Let $$[x]$$ denote the greatest integer $$\le x$$. Consider the function $$f(x) = \max \left\{ {{x^2},1 + [x]} \right\}$$. Then the value of the integral $$\int\limits_0^2 {f(x)dx} $$ is

Let $$A=\left\{(x, y) \in \mathbb{R}^{2}: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^{2}}\right\}$$ and

$$ B=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: 0 \leq y \leq \min \left\{2 x, \sqrt{4-(x-1)^{2}}\right\}\right\} \text {. } $$.

Then the ratio of the area of A to the area of B is

The area enclosed by the curves $${y^2} + 4x = 4$$ and $$y - 2x = 2$$ is :

The area of the region

$$\left\{(x, y):|x-1| \leq y \leq \sqrt{5-x^{2}}\right\}$$ is equal to :