Gaussian quadrature with a to $[0,1]$ reference domain instead of a $[-1,1]$ reference domain? For 1-d Gaussian quadrature with two points per element we have the following formula to transform an integral from an arbitrary domain $[a, b]$ to the reference domain $[-1,1]$ on which various Gaussian quadrature formulas for weights node points are defined -
$$\int_a^b f(x) dx \approx \frac{b-a}{2}\sum_{i=1}^n w_i f(\frac{b-a}{2}x_i + \frac{a+b}{2})$$
where, in the case of $n = 2$ points we have $x_i = \pm \frac{1}{\sqrt{3}}$ and $w_i = 1$. See Gaussian Quadrature
Now I am wondering what is the equivalent transformation if our reference domain is $[0,1]$ instead of $[-1,1]$. We will obviously have to change the transformation that gets applied to $x_i$ and the scaling factor outside the integral. Do we also have to transform the weights?
 A: Yes, we do have to transform the weights.
Consider first the mapping from $[-1,1]$ to $[a,b]$, that is,
$$
g_1(\widehat{x})=\frac{b-a}{2}\widehat{x}+\frac{b+a}{2},
$$
which clearly satisfies $g_1(-1)=a$ and $g_1(1)=b$. Then by the change of variables
$$
\int_a^b f(x)\,\mathrm{d}x = \int_{-1}^1 f(g_1(\widehat{x}))\frac{b-a}{2}\,\mathrm{d}\widehat{x} = \frac{b-a}{2}\int_{-1}^1 f(g_1(\widehat{x}))\,\mathrm{d}\widehat{x},
$$
where the integral on the right hand side can be computed using the quadrature rule
$$
\int_{-1}^1 f(g_1(\widehat{x}))\,\mathrm{d}\widehat{x} = \sum_{i=1}^n \widehat{w}_i f(g_1(\widehat{x}_i))
$$
with some points $\widehat{x}_i \in [0,1]$ and weights $\widehat{w}_i$, $i \in \{1,\cdots,n\}$.
Next consider the mapping from $[0,1]$ to $[a,b]$, that is,
$$
g_2(\widetilde{x})=(b-a)\widetilde{x}+a,
$$
which satisfies $g_2(0)=a$ and $g_2(1)=b$. Similarly as before
$$
\int_a^b f(x)\,\mathrm{d}x = \int_0^1 f(g_2(\widetilde{x}))(b-a)\,\mathrm{d}\widetilde{x} = (b-a)\int_0^1 f(g_2(\widetilde{x}))\,\mathrm{d}\widetilde{x},
$$
and
$$
\int_0^1 f(g_2(\widetilde{x}))\,\mathrm{d}\widetilde{x}=\sum_{i=1}^n \widetilde{w}_i f(g_2(\widetilde{x}_i))
$$
for some other points $\widetilde{x}_i \in [0,1]$ and weights $\widetilde{w}_i$, $i \in \{1,\cdots,n\}$.
Combining some of the previous equations we get
$$
(b-a)\int_0^1 f(g_2(\widetilde{x}))\,\mathrm{d}\widetilde{x}=\frac{b-a}{2}\int_{-1}^1 f(g_1(\widehat{x}))\,\mathrm{d}\widehat{x}
$$
or, equivalently,
$$
\sum_{i=1}^n \widetilde{w}_i f(g_2(\widetilde{x}_i)) = \frac{1}{2} \sum_{i=1}^n \widehat{w}_i f(g_1(\widehat{x}_i)).
$$
This holds true if
$$
\widetilde{w}_i = \frac{\widehat{w}_i}{2}
$$
and
$$
(b-a)\widetilde{x}_i+a = \frac{b-a}{2}\widehat{x}_i+\frac{b+a}{2} \Rightarrow \widetilde{x}_i = \frac{1}{2}\widehat{x}_i+\frac{1}{2}
$$
for every $i\in\{1,\cdots,n\}$. In conclusion, we can get the quadrature $(\widetilde{x}_i,\widetilde{w}_i)$ for the domain $[0,1]$ from the quadrature $(\widehat{x}_i,\widehat{w}_i)$ for the domain $[-1,1]$ by using the formulae
$$
\left\{
\begin{aligned}
\widetilde{w}_i &= \frac{\widehat{w}_i}{2}, \\
\widetilde{x}_i &= \frac{1}{2}\widehat{x}_i+\frac{1}{2}.
\end{aligned}
\right.
$$
A: Here me out, we already have a table for solvable values on [-1,1], we can adjust the range by doing +1 to get [0,2], then do /2 to get [0,1]. Why not modify our xi and wi with the above steps to get the new domain instead of modifying our function...
