How do I find the cdf of $X_1 + X_2$? $X_1$ uniform $(0,1)$ and $X_2$ uniform $(0,2)$
$$
\begin{cases}
f(x_1,x_2) = \frac{1}{2}, &\quad  \mbox{for} \  0<x_1<1, 0<x_2<2 \\
0, & \quad \mbox{otherwise}
\end{cases}
$$
The density of $X_1+X_2$:
$$
f(x_1,x_2) =
  \begin{cases}
    \displaystyle \frac{z}{2}       & \quad 0<z<1\\
    \displaystyle \frac{1}{2}  & \quad 1<z<2\\
\displaystyle \frac{3-z}{2} & \quad 2<z<3\\
  \end{cases}
$$
The question is how do I find the cdf of $X_1+X_2$?
I would do:
$$
f(x_1,x_2) =
  \begin{cases}
    \displaystyle \int_0^z \frac{z}{2} dx =\frac{z^2}{4},       & \quad 0<z<1\\
    \displaystyle \int_0^1 \frac{1}{2} dx =\frac{1}{2},    & \quad 1<z<2\\
  \end{cases}
$$
That just seems wrong.. Help

The PDF of $Y = X_1 + X_2$ is the convolution of the PDFs of $X_1$ and $X_2$
$$f_Y (y) = \begin{cases} \dfrac{y}{2} & \text{if } y \in [0,1]\\\\ \dfrac{1}{2} & \text{if } y \in [1,2]\\\\ \dfrac{3-y}{2} & \text{if } y \in [2,3]\\\\ 0 & \text{otherwise}\end{cases}$$
Integrating, we obtain the CDF
$$F_Y (y) = \begin{cases} 0 & \text{if } y < 0\\\\ \dfrac{y^2}{4} & \text{if } y \in [0,1]\\\\ \dfrac{1}{4} + \dfrac{y-1}{2} & \text{if } y \in [1,2]\\\\ \dfrac{3}{4} + \dfrac{3}{2} (y-2) - \dfrac{1}{4} (y^2 - 2^2) & \text{if } y \in [2,3]\\\\ 1 & \text{if } y \geq 3\end{cases}$$
