Let $X$ and $Y$ be identically distributed independent random variables with density function
\begin{align} f(t) = \begin{cases}\frac{1}{4} & 0 \leq t \leq 4\\0 & \text{otherwise}\end{cases} \end{align}
Find the density function of $Z=2X+Y$.
I am trying to solve this question as follow: $U=2X$ \begin{align} f_U(u) = \frac{1}{2}f_X(\frac {u}{2})=\begin{cases}\frac{1}{8} & 0 \leq u \leq 8\\0 & \text{otherwise}\end{cases} \end{align}
and
\begin{align} f_Y(y) = \begin{cases}\frac{1}{4} & 0 \leq y \leq 4\\0 & \text{otherwise}\end{cases} \end{align}
and now I am trying to solve $Z=U+Y$
\begin{align} f_Z(z) = \int_{-\infty}^{\infty} f_U(z-y)f_Y(y)dy \end{align}
$f_Y(y)=\frac{1}{4}$ if $0\leq y\leq4$ and $0$ otherwise, then this become:
\begin{align} f_Z(z) = \int_{0}^{8} f_U(z-y)dy \end{align}
now the integrated is $0$ unless $0 \leq z-y \leq 8 (i.e, z-8 \leq y \leq z)$ and then it is $\frac{1}{8}$
if $0 \leq z \leq 4$ then
\begin{align} f_Z(z) = \frac{1}{4}\int_{0}^{z} \frac {1}{8}dy = \frac{z}{32} \end{align}
if $4 < z \leq 12$ then
\begin{align} f_Z(z) = \frac{1}{4}\int_{z-8}^{8} \frac {1}{8}dy = \frac{16-z}{32} \end{align}
and $f_Z(z) = 0$ for values $z < 0$ and $z > 12$
\begin{align} f_Z(z) = \begin{cases} \frac{z}{32} & 0 \leq z \leq 4\\ \frac{16-z}{32} & 4 < z \leq 12\\ 0 & otherwise \end{cases} \end{align}