# The integral $\int_0^1 (-x^2+x)^{k}\cdot \lfloor kx \rfloor \, dx$ (BMT 2017)

Find the integral $$\int_0^1 (-x^2+x)^{k}\cdot \lfloor kx \rfloor \, dx$$ where $k = 2017.$

My attempt: By splitting up the integral and then using floor properties, I was able to transform the integral into $$\sum_{n=0}^{k-1} n\int_{n/k}^{(n+1)/k} (-x^2+x)^k\,dx$$

However I was unable to find a better representation for the sum. I did notice that the integrand was quite similar to a Beta function; however, the limits are wrong and I believe that the beta function cannot be applied.

Any suggestions?

• What is "BMT"? $\qquad$ – Michael Hardy Jun 18 '17 at 18:51
• @MichaelHardy Berkeley Math Tournament. – Teoc Jun 18 '17 at 18:55
• That's definitely something that should be explicit. My guess was something like "Basic Management Test" (although a test with that name would probably challenge students to solve difficult equations like $x+3=5$). I looked at the "BMT" disambiguation page on Wikipedia and it failed to shed any light. – Michael Hardy Jun 18 '17 at 18:59

This is $$I=\int_0^1 x^k(1-x)^k\lfloor kx\rfloor\,dx.$$ This just begs for the substitution $y=1-x$: $$I=\int_0^1 (1-y)^ky^k\lfloor k(1-y)\rfloor\,dy$$ so that $$2I=\int_0^1 x^k(1-x)^k\left( \lfloor kx\rfloor+\lfloor k(1-x)\rfloor\right)\,dx.$$ I wonder if the expression in brackets can be simplified?
• The terms in brackets simply equal $k-1$ if $x$ is not an integer, and $k$ if $x$ is an integer. – infinitylord Jun 18 '17 at 20:25