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Unformatted text preview: 2 . Solution: The region of integration is shown below: 1 By Fubini&s Theorem, Z 2 p ln(3) Z p ln(3) y= 2 e x 2 dx dy = Z p ln(3) Z 2 x e x 2 dy dx . The inner integral is Z 2 x e x 2 dy = ye x 2 & & & y =2 x y =0 = 2 xe x 2 . We now have Z p ln(3) Z 2 x e x 2 dy dx = Z p ln(3) 2 xe x 2 dx = e x 2 & & & x = p ln(3) x =0 = e ln(3) & e = 2 . 2...
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This note was uploaded on 02/05/2012 for the course MATH 2203 taught by Professor Ellermeyer during the Fall '10 term at FIU.
 Fall '10
 Ellermeyer
 Math

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