JEE Main 2003Annual, Paper 1MathematicsEasyIntegral Calculus369× since 2002If f(y)=ey,f\left( y \right) = {e^y},f(y)=ey, g(y)=y;y>0g\left( y \right) = y;y > 0g(y)=y;y>0 and F(t)=∫0tf(t−y)g(y)dy,F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,}F(t)=0∫tf(t−y)g(y)dy, then :AOption A. F(t)=te−tF\left( t \right) = t{e^{ - t}}F(t)=te−tBOption B. F(t)=1t−te−1(1+t)F\left( t \right) = 1t - t{e^{ - 1}}\left( {1 + t} \right)F(t)=1t−te−1(1+t)COption C. F(t)=et−(1+t)F\left( t \right) = {e^t} - \left( {1 + t} \right)F(t)=et−(1+t)DOption D. F(t)=tetF\left( t \right) = t{e^t}F(t)=tet.CheckSkip · show solution
