I am solving the following problem:
Find the quadrature of the following form: $Q(f) = Af(−1) + Bf(0) + > Cf(1)$, which has the highest degree and interpolates the integral: $\int_{-3}^{3} f(x) dx$. Is this an interpolation quadrature? Is this a Gaussian quadrature?
Here is how I am finding the degree of this quadrature:
Lagrange polynomials:
$L_0(x)=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}=\frac{x(x-1)}{2}=\frac{x^2-x}{2}\\ L_1(x)=\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}=\frac{(x+1)(x-1)}{-1}=1-x^2\\ L_2(x)=\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}=\frac{x^2+x}{2}\\ A=\int^{-3}_{3}\frac{1}{2}(x^2-x)dx=9\\ B=\int^{-3}_{3}(1-x^2)dx=-12\\ C=\int^{-3}_{3}\frac{x^2+x}dx=9$
So the quadrature we are looking for is: $Q(f)=9f(-1)-12f(0)+9f(1)$
$1: Q(1)=9-12+9=6\\ I(1)=6 \Rightarrow Q(1)=I(1)\\ x: Q(x)=9 (-1)+9=0 \\ %I(x^2)=\int_{-3}^{3} x^2 dx=\frac{1}{3}x^3\bigr|_{-3}^3=18\\ \Rightarrow Q(x)=I(x)\\ x^2: Q(x^2)=9( -1)^2+9=18\\ I(x^2)=\int_{-3}^{3} x^2 dx=\frac{1}{3}x^3\bigr|_{-3}^3=18\\ %I(x^2)=\int_{-3}^{3} x^3 dx=\frac{1}{4}x^4\bigr|_{-3}^3=0\\ \Rightarrow Q(x^2)=I(x^2)\\ x^3: Q(x^3)=9( -1)^3+9=0\\ I(x^3)=\int_{-3}^{3} x^3 dx=\frac{1}{4}x^4\bigr|_{-3}^3=0\\ \Rightarrow Q(x^3)=I(x^3) x^4: Q(x^4)=9( -1)^4+9=0\\ I(x^4)=\int_{-3}^{3} x^4 dx=\frac{1}{5}x^5\bigr|_{-3}^3=\frac{480}{5}\\ \Rightarrow Q(x^4)\neq I(x^4)$
So we get a 3rd quadrature.
How do I answer both of the questions stated in the problem?
I think it's easier to fit $A,B,C$ so that the formula correctly integrates the monomials of low degree: this is also a direct way to ensure we get the maximum degree possible. So,
From here $B=-12$ and $A=C=9$ fall out at once. These are the coefficients we must have to correctly handle degrees up to $2$. Then go on to check if there are bonus degrees: and indeed, $x^3$ is also integrated exactly (to $0$), by symmetry. For $x^4$ the formula gives $18$, which isn't right.
Gaussian quadrature with $n=3$ nodes integrates polynomials of degrees $\le 2n-1=5$ exactly. Hence, what we have is not Gaussian.
But it is an interpolation quadrature, although I did not use any interpolating polynomials in deriving it. Indeed, interpolating quadrature on $n=3$ points is exact for polynomials of degrees up to $n-1=2$, since these polynomials coincide with the interpolant. And we already saw that the exactness on such polynomials forces $A,B,C$ to be what we have.