The area (in square units) lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and the lines $x=0$ and $x=2$ is

The area bounded by the curve $x=4-y^{2}$ and the $Y$-axis is

If a straight line $y - x = 2$ divides the region $x^2 + y^2 \le 4$ into two parts,then the ratio of the area of the smaller part to the area of the greater part is

The value of $a$ $(a > 0)$ for which the area bounded by the curves $y = \frac{x}{6} + \frac{1}{x^2}$,$y = 0$,$x = a$ and $x = 2a$ has the least value,is

The area (in sq units) enclosed by the loop of the curve $ay^2 = x^2(a - x), (a > 0)$ is

Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y = f(x)$,the $x$-axis,and the ordinates $x = \frac{\pi}{4}$ and $x = \beta > \frac{\pi}{4}$ is given by $\left( \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta \right)$. Then,find the value of $f\left( \frac{\pi}{2} \right)$.