Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 4 / 6A / b

Table of contents
- Solution b

Solution

A two-node Gaussian rule is exact through degree three, the maximal degree

2 n - 1

. Its nodes are the zeros of the monic quadratic orthogonal to

1, x

x

on

[0, 1]

. Writing it as

p = x^{2} + ux + v

gives

\int_{0}^{1} p x d x = 0, \int_{0}^{1} p x^{2} d x = 0,

so

u = - 6/5

,

v = 3/10

. Therefore

x_{0} = \frac{3}{5} - \frac{6}{10}, x_{1} = \frac{3}{5} + \frac{6}{10} .

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