Ok so we know for integration with the Gauss-Legendre method : $\int_1^1f(x)dx = \sum_{i=0}^nc_i^nf(x_i)$
that both $x_i,c_i$ are fixed,depending on $n$ , when we work at the [-1,1] space. I know that for converting $x_i$ from [-1,1]->[a,b] I have to do the following:
$y_i=\frac{(b+a)}{2}+ \frac{(b-a)}{2}x_i$
But what about the weights for the G-L. How do I convert the $c_i$ weights from [-1,1] to [a,b]???
Given a function $g \colon [a, b] \to \mathbb{R},$ define the transformed function: $$ f \colon [-1, 1] \to \mathbb{R}, \ x \mapsto g\left(\frac{b+a}2 + \frac{b-a}2x\right). $$ The standard formula you have quoted, when applied to $f,$ gives: \begin{align*} \int_a^b g(y)\,dy & = \frac{b-a}2\int_{-1}^1f(x)\,dx \\ & \bumpeq \frac{b-a}2\sum_{i=0}^nc_i^nf(x_i) \\ & = \frac{b-a}2\sum_{i=0}^nc_i^ng(y_i), \end{align*} so the effect is that the weights $c_i^n$ are multiplied by $\frac{b-a}2.$