The area bounded by the curves $|x| + |y| \geq 1$ and $x^2 + y^2 \leq 1$ 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:

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

Let $f:[0,1] \rightarrow[0,1]$ be the function defined by $f(x)=\frac{x^3}{3}-x^2+\frac{5}{9} x+\frac{17}{36}$. Consider the square region $S=[0,1] \times [0,1]$. Let $G=\{(x, y) \in S: y>f(x)\}$ be called the green region and $R=\{(x, y) \in S: y(A)$ There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $L_{h}$ equals the area of the green region below the line $L_{h}$.
$(B)$ There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $L_{h}$ equals the area of the red region below the line $L_{h}$.
$(C)$ There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $L_{h}$ equals the area of the red region below the line $L_{h}$.
$(D)$ There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $L_{h}$ equals the area of the green region below the line $L_{h}$.

The area of the region in the first quadrant inside the circle $x^2+y^2=8$ and outside the parabola $y^2=2x$ is equal to:

The area (in sq. units) of the region $A = \{(x, y) : x^2 \le y \le x + 2\}$ is