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November 23, 2014

November 23, 2014

Posted by **Anonymous** on Tuesday, October 5, 2010 at 11:02pm.

y=x^2

y^2=x

- Calculus: Application of Integration -
**Reiny**, Tuesday, October 5, 2010 at 11:15pmIt is easy to see that the two curves intersect at (0,0) and (1,1)

I will integrate with respect to x

area = [integral] (x^(1/2) - x^2) dx from 0 to 1

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

= 2/3 - 1/3 - 0

= 1/3

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