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October 1, 2014

October 1, 2014

Posted by **Anonymous** on Thursday, March 10, 2011 at 10:22pm.

- calculus -
**Reiny**, Thursday, March 10, 2011 at 11:13pmDid you make a sketch?

first find their intersection.

I had them intersect at (0,1) and (3,4)

Area = [integral] ((3x)^(1/2) + 1 - x - 1) from 0 to 3

= ((2/9(3x)^(3/2) - x^2/2) from 0 to 2

= (2/9)(9)^(3/2) - 9/2 - (0-0))

= (2/9)(27) - 9/2

= 6 - 9/2 = 3/2

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