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.

Let $u = \int_{0}^{\infty} \frac{dx}{x^4 + 7x^2 + 1}$ and $v = \int_{0}^{\infty} \frac{x^2 dx}{x^4 + 7x^2 + 1}$. Then:

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

Let $\{x\}$ and $[x]$ denote the fractional part of $x$ and the greatest integer $\leq x$ respectively of a real number $x$. If $\int_{0}^{n}\{x\} dx$,$\int_{0}^{n}[x] dx$,and $10(n^{2}-n)$ $(n \in N, n > 1)$ are three consecutive terms of a $G.P.$,then $n$ is equal to

Value of the $\int_0^9 \sqrt{x} \,dx + \int_0^{\pi/2} (\cos x + \sin x) \,dx$ is: