# Math- integration

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Find the area bounded by x^(2/3) and x= y^2 in the first quadrant.

(I've worked out the x intercepts to be 0 and 1 in the first quadrant, but how would I find the area?)

thanks :)

• Math- integration -

Is your first function y = x^(2/3) ?
then your intersection is indeed at x-0 and x=1

in that domain x = y^2 or y = x^(1/2) is above y = x^(2/3)
so the effective height is x^(1/2) - x^(2/3)

area = ∫( x^(1/2) - x^(2/3) ) dx
= (2/3)x^(3/2) - (3/5)x^(5/3) | from 0 to 1
= (2/3 - 3/5 - 0)
= 1/15

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