$$f(t) = t^2 , t>=1$$ $$f(t) = 0, 0<t<1$$ what is the laplace transform of $f(t)$, It is solved In my sheet as
$$t^2 = (t-1)^2 + 2t -1 $$ $$ L(f(t)) = e^{-s} L(t^2 +2t +1) = e^{-s}(2/s^3 + 2/s^2 + 1/s) $$
why it does not use the simple rule $L(t^n)=n!/s^{n+1}$
$$f(t)=t^2H(t-1)=\left[(t-1)^2+2(t-1)+1\right]H(t-1)$$ $$\mathcal L \{f(t)\}=\mathcal L \left\{\left[(t-1)^2+2(t-1)+1\right]H(t-1)\right\}=\mathrm e^{-s}\left[\frac{2}{s^3}+2\cdot\frac{1}{s^2}+\frac{1}{s}\right]$$