Let U and V be independent and identically distributed uniformly on the interval [0,1]. Show that for 0<x<1: $\displaystyle P(x<V<U^2)=\frac{1}{3}-x+\frac{2}{3} x^{\frac{3}{2}}$.

So far I have worked out the cdf, and hence pdf of $\displaystyle U^2$. I thought about $\displaystyle P(x<V<U^2)=P(V<U^2)-P(V<x)$ but a) this doesn't give me the correct answer and b) I don't actually think P(V<U^2) can be calculated.

Could someone give me a hint as to where to go?

Thanks.