If the area (in $sq. units$) of the region $\{(x,y): y^2 \le 4x, x + y \le 1, x \ge 0, y \ge 0\}$ is $a\sqrt{2} + b$,then $a - b$ 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,

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 $C_{1}$ be the curve obtained by the solution of the differential equation $2xy \frac{dy}{dx} = y^{2} - x^{2}, x > 0$. Let the curve $C_{2}$ be the solution of $\frac{2xy}{x^{2} - y^{2}} = \frac{dy}{dx}$. If both the curves pass through $(1, 1)$,then the area enclosed by the curves $C_{1}$ and $C_{2}$ is equal to:

The area of the region enclosed between the parabolas $y^{2}=2x-1$ and $y^{2}=4x-3$ is

The parabola $y^2=4x+1$ divides the disc $x^2+y^2 \leq 1$ into two regions with areas $A_1$ and $A_2$. Then,$|A_1-A_2|$ equals