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Let f: \mathbb{R} \rightarrow \mathbb{R} be a differentiable function such that f^{\prime}(x)+f(x)=\int_0^2 f(t) d t. If f(0)=e^{-2}, then 2 f(0)-f(2) is equal to

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$$ \begin{aligned} & \text { Let } \int_0^2 \mathrm{f}(\mathrm{t}) \mathrm{dt}=\lambda \\ & \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})=\lambda \end{aligned}
is linear Differential equation
$$ \begin{aligned} & \text { I.f. }=\mathrm{e}^{\int \mathrm{dx}}=\mathrm{e}^{\mathrm{x}} \\ & \mathrm{f}(\mathrm{x}) \cdot \mathrm{e}^{\mathrm{x}}=\int \mathrm{e}^{\mathrm{x}} \lambda \mathrm{dx} \\ & \Rightarrow \mathrm{f}(\mathrm{x}) \cdot \mathrm{e}^{\mathrm{x}}=\lambda \mathrm{e}^{\mathrm{x}}+\mathrm{C} \\ & \Rightarrow \mathrm{f}(\mathrm{x})=\lambda+\mathrm{Ce}^{-\mathrm{x}} \\ & \text { put } \mathrm{f}(0)=\mathrm{e}^{-2} \\ & \mathrm{e}^{-2}=\lambda+\mathrm{C} \Rightarrow \mathrm{C}=\mathrm{e}^{-2}-\lambda \\ & \mathrm{f}(\mathrm{x})=\lambda+\left(\mathrm{e}^{-2}-\lambda\right) \mathrm{e}^{-\mathrm{x}} \\ & \lambda=\int_0^2 \mathrm{f}(\mathrm{t}) \mathrm{dt} \\ & =\int_0^2\left(\lambda+\left(\mathrm{e}^{-2}-\lambda\right) \mathrm{e}^{-\mathrm{t}}\right) \mathrm{dt} \\ & \Rightarrow \lambda=\lambda+\lambda \mathrm{e}^{-2}-\mathrm{e}^{-4}+\mathrm{e}^{-2} \\ & \Rightarrow \lambda=\mathrm{e}^{-2}-1 \\ & \mathrm{f}(\mathrm{x})=\mathrm{e}^{-2}-1+\mathrm{e}^{-\mathrm{x}} \end{aligned}

\begin{aligned} f(0)= & \mathrm{e}^{-2} \\ f(2)= & 2 \mathrm{e}^{-2}-1 \\ & 2 f(0)-f(2)=1 \end{aligned}

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