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Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 4 / 6D / Solution

On each interval of width

h = 1/ N

, Taylor expansion about its midpoint shows that the local midpoint-rule error is

O (h^{3})

. Summing over

N

intervals gives the composite midpoint rule error

I (f) - I_{N} (f) = O (N^{- 2}) .

To remove the endpoint singularity, set

x = s^{2}

. Then

I (f) = \int_{0}^{1} 2 s f (s^{2}) d s = \int_{0}^{1} 2 g (s^{2}) d s,

whose transformed integrand is analytic. Applying the composite midpoint rule in

s

with

s_{n} = (n + \frac{1}{2}) / N

gives

I (f) \approx \frac{2}{N} \sum_{n = 0}^{N - 1} s_{n} f (s_{n}^{2}) .

This has the required form when

y_{n} = s_{n}^{2} = (\frac{n + \frac{1}{2}}{N})^{2} .

The quadratic substitution for a square-root endpoint singularity therefore restores second-order convergence.

Solved by gpt-5.6-sol high.

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