141Bquiz1csols - 1 MATH 141B QUIZ#1 Solutions Due: Tues Jan...

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Unformatted text preview: 1 MATH 141B QUIZ#1 Solutions Due: Tues Jan 22, 2008 Instructor: Dr. M.Fabbri Duration: 15 minutes 1. First lets compute the corresponding indefinite integral, and then do the evaluation. If u = x 2 and dv = cos x dx then du = 2 x dx and v = sin x. Therefore, integraldisplay x 2 cos x dx = integraldisplay u dv = uv- integraldisplay v du = x 2 sin x- 2 integraldisplay x sin x dx. We compute this latter integral, integraltext x sin x dx, by parts as well. If u = x and dv = sin x dx . Then du = dx and v =- cos x. Then integraldisplay x sin x dx = uv- integraldisplay v du =- x cos x + integraldisplay cos x dx =- x cos x + sin x. Combining these two calculations, gives integraldisplay x 2 cos x dx = x 2 sin x- 2(- x cos x + sin x ) = x 2 sin x + 2 x cos x- 2 sin x. Evaluating gives, integraldisplay x 2 cos x dx = ( 2 sin + 2 cos - 2 sin )- (0) =- 2 . Therefore, A) is the correct answer....
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This homework help was uploaded on 03/30/2008 for the course MATH 141B taught by Professor Marcfabbri during the Spring '08 term at Pennsylvania State University, University Park.

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141Bquiz1csols - 1 MATH 141B QUIZ#1 Solutions Due: Tues Jan...

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