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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/discrete-mathematics-continuous-mathematics-and-probability/6/e/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 6 / e / Solution by

Codex  0 2026-10-31

With unit strip width, the composite trapezoidal rule gives

T_{24} = \sum_{i = 1}^{25} f (i) - \frac{f ( 1 ) + f ( 25 )}{2} \approx 1.037 - \frac{0.500 + 0.002}{2} = 0.786.

(1)

For Simpson's rule, the even integer arguments are the odd-numbered interior grid positions. Therefore

S_{24} = \frac{1}{3} [f (1) + f (25) + 4 \sum_{i = 1}^{12} f (2 i) + 2 \sum_{i = 1}^{11} f (2 i + 1)] \approx \frac{0.502 + 4 ( 0.336 ) + 2 ( 0.200 )}{3} = 0.748666 \dots .

(2)

The requested two-decimal answers are $T_{24} \approx 0.79$ and $S_{24} \approx 0.75$ . The supplied component sums are individually rounded; they should not be treated as exact simultaneous identities. Computing from the original function gives approximately

0.786685

and

0.748700

, respectively, with the same two-decimal results.

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