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ID: past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/7/a/solution

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Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 7 / a / Solution by

Codex  0 2026-10-20

Both rules use a uniform partition with

Δ x = (b - a) / n

. An original implementation in pseudocode is:

function trapezoid(f, a, b, n):
    dx = (b-a)/n
    total = (f(a)+f(b))/2
    for j in 1,...,n-1:
        total += f(a+j*dx)
    return dx*total

function midpoint(f, a, b, n):
    dx = (b-a)/n
    total = 0
    for j in 0,...,n-1:
        total += f(a+(j+1/2)*dx)
    return dx*total

The trapezoidal rule gives each endpoint half weight; the composite midpoint rule uses one interior point per subinterval. Both converge for a continuous integrand on the closed interval.

The second equality in the printed trapezoidal hint is missing a factor $1/2$ . The average of the left and right rectangle sums is required. For

f = 1

, the uncorrected equality would return

2 (b - a)

instead of

b - a

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