Question: Actually, I know that $\int\_0^n \lfloor t \rfloor dt$ is $\frac{n(n-1)}{2}$.
But I can't solve this simple problem. Can anybody help me?
P.S. I'm seventeen, I'm studying in last year of high school, and I'm preparing for the national exams.

Answer: 
\begin{eqnarray\*}
\int\_0^n 2x \lfloor x \rfloor dx = \sum\_{i=0}^{n-1} i \underbrace{\int\_{i}^{i+1} 2x \,dx}\_{x^2 \mid\_i^{i+1}=2i+1} = \underbrace{\sum\_{i=0}^{n-1} i (2i+1)}\_{\text{using} \sum\_{i=0}^{n-1} i^2=\frac{n(n-1)(2n-1)}{6} } = \frac{(n-1)n(n+1)}{3}. 
\end{eqnarray\*}
**hint:** $\displaystyle \int\_{0}^n = \displaystyle \int\_{0}^1 +\displaystyle \int\_{1}^2 +...+\displaystyle \int\_{n-1}^n$