If the area of the region bounded by the curves $y=4-\frac{x^2}{4}$ and $y=\frac{x-4}{2}$ is equal to $\alpha$,then $6 \alpha$ equals

Let the function $f(x) = \begin{cases} -3ax^2 - 2, & x < 1 \\ a^2 + bx, & x \geq 1 \end{cases}$ be differentiable for all $x \in R$,where $a > 1, b \in R$. If the area of the region enclosed by $y = f(x)$ and the line $y = -20$ is $\alpha + \beta \sqrt{3}$,where $\alpha, \beta \in Z$,then the value of $\alpha + \beta$ is . . . . .

The area (in sq. units) bounded by the parabola $y = x^2 - 1$,the tangent at the point $(2, 3)$ to it,and the $y$-axis is

The area (in sq. units) of the region $A = \{(x, y) : x^2 \le y \le x + 2\}$ is

The area between the parabolas $x^2 = \frac{y}{4}$ and $x^2 = 9y$ and the straight line $y = 2$ is:

If the area enclosed by the parabolas $P_1: 2y = 5x^2$ and $P_2: x^2 - y + 6 = 0$ is equal to the area enclosed by $P_1$ and the line $y = \alpha x$,where $\alpha > 0$,then $\alpha^3$ is equal to $......$.