Let the area of the region $\{(x, y) : |2x - 1| \leq y \leq |x^2 - x|, 0 \leq x \leq 1\}$ be $A$. Then $(6A + 11)^2$ is equal to $.......$.

Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2$, $x+y=8$ and the y-axis that lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y=x^2+2$, $y^2=x$, $x=2$, and the y-axis that lies in the first quadrant. Then $A_1-A_2$ is equal to

The area bounded by the curve $y=\cos x$,the line joining $(-\pi / 4, \cos (-\pi / 4))$ and $(0,2)$ and the line joining $(\pi / 4, \cos (\pi / 4))$ and $(0,2)$ is

The area bounded by the parabolas $y=4x^2$, $y=\frac{x^2}{9}$ and the line $y=2$ is

The area (in sq. units) of the region $\{(x, y): 0 \leq y \leq 2|x|+1, 0 \leq y \leq x^2+1, |x| \leq 3\}$ is

The area of the figure bounded by the parabolas $x = -2y^{2}$ and $x = 1 - 3y^{2}$ is