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Reading through some notes on 2-point gauss quadrature, I came across the following general formula. I'm currently doing an assignment with 3-point quadrature, and have gotten to a similar formula, but now am trying to work out how to evaluate it

$$\int_a^bf(x) \approx \frac{h}{2}\sum\limits_{j=0}^n[f(x_{j,0}) + f(x_{j,1})]$$

I'm mostly just uncertain what $f(x_{j,0})$ means in this context, could anyone give me a hand understanding what to do next?

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1 Answer 1

It seems that your integral is divided into n smaller integrals in the domain. $$\int_a^b f(x)dx=\int_{a_{0,0}}^{b_{0,1}} f(x)dx+\int_{a_{1,0}}^{b_{1,1}} f(x)dx+...+\int_{a_{n,0}}^{b_{n,1}} f(x)dx$$ Hence every small integral can be represented such as $$\int_{a_{j,0}}^{b_{j,1}} f(x)dx=\frac{h_j}{2}\bigg(f(x_{j,0})+f(x_{j,1})\bigg)$$ where $$h_j=b_{j,1}-a_{j,0}$$ $$x_{j,0}=\bigg(\frac{b_{j,1}-a_{j,0}}{2} \bigg)\frac{-1}{\sqrt{3}}+\frac{b_{j,1}+a_{j,0}}{2}$$ $$x_{j,1}=\bigg(\frac{b_{j,1}-a_{j,0}}{2} \bigg)\frac{1}{\sqrt{3}}+\frac{b_{j,1}+a_{j,0}}{2}$$ If your division is equal $h_j$ is equal everywhere such as $$b_{0,1}-a_{0,0}=b_{1,1}-a_{1,0}=...=b_{n,1}-a_{n,0}\Rightarrow h_0=h_1=...=h_n=h$$ As final result $$\int_a^b f(x)dx=\frac{h}{2} \sum_{j=0}^n\bigg(f(x_{j,0})+f(x_{j,1})\bigg)$$

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