TITLE: Gauss-Legendre quadrature rule question
QUESTION [0 upvotes]: In the working below, i understand the Gauss Legendre quadrature rule has degree of precision 3 and hence it is exact for polynomials of degree at most 3, which includes the Hermite interpolating polynomial $H_3(x)$.
Then, according to the quadrature rule, $\int_{-1}^{3} H_3(x) dx = c_{0}\cdot H_3(x_0) + c_1\cdot H_3(x_1)$.
But, in the working it is shown as just a sum of $H_3(x_o)$ and $H_3(x_1)$.
Why is it so?
And also, how should i properly expand an integral in the absolute function? 
Can someone show me the full workings for the expansion of the absolute error? 
Many thanks!

REPLY [1 votes]: We have $c_0=c_1=1$ because Gauss-Legendre quadratre is interpolatory. On the other hand, it follows immediately by its precision. To be exact on $1$ function we should have $c_0+c_1=2$. To be exact on $x$ we have $(-c_0+c_1)\frac{\sqrt{3}}{3}=0$. Hence $c_0=c_1=1$.
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