Mathematics - Differential Equation Question with Solution | TestHub

$f\left(x\right)=1-2x+{\int }_0^x{e}^{x-t}f\left(t\right)dt$$\Rightarrow e^{-x}f\left(x\right)=e^{-x}\left(1-2x\right)+{\int }_0^x{e}^{-t}f\left(t\right)dt$Differentiate w.r.t.$x$$e^{-x}f\left(x\right)+e^{+x}{f}^{\prime }\left(x\right)={-e}^{-x}\left(1-2x\right)+e^{-x}\left(-2\right)+e^{-x}f\left(x\right)$$\Rightarrow -f\left(x\right)+f^{\prime }\left(x\right)=-\left(1-2x\right)-2+f\left(x\right)$$\Rightarrow f^{\prime }\left(x\right)-2f\left(x\right)=2x-3$Integrating factor$=e^{-2x}$$f\left(x\right).e^{-2x}=\int e^{-2x}\left(2x-3\right)dx$$=\left(2x-3\right)\int e^{-2x}dx-\int \left(\left(2\right)\int e^{-2x}dx\right)dx$$= \frac{\left(2x-3\right)e^{-2x}}{-2}- \frac{e^{-2x}}{2}+c$$f\left(x\right)=\frac{2x-3}{-2}-\frac{1}{2}+c{e}^{2x}$$f\left(0\right)=\frac{3}{2}-\frac{1}{2}+c=1\Rightarrow c=0$$\therefore f\left(x\right)=1-x$Area$=\frac{\pi }{4}-\frac{1}{2}=\frac{\pi -2}{4}$