Q. If $f(y) = e^{y},\ g(y) = y;\ \ \ y > 0$ and $F(t) = \int_{0}^{t}{f(t - y)g(y)}\ dy$, then
A) $F(t) = 1 - e^{- t}(1 + t)$
B) $F(t) = e^{t} - (1 + t)$
C) $F(t) = te^{t}$
D) $F(t) = te^{- t}$
Explanation
(b)
$F(t) = \int_{0}^{1}{f(t - y)g(y)dy}$
$= \int_{0}^{t}{e^{t - y}.ydy = e^{t}\int_{0}^{t}e^{- y}ydy}$
$= e^{t}\lbrack\left( - ye^{- y} \right)_{0}^{t} - \left( e^{- y} \right)_{0}^{t}\rbrack$
$= e^{t}\left\lbrack - te^{- t} - e^{- t} + 1 \right\rbrack = e^{t} - (1 + t)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.