Let f : ℝ → ℝ be a differentiable function such that f ' x + f x = ∫ 0 2 f t d t .…

Let

${\int }_0^{2}f\left(t\right)dt=k$

$y=f\left(x\right)$

Now, we have

$f\left(x\right)+f^{\text{'}}\left(x\right)={\int }_0^{2}f\left(t\right)dt$

$\Rightarrow \frac{dy}{dx}+y=k$

This is a linear differential equation.

$\mathrm{I}.\mathrm{F}.=e^{\int dx}=e^x$

Solution of the given differential equation is

$y{e}^x=k\int e^{x}dx$

$\Rightarrow y{e}^x=k{e}^x+C$

Now, $f\left(0\right)=e^{-2}$, so

$C=e^{-2}-k$

Hence,

$y{e}^x=k{e}^x+\left(e^{-2}-k\right)$

$\Rightarrow y=f\left(x\right)=k+\left(e^{-2}-k\right)e^{-x} ....\left(\mathrm{i}\right)$

Now,

$k={\int }_0^{2}f\left(t\right)dt$

$\Rightarrow k={\int }_0^2\left[k+\left(e^{-2}-k\right)e^{-t}\right]dt$

$\Rightarrow k={\left[kt-\left(e^{-2}-k\right)e^{-t}\right]}_0^2$

$\Rightarrow k=2k-\left(e^{-2}-k\right)\left(e^{-2}-1\right)$

$\Rightarrow \left(e^{-2}-k\right)\left(e^{-2}-1\right)=k$

$\Rightarrow k=\left(e^{-2}-1\right)$

So,

$f\left(x\right)=\left(e^{-2}-1\right)+e^{-x}$

$\therefore f\left(2\right)=2e^{-2}-1$

$f\left(0\right)=e^{-2}$

So,

$2f\left(0\right)-f\left(2\right)=1$