I want to obtain an integration formula with 3 equally spaced nodes in the interval $[-1; 1]$ exact for polynomials of degree $\le $ 3 and it's truncation error knowing that $ET = Kf^4(\xi)$.
I assume $\xi$ is the value for which $f^4(\xi)$ is maximum in $[-1; 1]$. But I am lost since I thought that you needed $n+1$ points (in this case 4) to work with polynomials of degree 3. I also tried to get information from the truncation error, but it only supports my thoughts above $f^{n+1}(\xi)$.
Any comment or hint would be helpful.
Assume your formula is of the form $Q(f) = w_0 f(-1) + w_1 f(0) + w_2 f(1)$. Then we require $Q$ to be exact for the polynomials $p_0(x)=1, p_1(x)=x,p_2(x)=x^2$, so \begin{align*} w_0 p_0(-1) + w_1 p_0(0) + w_2 p_0(1) &= Q(p_0) = \int_{-1}^1 p_0(x)dx \\ w_0 p_1(-1) + w_1 p_1(0) + w_2 p_1(1) &= Q(p_1) =\int_{-1}^1 p_1(x)dx \\ w_0 p_2(-1) + w_1 p_2(0) + w_2 p_2(1) &= Q(p_2) =\int_{-1}^1 p_2(x)dx \end{align*} From this you can determine $w_0,w_1,w_2$. Then applying your formula to $p_3(x)=x^3$ should give $Q(p_3)=0$ which shows that your formula is exact for polynomials of degree at most 3.
Given $n+1$ distinct nodes, any interpolatory quadrature formula $\sum_{i=0}^n w_i f(x_i)$ has degree of exactness $ \geq n$. Taking the nodes to be parameters, instead of fixing them to be equidistant, it's possible for the degree of exactness to be as large as $2n+1$ (Gaussian Quadrature). Newton-Cotes is the keyword to google if you want to learn more about quadrature formulas with equally spaced nodes.