The main theme of this post is painted with red below. Please scroll down a bit to see it.
$$X,Y:=\text{2 independent continuous random variables}\tag{1}$$
$$\text{These random variables follow the below probabilty density. }\tag{2}$$
$$f_{XY}(x,y)=\begin{cases}\frac{1}{9}&&(0\leq x\leq 3~~0\leq y\leq 3)\\0&&(\text{otherwise})\end{cases}\tag{3}$$
$$T:=X+Y\tag{4}$$
$$f_{XY}(x,y)=\underbrace{f_{X}(x)f_{Y}(y)}_{\text{since independent}}\tag{5}$$
$$\int_{0}^{3}f_{X}(x)\,\mathrm{dx}=1\tag{6}$$
$$=\int_{0}^{3}\left(\int_{-\infty}^{\infty}f_{XY}(x,y)\,\mathrm{dy}\right)\,\mathrm{dx}\tag{7}$$
$$=\int_{0}^{3}\left(\int_{0}^{3}f_{XY}(x,y)\,\mathrm{dy}\right)\,\mathrm{dx}=1~~\leftarrow~~\text{total probability}\tag{8}$$
$$f_{X}(x)=\int_{-\infty}^{\infty}f_{XY}(x,y)\,\mathrm{dy}~~~~\text{where}~~0\leq x\leq 3~~\text{is satisfied}\tag{9}$$
$$=\int_{0}^{3}f_{XY}(x,y)\,\mathrm{dy}\tag{10}$$
$$=\int_{0}^{3}\frac{1}{9}\,\mathrm{dy}\tag{11}$$
$$=\frac{1}{9}\int_{0}^{3}\,\mathrm{dy}=\frac{1}{3}\tag{12}$$
$$\therefore~~f_{X}(x)=\frac{1}{3}\underbrace{=f_{Y}(y)}_{\because~\text{symmetry}}\tag{13}$$
$$t=x+y\tag{14}$$
$$0\leq t\leq6~~\text{is held since}~~0\leq x,y\leq 3~~\text{should be satisfied}\tag{15}$$
$$f_{T}(t):=\underbrace{\int_{-\infty}^{\infty}f_{X}(x)\cdot f_{Y}(t-x)\,\mathrm{dx}}_{\text{formula of convolution}}\tag{16}$$
$$\text{As}~~0\leq t\leq 3~~\text{is held,}\tag{17}$$
$$=\underbrace{\color{red}{\int_{-\infty}^{0}0\,\mathrm{dx}+\int_{0}^{t}f_{X}(x)f_{Y}(t-x)\,\mathrm{dx}+\int_{t}^{\infty}0\,\mathrm{dx}}}_{\text{The current my problem}}\tag{18}$$
About the first term$~\int_{-\infty}^{0}0\,\mathrm{dx}~$, I can easiely get the meaning of it since the probability density function can only return$~\frac{1}{9}~$as only with$~0\leq x,y\leq 3~$
About the second term$~\int_{0}^{t}f_{X}(x)\cdot f_{Y}(t-x)\,\mathrm{dx}~$, it is also really clear for me since$~0\leq t\leq 3~$with$~y=t-x~$are satisfied.
The third term's range of integation$~[t,\infty]~$is really making me confused currently.
How should I interpret it?
For instance as$~t=2~$is set, then the range of$~[t,\infty]=[t,2]+(2+\infty]~$is held and$~[t,2]~$can generate$~f_{X}(x)=f_{Y}(t-x)=\frac{1}{3}~$so$~\int_{t}^{\infty}0\,\mathrm{dx}~$seems wrong.
Please see if the below helps explain. We have $f_Y(t-x) = \frac 13$ when $0 \lt t-x \lt 3$ and $0$ otherwise. We know support of $T$ is $0 \lt t \lt 6$.
$\begin{align} \text{So, } ~f_{T}(t) &= \int_{-\infty}^{\infty}f_{X}(x)\cdot f_{Y}(t-x)\,dx \\ &= \int_0^3f_{X}(x)\cdot f_{Y}(t-x)\,dx\\ &= \frac 19 \int_0^3~\mathbf 1_{0\lt t\lt 6~,~ t-3\lt x\lt t}~ dx \\ &= \mathbf 1_{0\lt t\le 3} \int_0^t \frac 19 ~dx ~ + ~ \mathbf 1_{3\lt t\lt 6} \int_{t-3}^3 \frac 19 ~dx\\ \end{align} $