soln1 - Solutions to Homework 1 1(a For 0&lt...

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Unformatted text preview: Solutions to Homework 1 1 . (a) For 0 < t < 1 use integration by parts in: Z 1 t log ( x ) dx = xlog ( x ) | 1 t- Z 1 t x ( log ( x )) dx =- tlog ( t )- Z 1 t 1 dx =- tlog ( t )- 1+ t Then Z 1 log ( x ) dx = lim t & Z 1 t log ( x ) dx = lim t & (- tlog ( t ) + t )- 1 =- 1 Note: lim t & tlog ( t ) = lim t & log ( t ) 1 t L Hospital = lim t & 1 t- 1 t 2 = lim t & (- t ) = 0 (b) Fix an a > 1, and M > 1 Z M 1 1 x a dx = x- a +1- a + 1 | M 1 = M 1- a 1- a- 1 1- a Since 1- a < 0 we get Z 1 1 x a dx = lim M Z M 1 1 x a dx =- 1 1- a = 1 a- 1 For a = 1 we obtain Z M 1 1 x dx = log ( x ) | M 1 = log ( M ) Thus Z 1 1 x dx = lim M log ( M ) = 2 . f : R R , f ( x ) = xlog ( x 2 ) if x 6 = 0 if x = 0 At every point x > 0 the function is continuous, differentiable, and infinitely many times differentiable, since f ( x ) = 2 log ( x ) + 2 , f 00 ( x ) = 2 x , f ( k ) ( x ) = (- 1) k 2( k- 2)!...
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This note was uploaded on 05/05/2010 for the course MATH 464 taught by Professor Staff during the Spring '08 term at Maryland.

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soln1 - Solutions to Homework 1 1(a For 0&lt...

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