If $$f(x) = \int_{ - 1}^x {|t|} \,dt$$, then for any $$x \ge 0,\,f(x)$$ is equal to

Let $$I = \int_0^{100\pi } {\sqrt {(1 - \cos 2x)} } \,dx$$, then

$$\int\limits_0^1 {\log \left( {{1 \over x} - 1} \right)} dx$$ is equal to

If [x] denotes the greatest integer less than or equal to x, then the value of the integral $$\int\limits_0^2 {{x^2}[x]\,dx} $$ equals