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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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- Spring '08
- Johnson
- Math, Calculus, Logic, dt u=t dv, te dt
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