Evaluate the definite integral $\int_0^{2016} x (x-1)(x-2)(x-3)... (x-2016)\,dx$ 
$$\int_0^{2016} x (x-1)(x-2)(x-3)... (x-2016)\,dx$$ 

I have tried integration by parts using calling $f (x)=(x-1)(x-2)... (x-2016)$ but it doesn't help. I can't think of anything else. Please help.
 A: Hint Shift the variable of integration so that the interval is centered at $0$, that is, set $$x = u + 1008 , \qquad dx = du .$$
Alternatively, substitute $$x = 2016 - v, \qquad dx = -dv ,$$ and compare the result with the given integral.
A: This Definite integral can be evaluated using the property  $$\int_a^b f(x) dx= \int_a^b f(a+b-x) dx$$
Lets say that $$I= \int_0^{2016} x(x-1)(x-2) \cdots (x-2016) dx$$ 
Applying the property above,
$$I= \int_0^{2016} (2016-x)(2015-x) \cdots (-x) $$
$$\implies I= - \int_0^{2016} x(x-1)(x-2) \cdots (x-2016) dx$$
Adding our initial equation to the one we have obtained, we can see that
$$ 2I=0 \implies I=0$$
A: Note that 
$$\scriptsize\begin{align}
\int_0^2 x(x-1)(x-2)dx&=\int_{-1}^1 (x-1)x(x+1)dx&=0\\
\int_0^4 x(x-1)(x-2)(x-3)(x-4)dx&=\int_{-2}^2 (x-2)(x-1)x(x+1)(x+2)dx&=0\\
&\vdots\\
\int_0^{2016} x(x-1)(x-2)(x-3)\cdots (x-2016)dx&=\int_{-1008}^{1008} (x-1008)(x-1007)\cdots x\cdots (x+1007)(x+1008)dx&=\color{red}0\\
\end{align}$$
The result follows from the antisymmetric nature of the curve
$\scriptsize(x-a)(x-a+1)\cdots x\cdots (x+a-1)(x+a)$.
