Two Gauss nodes integrate cubics exactly and miss a concrete quartic

Work throughout with the standard $L^2[-1,1]$ inner product. Let $P_0\equiv 1$, $P_1(x)=x$, and $P_2(x)=(3x^2-1)/2$ denote the first three Legendre polynomials, which are orthogonal on $[-1,1]$. 1. Show that the roots of $P_2$ are $x_\pm=\pm 1/\sqrt{3}$. Determine the unique weights $w_\pm$ that make the interpolatory rule \[ Q_2(f)=w_-f(x_-)+w_+f(x_+) \] exact for $f(x)=1$ and $f(x)=x$. Equivalently, evaluate the classical weight formula \[ w_i=\frac{2}{(1-x_i^2)\bigl(P_2'(x_i)\bigr)^2}. \] Record $Q_2$ explicitly. 2. Prove that $Q_2$ is exact for every polynomial of degree at most $3$. You may use either the interpolatory error formula or orthogonality of $P_2$ to $\mathbb P_1$. Give a one-line counterexample showing that exactness can fail at degree $4$. 3. Let $I(f)=\int_{-1}^{1}f(x)\,dx$. Compute $I(x^4)$ and $Q_2(x^4)$ as rational numbers, and record the signed error $I(x^4)-Q_2(x^4)$. Deduce the constant $c$ in the representation $I(p)-Q_2(p)=c\,p^{(4)}$ for every quartic $p$. Then, using the three-point Gauss–Legendre nodes $0,\pm\sqrt{3/5}$ and weights $8/9,\,5/9,\,5/9$, verify that the three-point rule is exact on $x^4$. 4. Transfer $Q_2$ to $[0,2]$ by the affine map $t=x+1$. Write the resulting nodes and weights, evaluate the mapped rule on $t\mapsto t^3$ and on $t\mapsto t^4$, and compare with the exact integrals $\int_0^2 t^3\,dt$ and $\int_0^2 t^4\,dt$. Explain why the cubic is recovered exactly while the quartic error equals the reference error from part 3. Do not replace Gauss nodes by Chebyshev extrema or by equispaced Newton–Cotes nodes, and do not invoke an adaptive panel-halving estimator.

The even/odd split of the moment system forces $w_+=w_-$ as soon as the nodes are symmetric, and the constant-function moment then gives each weight equal to $1$.