Math1B Quiz2Solutions

# Math1B Quiz2Solutions - Z ln x 2 2 x 2 dx = x ln x 2 2 x...

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QUIZ 2 SOLUTION Question 1 x 3 + x 2 + x = x ( x 2 + x + 1) , is the decomposition of the denominator into irreducible factors. 3 x 2 + 4 x + 3 x ( x 2 + x + 1) = A x + Bx + C x 2 + x + 1 . Forming a common denominator and then multiplying both sides, we get 3 x 2 + 4 x + 3 = ( A + B ) x 2 + ( A + C ) x + A. Therefore A = 3 , A + C = 3 , A + B = 3 = A = 3 ,B = 0 ,C = 1 . Z 3 x 2 + 4 x + 3 x ( x 2 + x + 1) dx = Z 3 x dx + Z 1 x 2 + x + 1 dx. Z 1 x 2 + x + 1 dx = 2 3 tan - 1 ( 2 3 x + 1 3 ) . Question 2 Make substitution u = x + 1. Then du = 1 2 x +1 dx . Therefore: 8 Z 3 1 x x + 1 dx = 1 2 3 Z 2 1 ( u - 1) 2 du = = - 1 2( u - 1) ± ± ± 3 2 = 1 4 . Question 3 Integration by parts f ( x ) = ln( x 2 + 2 x + 2) ,f 0 ( x ) = 2 x + 2 x 2 + 2 x + 2 , g 0 ( x ) = 1 ,g ( x ) = x. Therefore Z ln( x 2 + 2 x + 2) dx = x ln( x 2 + 2 x + 2) - 2 Z x 2 + x x 2 + 2 x + 2 dx. x 2 + x x 2 + 2 x + 2 = 1 - x + 2 x 2 + 2 x + 2 = 1 - x + 1 x 2 + 2 x + 2 - 1 ( x + 1) 2 + 2 . We integrate each term in the last equation to get:

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Unformatted text preview: Z ln( x 2 + 2 x + 2) dx = x ln( x 2 + 2 x + 2)-2 x + ln | x 2 + 2 x + 2 | + 2 tan-1 ( x + 1) . Date : February 6,2008. 1 2 QUIZ 2 SOLUTION Question 4 The second derivative on ln( x ) is-1 x 2 therefore, the maximum of the modulus on [1 ,T ] is 1. Therefore, we want ( T-1) 3 12 · 10 6 ≤ 10-3 . Therefore ( T-1) 3 ≤ 12 · 10 3 . Therefore T ≤ 1 + 10 3 √ 12 ....
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## This note was uploaded on 07/10/2008 for the course MATH 1B taught by Professor Reshetiken during the Spring '08 term at Berkeley.

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Math1B Quiz2Solutions - Z ln x 2 2 x 2 dx = x ln x 2 2 x...

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