If $$\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } $$, then I equals

If y = y(x) is the solution of the differential equation $$\left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0$$ and y (0) = 0, then $$6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)$$ is equal to

Let a triangle ABC be inscribed in the circle $${x^2} - \sqrt 2 (x + y) + {y^2} = 0$$ such that $$\angle BAC = {\pi \over 2}$$. If the length of side AB is $$\sqrt 2 $$, then the area of the $$\Delta$$ABC is equal to :

The distance of the origin from the centroid of the triangle whose two sides have the equations $$x - 2y + 1 = 0$$ and $$2x - y - 1 = 0$$ and whose orthocenter is $$\left( {{7 \over 3},{7 \over 3}} \right)$$ is :