# Determine the Laplace transform of f(t) below: f(t)={(0",",if t<2),((t-2)^2 ",", if t >=2):}

Determine the Laplace transform of f(t) below:

$2{e}^{-2s}/{s}^{3}$
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With problems of this form, we can simplify the integration by recognizing the derivative.
$\begin{array}{rl}{\int }_{2}^{\mathrm{\infty }}\left(t-2{\right)}^{2}{e}^{-st}dt& ={\int }_{2}^{\mathrm{\infty }}{t}^{2}{e}^{-st}dt-4{\int }_{2}^{\mathrm{\infty }}t{e}^{-st}dt+4{\int }_{2}^{\mathrm{\infty }}{e}^{-st}dt\\ & =\frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }{s}^{2}}{\int }_{2}^{\mathrm{\infty }}{e}^{-st}dt+4\frac{\mathrm{\partial }}{\mathrm{\partial }s}{\int }_{2}^{\mathrm{\infty }}{e}^{-st}dt+4{\int }_{2}^{\mathrm{\infty }}{e}^{-st}dt\\ & =\frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }{s}^{2}}\frac{{e}^{-2s}}{s}+\frac{\mathrm{\partial }}{\mathrm{\partial }s}\frac{4{e}^{-2s}}{s}+\frac{4{e}^{-2s}}{s}\end{array}$
From this, we obtain the answer
###### Did you like this example?
duandaTed05
Given the Laplace transform as
$\begin{array}{r}f\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt\end{array}$
then for

it is seen that
$\begin{array}{rl}f\left(s\right)& ={\int }_{0}^{2}{e}^{-st}\phantom{\rule{thinmathspace}{0ex}}\left(0\right)\phantom{\rule{thinmathspace}{0ex}}dt+{\int }_{2}^{\mathrm{\infty }}{e}^{-st}\left(t-2{\right)}^{2}\phantom{\rule{thinmathspace}{0ex}}dt\\ & =0+{\int }_{0}^{\mathrm{\infty }}{e}^{-s\left(u+2\right)}{u}^{2}\phantom{\rule{thinmathspace}{0ex}}du\\ & =\frac{2{e}^{-2s}}{{s}^{3}}.\end{array}$