Let the area of the region $\{(x, y): 0 \leq x \leq 3, 0 \leq y \leq \min \{x^2+2, 2x+2\}\}$ be $A$. Then $12A$ is equal to

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 of the region bounded by the curve $y = x^2 + 2$ and the lines $y = x$,$x = 0$,and $x = 3$ is

The area (in $sq. \, units$) of the region bounded by the curves $x^{2}+2y-1=0$,$y^{2}+4x-4=0$,and $y^{2}-4x-4=0$ in the upper half plane is $....$

The area enclosed by the curves $y = \ln x$,$y = \ln |x|$,$y = |\ln x|$,and $y = |\ln |x||$ is equal to:

The area of the region,bounded by the parabola $y=x^2+2$ and the lines $y=x, x=0$ and $x=3$,is