I've come across this sort of exercises a few times now and I always feel so unsure about how to solve them. This is one example:
Determine the density function of $X + Y$ if $X$ and $Y$ are i.i.d random variables in $[-1,1]$.
This is my solution so far:
$f_X(x) = -\frac{1}{2}$ for $x$ that belongs to $[-1,1],$ $f_Y(y) = -\frac{1}{2}$ for $y$ that belongs to $[-1,1]$
$P(X+Y \leq u) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_X(x) f_Y(y) dxdy \Rightarrow$
$f_{X+Y}(u) = \int_{-\infty}^{\infty} f_X(u-y) f_Y(y) dy $
$X + Y$ belongs to $[-2, 2]$
Now, after this, I'm lost. I don't know what the limits of the integral should be, and this is where I always have problems when doing this kind of exercise. The solution manual says to put it as $\int_{u-1}^{1} \frac{1}{4} dy$, I don't "see" it.
Your equation $P(X+Y \leq u) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_X(x) f_Y(y) dxdy $ is wrong. The integral must be done inside an area that depends on $u$, where $X+Y\le u$ holds (pink region in the graphic)
In the range $0<u<1$ it's a little easier to compute the complementary probability (green zone). The integral is
$$P(X+Y \ge u)=\int_{Green} f_{X,Y}(x,y) dx dy = \int_{Green} \frac 12 \frac 12dx dy = \frac14 Area_{Green}=\frac14 \frac{(2-u)^2}{2}$$
I think you can go on from here.