$X$ and $Y$ are independent and uniformly distributed over $[-1,1]$.
I was thinking about splitting the problem in 2:
$ Z = X+Y $ and $V = Z^2$
The density for the second one I can find by using:
$ F_v(v) = P(V \leq v) = P ( Z^2 \leq v) = F_z(\sqrt{v} ) - F_z(- \sqrt{v}) + P(Z=- \sqrt{v}) $
and taking the derivative.
But I have some problems finding the correct $F_z$ and $f_z$. I think the bound for $z$ will be $[-2,2]$, but I don't know how to correctly calculate the integrals with the right bounds. Can anybody help?
I know the final solution is $ f_v(v) = \sqrt{v} -v/4$
We find the density of the sum by pdf convolution: $$4\cdot f_{X+Y}(z)=\int_{\mathbb{R}}\mathbb{I}_{[-1,1]}(z-y)\mathbb{I}_{[-1,1]}(y)dy=\int_{[-1,1]}\mathbb{I}_{[-1,1]}(z-y)dy=\\ =\int_{[-1,1]}\mathbb{I}_{[z-1,z+1]}(y)dy$$ In $z \in [-2,0]$ $$\int_{[-1,z+1]}dy=(z+1)+1=z+2$$ In $z \in (0,2]$ $$\int_{[z-1,1]}dy=1-(z-1)=2-z$$ Therefore $$f_{X+Y}(z)=\begin{cases} \frac{z+2}{4} & z \in [-2,0] \\ \frac{2-z}{4} & z \in (0,2] \end{cases}$$ Now set $X+Y=Z$ $$P_Z(Z^2\leq u)=P_Z(Z \in [-\sqrt{u},\sqrt{u}])=\int_{[-\sqrt{u},\sqrt{u}]}f_{X+Y}(z)dz=\\ =\int_{[-\sqrt{u},0]}\frac{z+2}{4}dz+\int_{(0,\sqrt{u}]}\frac{-z+2}{4}dz=$$ $$=\frac{1}{4}\bigg(\int_{[-\sqrt{u},0]}(z+2)dz+\int_{(0,\sqrt{u}]}(-z+2)dz\bigg)= \\ =\frac{1}{4}\bigg(-\frac{u}{2}+2\sqrt{u}-\frac{u}{2}+2\sqrt{u}\bigg)=\sqrt{u}-\frac{u}{4} $$