What is the quickest way to show that the integral equals zero? Let $a=x_0<\ldots<x_n=b$ be equidistant sampling points with $n$ being an even number.
How can one show that $$\int_a^b \prod_{k=0}^n(x-x_k) \ \text{d}x=0$$ in a fast way? I showed it by proving that $\prod_{k=0}^n(x-x_k)$ is an odd function w.r.t. $\frac{a+b}{2}$ and then substituting to shift the boundaries of integration. This was quite tedious and I thought maybe there is a quick and smart argument to show it's zero.
 A: Take an arbitrary monic polynomial $f(x)$ of degree $n+1$ and consider the $n$-th degree interpolating polynomial $p(x)$ through the $n+1$ points $$(x_0,f(x_0)),\ldots,(x_n,f(x_n)).$$ We have $f^{(n+1)}(x)\equiv (n+1)!$ so by the interpolation error formula
$$f(x)-p(x)=\prod_{k=0}^n(x-x_k).$$ The function $f(x)-p(x)$ has $n+1$ zeros evenly spaced along the interval $[a,b]$. We can take the Newton-Cotes evenly-spaced approximation with those $n+1$ sample points for the integral of this function, which is exact for polynomials up to degree at least $n+1$. But $f-p$ is a polynomial of degree at most $n+1$, and is zero at all sample points, so $$\int_a^b \prod_{k=0}^n(x-x_k)\,\mathrm{d}x=\int_a^b f(x)-p(x)\;\mathrm{d}x=0$$ QED.
A: You can shift by $(a+b)/2$ first. Then the product of the first and last factors is of the form $(x - c)(x + c) = x^2 - c^2$, as is the product of the second and second to last factors, and so on. But in addition, there's one "middle" factor which is just $x$, so the overall product is odd. Since the domain of integration is now from $-(a+b)/2$ to $(a+b)/2$, the integral of this odd function is zero.
A: Your way was actually quite smart. Why was it long and tedious? I think after an appropriate choice of notation it's going to sound quite trivial. In any case, your idea is the best approach, that's for sure. 
Hope that helps,
