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Unformatted text preview: HOMEWORK 5: MATH 265 Due in class on Oct 27 Problem 1: Improper integrals. (a) Consider the integral I = R 1 1 t p dt . Show that this improper integral converges for p > 1 and find its value. Show that it diverges for p = 1. What happens when p < 1? (b) Consider the integral I = R e- (5- k ) t dt . Does this integral converge? To what value and under what condition(s) if any. (c) Consider the integral I = R x 10 5 dx . Determine whether this integral converges or diverges and explain why. Solution to Problem 1: (a) Provided p 6 = 1 we can write I = Z 1 1 t p dt = Z 1 t- p dt = 1- p + 1 t (- p +1) 1 = 1 1- p [ lim t t (1- p )- 1]. If p > 1 then the limit in the above expression is 0 and we get I = 1 1- p [0- 1] = 1 p- 1 . If p < 1 then the power of t ( 1- p ) is positive, and hence the limit lim t t (1- p ) = so the integral diverges. If p = 1 then the integral is different, indeed I = Z 1 1 t dt = ln( t ) 1 . As discussed in class, this integral diverges since ln(...
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