Let f : R $\to$ R be a twice differentiable function such that f(2) = 1. If F(x) = xƒ(x) for all x $\in$ R,  ${\int }_0^2 x{F}^{\prime }\left(x\right)dx=6\text{ and }{\int }_0^2 x^2{F}^{\prime \prime }\left(x\right)dx=40,$ then $F^{\prime }\left(2\right)+{\int }_0^2 F\left(x\right)dx$ is equal to :

$\begin{array}{l}{\int }_0^2 x{F}^{\prime }\left(x\right)dx=6\\ ={\left.xF\left(x\right)\right|}_0^2-{\int }_0^2 f\left(x\right)dx=6\\ =2F\left(2\right)-{\int }_0^2 xF\left(x\right)dx=6[\therefore f(2)=2F(2)=2]\\ {\int }_0^2 xF\left(x\right)dx=-2\\ \Rightarrow {\int }_0^2 F\left(x\right)dx=-2\end{array}$

Also

$\begin{array}{l}{\int }_0^2 x^2{F}^{\prime \prime }\left(x\right)dx={\left.x^2{F}^{\prime }\left(x\right)\right|}_0^2-2{\int }_0^2 x{F}^{\prime }\left(x\right)dx=40\\ =4F^{\prime }\left(2\right)-2\times 6=40\\ F^{\prime }\left(2\right)=13\\ \therefore F^{\prime }\left(2\right)+{\int }_0^2 F\left(x\right)=13-2=11\end{array}$