Posted by Bjorn on Wednesday, January 2, 2008 at 3:50pm.
Calculate the area above y=x^2 and to the right of x=y^2 using integrals.

Math  Damon, Wednesday, January 2, 2008 at 4:09pm
y = 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 4:11pm
The 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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