Given $(X ,Y)$ uniformly distributed over the parallelogram with vertices $(-1,0)$, $(1,0)$, $(2,1)$, $(0,1)$, I'm intrested in the distribution of $Z = X+Y$.
Solution
The contours of (X, Y) look like this 
Now \begin{eqnarray}\label{myeq} f_Z(z) = \int_{-\infty}^{+\infty} f(x,z-x)\,dx\,, \end{eqnarray}
since $x\in[-1,2]$. Now, since $y \in [0,1]$, $0 \leq z-x \leq 1$, hence the region of integration for computing $f_Z(z)$ is something like
Therefore, for $-1\leq z < 0$ (the blue triangle) $$f_Z(z) = \int_{-1}^{z}\frac{1}{2} dx = \frac{z+1}{2}\,,$$ for $0\leq z < 2$ (the red parallelogram) $$f_Z (z) = \int_{z-1}^{z}\frac{1}{2}\,dx = \frac{1}{2}\,,$$ and for $2 \leq z\leq 3$ (the green triangle) $$f_Z(z) = \int_{z-1}^{2}\frac{1}{2}\,dx = \frac{3-z}{2}.$$
So the density of $Z$ is trapezoid when it should be a triangle with vertex at 1. What am I missing?