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

If the area enclosed between the curves $y^2 = 4kx$ and $y = kx$ for $k > 0$ is $\frac{2}{3}$ sq. units, then $k =$ ?

The area of the region $\{(x, y): x^2 \leq y \leq |x^2-4|, y \geq 1\}$ is

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:

Let $A_1, A_2$ and $A_3$ be the regions on $\mathbb{R}^2$ defined by:
$A_1 = \{(x, y) : x \geq 0, y \geq 0, 2x + 2y - x^2 - y^2 > 1 > x + y\}$
$A_2 = \{(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^2 + y^2\}$
$A_3 = \{(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^3 + y^3\}$
Denote by $|A_1|, |A_2|$ and $|A_3|$ the areas of the regions $A_1, A_2$ and $A_3$ respectively. Then,

If the area of the region ${(x, y) : -2x + 1 \le y \le 4 - x^2, x \ge 0, y \ge 0}$ is $\frac{\alpha}{\beta}$, where $\alpha, \beta \in N$ and $\gcd(\alpha, \beta) = 1$, then the value of $(\alpha + \beta)$ is: