Difference between 2 quadrature formulas $I_{2n}(f(x))$ and $I(f(x))$

Question

Let $h = \frac{1}{n}$, $x_i = -1 + ih$ , $ i = 0, 1, ... , 2n,$ where $n$ is a positive integer. Let $p_{2n} (x)$ be a polynomial of degree at most $2n$ interpolating the function $f(x)$ at $x_i$ $(i = 0, 1, ..., 2n)$.

Let

$$ I(f(x)) = \int^{1}_{-1} f(x) dx, $$ $$ I_{2n} (f(x)) = \int^{1}_{-1} p_{2n} (x) dx. $$

Then compute

$$ I_{2n}(\sum_{i=0}^{2n+1}(ix^i)) - I (\sum_{i=0}^{2n+1}(ix^i)).$$

I thought of using Gaussian quadrature methods to solve this question but I am clueless as to how I can proceed. Also, is the answer indeed zero? Some hints and/or clarifications will be deeply appreciated.

Answer 1

You get $p_{2n}=f$ if $f$ is any polynomial of degree at most $2n$.

You only have to care what happens to $f(x)=x^{2n+1}$, but as this is an odd function...