Finding Interval of Integration If we let :
$$
x(t)=\begin{cases}
1&amp;\text{if $0&lt;t&lt;1$}\\
0&amp;\text{if otherwise}
\end{cases}
$$
and
$$
h(t)=x(t/a)=\begin{cases}
1&amp;\text{if $0&lt;t&lt;a$}\\
0&amp;\text{if otherwise}\end{cases}
$$
where $0&lt;a\leq 1$
I wish to find convolution of $(x*h)(t)$ without graphing. We have that :
$$
\max(0,t-a)&lt;\tau&lt;\min(1,t)
$$

Is the following assertion correct?
$$ 
\max(0,t-a)=\begin{cases}
0&amp;\text{if $0&lt;t&lt;a$}\\
t-a&amp;\text{if $a&lt;t&lt;1$}
\end{cases}
\quad\text{and}\quad
\min(t,1)=\begin{cases}
1&amp;\text{if $1&lt;t&lt;1+a$}\\
t&amp;\text{if $0&lt;t&lt;1$}
\end{cases}
$$
By definition:
$$x(t)*h(t)=\int_{-\infty}^\infty x(\tau)h(t-\tau) \ \mathrm{d}\tau$$
If any of the functions inside the integral equals 0, then the product does as well. We know that:
$$x(\tau)\neq0  \iff \tau\in(0,1)$$
$$h(t-\tau)\neq0  \iff t-\tau\in(0,a) \iff \tau \in (t-a,t)$$
So the convolution integral can be simplified:
$$x(t)*h(t)=\int_{-\infty}^\infty x(\tau)h(t-\tau) \ \mathrm{d}\tau
= \int_{\max(0, t-a)}^{\min(1,t)} x(\tau)h(t-\tau) \ \mathrm{d}\tau
$$
Just as you had found out.
Notice that this works just because $0&lt;a\leq 1$. If that was not the case, then some other things should be taken into account.