The area of the region bounded by the curve $y = x^2 + 2$ and the lines $y = x$,$x = 0$,and $x = 3$ is

Find the area (in square units) of the region bounded by the circle $x^2 + y^2 = 9$ and the parabola $y^2 \leq 8x$.

The area (in square units) bounded by the curves $x = -2y^2$ and $x = 1 - 3y^2$ 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 $L_{h}$.

The area (in square units) enclosed between the curves $y=\sin x$ and $y=\cos x$ for $\frac{\pi}{4} \leq x \leq \frac{5 \pi}{4}$ is

Let $T$ be the tangent to the ellipse $E: x^{2}+4 y^{2}=5$ at the point $P(1,1)$. If the area of the region bounded by the tangent $T$,ellipse $E$,lines $x=1$ and $x=\sqrt{5}$ is $\alpha \sqrt{5}+\beta+\gamma \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)$,then $|\alpha+\beta+\gamma|$ is equal to $....$