Past exam of the options I course of the University of Oxford / 2013 / P / 1

The Lagrange interpolation polynomial isThe distinct nodes make all denominators nonzero. Each Lagrange basis polynomial has degree $n$ and satisfies ${\ell }_j(x_i)={\delta }_{ij}$, so the sum has degree at most $n$ and interpolates every specified value.

If $p$ and $q$ both interpolate the data and have degree at most $n$, the polynomial $r=p-q$ has the $n+1$ distinct roots $x_0,\dots ,x_n$. A nonzero polynomial of degree at most $n$ has at most $n$ roots: repeated use of the factor theorem would otherwise make a product of $n+1$ linear factors divide it. Hence $r=0$, proving the interpolating polynomial is unique.

Integrate the Lagrange interpolation polynomial and exchange the integral with the finite sum:Therefore the interpolatory quadrature weights areThe resulting quadrature rule is exact for every polynomial of degree at most $n$, since its interpolant is itself. For a general continuous $f$, replacing $f$ by its interpolant supplies the approximation.