Finding Interval of Integration
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.