Question

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

The correct option is A $F\left(t\right)=e^t-(1+t)$

We have $f\left(y\right)=e^y,g\left(y\right)=y:y>0$

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

$={\int }_0^t{e}^{t-y}ydy$

$=e^t{\int }_0^t{e}^{-y}ydy$

using by-parts theorem

$=e^t({[-y{e}^{-y}]}_0^t+{\int }_0^t{e}^{-y}dy)$$=e^t(-t{e}^{-t}-{\left[e^{-y}\right]}_0^t)$

$=e^t(-t{e}^{-t}-e^{-t}+1)$

$=e^t-(1+t)$