The area bounded between the parabola $y^2 = 4x$ and the line $2x + y - 4 = 0$ is

Using the method of integration,find the area of the region bounded by the lines: $2x + y = 4$,$3x - 2y = 6$,and $x - 3y + 5 = 0$.

The area (in sq. units) of the region $\{(x, y) \in R^{2}: x^{2} \leq y \leq 3-2x\}$ is

The area of the region $\{(x, y): x^2+4x+2 \leq y \leq |x+2|\}$ is equal to

The area of the region enclosed by the parabolas $y=x^2-5x$ and $y=7x-x^2$ 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 $L_{h}$.