$\text{ Let }f:R^{+}\to R\text{ be a differentiable function satisfying }$$f\left(x\right)=e+(1-x)\ln \left(\frac{x}{e}\right)+{\int }_1^{x}f\left(t\right)dt,\text{ for all }x\epsilon R^{+}\text{. If the area enclosed by the curve }$$g\left(x\right)=x\left(f\left(x\right)-e^x\right)$ $lying in the fourth quadrant is A, then A=$

$f^{\text{'}}\left(x\right)=0+\left(1-x\right)\left(\frac{1}{x}\right)+\ln \left(\frac{x}{e}\right)\left(-1\right)+f\left(x\right) \left(by newton\text{'}s Leibnitz rule\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{1}{x}-1+\left(\ln x-1\right)\left(-1\right)+f\left(x\right)$

$\Rightarrow \frac{dy}{dx}-y=\frac{1}{x}-\ln x$

$\Rightarrow y e^{-x}=\int \left(\frac{1}{x}-\ln x\right)e^{-x}dx \left(solving of linear differential equation\right)$

$\Rightarrow y{e}^{-x}=e^{-x}\left(\log x\right)+c$

$\sin ce f\left(1\right)=e\Rightarrow e{e}^{-1}=0+c\Rightarrow c=1$

$y=\log x+e^x$

$f\left(x\right)=e^x+\ln x \Rightarrow g\left(x\right)=x\left(f\left(x\right)-e^x\right)=x\ln x$

$Area=A={\int }_0^1 x\ln xdx=\frac{1}{4}$