By definition, you have
[tex]\displaystyle(f*g)(t)=\int_{-\infty}^\infty f(\tau)g(t-\tau)\,\mathrm d\tau[/tex]
Take [tex]f(t)=u(t-1)[/tex] and [tex]g(t)=t^2[/tex]. Then
[tex]\displaystyle(f*g)(t)=\int_{-\infty}^\infty u(\tau-1)(t-\tau)^2\,\mathrm d\tau[/tex]
The Heaviside step function is defined as
[tex]u(t)=\begin{cases}0&\text{for }t<0\\1&\text{for }t\ge0\end{cases}[/tex]
and adjusting its argument by 1, you have
[tex]u(t-1)=\begin{cases}0&\text{for }t<1\\1&\text{for }t\ge1\end{cases}[/tex]
which means the integral reduces to
[tex](f*g)(t)=\displaystyle\int_1^\infty (t-\tau)^2\,\mathrm d\tau[/tex]
This integral doesn't converge, but hopefully it's a simple enough demonstration of how you might go about computing such an integral.
