Let f: ℝ → ℝ be a differentiable function such that f ^ ' (x) + f...

Let

${∫}_{0}^{2}f\left(t\right)dt = k$

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

Now, we have

$f\left(x\right) + {f}^{'}\left(x\right) = {∫}_{0}^{2}f\left(t\right)dt$

$⇒ \dfrac{dy}{dx} + y = k$

This is a linear differential equation.

$I . F . = {e}^{∫ dx} = {e}^{x}$

Solution of the given differential equation is

$y{e}^{x} = k ∫ {e}^{x}dx$

$⇒ y{e}^{x} = k{e}^{x} + C$

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

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

Hence,

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

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

Now,

$k = {∫}_{0}^{2}f\left(t\right)dt$

$⇒ k = {∫}_{0}^{2}\left[k + \left({e}^{- 2} - k\right){e}^{- t}\right]dt$

$⇒ k = {\left[kt - \left({e}^{- 2} - k\right){e}^{- t}\right]}_{0}^{2}$

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

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

$⇒ k = \left({e}^{- 2} - 1\right)$

So,

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

$∴ f\left(2\right) = 2{e}^{- 2} - 1$

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

So,

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