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September 16, 2014

September 16, 2014

Posted by **Corey** on Monday, December 17, 2012 at 9:57am.

- Calculus -
**Steve**, Monday, December 17, 2012 at 10:04amThe curves intersect at (0,4) and (3,0)

area = ∫[0,3] (-x^2+2x+3) - (x^2-4x+3) dx

= ∫[0-3] 6x - 2x^2 dx

= 3x^2 - 2/3 x^3 [0,3]

= 9

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