asst4_solns

# asst4_solns - MATH 138 ASSIGNMENT 4 SOLUTIONS WINTER 2010...

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MATH 138 ASSIGNMENT 4 SOLUTIONS WINTER 2010 1. (a) Let u = 1 + x so that du = 1 2 x dx = 1 2( u - 1) dx . Then Z 0 1 1 + x dx = Z 1 2( u - 1) u du = 2 Z 1 ± 1 - 1 u ² du = 2 lim t →∞ h u - ln | u | i t 1 = 2 lim t →∞ h t - ln | t | - 1 i = 2 lim t →∞ t ³ 1 - ln | t | t - 1 t ´ = (1 - 0 - 0) = . Hence the integral is divergent. (b) See the solution to Question 6(b) in Assignment 3. (c) Note that x + 1 x 4 - x is not continuous at x = 1. Hence our improper integral is of both Type 1 and Type 2. We break the integral into Z 1 x + 1 x 4 - x dx = Z c 1 x + 1 x 4 - x dx + Z c x + 1 x 4 - x dx for some ﬁxed constant c > 1. (For concreteness, we can take c = 2.) If x c > 1, then x 4 - x < x 4 = x 2 . Thus x + 1 x 4 - x > x + 1 x 2 > 0. By Comparison Theorem, Z c x + 1 x 4 - x dx Z c x + 1 x 2 dx = Z c 1 x dx + Z c 1 x 2 dx. Since Z c 1 x dx = lim t →∞ ln | t | - ln | c | = , our integral must be divergent. 2. (a) g ( - x ) = Z - x 0 e - t 2 dt = Z x 0 e - ( - u ) 2 ( - du ) = - Z x 0 e - u 2 du = - g ( x ) . Thus g is an odd function. (b) By Fundamental Theorem of Calculus, g 0 ( x ) = e - x 2 . Since e - x 2 is always positive, g has to be increasing. 1

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2 MATH 138 ASSIGNMENT 4 SOLUTIONS (c) Let f ( t ) = e t - (1 + t ). If t 0, then f 0 ( t ) = e t - 1 0 and f 0 ( t ) = 0 only when t = 0. Hence f is increasing when t 0. If t > 0, then f ( t ) > f (0) = e t - (1 + t ) > e 0 - (1 + 0) = e t - (1 + t ) > 0 = e t > 1 + t.
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asst4_solns - MATH 138 ASSIGNMENT 4 SOLUTIONS WINTER 2010...

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