# Given the function begin{cases}e^{-t}& text{if } 0leq t<2 0&text{if } 2leq tend{cases}Express f(t) in terms of the shifted unit step function u(t -a)F(t)

Given the function
Express f(t) in terms of the shifted unit step function u(t -a)
F(t) - ?
Now find the Laplace transform F(s) of f(t)
F(s) - ?

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Step 1
Since f(t)=0 for so create step function: $u\left(t\right)-u\left(t-2\right)$ which will give those zero values.
Thus,
Step 2
Calculate Laplace transform:
$F\left(s\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt$
$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left({e}^{-t}u\left(t\right)-{e}^{-t}u\left(t-2\right)\right){e}^{-st}dt$
$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left({e}^{-t}u\left(t\right)\right){e}^{-st}dt-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left({e}^{-t}u\left(t-2\right)\right){e}^{-st}dt$
$={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+1\right)t}dt-{\int }_{2}^{\mathrm{\infty }}{e}^{-\left(s+1\right)t}dt$
$=\frac{1}{\left(s+1\right)}-\frac{{e}^{-2\left(s+1\right)}}{\left(s+1\right)}$
$=\frac{1-{e}^{-2\left(s+1\right)}}{\left(s+1\right)}$
Step 3
Thus, Laplace transform is $F\left(s\right)=\frac{1-{e}^{-2\left(s+1\right)}}{\left(s+1\right)}$