# For the function f(t)=e^t g(t)=e^{-2t} 0leq t < infty compute in two different ways: a) By directly evaluating the integral in the defination of f cdot g b) By computing L^{-1}left{F(s)G(s)right} text{ where } F(s)=Lleft{f(t)right} text{ and } G(s)=Lleft{g(t)right}

For the function
$f\left(t\right)={e}^{t}$
$g\left(t\right)={e}^{-2t}$
$0\le t<\mathrm{\infty }$
compute in two different ways:
a) By directly evaluating the integral in the defination of $f\cdot g$
b) By computing
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$\text{Then the convolution of f with g is given below:}$
$\left(f\ast g\right)\left(t\right)={\int }_{0}^{t}f\left(\tau \right)g\left(t-\tau \right)d\tau$
$={\int }_{0}^{t}\left({e}^{w}\right)\left({e}^{-2\left(t-w\right)}\right)dw$
${\int }_{0}^{t}\left({e}^{3w}\right)\left({e}^{-2w}\right)dw$
$=\frac{1}{3}{e}^{-2t}{\left[{e}^{3w}\right]}_{0}^{t}$
$=\frac{1}{3}{e}^{-2t}\left[{e}^{3t}-1\right]$
$\text{Step 2}$

$\left(f\ast g\right)\left(t\right)={L}^{-1}\left\{F\left(s\right)G\left(s\right)\right\}$
$={L}^{-1}\left\{\frac{1}{s-1}×\frac{1}{s+2}\right\}$
$={L}^{-1}\left\{\frac{1}{s-1}\right\}\ast {L}^{-1}\left\{\frac{1}{s+2}\right\}$
$={e}^{t}\ast {e}^{-2t}$
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