Meaning of limits, $\int_{\max(0, t-5)}^{\max(0,t-3)} e^{-3s} \, ds $?

Question

What does it mean to have $\max(0,t-3)$ and $\max(0,t-5)$ in the limits? Is it a abbreviation? $$ \int_{t-5}^{t-3} e^{-3s}u(s) \, ds = \int_{\max(0, t-5)}^{\max(0,t-3)} e^{-3s} \, ds $$

Source of question:

Calculate the convolution $y(t)=(x\ast h)(t)$ where $x(t)=u(t-3)-u(t-5)$ and $h(t)=e^{-3t}u(t)$. \begin{align} y=\cdots &=\int_{t-5}^{t-3} e^{-3s}u(s) \, ds = \int_{\max(0, t-5)}^{\max(0,t-3)} e^{-3s} \, ds \\ &= \left[-\frac{1}{3}e^{-3s}\right]_{\max(0, t-5)}^{\max(0,t-3)} \\ &= \frac{1}{3}(e^{-3\max(0,t-5)}-e^{-3\max(0,t-3)}) \end{align}

Answer 1

Well, $\max(a,b)$ is simply a function that returns the largest value of $a$ and $b$.

So $\max(0, t-3)$ returns the largest value of $0$ and $t-3$, which is $0$ if $t\leq 3$ and $t-3$ if $t>3$, or put in symbols $$\max(0,t-3) = \begin{cases} 0&\mbox{if }\, t\leq 3\\ t-3&\mbox{if }\, t > 3\end{cases}. $$