find the area enclosed by y=x^2 and y=sinx in the first quadrant.

Answer 1

The hardest part is to find the intersection of the two curves.
I don't know what level of Calculus this is but the equation is nasty to solve.

WolfFram gave me
http://www.wolframalpha.com/input/?i=x%5E2+%3D+sin%28x%29

x = ..876726 and of course the obvious x = 0

so area = ∫(sinx - x^2) dx from 0 to .876726
= [ -cosx - x^3/3 ]
= -.63967 - .22463 - (-1 - 0)
= .1356975

Answer 2

x^2 = sinx at x=.876
So integrate

sin(x)-x^2 from 0 to .876