After Laplace Transformations, how do initial conditions come into play?

Question

The Laplace transform is relatively simple to derive but my solution was wrong. I am positive it is because of the initial conditions.

Find the laplace transform

f(t)={0 for t < 3, (t-3)^2 for t greater than or equal to 3}

My solution was 2/(s^3)-6/(s^2)+9/s

Answer 1

As you will no doubt be aware:

$$\mathcal{L}[\operatorname{f}](s) = \int_0^{\infty} \operatorname{f}(t)\operatorname{e}^{-st} \, \operatorname{d}\!t $$

Since $\operatorname{f}(t) = 0$ for all $t<3$ and $\operatorname{f}(t) = (t-3)^2$ for all $t \ge 3$ we have

$$\begin{eqnarray*} \int_0^{\infty} \operatorname{f}(t)\operatorname{e}^{-st}\, \operatorname{d}\!t &=& \int_0^3 0\operatorname{e}^{-st} \, \operatorname{d}\!t +\int_3^{\infty}(t-3)^2\operatorname{e}^{-st} \, \operatorname{d}\!t \\ \\ \\ &=& \int_3^{\infty}(t-3)^2 \, \operatorname{e}^{-st} \, \operatorname{d}\!t \\ \\ \\ &=& \frac{2}{s^3}\operatorname{e}^{-3s}\end{eqnarray*}$$