Evaluate $I$:
$$I = \int_0^{10} \lfloor x \rfloor^3 \{x\}\,dx.$$
[$\{.\}$ represents the fractional part of $x$]
[ $\lfloor x \rfloor$ represents the greatest integer function / floor function of $x$]
PS- Please cite any sources that uses the proper method of evaluating floor function integrals.
I started off with re-writing $I$ as
$$I = \int_0^{10} \lfloor x \rfloor^3 (x-\lfloor x \rfloor) \, dx,$$
which I suppose can written as
$$I = I_1 + I_2 = \int_0^{10} \lfloor x \rfloor^3 x \, dx-\int_0^{10}\lfloor x \rfloor^4 \, dx.$$
However, on evaluating $I_1$ and $I_2$ individually, I ended up the value $I = 8362.5$, which did not match with options from where I have taken this question from.
Now I believe I have made a mistake with $I_2$ on the part of writing it as $\lfloor x \rfloor^4$. It would be great to learn how to properly evaluate $I$. (or any other faster methods)
Best Answer
\begin{align*} \int_0^{10} {\left\lfloor x \right\rfloor ^3 \left\{ x \right\}{\rm d}x} & = \sum\limits_{k = 0}^9 {\int_k^{k + 1} {\left\lfloor x \right\rfloor ^3 \left\{ x \right\}{\rm d}x} } = \sum\limits_{k = 0}^9 {\int_k^{k + 1} {k^3 \left\{ x \right\}{\rm d}x} } \\ & = \sum\limits_{k = 0}^9 {k^3 \int_k^{k + 1} {\left\{ x \right\}{\rm d}x} } = \sum\limits_{k = 0}^9 {k^3 \frac{1}{2}} = \frac{1}{2}\sum\limits_{k = 0}^9 {k^3 } = \frac{{2025}}{2}. \end{align*}