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February 1, 2015

February 1, 2015

Posted by **tiffany** on Wednesday, December 11, 2013 at 6:35pm.

Enclosed by y = x^2 − 4x + 1 and y = −x^2 + 4x − 5

- Calculus -
**Steve**, Wednesday, December 11, 2013 at 6:44pmthe curves intersect at x=1,3 so the area is

∫[1,3] (−x^2 + 4x − 5 )-(x^2 − 4x + 1) dx

= ∫[1,3] -2x^2 + 8x - 6 dx

= -2/3 x^3 + 4x^2 - 6x [1,3]

= 8/3

- Calculus -
**tiffany**, Wednesday, December 11, 2013 at 6:56pmthank you so much

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