Fall 2007 - Hohnhold's Class - Quiz 1 (Version A)

Fall 2007- - u 7 2 6 2 5 u 5 2 9 2 3 u 3 2 | 3 1 = 2 7 3 7 2 12 5 3 5 2 6 3 3 2-2 7-12 5-6(b We use integration by parts with u = x 2 and v

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Quiz 1 for Math 20B Fall Quarter 2007, UCSD Henning Hohnhold Name: Section: #1: #2: Total: (1) (3 points) Compute the derivative of the function g defined by g ( x ) = Z e x 1 t 2 t 3 + 1 dt. Solution: Using the FTC, part 1, and the chain rule, we compute d dx Z e x 1 t 2 t 3 + 1 dt = ( e x ) 2 ( e x ) 3 + 1 · e x = e 2 x e 3 x + 1 · e x = e 3 x e 3 x + 1 where the second factor in the second expression from the left comes from the interior derivative: d dx e x = e x .
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(2) (7 points) Evaluate the following integrals: ( a ) Z 6 4 x 2 · x - 3 dx and ( b ) Z x 2 · sin ± 2 x 3 ² dx. Solution: (a) We substitute u = x - 3. Then du = dx , x 2 = ( u + 3) 2 , so that Z 6 4 x 2 · x - 3 dx = Z 3 1 ( u + 3) 2 · u 1 2 du = Z 3 1 u 5 2 + 6 u 3 2 + 9 u 1 2 du = 2 7 ·
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Unformatted text preview: u 7 2 + 6 · 2 5 u 5 2 + 9 · 2 3 u 3 2 | 3 1 = 2 7 · 3 7 2 + 12 5 3 5 2 + 6 · 3 3 2-2 7-12 5-6 (b) We use integration by parts with u = x 2 and v = sin ( 2 x 3 ) : Z x 2 · sin ± 2 x 3 ² dx = x 2 · ±-3 2 ² · cos ± 2 x 3 ²-Z 1 2 · ±-3 2 ² · cos ± 2 x 3 ² dx =-3 x 4 · cos ± 2 x 3 ² + 3 4 · ± 3 2 ² · sin ± 2 x 3 ² + C =-3 x 4 · cos ± 2 x 3 ² + 9 8 · sin ± 2 x 3 ² + C 2 Good Luck! 3...
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This note was uploaded on 04/29/2008 for the course MATH 20B taught by Professor Justin during the Fall '08 term at UCSD.

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Fall 2007- - u 7 2 6 2 5 u 5 2 9 2 3 u 3 2 | 3 1 = 2 7 3 7 2 12 5 3 5 2 6 3 3 2-2 7-12 5-6(b We use integration by parts with u = x 2 and v

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