Find the Laplace transform of the following: $$ f(t) = \begin{cases} 0, & t<1,\\ t^2-2t+2, & t \ge 1\end{cases} $$
My solution:
$ u_1f(t-1)=u_1(t)(t^2-2t+2) $
$ f(t-1)=t^2-2t+2$
$f(t)=(t+1)^2-2(t+1)+2=t^2+2t+1-t-2+2=t^2+1$
$ f(s)=Laplace(t^2+1)= 2/s^3+ 1/s $
Final answer:
$ e^{-s} \left( 2/s^3+1/s\right) $