Answer
\[ = \frac{1}{{{s^2}}}\]
Work Step by Step
\[\begin{gathered}
f\,\left( t \right) = t\,\,\,\,\, \to \,\,\,\,f\,\left( s \right) = \frac{1}{{{s^2}}} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\mathcal{L}\,\,\,\left[ t \right]\left. { = \frac{{ - t{e^{ - st}}}}{s}} \right|_0^\infty + \frac{1}{s}\int_0^x {{e^{ - st}}dt} \hfill \\
\hfill \\
integrate\,\,and\,\,evaluate \hfill \\
\hfill \\
= \frac{1}{s}\mathcal{L}\,\,\,\left[ t \right] = \frac{1}{s}\,\left( {\frac{1}{s}} \right) \hfill \\
\hfill \\
solution \hfill \\
\hfill \\
= \frac{1}{{{s^2}}} \hfill \\
\end{gathered} \]
