Considering the following integral of the function $\rho(t-u)$:
\begin{equation} \int_0^L\int_0^L\rho(t-u)\,\mathrm{d}u\,\mathrm{d}t \end{equation}
Making the substitution $s = t - u$:
\begin{equation} \int_0^L\int_0^L\rho(t-u)\,\mathrm{d}u\,\mathrm{d}t = \int_0^L\int_{t-L}^t \rho(s)\,\mathrm{d}s\,\mathrm{d}t \end{equation}
My question is how are the new limits $s = t-L$ and $s = t$ obtained? I am unable to understand this.
$\begin{align}\int_{0\leqslant t\leqslant L}\int_{0\leqslant u\leqslant L} \rho(t-u)~\mathsf d u~\mathsf d t ~&=~ \int_{0\leqslant t\leqslant L}\int_{0\leqslant t-s\leqslant L} \rho(s)~\lvert\frac{\partial (t-s)}{\partial s}\rvert\mathsf ds~\mathsf dt && u\gets t-s\\[1ex]&=~\int_{0\leqslant t\leqslant L}\int_{t-L\leqslant s\leqslant t}\rho(s)~\mathsf d s~\mathsf dt\\[3ex]\therefore\qquad\int_0^L\int_0^L \rho(t-u)~\mathsf d u~\mathsf dt~&=~\int_0^L\int_{t-L}^t\rho(s)~\mathsf d s~\mathsf d t \end{align}$