Distribution of function of a Random Variable

Question

If $X$ is uniform on $(0,1)$, how would I go about finding the CDF of $Y=(X-X^2)^2$ ?Thanks.

Answer 1

A sketch of $y=(x-x^2)^2$ will reveal that if $X\in (0;1)$ then $Y\in (0;0.0625]$, and that the section $0\leq Y\leq y$ will correspond to two intervals on the support of $X$.

$\begin{align} \mathsf P(Y\leq y) & =\mathsf P(0\leq (X-X^2)^2\leq y) & :\text{for } y\in(0;1/16] \\ & = \mathsf P\left(\left\{0\leq X\leq \Box\right\}\cup \left\{\Box\leq X\leq 1\right\}\right) \end{align}$

Can you determine what these boxes contain?