Application of ...

Area common to the curves $$y^{2} = ax$$ and $$x^{2} + y^{2} = 4ax$$ is equal to

The area bounded by $$x^2+y^2-2x=0$$ & $$y=\sin\displaystyle\frac{\pi x}{2}$$ in the upper half of the circle is?

On the real line R, we define two functions f and g as follows:
$$f(x) = min [x - [x], 1 - x + [x]]$$,
$$g(x) = max [x - [x], 1 - x + [x]]$$,
where [x] denotes the largest integer not exceeding x. 

The area in sq.units bounded by the hyperbola $$xy={ c }^{ 2 }$$, the x-axis and the ordinates at $$x=a$$ and $$x=b$$ ($$0< a< b)$$ is

The parabola $$y^2=4x+1$$ divides the disc $$x^2+y^2\leq 1$$ into two regions with areas $$A_1$$ and $$A_2$$. Then $$|A_1-A_2|$$ equals.

The area bounded by the curves $$y = \sin x, y = \cos x$$ and x-axis from $$x = 0$$ to $$x = \pi /2$$ is

The area bounded by min (|x|, |y|) = 2 and max (|x|, |y|) = 4 is

The area bounded by the curves $$y = \dfrac {1}{4} |4 - x^{2}|$$ and $$y = 7 -|x|$$ is

Consider two curves $$C_1 : (y - \sqrt 3)^2 = 4 ( x - \sqrt2) $$ and $$ C_2 : x^2 + y^2 = ( 6 + 2 \sqrt2 ) x + 2 \sqrt{3y} - 6 ( 1 + \sqrt2)$$ then

The area of the region $$\left\lfloor x \right\rfloor +\left\lfloor y \right\rfloor =1,-1\le x\le 1$$ and $$xy\le 1/2$$