 # Find the Laplace transform of the given functionbegin{cases}t & 0,4leq t<infty 0 & 4leq t<infty end{cases}Lleft{f(t)right} - ? remolatg 2020-11-22 Answered

Find the Laplace transform of the given function
$\left\{\begin{array}{ll}t& 0,4\le t<\mathrm{\infty }\\ 0& 4\le t<\mathrm{\infty }\end{array}$
$L\left\{f\left(t\right)\right\}-?$

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Step 1
The Laplace transform of a function is given as follows.
$L\left\{f\left(t\right)\right\}={\int }_{0}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt$
Here the given function is as follows.
$\left\{\begin{array}{ll}t& 0,4\le t<\mathrm{\infty }\\ 0& 4\le t<\mathrm{\infty }\end{array}$
Step 2
The Laplace transform of the given function is computed as follows.
$L\left\{f\left(t\right)\right\}={\int }_{0}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt$
$={\int }_{0}^{4}f\left(t\right){e}^{-st}dt+{\int }_{4}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt$
$={\int }_{0}^{4}t{e}^{-st}dt+{\int }_{4}^{\mathrm{\infty }}\left(0\right){e}^{-st}dt$
$={\int }_{0}^{4}t{e}^{-st}dt$
$=\frac{-t{e}^{-st}}{s}+{\int }_{0}^{4}\frac{{e}^{-st}}{s}\left[\begin{array}{cc}u=t& dv={e}^{-st}\\ du=1& v=\frac{-{e}^{-st}}{s}\\ uv-\int vdu\end{array}\right]$
$={\left[\frac{-t{e}^{-st}}{s}-\frac{{e}^{-st}}{{s}^{2}}\right]}_{0}^{4}$
$=\frac{-4{e}^{-4s}}{s}-\frac{{e}^{-4s}}{{s}^{2}}-\left[0-\frac{1}{{s}^{2}}\right]$
$=\frac{1-{e}^{-4s}}{{s}^{2}}-\frac{4{e}^{-4s}}{s}$