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

The area (in sq. units) of the region enclosed by the curves $y=x^{2}-1$ and $y=1-x^{2}$ is equal to

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 enclosed by the curves $y = \cos x$,$y = 1 + \sin 2x$ and $x = \frac{3\pi}{2}$ in the first and fourth quadrants (as shown in the figure) is:

Let $S$ be the region bounded by the curves $y=x^{3}$ and $y^{2}=x$. The curve $y=2|x|$ divides $S$ into two regions of areas $R_{1}$ and $R_{2}$. If $\max \{R_{1}, R_{2}\}=R_{2}$,then $\frac{R_{2}}{R_{1}}$ is equal to

The area (in sq. units) bounded by the curves $y=\sqrt{x}$,$2y-x+3=0$,$X$-axis and lying in the first quadrant,is