The area (in $sq. units$) of the region $A = \{(x,y) : \frac{y^2}{2} \le x \le y + 4\}$ is

The area of the region (in sq. units) enclosed between the curves $y=|x|$,$y=[x]$ and the ordinates $x=-1$,$x=0$,$x=1$ is

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

The area bounded by $y = x^2 + 2$ and $y = 2|x| - \cos(\pi x)$ is equal to

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:

The area bounded by the curve $y = |x^{2}-9|$ and the line $y = 3$ is