Probability that $xI have a question asking for $P(x<V<U^2)$ when $U$ and $V$ are independent, identically distributed uniform variables on $[0,1]$.
I started with:
$$
\begin{align}
P(x<V<U^2) &= P(V<U^2) - P(V<x)\\
&= P(U^2-V>0) - x
\end{align}
$$
But I don't know where to go from here. Any advice appreciated.
 A: You need to integrate the constant function $\mathbf{1}$ over the set whose probability you want to determine. In your case, that is the set $$
    A_x
  = \{(u,v) \in [0,1]^2 \,:\, x < v < u^2 \}
  = \{(u,v) \in [0,1]^2 \,:\, v > x \:\:\text{and}\:\: u > \sqrt{v}\}
$$
and you therefore need to determine $$
  P(A_x)
  = \int_A \mathbf{1} \,d(\mu_u\times\mu_v)
  = \int_{[x,1)} \int_{\left(\sqrt{v},1\right]} \,d\mu_u(u)\,d\mu_v(v)
  = \int_x^1 \int_{\sqrt{v}}^1 \,du\,dv \text{.}
$$
(The last identity holds because both probability measures $\mu_v$ and $\mu_v$ have constant density $1$)
A: First of all we have
$$ P(x<V<U^2) = P(x<V)P(V<U^2|x<V) = (1-x)P(V<U^2|x<V).$$
Secondly 
$$P(V<U^2|x<V) = \frac{\int_x^1 P(V<U^2)\,\mathrm{d}V}{\int_x^1 1\,\mathrm{d}V} = \frac{1}{1-x}\int_x^1P(U>\sqrt{V})\,\mathrm{d}V.$$
Thus
$$P(x<V<U^2) = \int_x^1P(U>\sqrt{V})\,\mathrm{d}V = \int^1_x 1-\sqrt{V}\,\mathrm{d}V,$$
which evaluates to
$$P(x<V<U^2) = \left[V-\frac{2}{3}V^\frac{3}{2}\right]_x^1 = \frac{1}{3}-x+\frac{2}{3}x^\frac{3}{2}.$$
