Let $[t]$ denote the greatest integer less than or equal to $t.$ Then,the value of the integral $\int\limits_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x$ is equal to

Let the function $f: R \rightarrow R$ be defined by
$f(t)=\begin{cases} (-1)^{n+1} 2, & \text{if } t=2n-1, n \in N \\ \frac{(2n+1-t)}{2} f(2n-1) + \frac{(t-(2n-1))}{2} f(2n+1), & \text{if } 2n-1 < t < 2n+1, n \in N \end{cases}$
Define $g(x) = \int_1^x f(t) dt, x \in (1, \infty)$. Let $\alpha$ denote the number of solutions of the equation $g(x) = 0$ in the interval $(1, 8]$ and $\beta = \lim_{x \rightarrow 1^+} \frac{g(x)}{x-1}$. Then the value of $\alpha + \beta$ is equal to.

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

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:

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

Which of the following inequalities is/are $TRUE$?
$(A)$ $\int_0^1 x \cos x \, dx \geq \frac{3}{8}$
$(B)$ $\int_0^1 x \sin x \, dx \geq \frac{3}{10}$
$(C)$ $\int_0^1 x^2 \cos x \, dx \geq \frac{1}{2}$
$(D)$ $\int_0^1 x^2 \sin x \, dx \geq \frac{2}{9}$