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October 1, 2014

October 1, 2014

Posted by **Bjorn** on Wednesday, January 2, 2008 at 3:50pm.

- Math -
**Damon**, Wednesday, January 2, 2008 at 4:09pmy = x^2 is 0 at x = 0, .25 at x = 1/2 and 1 at x = 1

y = +/- sqrt x is 0 at x=0, .707 at x = 1/2 and 1 at x = 1

sketch that and you will see that what you want is the integral from x = 0 to x = 1 of x^(1/2)dx minus the integral over the same interval of x^2 dx

- Math -
**drwls**, Wednesday, January 2, 2008 at 4:11pmThe two cuves intersect at (x=0, y=0) and (x=1, y=1). The area between them is

(Integral of) x^1/2 - x^2 dx

0->1

Evaluate x^(3/2)/(3/2) - x^3/3.

at the two limits and take the difference

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