The area enclosed between the curves $y = \sin x$,$y = \cos x$ and the $x$-axis for $0 \le x \le \frac{\pi}{2}$ is:

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,

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

If the area of the larger portion bounded between the curves $x^2+y^2=25$ and $y=|x-1|$ is $\frac{1}{4}(b \pi+c)$,where $b, c \in N$,then $b+c$ is equal to $ . . .. .. $

If the area of the region $\{(x, y): x^{2/3} + y^{2/3} \leq 1, x + y \geq 0, y \geq 0\}$ is $A$,then find the value of $\frac{256A}{\pi}$.

The area (in sq. units) of the region $A=\{(x, y) : |x|+|y| \leq 1, 2y^{2} \geq |x|\}$ is