Question

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

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

Answer 1

Answer: (C)

$F\left(t\right)=\underset{0}{\overset{t}{\int }}f\left(t-y\right)g\left(y\right)dy$
$=\underset{0}{\overset{t}{\int }}{e}^{t-y}ydy=e^t\underset{0}{\overset{t}{\int }}{e}^{-y}ydy$
$=e^t{\left[-y{e}^{-y}-e^{-y}\right]}_0^t=-e^t{\left[y{e}^{-y}+e^{-y}\right]}_0^t$
$=-e^t\left[t{e}^{-t}+e^{-t}-0-1\right]=-e^t\left[\frac{t+1-e^t}{e^t}\right]$
$=e^t-\left(1+t\right)$