Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a twice differentiable function such that $f(2)=1$. If $\mathrm{F}(\mathrm{x})=\mathrm{x} f(\mathrm{x})$ for all $\mathrm{x} \in \mathrm{R}$, $\int\limits_0^2 x F^{\prime}(x) d x=6$ and $\int\limits_0^2 x^2 F^{\prime \prime}(x) d x=40$, then $F^{\prime}(2)+\int\limits_0^2 F(x) d x$ is equal to :

<p>$$\begin{aligned} &\begin{aligned} & \int_0^2 \mathrm{xF}^{\prime}(\mathrm{x}) \mathrm{dx}=6 \\ & =\left.\mathrm{xF}(\mathrm{x})\right|_0 ^2-\int_0^2 \mathrm{f}(\mathrm{x}) \mathrm{dx}=6 \\ & =2 \mathrm{~F}(2)-\int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=6[\therefore \mathrm{f}(2)=2 \mathrm{~F}(2)=2] \\ & \int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=-2 \quad \ldots(1) \\ & \Rightarrow \int_0^2 \mathrm{~F}(\mathrm{x}) \mathrm{dx}=-2\quad \ldots(2) \end{aligned}\\ &\text { Also }\\ &\begin{aligned} & \int_0^2 x^2 F^{\prime \prime}(x) d x=\left.x^2 F^{\prime}(x)\right|_0 ^2-2 \int_0^2 x F^{\prime}(x) d x=40 \\ & =4 F^{\prime}(2)-2 \times 6=40 \\ & F^{\prime}(2)=13 \\ & \therefore F^{\prime}(2)+\int_0^2 F(x)=13-2=11 \end{aligned} \end{aligned}$$</p>