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:

Let the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=e^{x-1}-e^{-|x-1|}$ and $g(x)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)$. Then the area of the region in the first quadrant bounded by the curves $y=f(x)$,$y=g(x)$ and $x=0$ is

The area of the region $R = \{(x, y) : 5x^2 \leq y \leq 2x^2 + 9\}$ is ........ $\text{square units}$. (in $\sqrt{3}$)

The area of the shorter region bounded by $|y| = 4 - x^2$ and $|y| = 3x$ is given by $\left( 3K + \frac{1}{3} \right)$ sq-unit,where $K$ is equal to:

The area (in sq. units) of the region $\{(x, y): 0 \leq y \leq x^{2}+1, 0 \leq y \leq x+1, \frac{1}{2} \leq x \leq 2\}$ is

The area of the region bounded by the curves $y=e^x, y=\log x$ and lines $x=1, x=2$ is