The area (in square units) of the region bounded by the parabola $y^2=4(x-2)$ and the line $y=2x-8$ 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 $A = \{(x, y) \in R^2 : y \geq 0, 2x \leq y \leq \sqrt{4-(x-1)^2}\}$ and $B = \{(x, y) \in R \times R : 0 \leq y \leq \min \{2x, \sqrt{4-(x-1)^2}\}\}$. Then the ratio of the area of $A$ to the area of $B$ is

If the area of the bounded region $R=\{(x, y): \max \{0, \log _{e} x\} \leq y \leq 2^{x}, \frac{1}{2} \leq x \leq 2\}$ is $\alpha(\log _{e} 2)^{-1}+\beta(\log _{e} 2)+\gamma$,then the value of $(\alpha+\beta-2 \gamma)^{2}$ is equal to:

The area bounded by $x^2 + y^2 - 2x = 0$ and $y = \sin \frac{\pi x}{2}$ in the upper half of the circle is:

What is the area bounded by the curves $x^2 + y^2 = 9$ and $y^2 = 8x$?