Question

In Problems 1 and 2, use the definition of the Laplace transform to determine $\mathit{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right\}}$.

$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{,}\mathbf{0}\mathbf{\le }\mathbf{t}\mathbf{\le }\mathbf{5}\\ \mathbf{-}\mathbf{1}\mathbf{,}\mathbf{5}\mathbf{<}\mathbf{t}\end{array}$

Short answer

$L\left\{f\right(t\left)\right\}\left(s\right)=\frac{1}{s+1}(1-e^{-5(s+1)})-\frac{1}{s}{e}^{-5s}$

Step-by-step solution

Step 1:Given Information

The given function is $f\left(t\right)=\{\begin{array}{l}{e}^{-t},0\le t\le 5\\ -1,5<t\end{array}$.

Determining the

Using the Laplace transform definition, we get

$\begin{array}{c}L\left\{f\right(t\left)\right\}\left(s\right)={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt\\ ={\int }_0^5{e}^{-st}{e}^{-t}dt+{\int }_5^{\infty }(-1)e^{-st}dt\\ ={\int }_0^5{e}^{-t(s+1)}dt-\underset{N\to \infty }{\lim }{\int }_5^N{e}^{-st}dt\end{array}$

Simplify further as:

$\begin{array}{c}L\left\{f\right(t\left)\right\}={[-\frac{e^{-t(s+1)}}{s+1}]}_0^5-\underset{N\to \infty }{\lim }{[-\frac{e^{-st}}{s}]}_5^N\\ =\frac{1}{s+1}(1-e^{-5(s+1)})-\underset{N\to \infty }{\lim }\left(\frac{1}{s}(e^{-5s}-e^{-Ns})\right)\\ =\frac{1}{s+1}(1-e^{-5(s+1)})-\frac{1}{s}{e}^{-5s}\end{array}$

Determining the Result

Thus, the required Laplace transform is$L\left\{f\right(t\left)\right\}\left(s\right)=\frac{1}{s+1}(1-e^{-5(s+1)})-\frac{1}{s}{e}^{-5s}$