Elleanor Mckenzie

2020-12-27

Let f(t) be a function on $\left[0,\mathrm{\infty }\right)$. The Laplace transform of fis the function F defined by the integral $F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$ . Use this definition to determine the Laplace transform of the following function.
$f\left(t\right)=\left\{\begin{array}{cc}1-t& 0

grbavit

Step 1
We know that , $F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
Here $f\left(t\right)=\left\{\begin{array}{cc}1-t& 0
from $F\left(s\right)={\int }_{0}^{\mathrm{\infty }}\left(f\left(t\right)\right){e}^{-st}dt={\int }_{0}^{1}\left(1-t\right){e}^{-st}dt+{\int }_{1}^{\mathrm{\infty }}0\cdot {e}^{-st}dt$
$={\int }_{0}^{1}\left(1-t\right){e}^{-st}dt+0$
$F\left(s\right)={\left[\frac{\left(1-t\right){e}^{-st}}{-s}-\frac{\left(-1\right){e}^{-st}}{{s}^{2}}\right]}_{0}^{1}$
$F\left(s\right)=\frac{1}{s}+\frac{{e}^{-s}}{s}-\frac{1}{{s}^{2}}$
Step 2
This is required Laplace Transform.

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