Question

# Let f(t) be a function on. The Laplace transform of fis the function F defined by the integral . Use this definition to determine the Laplace transform of the following

Laplace transform

Let f(t) be a function on $$\displaystyle{\left[{0},\infty\right)}$$. The Laplace transform of fis the function F defined by the integral $$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}$$ . Use this definition to determine the Laplace transform of the following function.
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}$$

2020-12-28

Step 1
We know that , $$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}$$
Here $$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}$$
from $$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{\left( f{{\left({t}\right)}}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}={\int_{{0}}^{{1}}}{\left({1}-{t}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}+{\int_{{1}}^{\infty}}{0}\cdot{e}^{{-{s}{t}}}{\left.{d}{t}\right.}$$
$$\displaystyle={\int_{{0}}^{{1}}}{\left({1}-{t}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}+{0}$$
$$\displaystyle{F}{\left({s}\right)}={{\left[\frac{{{\left({1}-{t}\right)}{e}^{{-{s}{t}}}}}{{-{s}}}-\frac{{{\left(-{1}\right)}{e}^{{-{s}{t}}}}}{{{s}^{2}}}\right]}_{{0}}^{{1}}}$$
$$\displaystyle{F}{\left({s}\right)}=\frac{1}{{s}}+\frac{{{e}^{{-{s}}}}}{{s}}-\frac{1}{{s}^{2}}$$
Step 2
This is required Laplace Transform.