Question

ƒ(x) =$$\begin{gathered} 1-2x+{\int }_0^x{e}^{x-t}f\left(t\right)dt \\ \Rightarrow e^{-x}f\left(x\right)=e^{-x}(1-2x)+{\int }_0^x{e}^{x-t}f\left(t\right)dt \end{gathered}$$
Differentiate w.r.t. x.
$\Rightarrow$e^-x ƒ(x) + e^-x ƒ '(x) =-e^-x (1- 2x) + e^-x (-2) + e^-x ƒ(x)
$\Rightarrow$ –ƒ(x) + ƒ'(x) = –(1 – 2x) – 2 + ƒ(x).
$\Rightarrow$ƒ'(x) – 2ƒ(x) = 2x – 3
Integrating factor = e^–2x.
ƒ(x).e^–2x =$\int e^{-2x}$(2x-3)dx
=(2x-3)$\int e^{-2x}$(2x-3)dx
=$$\begin{gathered} \frac{\left(2x-3\right)e^{-2x}}{-2}-\frac{{\displaystyle e^{-2x}}}{{\displaystyle 2}}+c \\ ƒ\left(x\right) =\frac{2x-3}{-2}-\frac{1}{2}+c{e}^{2x} \\ ƒ\left(0\right) =\frac{{\displaystyle 3}}{{\displaystyle 2}}-\frac{{\displaystyle 1}}{{\displaystyle 2}}+c=1\Rightarrow c=0 \\ \therefore f\left(x\right)=1-x \\ Area=\frac{\pi }{4}-\frac{1}{2}=\frac{\pi -2}{4} \end{gathered}$$