An extremum value of $y = \int_{0}^{x} (t - 1)(t - 2) dt$ is

Let $u = \int_0^1 \frac{\ln(x + 1)}{x^2 + 1} \, dx$ and $v = \int_0^{\frac{\pi}{2}} \ln(\sin 2x) \, dx$,then:

If for a real number $y$,$[y]$ is the greatest integer less than or equal to $y$,then the value of the integral $\int_{\pi /2}^{3\pi /2} [2\sin x] \, dx$ is

The tangent to the graph of the function $y = f(x)$ at the point with abscissa $x = a$ forms with the $x$-axis an angle of $\pi/3$ and at the point with abscissa $x = b$ at an angle of $\pi/4$. Then the value of the integral $\int_{a}^{b} f(x) \cdot f''(x) \, dx$ is equal to (assume $f''(x)$ to be continuous).

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.

$A$ continuous and differentiable function $f$ satisfies the condition $\int_{0}^{x} f(t) dt = f^2(x) - 1$ for all real $x$. Then: