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

The area bounded by the parabola $y^2 = 4x$ and the line $2x - 3y + 4 = 0$,in square units,is

Let the line $x=-1$ divide the area of the region $\{(x,y):1+x^{2}\le y\le3-x\}$ in the ratio $m:n$,where $\gcd(m,n)=1$. Then $m+n$ is equal to

Let $y=p(x)$ be the parabola passing through the points $(-1,0), (0,1)$ and $(1,0)$. If the area of the region $\{(x, y) : (x+1)^2+(y-1)^2 \leq 1, y \leq p(x)\}$ is $A$,then $12(\pi-4A)$ is equal to $.........$.

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.

The area of the region $\{(x, y): x^2+4x+2 \leq y \leq |x+2|\}$ is equal to