# Derive the quadrature rule of the form $\int_{-1}^1 f(x) dx \approx w_0 f(\alpha) + w_1 f'(\beta)$

From a practice problem

## Problem Statement

Let the interval $$[a,b]$$ be divided into $$n$$ subintervals by $$n+1$$ points $$x_0 = a , x_1 = a+h$$, $$x_2 = a+2h , \dots ,x_n = a+nh =b$$. Derive the quadrature rule of the form
$$\int_{-1}^1 f(x) \mathrm{d} x \approx w_0 f(\alpha) + w_1 f'(\beta) ,\quad -1 \le \alpha < \beta \le 1 \tag{1} ,$$
that is exact for polynomials of as high degree as possible.

### Attempt at solution

Let $$p(x)$$ be the lagrange interpolation polynomial interoplating the points $$(x_0, f(x_0) ), \dots , (x_n , f(x_n))$$. Then

$$p(x) = \sum_{i=0}^n l_i(x) f(x_i) \qquad \text{for} , \ l_i (x) = \prod_{j=0, j\neq i}^n \left( \frac{x-x_j}{x_i – x_j} \right)$$
it follows that
$$\int_{-1}^1 f(x) \mathrm{d} x \approx \int_{-1}^1 p(x) \mathrm{d} x$$
where
\begin{align} \int_{-1}^1 p(x) \mathrm{d} x &= \int_{-1}^1 \sum_{i=0}^n l_i(x) f(x_i) \mathrm{d} x \\ &= \sum_{i=0}^n \left( \int_{-1}^1 l_i(x) \mathrm{d} x \right) f(x_i) \mathrm{d} x \\ &:= \sum_{i=0}^n w_i f(x_i) \end{align}
We conclude
$$\therefore \int_{-1}^1 f(x) \mathrm{d} x = \sum_{i=0}^n w_i f(x_i)$$

## Questions

From here I am stuck. In $$(1)$$ we have a function derivative $$f'(\beta)$$, but in the formula derived above, there's no such thing. Moreover, I don't know how many terms in this sum to take as well. Does anyone know how I can complete this derivation?

The reference to the sub-intervals is useless... We just want to determine the quadrature nodes ($$\alpha, \beta)$$ and weights ($$\omega_0, \omega_1$$) that make the rule exact for polynomial of degree as high as possible. So, forcing the rule to be exact for $$f(x)=1, f(x) = x, f(x)=x^2, \cdots$$, you get a system of equations that allows the calculation of $$\alpha, \beta, \omega_0, \omega_1$$:
$$\omega_0 \cdot 1 + \omega_1 \cdot 0 = 2 \Leftrightarrow \omega_0=2$$
$$\omega_0 \cdot \alpha + \omega_1 \cdot 1 = 0\Leftrightarrow 2 \alpha+\omega_1=0$$
$$\omega_0 \alpha^2 + \omega_1 \cdot 2 \beta = \frac 23$$
$$\vdots$$