Help to solve this integral :) $$ g(t) = A \int_{-\infty}^{+\infty} h(\omega) f(t-\omega) d\omega$$

$$h(t) = \left\{\begin{array}{ll} A &amp; 0\leq t\leq {T\over2}\\ 0 &amp; \mbox{otherwise} \end{array}\right.     $$

$$f(t) = \left\{\begin{array}{ll} A &amp; {T\over2}\leq t\leq T \\ 0 &amp; \mbox{otherwise} \end{array}\right.     $$

I didn't find a way to solve it :(, please help
The given expression is the convolution between $h(t)$ and $f(t)$ 

$$
 g(t) = \left\{
\begin{array}{ll} 0 &amp; t\leq {T\over2}  \ \mbox{or}\  t\geq {3T\over2}\\
A^3\left(t-{T\over2}\right) &amp; {T\over2}\leq t\leq {T}\\ 
A^3\left({3T\over2}-t\right) &amp; {T}\leq t\leq {3T\over2} \end{array} \right.
$$
A correct result has already been given; the following just examines some of the calculations in detail.

The expression $f(t - \omega)$ in the integrand is non-zero precisely when
$$\frac T 2 \le t - \omega \le T.$$
Multiplying each part by $-1$ reverses the direction of the inequalities:
$$-T \le \omega - t \le -\frac T 2,$$
so
$$t - T \le \omega \le t - \frac T 2.$$

So this part of the integrand is non-zero just when $\omega$ is in the interval
$\left[t - T, t - \frac T 2\right].$
The other part of the integrand is non-zero when $\omega$ is in 
$\left[0, \frac T 2\right].$
The three cases arise from whether one interval overlaps the other on the left, on the right, or not at all.