If the area lying in the first quadrant and bounded by the circle $x^2+y^2-4x=0$,the parabola $y^2=x$ and the $X$-axis is $A$,then $6A-9\sqrt{3}=$

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

The polynomial $f(x)$ satisfies the condition $f(x + 1) = x^2 + 4x$. The area enclosed by $y = f(x - 1)$ and the curve $x^2 + y = 0$ is

The area enclosed by $y=\sqrt{5-x^{2}}$ and $y=|x-1|$ 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 $R$ denote the set of all real numbers. Then the area of the region $\{(x, y) \in R \times R : x > 0, y > \frac{1}{x}, 5x - 4y - 1 > 0, 4x + 4y - 17 < 0\}$ is