Convolution, sum of two independent random variables $X+Y$ 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?
 A: I think you are correct. An alternate proof (with many steps omitted for concision) to sanity check:
\begin{align}
f_Z(z)&=\int_0^1f_Y(z-x)f_X(x)dx\tag{1}
\end{align}
For $0\leq z \leq1$,
$$
f_Z(z)=\int_{0}^{z}2xdx=z^2\tag{2}.
$$
For $1<z\leq 2$,
$$
f_Z(z)=\int_{z-1}^{1}2xdx=2z-z^2\tag{3}.
$$
As such,
$$
f_Z(z)=\left\{
\begin{aligned} 
z^2 &\ \ \ \text{ for }0\leq z \leq 1,\\ 
2z-z^2&\ \ \ \text{ for }1 < z \leq 2,\\ 
0 &\ \ \ \text{ otherwise. }
\end{aligned} 
\right. 
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{{\displaystyle #1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Hereafter, $\ds{\bracks{\cdots}}$ is an Iverson's Bracket.
The $\ds{\underline{answer}}$ is given by
\begin{align}
& \color{#44f}{\int_{0}^{1}2x\bracks{0 < z - x < 1}\dd x} =
2\int_{0}^{1}x\bracks{z - 1 < x < z}\dd x
\\[5mm] = & \
2\bracks{0 < z < 1}\int_{0}^{z}x\,\dd x\quad +\quad
2\bracks{0 < z - 1 < 1}\int_{z - 1}^{1}x\,\dd x
\\[5mm] = & \
\bracks{0 < z < 1}z^{2}\quad +\quad
\bracks{1 < z < 2}\pars{2z - z^{2}}
\\[5mm] = & \
\bbx{\color{#44f}{ \color{black}{=}
\left\{\begin{array}{lcl}
\ds{0} & \mbox{if} & \ds{z \leq 0\ \mbox{or}\ z \geq 2}
\\[2mm]
\ds{z^{2}} & \mbox{if} & \ds{0 < z \leq 1}
\\[2mm]
\ds{2z - z^{2}} & \mbox{if} & \ds{1 < z < 2}
\end{array}\right.}}
\\[5mm] & \
\pars{~z = 0, 1, 2~}\mbox{-results}\ \mbox{are found as}\ limiting\ cases\
\mbox{in this approach.}
\end{align}

