 # Compute the Laplace transform of f(t)={(8 ,if 0 <= x <14),(9t,if 14 <= x <infty):} chillywilly12a 2021-09-09 Answered

Compute the Laplace transform of

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Step 1
Consider the given information.

The Laplace transform is defined as,
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
Put the value in the integral.
$L\left\{f\right\}\left(s\right)={\int }_{0}^{14}{e}^{-st}8dt+{\int }_{14}^{\mathrm{\infty }}{e}^{-st}\left(9t\right)dt$
$=8{\int }_{0}^{14}{e}^{-st}dt+9{\int }_{14}^{\mathrm{\infty }}t{e}^{-st}dt$
Step 2
Consider the values for integration as,
$u=t,dv={e}^{-st}$
$du=1,v=\frac{{e}^{-st}}{-s}$
Put the values in the function.
$L\left\{f\right\}\left(s\right)=8{\int }_{0}^{14}{e}^{-st}+9{\int }_{14}^{\mathrm{\infty }}t{e}^{-st}$
$=8{\left[\frac{{e}^{-st}}{-s}\right]}_{0}^{14}+9{\left[\frac{t{e}^{-st}}{-s}\right]}_{14}^{\mathrm{\infty }}+\frac{1}{s}{\int }_{14}^{\mathrm{\infty }}{e}^{-st}dt$
$=-8\frac{\left({e}^{-14s}-1\right)}{s}-9{\left[\frac{t{e}^{-st}}{s}\right]}_{14}^{\mathrm{\infty }}-9\frac{1}{s}{\left[\frac{{e}^{-st}}{s}\right]}_{14}^{\mathrm{\infty }}$
$=-8\frac{\left({e}^{-14s}-1\right)}{s}+\frac{\left(126s+9\right){e}^{-14s}}{{s}^{2}}$
$=\frac{118{e}^{-14s}s+8s+9{e}^{-14s}}{{s}^{2}}$