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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions to Homework 1 1 . (a) For 0 &amp;lt; t &amp;lt; 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 &amp;amp; Z 1 t log ( x ) dx = lim t &amp;amp; (- tlog ( t ) + t )- 1 =- 1 Note: lim t &amp;amp; tlog ( t ) = lim t &amp;amp; log ( t ) 1 t L Hospital = lim t &amp;amp; 1 t- 1 t 2 = lim t &amp;amp; (- t ) = 0 (b) Fix an a &amp;gt; 1, and M &amp;gt; 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 &amp;lt; 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 &amp;gt; 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)!...
View Full Document

## This note was uploaded on 05/05/2010 for the course MATH 464 taught by Professor Staff during the Spring '08 term at Maryland.

### Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online