b, 52. (a) Since f is even, f ( x) b f (x) dx is as small as we want. f (x). Let u x, du dx. 0 f (x) dx f (x) dx f (x) dx 0 0 f ( u)( 1) du f (x) dx 0 f (u) du f (x) dx 0 0 2 f (x) dx 0 (b) Since f is odd, f ( x) f (x). Let u 0 x, du 0 f (x) dx f (x) dx f (x) dx 0 f ( u)( 1) du f (x) dx 0 f (u) du 0 f (x) dx 0 0 dx Section 8.3 b 2x dx x2 1 53. (a) 0 lim b→ 0 55. Suppose 0 2x dx x2 1 1) b→ b Thus the integral diverges. 0 2x dx 2x dx and must converge in order x2 1 x2 1 2x dx to converge. x2 1 (b) Both 0 for b (c) lim b→ b b 2x dx x2 1 lim ln (x 2 1) b→ lim [ln (b 1) b→ lim 0 2x Note that x2 ln (b If the infinite integral of g converges, then taking the limit in the above inequality as b→ shows that the infinite integral of f is bounded above by the infinite integral of g. Therefore, the infinite integral of f must be finite and it converges. If the infinite integral of f diverges, it must grow to infinity. So taking the limit in the above inequality as b→ shows that the infinite integral of g must also diverge to infinity. 1)] 0: b x dx e x e x lim [ e e b lim b→ 0 b b b→ 2x dx = 0. x2 1 (d) Because the determination of convergence is not made using the method in part (c). In order for the integral to converge, there must be finite areas in both directions (toward and toward ). In this case, there are infinite areas in both directions, but when one computes the integral over an interval [ b, b], there is cancellation which gives 0 as the result. For n 1: ux du dx xe x b dx 0 lim b→ xe y 2/3 xe y 1 x dy dx 3 (1 2 x dx x 1/3 (1 x 2/3)1/2 x2 e 2/3 x 2/3) (1 2/3 (x x b dx lim b→ 1) b→ dy 2 1/3 2/3 x dx 1/3 lim b→0 lim b→0 lim b→0 3 2/3 x 2 Thus, the perimeter is 4 1 x2 e x dx 0 x2 e x b b b b 2 eb 2 3 2/3 b 2 3 2 3 2 6. b→ x 2x e dx 0 0 lim 1 dx e x dx ex 2(1) b→ x 1 2b eb dx dx 0 lim 1/3 b 3 2 e b→ 2 lim b→ x b lim b2 eb lim 1 dx x x 0 dv v lim 1 0 x 0 dy 2 1 For n 2: u x2 du 2x dx 2 1/3 x 3 b e 1 eb lim x 2/3)1/2 b x b eb b→ x 2/3)3/2 (1 dx 0 b→ 2/3 x 0 b→ 0. 1 e x dx ex dv v lim 1, y 1] b→ 54. By symmetry, find the perimeter of one side, say for x b 0 = lim 0 dx 0 lim is an odd function so 1 2 0. b→ a 56. (a) For n b 2 a, g(x) dx. a 1) b→ a. b f (x) 0 lim ln (b 2 g(x) for all x From the properties of integrals, for any b b lim ln (x 2 f (x) 343 b→ xe 0 2 x dx 344 Section 8.4 56. continued c n x (b) Evaluate x e u xn du nx n 1 (b) dx using integration by parts dv e x dx v ex 0 f (x) dx c f (x) dx f (x) dx 0 0 f (x) dx f (x) dx c c f (x) dx 0 Thus, n x xe n dx x xe nx n1 x e dx c f (x) dx f (x) dx c f (n x ne 1) x 0 dx b→ x ne b x nx n 1e 0 lim b→ x n 1e (c) Since f (n f (n x n xe 0 0 nf (n), 1) … f (1) n!; thus dx converges for all integers n 0. 0 sin x , x, 0, x or create a table x 57. (a) On a grapher, plot NINT of values. For large values of x, f (x) appears to approach approximately 1.57. (b) Yes, the integral appears to converge. 