$$\because f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t$$

or, $f(x)=1-2 x+e^x \int_0^x e^{-t} f(t) d t$

on differentiating both sides w.r.t. $x$ we get

$$\begin{aligned} & f^{\prime}(x)=-2+e^x \int_0^x e^{-t} f(t) d t+e^x \cdot e^{-x} f(x) \\ & f^{\prime}(x)=-2+f(x)+2 x-1+f(x)\{\text { from eq. (1) \}} \end{aligned}$$

$$\begin{aligned} & \therefore \quad f(x)-2 f(x)=2 x-3 \\ & \text { I.F. }=e^{\int-2 d x}=e^{-2 x} \\ & \therefore \quad e^{-2 x} \cdot f(x)=\int e^{-2 x}(2 x-3) d x \\ & e^{-2 x} \cdot f(x)=(2 x-3) \cdot \frac{e^{-2 x}}{-2}-\int 2 \cdot \frac{e^{-2 x}}{-2} d x \\ & e^{-2 x} \cdot f(x)=\frac{(2 x-3) e^{-2 x}}{-2}+\frac{e^{-2 x}}{-2}+c \\ & f(x)=-x+1+c^{\prime} e^{2 x} \\ & \because \quad f(x)=1 \text { from eq. (1) } \\ & \therefore \quad 1=0+1+c^{\prime} \Rightarrow c^{\prime}=0 \\ & \therefore \quad f(x)=-x+1 \end{aligned}$$

$\Rightarrow \quad$ Area $=\frac{1}{2}$