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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 tet dt 2 = lim T tet dt 2 T u = t dv = et du = dt v = et = lim Ttet 2 T + et dt 2 T [ ] = lim Ttetet [ ] 2 T = lim TTeTeT [ ] 2 e 2e 2 [ ] [ ] = lim TTeTeT + 2 e 2 + e 2 [ ] = 3 e 2 or 3 e 2...
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This note was uploaded on 02/21/2008 for the course MATH 155 taught by Professor Johnson during the Spring '08 term at Colorado State.
 Spring '08
 Johnson
 Math, Calculus, Logic

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