Let f:R+→R be a differentiable function satisfying f(x)=e+(1-x)lnxe+∫1xf(t)dt, for all xεR+. If the area enclosed by the curve gx=xfx-ex lying in the fourth quadrant is A, then A=

f'x=0+1-x1x+lnxe-1+fx by newton's Leibnitz rule   ⇒f'x=1x-1+lnx-1-1+fx ⇒dydx-y=1x−ln⁡x ⇒y e-x=∫1x−ln⁡xe-xdx    solving of linear differential equation ⇒ye-x=e-xlogx+c   since f1=e⇒ee-1=0+c⇒c=1     y=logx+exf(x)=ex+ln⁡x  ⇒gx=xfx-ex=xlnxArea=A=∫01 xln⁡xdx=14