Question regarding $\mathcal{L}\{t*\mathcal{U}(t-2)\}$

Question

I'm working on a problem for homework (* is multiplication not convolution): $\mathcal{L}\{t*\mathcal{U}(t-2)\}$

I understand that $\mathcal{L}\{(t-a)\mathcal{U}(t-a)\}=e^{-as}F(s)$

The first step of the solution shows that this becomes: $e^{-2s}*\mathcal{L}\{t+2\}$

Why do we take the Laplace transform of $t+2$ instead of just $t$?

Answer 1

Note $t=(t-2)+2$. Then,

$$\begin{align} \mathscr{L}\{tu(t-2)\}&=\mathscr{L}\{(t-2+2)u(t-2)\}\\ &=\mathscr{L}\{(t-2)u(t-2)\}+2\mathscr{L}\{u(t-2)\} \end{align}$$

Can you finish?