$X \sim uniform[0,4]$
Another probability variable $Y$ is defined as $Y:=(X+1)^2$. I'm searching for the CDF of $Y$.
Thing's I already know:
If $W:=X+1$ then $W \sim uniform[1,5]$
$Y=W^2$, so $P(Y\le y) = P(W^2\le y) = P(|W|\le \sqrt y)$
Is $P(|W|\le \sqrt(y)) = \frac{\sqrt z+1}{4}$??
$1<Y<25$ a.s. so $P(Y<y)=0$ for $y<1$ and $P(Y<y)=1$ for $y>25$. Now let $1\leq y \leq 25$.
$$ P(Y<y) = P((X+1)^2<y) = P(X+1 < \sqrt{y}) = P(X<\sqrt{y}-1) = \frac{1}{4}({\sqrt{y}-1}) $$