Quiz1 v2 Solutions

Quiz1 v2 Solutions - MAC 2312 Quiz 1 January 17, 2012...

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Unformatted text preview: MAC 2312 Quiz 1 January 17, 2012 Solutions 1. Evaluate the indefinite integral sin3 x dx. (cos x)3/2 Solution: First, write sin3 x = sin2 x sin x = (1 - cos2 x) sin x so the given integral may be rewritten as sin3 x dx = (cos x)3/2 1 - cos2 x sin x dx. (cos x)3/2 (1) Next, use the change of variable w = cos x, dw = - sin x dx to rewrite the right-hand side of equation (1) as - 1 - w2 dw = w 3/2 = = w2 - 1 dw w 3/2 w 1/2 - w -3/2 dw 2 3/2 w + 2w -1/2 + C. 3 Finally, use w = cos x on the right-hand side of the above string of equalities to obtain 2 sin3 x dx = (cos x)3/2 + 2(cos x)-1/2 + C. 3/2 (cos x) 3 2. Evaluate the definite integral. Simplify your answer. e3 ln(x1/3 ) dx 1 Solution: Observe first that ln(x1/3 ) = 1 3 e3 1 1 3 ln x, so the given integral is equal to e3 1 ln x dx. Therefore, it suffices to first compute ln x dx then multiply the result by 1 . 3 To compute e3 1 ln x dx, use integration by parts with u = ln x 1 du = x dx dv = dx v = x. This gives e3 1 ln x dx = x ln x 1 dx x 1 1 = 3e3 - 0 - (e3 - 1) = 2e3 + 1. - x e3 e3 Multiplying answer 1 3 to the right-hand side of the above string of equalities gives the final e3 1 1 ln(x1/3 ) dx = (2e3 + 1). 3 ...
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Quiz1 v2 Solutions - MAC 2312 Quiz 1 January 17, 2012...

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