The parabolas $${y^2} = 4x$$ and $${x^2} = 4y$$ divide the square region bounded by the lines $$x=4,$$ $$y=4$$ and the coordinate axes. If $${S_1},{S_2},{S_3}$$ are respectively the areas of these parts numbered from top to bottom ; then $${S_1},{S_2},{S_3}$$ is :

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

If $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha ,$$ then $$4{x^2} - 4xy\cos \alpha + {y^2}$$ is equal to :

The value of integral, $$\int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx $$ is