Answer
$$F\left( s \right) = \frac{2}{{{s^3}}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( t \right) = {t^2} \cr
& {\text{Find the Laplace Transform of the function }} \cr
& F\left( s \right) = \int_0^\infty {{e^{ - st}}f\left( t \right)} dt, \cr
& {\text{Substitute }}f\left( t \right) = {t^2} \cr
& F\left( s \right) = \int_0^\infty {{e^{ - st}}\left( t \right)} dt \cr
& F\left( s \right) = \int_0^\infty {{t^2}{e^{ - st}}} dt \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ then}} \cr
& F\left( s \right) = \mathop {\lim }\limits_{b \to \infty } \int_0^b {{t^2}{e^{ - st}}} dt \cr
& {\text{Integrating by tabulation we obtain}} \cr
& F\left( s \right) = \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{s}{t^2}{e^{ - st}} - \frac{2}{{{s^2}}}t{e^{ - st}} - \frac{2}{{{s^3}}}{e^{ - st}}} \right]_0^b \cr
& F\left( s \right) = \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{1}{s}{b^2}{e^{ - sb}} - \frac{2}{{{s^2}}}b{e^{ - sb}} - \frac{2}{{{s^3}}}{e^{ - sb}}} \right] - \mathop {\lim }\limits_{b \to \infty } \left[ {0 - \frac{2}{{{s^3}}}{e^0}} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty ,{\text{ }}s > 0 \cr
& F\left( s \right) = \left[ { - \frac{1}{s}\left( \infty \right){e^{ - \infty }} - \frac{2}{{{s^2}}}\left( \infty \right){e^{ - \infty }} - \frac{2}{{{s^3}}}{e^{ - \infty }}} \right] - \left[ {0 - \frac{2}{{{s^3}}}{e^0}} \right] \cr
& F\left( s \right) = \left[ 0 \right] - \left[ {0 - \frac{2}{{{s^3}}}} \right] \cr
& F\left( s \right) = \frac{2}{{{s^3}}} \cr} $$
