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

The area of the region bounded by the curve $y = x^2 + 2$ and the lines $y = x$,$x = 0$,and $x = 3$ is

The area of the region bounded by the curves $x + 3y^2 = 0$ and $x + 4y^2 = 1$ is equal to: (in $/3$)

The area enclosed by the curves $y = \sqrt{x}$ and $x = -\sqrt{y}$,and the circle $x^2 + y^2 = 2$ above the $x$-axis,is

The area of the region common to the parabolas $4y^2 = 9x$ and $3x^2 = 16y$ is...

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 . . . . .