The area of the region $\{ (x, y) : x^2 + y^2 \le 1 \le x + y \}$ is

Let $g(x) = \cos(x^2)$,$f(x) = \sqrt{x}$,and $\alpha, \beta$ (where $\alpha < \beta$) be the roots of the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$. Then,the area (in sq. units) bounded by the curve $y = (g \circ f)(x)$ and the lines $x = \alpha$,$x = \beta$,and $y = 0$ is:

Area under the curve $y = \sin 2x + \cos 2x$ between $x = 0$ and $x = \frac{\pi}{4}$ is ......... $sq. \text{ } unit$.

The area bounded by the curve $y = \ln(x)$ and the lines $y = 0$,$y = \ln(3)$,and $x = 0$ is equal to

The area of the region bounded by the curve $y = x^2 - x - 6$,the $x$-axis $(y = 0)$,and the lines $x = -1$ and $x = 1$ is . . . . . . sq. units.

Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x=0, x=1, y^{2}=x$ and $y=|\alpha x-5|-|1-\alpha x|+\alpha x^{2}$. Then $f(0)+f(1)$ is equal to