Let $f$ is a differentiable function such that $f^{\prime}(x)=f(x)+\int_0^2 f(x) d x, f(0)=\frac{4-e

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Let $f$ is a differentiable function such that $f^{\prime}(x)=f(x)+\int_0^2 f(x) d x, f(0)=\frac{4-e^2}{3}$, then:

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It is given that $f^{\prime}(x)=f(x)+\int_{0}^{2}f(x)dx$ $\Rightarrow f^{\prime}(x)=f(x)+c[:..c=\int_{0}^{2}f(x)dx]$ $\Rightarrow\frac{f^{\prime}(x)}{f(x)+c}=1$ On integrating both the sides of above expression, $log_{e}\{f(x)+c\}=k\Rightarrow f(x)+c=ke^{x}$ [k being constant of integration] Since, $f(0)=\frac{4-e^{2}}{3}$ and $f(0)+c=k$ kk=- 4-6 +cc=k 3 4-e 3 From Eqs i and get $t~f(x)+k-(\frac{4-e^{2}}{3})=k(e^{x})\Rightarrow f(x)=k(e^{k}-1)+(\frac{4-e^{2}}{3}).$ Now, to find constant of integration k. On integrating both the sides form 0 to 2, we get $\int_{0}^{2}f(x)dx=k\int_{0}^{2}(e^{x}-1)dx+\frac{4-e^{2}}{3}\int_{0}^{2}dx$ $\Rightarrow c=k(e^{2}-2-1)+\frac{4-e^{2}}{3}(2)\Rightarrow c=\frac{2}{3}(4-e^{2})+k(e^{2}-3)$ From Eqs. ( ) and (iv), we get On putting $k-(\frac{4-e^{2}}{3})=\frac{2}{3}(4-e^{2})+k(e^{2}-3)\Rightarrow k=1$ $k=1$ in Eq (ii), we get $f(x)=(e^{x}-1)+\frac{4-e^{2}}{3}$ Hence, $f(x)=e^{x}-\frac{(e^{2}-1)}{3}$