The area bounded by $y-1=-|x|$ and $y+1=|x|$ is

The area (in sq. units) of the region $A = \{(x,y) \in R \times R | 0 \le x \le 3, 0 \le y \le 4, y \le x^2 + 3x\}$ is

Area lying in the first quadrant between the curves $x^2 + y^2 = \pi^2$ and $y = \sin x$ is equal to :-

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:

Given: $f(x) = \begin{cases} x, & 0 \leq x < \frac{1}{2} \\ \frac{1}{2}, & x = \frac{1}{2} \\ 1-x, & \frac{1}{2} < x \leq 1 \end{cases}$ and $g(x) = (x-\frac{1}{2})^2, x \in R$. Then the area (in sq. units) of the region bounded by the curves $y=f(x)$ and $y=g(x)$ between the lines $2x=1$ and $2x=\sqrt{3}$ is:

The area (in sq. units) bounded between the parabolas $x^2 = \frac{y}{4}$ and $x^2 = 9y$ and the line $y = 2$ is