The area enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ over the interval $\left[0, \frac{\pi}{2}\right]$ is

In the interval $(0, \pi / 2)$,the area lying between the curves $y = \tan x$ and $y = \cot x$ and the $X$-axis is:

Let $S$ be the region bounded by the curves $y=x^{3}$ and $y^{2}=x$. The curve $y=2|x|$ divides $S$ into two regions of areas $R_{1}$ and $R_{2}$. If $\max \{R_{1}, R_{2}\}=R_{2}$,then $\frac{R_{2}}{R_{1}}$ 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 square unit) of the region enclosed by the curves $y=x^2$ and $y=x^3$ is

Let $S = \{(x, y) \in R \times R : x \geq 0, y \geq 0, y^2 \leq 4x, y^2 \leq 12 - 2x \text{ and } 3y + \sqrt{8}x \leq 5\sqrt{8}\}$. If the area of the region $S$ is $\alpha \sqrt{2}$,then $\alpha$ is equal to