If f^ ' (x) = f(x) + ∫ _ 0 ^ 1 f(x)dx, f(0) = 1 , then f(x) =

Given,

$f^{'}\left(x\right) = f\left(x\right) + {∫}_{0}^{1}f\left(x\right)dx, f\left(0\right) = 1... ( i )$

⇒ f^{' '}\left(x\right) = f^{'}\left(x\right) + 0

⇒ \dfrac{f^{' '}\left(x\right)}{f^{'}\left(x\right)} = 1

⇒ ∫ \dfrac{f^{' '}\left(x\right)}{f^{'}\left(x\right)}dx = ∫ dx

$⇒ logf^{'}\left(x\right) = x + C$

$⇒ f^{'}\left(x\right) = Ae^{x}$

$⇒ f\left(x\right) = Ae^{x} + K$

$f\left(0\right) = A + K = 1... ( ii )$

$∴ Ae^{x} = Ae^{x} + K + {∫}_{0}^{1}\left(Ae^{x} + k\right)dx$

$⇒ k + {Ae^{x} + Kx}_{0}^{1} = 0$

$⇒ k + Ae - A + K = 0$

$⇒ A ( e - 1 ) + 2k = 0... ( iii )$

From Eqs. $( ii )$ and $( iii )$, we get

$A = \dfrac{2}{3 - e}, k = \dfrac{1 - e}{3 - e}$

$∴ f\left(x\right) = \dfrac{2e^{x}}{3 - e} + \dfrac{1 - e}{3 - e}$