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 :

Let $f$ be a real valued continuous function defined on the positive real axis such that $g(x)=\int\limits_0^x t f(t) d t$. If $g\left(x^3\right)=x^6+x^7$, then value of $\sum\limits_{r=1}^{15} f\left(r^3\right)$ is :

If $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} \mathrm{d} x=\pi\left(\alpha \pi^2+\beta\right), \alpha, \beta \in \mathbb{Z}$, then $(\alpha+\beta)^2$ equals

If $I(m, n)=\int_0^1 x^{m-1}(1-x)^{n-1} d x, m, n>0$, then $I(9,14)+I(10,13)$ is