Finding the Laplace transform of functions with finitely many jump discontinuities Find the Laplace transform of the given function.
$f(t)=\left\{
\begin{array}{lr}
       0, &  \hspace{2mm} t<2 \\
       (t-2)^2, & \hspace{2mm} t \ge 2\\
     \end{array}
   \right.$
My attempt
$f(t)=(t-2)^2u_2(t)=(t^2-4t+4)u_2(t)$, where $u_2(t)=\left\{
\begin{array}{lr}
       0, &  \hspace{2mm} t<2 \\
       1, & \hspace{2mm} t \ge 2\\
     \end{array}
   \right.$
$\mathcal{L}(f(t))=\mathcal{L}(u_2(t)(t^2-4t+4))$
Because we have a $u_2(t)$ out front, we want to express $t^2-4t+4$ as a function of $t-2$, so let $u=t-2 \Rightarrow t=u+2$.
$f(u+2)=(u+2)^2-4(u+2)+4=u^2=(t-2)^2$.
Thus, $\mathcal{L}(u_2(t)(t^2-4t+4))=\mathcal{L}(u_2(t)f(t-2))=e^{-2s}\mathcal{L}(t^2)=e^{-2s}(\frac{2}{s^3})=\frac{2e^{-2s}}{s^3},s>0$
However, my book says the correct answer is $F(s)=\frac{2e^{-s}}{s^3}$
Would you please let me know what I'm missing here thanks!

You already started with $$f(t) = (t-2)^2 u_2(t)$$ Why go through all that extra error prone algebra?
Let's use the definition of the Laplace Transform to find the result and compare $$ \int_2^{\infty } \frac{(t-2)^2}{e^{s t}} \, dt = \frac{2 e^{-2 s}}{s^3}$$
Let's also use the Laplace Transform Table approach, items $3$ and $27$.
The Laplace transform of $u_c(t) f(t-c) = e^{-c s}F(s)$ and of $f(t-c) = \dfrac{n!}{s^{n+1}}$.
Putting it all together
$$e^{-2s} \dfrac{2!}{s^{2+1} } = \dfrac{2 e^{-2s}}{s^3} $$
Lastly, let's use WA.
It appears there is a typo in the book.
Great work!
