I started along these lines:
Let $Z = X + Y$ where $\frac{L}{2}< z < \frac{3L}{2}$, then,
$$f_{X+Y}(z)=f_{Z}(z) = \int f_{X}(x)f_{Y}(z-x)dx$$
However, I am not sure how to fill in the bounds of the integral to properly account for the domains of $X$ and $Y$. Also, I think $f_{Z}(z)$ needs to computed via 2 cases:
1) $\frac{L}{2}< z < L$
2) $L< z < \frac{3L}{2}$
Since in case (1) $X$ can take any value in its domain, whereas in case (2) $X$ must take the value $\frac{L}{2}$.
Regardless, I am having a hard time properly setting up the integrals. Any insight would be appreciated.
Update: After some more thinking, for case (1), I think this is the integral:
$$f_{Z}(z) = \int_{0}^{z} f_{X}(y-z)f_{Y}(y)dy$$
And for case (2):
$$f_{Z}(z) = \int_{z}^{\frac{3L}{2}} f_{X}(X=\frac{L}{2})f_{Y}(z-\frac{L}{2})dy$$
However, I am unsure of the above.
$P(X+Y \le z)=P(X\le z-Y)$
Therefore,
$F(z) = \int_{L/2}^L P(X \le z-y) \frac{2}{L} dy$
Now, it is obvious that:
$$P(X \le z-y) = 0 \text{ when } z-y<0$$
$$P(X \le z-y) = \frac{2(z-y)}{L} \text{ when } 0\le z-y \le \frac{L}{2}$$
$$P(X \le z-y) = 1 \text { when } z-y \ge \frac{L}{2}$$
Therefore, it follows that:
$$F(z) = 0 \text{ when } z \le \frac{L}{2}$$
$$F(z) = \int_{L/2}^L \frac{2(z-y)}{L} \frac{2}{L} dy \text{ when } \frac{L}{2} \le z \le L$$
$$F(z) = \int_{L/2}^L \frac{2}{L} dy \text{ when } L \le z \le \frac{3L}{2}$$