Let f(x) be an integrable function in the interval $[a,b]$, and let $x_0, x_1, …, x_n $ be distinct points in $[a,b]$ (not necessarily uniformly spaced). Integrating the polynomial of degree n that interpolates f(x) at the points $x_i$, we obtain a quadrature formula of the form $$I_n(f) = \sum_{i=0}^n a_{i}f(x_i)$$ Prove that $ \sum_{i=0}^n a_{i} = b – a$.
Attempt: I believe that the $a_i$ are the quadrature weights? One can obtain $a_i$ by setting $$a_i = \int_a^b L_i(x)dx $$ and computing the Lagrange polynomials. For example: for $n=1, x_0 = a,$ and $ x_1 = b $: $L_o(x) = \dfrac {x-b}{a-b}$ and $L_1 = \dfrac {x-a}{b-a}$. Then we have the weights $$a_o = \int_a^b \dfrac {x-b}{a-b}dx = \dfrac {b-a}{2}$$ $$a_1 = \int_a^b \dfrac {x-a}{b-a}dx = \dfrac {b-a}{2}$$
Taking the sum for $n = 1$ we get: $$ \sum_{i=0}^1 a_{i} = b – a$$
Is this a correct proof?
Best Answer
As polynomials of degrees up to $n$ are interpolated exactly, you know from interpolating $p(x)=1$ that $$ 1=\sum_{k=0}^nL_k(x)~\text{ for all }x\in\Bbb R. $$ Now integrate this over $[a,b]$.