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December 19, 2014

December 19, 2014

Posted by **Josh** on Tuesday, November 8, 2011 at 11:55pm.

- Math - Calculus III -
**Steve**, Wednesday, November 9, 2011 at 5:38amThe curves intersect at (0,0) and (3/2,9/4)

So, we want

Int[0,3/2] Int[x^2,3x-x^2] (x^7 y) dy dx

= Int[0,3/2] (x^7 y^2)/2 [x^2,3x-x^2] dx

= Int[0,3/2] (1/2)(9x^9 - 6x^10) dx

= 1/2(9/10 x^10 - 6/11 x^11)[0,3/2]

= 1/2 (3/2)^10 (9/10 - 6/11 * 3/2)

= 1/2 (3/2)^10 (9/110)

= 531441/225280 = 2.359

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