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

If the area of the region $\{(x, y): |4-x^2| \leq y \leq x^2, y \leq 4, x \geq 0\}$ is $\left(\frac{80 \sqrt{2}}{\alpha}-\beta\right)$,where $\alpha, \beta \in \mathbb{N}$,then $\alpha+\beta$ is equal to . . . . . . .

Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2$,$x+y=8$ and the y-axis that lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y=x^2+2$,$y^2=x$,$x=2$,and the y-axis that lies in the first quadrant. Then $A_1-A_2$ is equal to

The area (in sq. units) of the region enclosed by the curves $y=x^{2}-1$ and $y=1-x^{2}$ is equal to

The area of the region (in square units) bounded by the curves $y=x^3$,$y=x$ and $-1 \leq x \leq 1$ is:

The area (in sq. units) bounded by the curves $x^2=9y$,$(x-6)^2=9y$ and the $X$-axis is