Convolution of two independent uniform variables Question: Let's say $X$ and $Y$ are two independent random variables with $Uniform (0,3)$ distribution. What is the probability density function of $X+Y$?
In solution, it is told to use convolution method by many sources since the pdf of the sum of two independent random variables $(X+Y)$ is the convolution of the individual pdfs.
However, I couldn't solve this problem by looking at $Uniform (0,1)$ version solutions of this question. What I've done by now is; 
$fx+y(a) = ∫_0^3 fx(a-y)*fy(y)dy$

$= 1/3 *∫_0^3 fx(a-y)dy$

If what I've found until here is true (I have doubts about upper and lower limits of the integral), how should I determine the limits of pdf of X+Y? Some sources assume $a=.5$ and $a=1.5$ and solve it that way; while other sources assume $0<a<1$ and $1<a<2$ and solve it that way. In both solutions, they don't explain how they find upper and lower limit of the integral. Fully solve the problem given above to clearly explain.
 A: Let $Z = X + Y$, then probability distributionn function of $Z$ is defined as
$$
\begin{aligned}
F_Z(z) &= \text{Pr}(Z \leq z) = \text{Pr}(X + Y \leq z) = \text{Pr}(X \leq z - Y) = \int\limits_{-\infty}^{+\infty}F_X(z-y)f_Y(y)dy = \\
&= \frac{1}{3}\int\limits_{0}^{3}F_X(z-y)dy = 
\left|
\begin{array}{l}
u = z-y \Leftrightarrow du = -dy \\
y = 0 \Leftrightarrow u = z \\
y = 3 \Leftrightarrow u = z-3
\end{array}
\right| = \frac{1}{3}\int\limits_{z-3}^zF_X(u)du = \\ 
& = \left\{
\begin{array}{rl}
0, & z \leq 0 \\
\frac{1}{3}\int\limits_{0}^z\frac{u}{3}du = \frac{z^2}{18}, & 0 < z \leq 3 \\
\frac{1}{3}\int\limits_{z-3}^3\frac{u}{3}du + \frac{1}{3}\int\limits_{3}^zdu=  
\frac{1}{2}-\frac{(z-3)^2}{18} + \frac{z-3}{3} = -\frac{z^2-12z+18}{18}, & 3 < z \leq 6 \\
1, & z > 6
\end{array}
\right.
\end{aligned}
$$
So, we have that
$$
F_Z(z)=\left\{
\begin{array}{rl}
0, & z \leq 0 \\
\frac{z^2}{18}, & 0 < z \leq 3 \\
-\frac{z^2-12z+18}{18}, & 3 < z \leq 6 \\
1, & z > 6
\end{array}
\right. \Rightarrow f_Z(z) = F_Z'(z) = 
\left\{\begin{array}{rl}
0, & z \leq 0 \\
\frac{z}{9}, & 0 < z \leq 3 \\
-\frac{z-6}{9}, & 3 < z \leq 6 \\
0, & z > 6
\end{array}
\right.
$$
