TITLE: Evaluate the definite integral $\int_0^{2016} x (x-1)(x-2)(x-3)... (x-2016)\,dx$
QUESTION [9 upvotes]: $$\int_0^{2016} x (x-1)(x-2)(x-3)... (x-2016)\,dx$$ 

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.

REPLY [6 votes]: 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
$$ 2I=0 \implies I=0$$