If $f^{\mathrm{\prime }}\left(x\right)=f\left(x\right)+{\int }_0^1 f\left(x\right)dx$ and given f(0)=1 Then f(x)=

Given $f^{\text{'}}\left(x\right)=f\left(x\right)+{\int }_0^1 f\left(x\right)dx$  ……….(1)

Differentiating we get, $\frac{f^{\text{'}\text{'}}\left(x\right)}{f^{\text{'}}\left(x\right)}=1$

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

$\Rightarrow \ln f^{\text{'}}\left(x\right)=x+c\Rightarrow f^{\text{'}}\left(x\right)=e^{x+c}=k{e}^x$

where k = e^c, a constant Again integrate 

$\int f^{\text{'}}\left(x\right)dx=\int k{e}^{x}dx\Rightarrow f\left(x\right)=k{e}^x+D\dots .......\left(2\right)$

Put x = 0 in (2), f (0) = ke⁰+D $\Rightarrow$ k + D = 1     (given f(0) = 1)    ……….(3)

Also from (1), $f^{\text{'}}\left(x\right)=f\left(x\right)+{\int }_0^1 f\left(x\right)dx$

$\begin{array}{l}\Rightarrow {\mathrm{ke}}^{\mathrm{x}}={\mathrm{ke}}^{\mathrm{x}}+\mathrm{D}+{\int }_0^1 \left({\mathrm{ke}}^{\mathrm{x}}+\mathrm{D}\right)\mathrm{dx}\\ \Rightarrow {\left[{\mathrm{ke}}^{\mathrm{x}}+\mathrm{Dx}\right]}_0^1+\mathrm{D}=0\Rightarrow [\mathrm{ke}+\mathrm{D}-\mathrm{k}]+\mathrm{D}=0\end{array}$

$\Rightarrow \mathrm{k}(\mathrm{e}-1)+2\mathrm{D}=0$    .………. (4)

Solving (3) and ( 4), we get

$\mathrm{k}=\frac{2}{3-\mathrm{e}}$ and $\mathrm{D}=\frac{1-\mathrm{e}}{3-\mathrm{e}}$.

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