Given,

$f^{\text{'}}\left(x\right)=f\left(x\right)+{\int }_0^{1}f\left(x\right)dx,f\left(0\right)=1$   $...\left(\mathrm{i}\right)$

$\Rightarrow f^{\text{'}\text{'}}\left(x\right)=f^{\text{'}}\left(x\right)+0$

$\Rightarrow \frac{f^{\text{'}\text{'}}\left(x\right)}{f^{\text{'}}\left(x\right)}=1$

$\Rightarrow \int \frac{f^{\text{'}\text{'}}\left(x\right)}{f^{\text{'}}\left(x\right)}dx=\int dx$

$\Rightarrow \log f^{\text{'}}\left(x\right)=x+C$

$\Rightarrow f^{\text{'}}\left(x\right)=A{e}^x$

$\Rightarrow f\left(x\right)=A{e}^x+K$

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

$\therefore A{e}^x=A{e}^x+K+{\int }_0^1\left(A{e}^x+k\right)dx$

$\Rightarrow k+{\left[A{e}^x+Kx\right]}_0^1=0$

$\Rightarrow k+Ae-A+K=0$

$\Rightarrow A(e-1)+2k=0$   $...\left(\mathrm{iii}\right)$

From Eqs. $\left(\mathrm{ii}\right)$ and $\left(\mathrm{iii}\right)$, we get

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

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