posted by Candice .
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?)
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)