Question

If $f\left(y\right)=e^y,g\left(y\right)=y;y>0$ and $F\left(t\right)={\int }_0^{t}f(t-y)g\left(y\right)dy$, then which of the following is true?

• 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 $F\left(t\right)=e^t-(1+t)$

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

Since, ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

Here we replace $y$ by $t-y$, so we get:-

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

It is given that $f\left(y\right)=e^y$ and $g\left(y\right)=y$ for $y>0$

Hence, we get:-

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

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

$=t{\left[e^y\right]}_0^t-{[y{e}^y-e^y]}_0^t$

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

Hence, answer is option-(A).