31B pfinal-sols

31B pfinal-sols - 31/B - Final - Solutions December 9, 2011...

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Unformatted text preview: 31/B - Final - Solutions December 9, 2011 1. (20 points) Calculate g (1) and g (1), where g ( x ) is the inverse of f ( x ) = x + ln x . Solution First we solve, 1 = x + ln x , and we see that x = 1. Thus, g (1) = 1. Now, f ( x ) = 1 + 1 x . So, g (1) = 1 f ( g (1)) = 1 1 + 1 g (1) = 1 1 + 1 1 = 1 2 . 2. (20 points) Evaluate the integral Z x 9- x 2 dx using trigonometric substitution. Solution We substitute x = 3 sin . Then, dx = 3 cos d . So, the integral becomes Z x 9- x 2 dx = Z 3 sin p 9- 9 sin 2 3 cos d = 27 Z sin cos 2 d. Now we substitute u = cos . Then, du =- sin d , so the integral becomes 27 Z sin cos 2 d =- 27 Z u 2 du =- 27 u 3 3 =- 9 cos 3 . Using triangles, we see that cos = 9- x 2 3 . 1 Thus, the final answer is Z x 9- x 2 dx =- 9 (9- x 2 ) 3 / 2 27 =- (9- x 2 ) 3 / 2 3 . 3. (20 points) Evaluate the integral Z x 5 + 2 x 2 ( x + 1) dx. Solution First, dividing x 5 +2 by x 3 + x 2 we see that x 5 +2 = ( x 2- x +1)( x 3 + x 2 )- x 2 +2. So, Z x 5 + 2 x 2 ( x + 1) dx = Z ( x 2- x + 1)( x 3 + x 2 )- x 2 + 2 x 3 + x 2 dx = Z ( x 2- x + 1 ) dx- Z x 2- 2 x 3 + x 2 dx = x 3 3- x 2 2 + x- Z x 2- 2 x 2 ( x + 1) dx....
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This note was uploaded on 04/06/2012 for the course MATH 31B taught by Professor Valdimarsson during the Winter '08 term at UCLA.

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31B pfinal-sols - 31/B - Final - Solutions December 9, 2011...

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