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 area of the region $\{(x, y): 0 \leq x \leq \frac{9}{4}, 0 \leq y \leq 1, x \geq 3y, x+y \geq 2\}$ is

Using integration,find the area of the region bounded by the triangle whose vertices are $(-1, 0)$,$(1, 3)$,and $(3, 2)$.

The smaller area enclosed by the circle $x^{2}+y^{2}=4$ and the line $x+y=2$ is

Let $A$ be the area bounded by the curve $y = \cos^{-1}\sqrt{1 - x^2}$,the tangent to the curve $y = \sin^{-1}x$ at $x = 0$,and the line $x = 1$. Then the value of $2(\{A\} + \text{sgn}(A))$ is (where $\{.\}$ is the fractional part function and $\text{sgn}(x)$ is the signum function).

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