Past exam of the options I course of the University of Oxford / 2011 / P / 2 / b / Solution

Since $T_2=2x^2-1$, take $x_1=-1/\sqrt{2}$, $x_2=1/\sqrt{2}$. Exactness for $T_0$ and $T_1$ requires $A+B=\pi$ and $(-A+B)/\sqrt{2}=0$. ThusFor $T_2$, both node values and its weighted integral vanish. For $T_3=4x^3-3x$, oddness and symmetric nodes likewise give zero on both sides. Therefore $R(T_j)=0$ for $j=0,1,2,3$. These four Chebyshev polynomials have successive degrees zero through three and form a basis of the degree-at-most-three polynomial space. By linearity, the rule is exact on that whole space: this is two-node Gauss–Chebyshev quadrature.