Question

If $f\left(y\right)=e^y$ and $g\left(y\right)=y,y>0$ and $F\left(t\right)={\int }_0^{t}f(t-y)g\left(y\right)dy$ then

• A. $F\left(t\right)=e^t-(1+t)$
• B. $F\left(t\right)=t{e}^t$
• C. $F\left(t\right)=t{e}^{-t}$
• D. $F\left(t\right)=1-e^t(1+t)$

Answer 1

Answer: (A)

$F\left(t\right)=e^t-(1+t)$

Given $F\left(t\right)={\int }_0^{1}f(t-y)g\left(y\right)dy$

$={\int }_0^t{e}^{t-y}ydy$ (By replacing $y\to t-y$ in $f\left(y\right))$
$F\left(t\right)=-{\int }_t^0(t-\theta )e^{\theta }d\theta ={\int }_0^t(t-\theta )e^{\theta }d\theta$
$={\left(t{e}^{\theta }\right)}_0^t-{\left[(\theta -1)e^{\theta }\right]}_0^t$
$=t(e^t-1)-(t-1)e^t-1$
$=e^t(t-t+1)-t-1$
$=e^t-(t+1)$