Math.112HW6Solns.(2008-9,Spring)

Math.112HW6Solns.(2008-9,Spring) - Â ï¬‚ ï¬‚ ï¬‚ c 1 We see...

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Homework 6 due on Monday, 30 March 2009 by 12:50 PM 1. Determine whether the integral Z π/ 2 0 x 3 / 2 x - sin x dx converges. Since sin x x we see that the integrand is positive. Using limit comparison with the function x - 3 / 2 we see that lim x 0+ x - 3 / 2 x 3 / 2 x - sin x = lim x 0+ x - sin x x 3 LR z}|{ = lim x 0+ 1 - cos x 3 x 2 LR z}|{ = lim x 0+ sin x 6 x = 1 6 . Hence the ﬁrst integral behaves similar to Z π/ 2 0 x - 3 / 2 dx and hence diverges. 2. Find lim x 0 + x Z 1 x cos t t 2 dt . Note that cos x cos1 for all x (0 , 1]. Hence we have Z 1 x cos t t 2 dt cos1 · ± 1 x - 1 → ∞ as x 0 + . Therefore the limit in question is in the form 0 · ∞ . We then use L’hospital’s rule to get lim x 0 + x Z 1 x cos t t 2 dt = lim x 0 + R 1 x cos t t 2 dt x - 1 = lim x 0 + - cos x x 2 - x - 2 = 1 . 3. Find all a for which the integral Z 1 ± ax 2 x 3 + 5 - 2 3 x dx converges. We have Z c 1 ± ax 2 x 3 + 5 - 2 3 x dx = ± a 2 ln( x 3 + 5) - 2 3 ln x c 1 = ±± 3 a 2 - 2 3 ln x + a 2 ln(1 + 5 x - 3 )
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Unformatted text preview: Â¶ ï¬‚ ï¬‚ ï¬‚ c 1 . We see that for the integral to converge we need a = 4 / 9. 4. For what values of p and q does the integral Z âˆž 1 x p + x q dx converge? The integral in question converges if both Z 1 1 x p + x q dx and Z âˆž 1 1 x p + x q dx converge. Assume, without loss of generality, that p â‰¥ q . Note that the former integral behaves as the integral Z 1 1 x q dx by LCT, since lim x â†’ + x q x p + x q = 1 . By the same reasoning, it is easily seen that the latter integral behaves as the integral Z âˆž 1 1 x p dx . Hence we require that q < 1 and p > 1....
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