140Bquiz14asols

140Bquiz14asols - 1 MATH 140B QUIZ#14 Solutions 1 Let u = 3...

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Unformatted text preview: 1 MATH 140B QUIZ#14 Solutions Dec 14, 2007 1. Let u = 3 - x so du = -dx. When x = -1, u = 4 and when x = 1, u = 2. Therefore, 1 2 dx 1 2 =- du = - ln u 4 = -(ln 2 - ln 4) = ln 4 - ln 2 = 2 ln 2 - ln 2 = ln 2. -1 3 - x 4 u Therefore, C) is the answer. 2. Let u = x3 - 1 so du = 3x2 dx. When x = 0, u = -1 and when x = 1, u = 0. Therefore, 0 1 0 1 1 1 = 0 - (-1)6 = - . 9x2 (x3 - 1)5 dx = 3 u5 du = u6 2 2 2 0 -1 -1 Therefore, E) is the answer. 3. Let u = x2 -2 so du = 2x dx and 2 du = 4x dx. When x = 0, u = -2 and when x = 1, u = -1. Hence -1 1 -1 1 1 4x -3 =- 1- = -3/4. dx = 2 u du = - 2 2 - 2)3 u -2 4 0 (x -2 Therefore, B) is the answer. 1 -1 4. Let u = so du = 2 dx. When x = 1/, = and when x = 2/, u = /2. Therefore, x x 2/ 1/ 1 sin x2 1 x /2 dx = - sin u du = cos u /2 = (0 - (-1)) = 1. Therefore, D) is the answer. ...
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This note was uploaded on 03/30/2008 for the course MATH 140B taught by Professor Fabbri,marc during the Fall '07 term at Penn State.

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