Find the area of the region: Above y=x^2 and to the right of x=y^2

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`y= x^2` (Black curve)

`x = y^2` ( Red curve)

We need to find the area between both curves.

First we will find the intersection points.

==> `y= x^2 ==gt y= +-sqrt(x).`

`` ==> `x^2 = +-sqrtx`

`` ==> `x^4 = x` ==> `x^4 - x =` 0

==> `x(x^3 -1) =` 0

==> `x (x-1)(x^2 +x +1) = ` 0

==> x = 0

==> x = 1

Then we need to find the bounded area between `x^2` and `sqrtx` from 0 to 1.

==> `int_0^1 (x2 -sqrtx) dx = int_0^1 (x^2 - x^(1/2) )` dx

==> `int_0^1 (x^2 -x^(1/2)) dx = x^3/3 - x^(3/2)/(3/2) = (1/3)x^3 - (2/3)x^(3/2)`

`` ==> (`1^3 - (2/3)(1^(3/2)) = 1- 2/3 = 1/3`

`` ==> `0^3 - (2/3)0^(3/2) =` 0

==> Then, the area is `1/3.`

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