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quizsoln13 - Area = 1 2 x 3 2 4 dx u = 2 x 3 x = 2 then u...

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Name:___________________________ Math 155 Quiz 13 6 December 2006 Instructions : This quiz is closed book and closed notes. You may use a calculator. Label your answers and SHOW ALL WORK! If substitution is needed, show your substitution. 1. Calculate the integral: 72 t + 5 ( ) dt - 1 2 = 36 t 2 + 5 t - 1 2 = 36(2) 2 + 5(2) [ ] - 36 - 5 [ ] =123 2. Calculate the integral: x cos( x ) dx 0 p (Hint: Use integration by parts) u = x dv = cos( x ) dx du = dx v = sin( x ) x cos( x ) dx = x sin( x ) 0 p - sin( x ) dx 0 p 0 p = x sin( x ) + cos( x ) 0 p = 0 - 1 [ ] - 0 +1 [ ] = - 2 3. Find the area under the curve f ( x ) = 1 2 x - 3
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Unformatted text preview: Area = 1 2 x- 3 2 4 dx u = 2 x- 3 x = 2 then u =1 du = 2 dx x = 4 then u = 5 1 2 1 u du 1 5 = 1 2 ln u 1 5 = 1 2 ln(5) -1 2 ln(1) = 1 2 ln(5) EXTRA CREDIT (5 pts) Evaluate the following integral te-t dt 2 = lim T te-t dt 2 T u = t dv = e-t du = dt v = -e-t = lim T-te-t 2 T + e-t dt 2 T [ ] = lim T-te-t-e-t [ ] 2 T = lim T-Te-T-e-T [ ]-- 2 e- 2-e- 2 [ ] [ ] = lim T-Te-T-e-T + 2 e- 2 + e- 2 [ ] = 3 e- 2 or 3 e 2...
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