<s>[INST] Density of a sum Let $X, Y$ be random variables having the joint density $f(x,y) = (a+1)(a+2)(y-x)^a, 0 \leq x < y \leq 1$, and $f(x,y) = 0$ elsewhere. Compute the density of $Z = X+Y$.
My solution doesn't match the textbook's and I'm not sure why. 
If $0 < z \leq 1$,
$$
\begin{align}
F_Z(z) & = \int_0^z \int_x^{z-x} (a+1)(a+2)(y-x)^a \mathrm{d}y\, \mathrm{d}x \\[8pt] & = \left. \int_0^z  (a+2)(y-x)^{a+1} \right|_x^{z-x}\mathrm{d}y\, \mathrm{d}x \\[8pt]
&= \int_0^z (a+2)(z-2x)^{a+1} \mathrm{d}x \\[8pt]
&= \left. \frac{(z-2x)^{a+2}}{-2} \right|_0^z \\[8pt]
&= \frac{z^{a+2}}{2} - \frac{(-z)^{a+2}}{2}
\end{align}
$$
Therefore,
$$
\begin{align}
f_Z(z) = \frac{\mathrm{d}}{\mathrm{d}z}F_Z(z) = \frac{a+2}{2}\left(z^{a+1} + (-z)^{a+1} \right)
\end{align}
$$
My setup for the case $1 < z < 2$,
$$
\begin{align}
F_Z(z) &= \int_0^{z-1} \int_0^1  (a+1)(a+2)(y-x)^a \mathrm{d}y\, \mathrm{d}x  + \int_{z-1}^1 \int_0^{z-x}  (a+1)(a+2)(y-x)^a \mathrm{d}y\, \mathrm{d}x 
\end{align}
$$
The solution for the first case is $\frac{a+2}{2} z^{a+1}$, so there is something with the term $(-z)^{a+1}$ but I can't figure out.
 [/INST] To find the bounds, it is useful to draw a picture. The joint density "lives" in the part of the square that is above the line $y=x$.
The interval of integration in the $0\lt z\le 1$ part is not right. The minimum value of the variable $y$ should be $x$. You have $y$ going instead from $0$ to $z-x$. 
For the range of $x$, the largest possible value is where $x+y=z$ meets the line $x=y$. Thus $x$ should go from $0$ to $z/2$.
</s>