If the area of the region $\{(x, y) : |x-5| \leq y \leq 4 \sqrt{x}\}$ is $A$,then $3A$ is equal to . . . . . . .

The area between the parabolas $x^2 = \frac{y}{4}$ and $x^2 = 9y$ and the straight line $y = 2$ is:

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}+\frac{2 x}{\left(1+x^2\right)^2} y=x e^{\frac{1}{\left(1+x^2\right)}}$ with $y(0)=0$. Then the area enclosed by the curve $f(x)=y(x) e^{-\frac{1}{\left(1+x^2\right)}}$ and the line $y=x/4+2$ is:

The area (in sq units) bounded by the curves $y^2=4x$ and $x^2=4y$ is

The area (in sq. units) of the region consisting of points $(x,y)$ on the $X-Y$ plane which satisfy $|x| \le 1 + |y|$ and $|y| \le 1$ is:

Find the area bounded by the curves $y = ||x - 1| - 2|$ and $y = 2$.