Find and prove c.d.f of a piece-wise defined random variable

Question

I have a random variable $X$ defined as:

$$X = \begin{cases} U, & \text{if }\text{U} >\text{V}^\text{2} \\ V & \text{if }\text{U} \leq\text{V}^\text{2} \\ \end{cases}$$

where $U$ and $V$ are i.i.d $\mathcal{U}[0,1]$. In my simulation, I figured out that c.d.f of $X$ is $F(x) = x^2$, but I have no idea how to demonstrate it.

Answer 1

$$F_X(x)=\mathbb P(X<x)=\mathbb P(U<x, U>V^2)+\mathbb P(V<x, U<V^2) = \mathbb P(V^2<U<x)+P(\sqrt{U}<V<x)$$ $$ =\int\limits_0^x du \int\limits_0^{\sqrt{u}} dv +\int\limits_0^x dv \int\limits_0^{v^2} du. $$ The result is $F_X(x)=\frac{1}{3}x^3+\frac{2}{3}x^{3/2}$ for $0<x<1$.