Let f: R → R be a twice differentiable function such that f(2)=1. If F (x)=x f(x) for all x ∈ R, ∫_0^2 x F^′(x) d x=6 and ∫_0^2 x^2 F^′ ′(x) d x=40,…

Let $\displaystyle \mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that $\displaystyle f(2)=1$. If $\displaystyle \mathrm{F}(x)=x f(x)$ for all $\displaystyle x \in \mathbf{R}$, $\displaystyle \int_0^2 x \mathrm{~F}^{\prime}(x) \mathrm{d} x=6$ and $\displaystyle \int_0^2 x^2 \mathrm{~F}^{\prime \prime}(x) \mathrm{d} x=40$, then $\displaystyle \mathrm{F}^{\prime}(2)+\int_0^2 \mathrm{~F}(x) \mathrm{d} x$ is equal to :