Question

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

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

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