calculus

2. V=\int_{0}^{\frac{\pi }{2}}{\frac{x}{2}\left|2sin^{2}x-1 \right|}dx
=\int_{0}^{\frac{\pi }{2}}{\frac{x}{2}\left|cos2x \right|}dx
=\int_{0}^{\pi }{\frac{x}{8}\left|cosx \right|}dx
=\int_{0}^{\frac{\pi }{2}}{\frac{x}{8}cosx}dx + \int_{\pi /2}^{\pi }{\frac{x}{8}(-cosx)dx}
in second integral, put t = π - x , and later change the variable from t to x
\int_{\pi /2}^{\pi }{\frac{x}{8}(cosx)dx}= -\int_{\pi /2}^{0}{\frac{\frac{\pi }{2}-x}{8}cosx} =-( \int_{0}^{\pi /2}{\frac{\pi cosx}{16}} - \int_{0}^{\pi /2}{\frac{xcosx}{8}})

hence V=\int_{0}^{\pi /2}{\frac{\pi cosx}{16}}= \frac{\pi }{16}

1st one is pretty trivial.

If x = n an integer, then n f(n) = n f(0), so that f(n) = f(0) for all integers

If 0<x<1, then we get x f(x) = x f(0) so that f(x) = f(0) for 0<x<1.

From this we see that f([x]) = f(0) and f({x}) = f(0) for all x.

Hence the equation becomes f(x) = [x] f(0) + {x} f(0) = ([x]+{x}) f(0) = x f(0)

Hence f(x) = f(0) for all x.

We see on plugging back that f(0) can assume any value

Hence f(x) = c for any value of c satisfies the above equation.