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

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

The area (in sq. units) of the region bounded by the curve $x^2=4y$ and the straight line $x=4y-2$ is

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 $L_{h}$.

Let the area of the region bounded by the curve $y=\max\{\sin x, \cos x\}$, lines $x=0, x=\frac{3\pi}{2}$ and the x-axis be $A$. Then, $A+A^{2}$ is equal to:

$A$ line passing through the point $A(-2, 0)$ touches the parabola $P: y^2 = x - 2$ at the point $B$ in the first quadrant. The area of the region bounded by the line $AB$,the parabola $P$,and the $x$-axis is: