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 (in square units) of the region bounded by the parabola $y^2=4(x-2)$ and the line $y=2x-8$ is:

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

Area of the region enclosed between the curves $y^2=4(x+7)$ and $y^2=5(2-x)$ is

If the area of the region bounded by the curves $y=4-\frac{x^2}{4}$ and $y=\frac{x-4}{2}$ is equal to $\alpha$,then $6 \alpha$ equals

Consider the functions $f, g: R \rightarrow R$ defined by
$f(x)=x^2+\frac{5}{12}$ and $g(x)=\begin{cases} 2\left(1-\frac{4|x|}{3}\right), & |x| \leq \frac{3}{4} \\ 0, & |x|>\frac{3}{4} \end{cases}$
If $\alpha$ is the area of the region
$\{( x , y ) \in R \times R :| x | \leq \frac{3}{4}, 0 \leq y \leq \min \{f( x ), g( x )\}\}$,
then the value of $9 \alpha$ is.