1 58. (a) 1 dx 1 x 2 dx lim b→ x2 1 b tan lim (tan 1 b→ 4 1 1 1 lim b→ 1 lim [tan 1 b x b→ 1 b→ 2 1 x2 b) x2 1 3 4 1 dx 1 lim tan dx tan 3 4 2 b x x b b→ 2 1 1 lim dx 0 c f (x) dx c c f (x) dx 0 f (x) dx 0. 0 s Section 8.4 Partial Fractions and Integral Tables (pp. 444–453) Quick Review 8.4 n(n x f (x) dx, because f (x) dx dx 0. 1) 1) f (x) dx c bn eb f (x) dx 0 0 Note: apply L’Hôpital’s Rule n times to show that b→ f (x) dx c 0 dx 0 n x nf (n) bn eb 0 f (x) dx 0 lim lim c f (x) dx 0 4 b 4 4 tan 1 1] 1. Solving the first equation for B yields B 3A Substitute into the second equation. 2A 3( 3A 5) 7 2A 9A 15 7 11A 22 A 2 Substituting A 2 into B 3A 5 gives B solution is A 2, B 1. 5. 1. The 2. Solve by Gaussian elimination. Multiply first equation by 3 and add to second equation. Multiply first equation by 1 and add to third equation. A 2B C 0 7B 5C 1 B 2C 4 Multiply third equation by 7 and add to second equation. A 2B C 0 9C 27 B 2C 4 Solve the second equation for C to get C 3. Solve for B by substituting C 3 into the third equation. B 2(3) 4 B 2 B2 Solve for A by substituting B 2 and C 3 into the first equation. A 2(2) 3 0 A10 A 1 The solution is A 1, B 2, C 3. 2x 1 3. x 2 3x 4 2x 3 5x 2 10x 7 2x 3 6x 2 8x x2 2x 7 x2 3x 4 x3 2x 1 x x2 3 3x 4 Section 8.4 4. x 2 3x 2 5. x x2 3 11x 8x 3x x 6. y 7. 8. 5y 2 3 4 2 x x (x (y (y x 2 3t t2 2 3) 2 3)(x 2 (x 3) 1) 4)(y 1) 2)(y 2)(y A 1)(y x2 3. 2 (x 2)(x A A x 1) B(x 4. s 3 B)x (A x 7. 2 x B 0, B 2, B s2 C(t (B C 1) C)t 1, and 2, C C C 1. 1. 2 t 2 t 1 s(s 2 6s A s 4 s2 A( s 3)(s s 6s (A 1 Solving the system simultaneously yields A 2 1) B)t 2 A( s 2 Equating coefficients of like terms gives 3 C t2 Bt(t At 2 4 2B) x 1)2 (x B t 1 s3 2) 2 1 A t1 t 2(t 1) 1) A( x 7 4 x t 2, B Solving the system simultaneously yields s3 7 3x B) 2. 2 1 1 1 B B C 2 ,B 3 A 2 5x B 1 t2 s 6) s(s B s 2) C)s 2 2) C 3 s 2 B(s)(s B(s 2 6) 3)(s (A 2) 2s) 2B C(s)(s C(s 2 3) 3s) 3C)s 0, A 2B 3C 0, 6A Solving the system simultaneously yields (A x2 B 2 2x t1 t 2(t 1) A 3x 2B 1)2 (x Equating coefficients of like terms gives B 5 and A B 1 6A Equating coefficients of like terms gives A 2 A (A 3 7 B (A t 1 5x A 1) 2x Section 8.4 Exercises 3x A( x 2 and x2 1 (x 1) (x 1) 2(x 1)2 3(x 1) 1 (x 1)3 (x 1)3 (x 1)3 2x 2 4x 2 3x 3 1 (x 1)3 2 2x 7x 6 (x 1)3 5x 2 A Solving the system simultaneously yields A 1) (3t 4)(t 2 2) (t 1)(t 2 1) (t 2 2)(t 2 1) (t 2 2)(t 2 1) 3 2 t t t 1 3t 3 4t 2 6t 8 (t 2 2)(t 2 1) 2t 3 5t 2 5t 9 (t 2 2)(t 2 1) 2t 3 5t 2 5t 9 (t 2 2)(t 2 1) 4 1 3 1. x 2 x 2 Equating coefficients of like terms gives 2 2 x 1 2x 1 2 4x 5 x5 (x 1)(x 5) 2(x 2 4x 5) (x 5)(x 2 4x 5) (x 5)(x 2 4x 5) x 2 4x 5 2x 2 8x 10 (x 5)(x 2 4x 5) x2 12x 15 (x 5)(x 2 4x 5) 1 2 2x 2x x x2 (x x2 3(x 3) 2(x 2) (x 2)(x 3) (x 2)(x 3) 2x 4 3x 9 (x 2)(x 3) x 13 (x 2)(x 3) x 13 x2 x 6 3 3 t 9. 2 t 10. 2 1)2 1 5 2 (x 4 2x 4 4x 3x 2. x 2 2 06 10 04 Ax 5 2x 2 2x 2 4x 345 3, B 2. 4 s2 6s 4 ,C 15 2 3s 2 . 5 4 15(s 3) 2 5(s 2) 4 4. ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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