If the random variables $X$ and $Y$ have joint density function $$f_{XY}(x,y)= \begin{cases} \frac{xy}{96} & 0<x<4,1<y<5 \\ 0 & \text{otherwise} \end{cases} $$ , find the density function of $U=X+2Y$.
Ans: $$g_U(u)= \begin{cases} \frac{1}{2304}(u-2)^2(u-4) & 2<u<6 \\\\ \frac{1}{144}(3u-8) & 6<u<10 \\\\ \frac{1}{2304}(348u-u^3-2128) & 10<u<14 \\\\ 0 & \text{otherwise} \end{cases} $$
I managed to get the expressions right except for $10<u<14$. Here's what I did,
I let $u=x+2y$, $v=x$. Then
$$x=v, 0<v<4$$
$$y=\frac{u-v}{2}, v+2<u<v+10$$
with Jacobian $$J = \frac{1}{2}$$
Graphing this out, I saw that I could describe the region as three split regions, namely $R_1$, $R_2$ and $R_3$. $$R_1 = \{(u,v)\in \mathbb{R}^2:2<u<6, 0<v<u-2\}$$ $$R_2 = \{(u,v)\in \mathbb{R}^2:6<u<10, 0<v<4\}$$ $$R_3 = \{(u,v)\in \mathbb{R}^2:10<u<14, 0<v<u-10\}$$ $$R = R_1 \cup R_2 \cup R_3$$
Then, the CDF would be: $$F_U(u)= \begin{cases} \int_2^u \int_0^{u-2} \frac{1}{192}v(u-v)|J| \,dv\,du & 2<u<6\\ \int_6^u \int_0^{4} \frac{1}{192}v(u-v)|J|\,dv\,du & 6<u<10\\ \int_{10}^u \int_0^{u-10} \frac{1}{192}v(u-v)|J|\,dv\,du & 10<u<14\\ 0 & \text{otherwise} \end{cases} $$ $$F_U(u)= \begin{cases} \frac{1}{2304}(\frac{1}{4}u^4-6u^2+16u)-\frac{1}{96} & 2<u<6\\ \frac{1}{144}(\frac{3}{2}u^2-8u)-\frac{1}{12} & 6<u<10\\ \frac{1}{2304}(\frac{1}{4}u^4-150u^2+2000u) & 10<u<14\\ 0 & \text{otherwise} \end{cases} $$ Differentiating to get $f_U(u)$ gives, $$f_U(u)= \begin{cases} \frac{1}{2304}(u-2)^2(u+4) & 2<u<6\\ \frac{1}{144}(3u-8) & 6<u<10\\ \frac{1}{2304}(u^3-300u+2000) & 10<u<14\\ 0 & \text{otherwise} \end{cases} $$ I'm not sure where I've went wrong with the third expression. The integral of the answer provided across all values of $u$ gives 1, which means the answer isn't wrong.