Integral $ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ Question
$ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ = ?
we tried to bound it from both sides using $x$ and $(x-1)$, which yield nice estimation ($\frac {24}\pi$) - ($\frac {36}\pi$) but not a precise one. 
we also tried to split it to 6 integrals using $\lfloor x\rfloor  $ as a different constant each time. $(0\cdot (\cos(...)-\cos(...))+1(\cos(\frac \pi3)-\cos(\frac\pi6))+2\cdot...$ etc
it produced the right result ($\frac {30}\pi$), but the number is not important as achieving the general function.

Here is another solution using Riemann-Stieltjes integral. Integrating by parts,
\begin{align*}
\int_{0}^{6} \lfloor x \rfloor \sin \left(\frac{6x}{\pi}\right) \, dx
&= \int_{0^{+}}^{6} \lfloor x \rfloor \sin \left(\frac{6x}{\pi}\right) \, dx \\
&= \left[ - \lfloor x \rfloor \frac{\pi}{6} \cos \left(\frac{6x}{\pi}\right) \right]_{0^{+}}^{6} + \int_{0^{+}}^{6} \frac{\pi}{6} \cos \left(\frac{6x}{\pi}\right) \, d\lfloor x \rfloor \\
&= - \pi \cos \left(\frac{36}{\pi}\right) + \sum_{k=1}^{6} \frac{\pi}{6} \cos \left(\frac{6k}{\pi}\right).
\end{align*}
More generally, let $f$ be continuous on $[a, b]$ and $F$ be an anti-derivative of $f$. Then
\begin{align*}
\int_{a}^{b} f(x)\lfloor x \rfloor \, dx
&= \int_{a^{+}}^{b} f(x)\lfloor x \rfloor \, dx \\
&= \left[ F(x)\lfloor x \rfloor \right]_{a^{+}}^{b} - \int_{a^{+}}^{b} F(x) \, d\lfloor x \rfloor \\
&= \lfloor b \rfloor F(b) - \lfloor a \rfloor F(a) - \sum_{a < k \leq b} F(k).
\end{align*}
