Answer
$\frac{8}{s^3}$
Work Step by Step
Given: $f(t)=4t^2$
Obtain: $F(s)=\int^{\infty}_0 e^{-st}f(t)dt\\
=\int^{\infty}_0 e^{-st} (4t^2)dt\\
=4\int^{\infty}_0 t^2e^{-st}dt\\
=4\lim \{e^{-st}[\frac{t^2}{s}+\frac{2t}{s^2}+\frac{2}{s^3}]^n_0\\
=3\lim [\frac{2}{s^3}-e^{-sn}(\frac{n^2}{s}+\frac{2n}{s^2}+\frac{2}{s^3})]\\
=\frac{8}{s^3}$
