Let $$ f_{(X,Y)}(x,y) = 2x $$ for $x \in (0,1), y \in (0,1)$. I need to compute density of $X+Y$. So, I know that $X \perp Y$, because \begin{align} f_X(x) &= 2x, \ \ x\in(0,1)\\ f_Y(x) &= 1, \ \ \ \ y \in (0,1) \end{align} thus $f_{(X,Y)}(x,y) = f_X(x)f_Y(y)$.
Let $Z=X+Y$ and $z\in(0,2)$, then $$ f_Z(z) = \int\limits_{-\infty}^{+\infty}f_Y(y)f_X(z-y)dy $$ Then, for $z\in(0,1)$ I need $z-y\geq 0$, so $z \geq y$ $$ f_Z(z) = \int\limits_{0}^{z}1\cdot2(z-y)dy = 2z^2-z^2 = z^2 $$ And for $z\in(1,2)$ inequality $z \geq y$ holds, but $z-y \leq 1$ $$ f_Z(z) = \int\limits_{z-1}^{1}1\cdot2(z-y)dy = 2z-z^2 $$ Thus $$ f_Z(z) = \begin{cases} z^2, \ \ \ \ \ \ \ \ \ \ z \in (0,1)\\ 2z-z^2, \ \ \ z \in (1,2)\\ 0, \ \ \ \ \ \ \ \ \ \ \ \ \text{elsewhere} \end{cases} $$ Am I correct?