A4W08-138Sol

A4W08-138Sol - Math 138 Assignment 4 Solutions Fall 2008 1....

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Assignment 4 Solutions Fall 2008 1. Use Comparison to determine the convergence or divergence of the following: a) Z 1 x x x + x 2 dx Solution Try to show that the Type 1 improper integral diverges. For x 1, x + x 2 x 2 + x 2 1 x + x 2 1 x 2 + x 2 x x x + x 2 x x x 2 + x 2 = x x 2 x 2 = 1 2 x 1 2 Thus Z 1 x x x + x 2 dx Z 1 1 2 x 1 2 dx = lim b →∞ Z b 1 1 2 x 1 2 dx = lim b →∞ ± x ² b 1 = lim b →∞ h b - 1 i As b → ∞ so does b , and so the improper integral diverges. Then Z 1 x x x + x 2 dx diverges by comparison. Note: the result that Z 1 1 x p dx diverges for p 1 could be used here. b) Z π 0 sin 2 x x dx Solution Try to show that the Type 2 improper integral converges. For 0 < x π , sin 2 x x 1 x . Thus Z π 0 sin 2 x x dx Z π 0 1 x dx = lim a 0 + Z π a 1 x dx = lim a 0 + ± 2 x ² π a = lim a 0 + ± 2 π - 2 a ² = 2 π Thus the integral converges, and by comparison, the integral
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This note was uploaded on 04/07/2010 for the course MATH 138 taught by Professor Anoymous during the Fall '07 term at Waterloo.

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A4W08-138Sol - Math 138 Assignment 4 Solutions Fall 2008 1....

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