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

If $f(x)$ is a function satisfying $f^{\prime}(x)=f(x)$ with $f(0)=1$ and $g(x)$ is a function that satisfies $f(x)+g(x)=x^2$. Then the value of the integral $\int_0^1 f(x) g(x) d x$ is

$\int_{0}^{1/3} (\sum_{r=0}^{101} \{x + \frac{r}{3}\}) dx$ is equal to (where $\{.\}$ represents the fractional part function).

Let $I_n = \int_0^1 (\log x)^n dx$,where $n$ is a non-negative integer. Then,$I_{2011} + 2011 I_{2010}$ is equal to

Let $f: R \rightarrow R$ be a differentiable function such that its derivative $f^{\prime}$ is continuous and $f(\pi)=-6$. If $F:[0, \pi] \rightarrow R$ is defined by $F(x)=\int_0^{ x } f( t ) dt$,and if $\int_0^\pi\left(f^{\prime}( x )+ F ( x )\right) \cos x dx =2$,then the value of $f(0)$ is.

The value of $\int\limits_0^2 {\frac{{dx}}{{{{(1 - x)}^2}}}} $ is