Uniform distributed variable X:U(-9,9) is given. Find the CDF of Y if Y is.. $$Y= \begin{cases} 4X,\ \ \  |X| \leq 3 \\ 0,\ \ \ \ \ \ |X|>3 \end{cases}$$
My take on it:
$$F_Y(y)=0; y\leq-12;F_Y(y)=1; y\geq 12;$$
$$F_Y(y)=\{Y < y \}$$
In class in a similar task we drew the functions and integrated in relation to the graph with respect to dx and the given density function. What I am having trouble is finding the boundaries of the integral... 
 A: Notice that the pdf of $X$ is
$$f_X(x)=\begin{cases}
\frac{1}{18},&\text{ if } -9\le x \le 9\\
\ 0,& \text{ otherwise. }\end{cases}$$
We can treat $F_Y(y)=P(Y<y)$ as the sum of two probabilities: 
$$F_Y(y)=P(Y<y\cap |X|>3)+P(Y<y\cap |X|\le 3).$$
First,
$$P(Y<y\cap |X|>3)=P(0<y \cap |X|>3)=\begin{cases}
0,&\text{ if } y<0\\
P(|X|>3)=\frac{2}{3},&\text{ if } y\ge 0.
\end{cases}$$
Second,
$$P(Y<y\cap |X|\le 3)=P(X<\frac{y}{4}\cap -3\le X \le 3)=\begin{cases}
0,&\text{ if }& y<-12\\
\frac{1}{72}y+\frac{1}{6},&\text{ if }& -12\le y\le12\\
\frac{1}{3},&\text{ if }& y> 12.\\
\end{cases}$$
Actually your integral is hiding above behind the function
$$\frac{1}{72}y+\frac{1}{6}=\int_{-3}^{y/4}f_X(x)dx=\frac{1}{18}[x]_{-3}^{y/4}.$$ 
So, the sum of the two probabilities gives
$$F_Y(y)=\begin{cases}
0,&\text{ if }& y<-12\\
\frac{1}{72}y+\frac{1}{6},&\text{ if }& -12\le y< 0\\\frac{1}{72}y+\frac{5}{6},&\text{ if }& 0\le y\le 12\\
1,&\text{ if }& y> 12.\\
\end{cases}$$
Below is the graph of $F_Y(y)$

A: You are correct that $Y$ takes values in $(-12,12)$ and "mainly" uniformly. However,
this is a 'mixed' distribution with positive probability at the
point $0$. So you are not going to find a PDF for $Y$. Think
it through in terms of intervals of $X$ in order to get the CDF of $Y.$
If you have ready access to R, you might run the following code for
a (necessarily crude, possibly helpful) histogram.
 x = runif(10^5, -9, 9);  y = 4*x;  y[abs(x)>3]=0
 hist(y, prob=T, br=50)

