Laplace transform of $(t-2)^2u_2$

Question

The homework problem is $$f(t) = \begin{cases} 0 & t < 2\\ (t-2)^2 & t\geq 2\end{cases}$$ $f(t)$ as a step function $$f(t) = (t-2)^2u_2(t)$$ Using what we learned in class $$\mathcal{L}\{f(t)u_c(t)\} = e^{-cs}\mathcal{L}\{f(t+c)\}$$ $\mathcal{L}\{f(t)\}$ should be $$e^{-2s} \mathcal{L}\{f(t+c)\}= e^{-2s} \frac2{s^3}$$ but the answer in the back of the book is $$e^{-s} \frac2{s^3}$$

I can't figure out why the answer does not include the 2 in the exponent.

Answer 1

Just, work out the problem from scratch

$$ \int_{0}^{\infty}f(t)e^{-st}dt = \int_{0}^{\infty}(t-2)^2u_2(t)\,e^{-st} dt = \int_{2}^{\infty}(t-2)^2e^{-st} dt$$

$$ = \int_{0}^{\infty}u^2e^{-s(u+2)} du= e^{-2s}\frac{2!}{s^3}, $$

which matches your answer.

Answer 2

I figured out how to verify it with mathematica and I have concluded the answer in the textbook is wrong.

LaplaceTransform[Piecewise[{{0, t < 2}, {(t - 2)^2, t >= 2}}], t, s]