Let $\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that $f(2)=1$. If $\mathrm{F}(x)=x f(x)$ for all $x \in \mathbf{R}$, $\int_{0}^{2} x \mathrm{~F}^{\prime}(x) \mathrm{d} x=6$ and $\int_{0}^{2} x^{2} \mathrm{~F}^{\prime \prime}(x) \mathrm{d} x=40$, then $\mathrm{F}^{\prime}(2)+\int_{0}^{2} \mathrm{~F}(x) \mathrm{d} x$ is equal to :