Quadrature grade

Question

Please give me any advice how to solve that task:

Find node $c$ and coefficients $\alpha$ and $\beta$ such that quadrature: $$ Q(f) = \alpha f(a) + \beta f(c) $$ which is aproximating integral $\int_b^a f(x) dx $ has maximal grade.

I know that minimum grade for that quadrature is 2 and maximal is 4. Grade 2 is possible, using trapezoidal quadrature. The case is, I don't know how to find out quadrature with higher grade or how to proof that grade 2 is maximal. I know I can try to solve system of equations, where I insert monomials ($ 1, x, x^2 \ ...)$ into $f$ in $Q$ but then I am not able to solve it or proof it isn't countable.

Please help me!

Answer 1

For a given function $f$ let define a quadrature $$Q(f) = \alpha f(a) + \beta f(c),$$

and an integral: $$I(f) = \int_a^b f(x) \, dx$$

Equations $Q(f) = I(f)$ for monomials $f = 1,\,x,\,x^2$ are

$$\beta+\alpha = b-a, \quad \beta c+a \alpha = b^2/2-a^2/2, \quad \beta c^2+a^2 \alpha = b^3/3-a^3/3$$

The system has a solution: $$ \alpha = (b-a)/4, \quad \beta = 3(b-a)/4, \quad c = (2b+a)/3 \tag{*}\label{*} $$

Unfortunately $Q(x^3) = I(x^3)$: $$ \beta c^3+a^3 \alpha = b^4/4-a^4/4 $$

and \eqref{*} are inconsistent. For example for $a=0$ and $b=1$ it reduces to $2/9 = 1/4$. So the quadrature is accurate for polynomials up to degree $3$. For $a=0$ and $b=1$ it is: $$ Q_0(f) = \frac{f(0) + 3 f(2/3)}{4} $$