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.
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,
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.
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.
