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

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Math 138 Assignment 4 Solutions Fall 2008 1. Use Comparison to determine the convergence or divergence of the following: a) 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 1 x x x + x 2 dx 1 1 2 x 1 2 dx = lim b →∞ b 1 1 2 x 1 2 dx = lim b →∞ x b 1 = lim b →∞ b - 1 As b → ∞ so does b , and so the improper integral diverges. Then 1 x x x + x 2 dx diverges by comparison. Note: the result that 1 1 x p dx diverges for p 1 could be used here. b) 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 π 0 sin 2 x x dx π 0 1 x dx = lim a 0 + π a 1 x dx = lim a 0 + 2 x π a = lim a 0 + 2 π - 2 a = 2 π Thus the integral converges, and by comparison, the integral π 0 sin 2 x x dx also converges.
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