If the area of the region $\{(x, y) : |x-5| \leq y \leq 4 \sqrt{x}\}$ is $A$,then $3A$ is equal to . . . . . . .

The positive values of the parameter $a$ for which the area of the figure bounded by the curve $y = \cos ax$,$y = 0$,$x = \frac{\pi}{6a}$,and $x = \frac{5\pi}{6a}$ is greater than $3$ are:

The area of the region bounded by the curves $y = x^2$ and $y = |x|$ is

Let for $x \in R$; $f(x) = \frac{x+|x|}{2}$ and $g(x) = \begin{cases} x, & x < 0 \\ x^2, & x \geq 0 \end{cases}$. Then the area bounded by the curve $y = (f \circ g)(x)$ and the lines $y = 0$,$2y - x = 15$ is equal to $...........$.

Find the area enclosed by the parabola $4y = 3x^{2}$ and the line $2y = 3x + 12$.

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 . . . . . . .