Question

$\mathit{L}\left\{\underset{0}{\overset{t}{\int }}{e}^{aT}f\left(T\right)dT\right\}\mathbf{=}\underset{\mathbf{¯}}{}$ whereas$\mathit{L}\left\{e^{at}\underset{0}{\overset{t}{\int }}f\left(T\right)dT\right\}\mathbf{=}\underset{\mathbf{¯}}{}$

In Problems 25–28 use the unit step function to and an equation for each graph in terms of the function , whose graph is given in figure 7.R.1

Short answer

The solution is

$f_1\left(t\right)=f\left(t\right)U\left(t-t_0\right)$

Step-by-step solution

given information

The given value is:

$L\left\{\underset{0}{\overset{t}{\int }}{e}^{aT}f\left(T\right)dT\right\}=\underset{¯}{}$ whereas

$L\left\{e^{at}\underset{0}{\overset{t}{\int }}f\left(T\right)dT\right\}=\underset{¯}{}$

find graph

Equation for given graph is

$$\begin{gathered} f_1\left(t\right)=\left\{0,0⩽t<t_0 \\ f\left(t\right),t⩾t_0\right. \end{gathered}$$

A piecewise function as

$$\begin{gathered} f\left(t\right)=\left\{g\left(t\right),0⩽t<a \\ h\left(t\right),t⩾a\right. \end{gathered}$$

Can be expressed as:

$f\left(t\right)=g\left(t\right)-g\left(t\right)U(t-a)+h\left(t\right)U(t-a)$

The given function$f_1\left(t\right)$can be expressed as:

$f_1\left(t\right)=f\left(t\right)U\left(t-t_0\right)$

The final solution is

$f_1\left(t\right)=f\left(t\right)U\left(t-t_0\right)$