The area bounded by the curves $y=\frac{1}{4}\left|4-x^2\right|$ and $y=7-|x|$ 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

The area enclosed by the curves $y=8x-x^2$ and $8x-4y+11=0$ is

Let $A_1, A_2$ and $A_3$ be the regions on $\mathbb{R}^2$ defined by:
$A_1 = \{(x, y) : x \geq 0, y \geq 0, 2x + 2y - x^2 - y^2 > 1 > x + y\}$
$A_2 = \{(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^2 + y^2\}$
$A_3 = \{(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^3 + y^3\}$
Denote by $|A_1|, |A_2|$ and $|A_3|$ the areas of the regions $A_1, A_2$ and $A_3$ respectively. Then,

The area bounded by the curves $y = x(x - 3)^2$ and $y = x$ is (in sq. units) :

The area (in sq. units) of the region $\{(x, y) \in R^{2}: x^{2} \leq y \leq 3-2x\}$ is