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

The area enclosed between the curves $y^2 = 4x$ and $y = |x|$ is

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 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

The area bounded between the parabola $y^2 = 4x$ and the line $2x + y - 4 = 0$ is

The area (in square units) bounded by the curves $y=2x^2$ and $y=\max \{x-[x], x+|x|\}$ in between the lines $x=0$ and $x=2$ is