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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/20/a/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 20 / a / Solution by

Codex  0 2026-10-18

Write

d = b - a

and

h = d / n

. The required Riemann sum is

S_{n} = h \sum_{k = 1}^{n} (a + kh)^{2} = h [n a^{2} + 2 ah \sum_{k = 1}^{n} k + h^{2} \sum_{k = 1}^{n} k^{2}] .

Using

\sum k = n (n + 1) /2

and the given sum of squares,

S_{n} = a^{2} d + a d^{2} \frac{n + 1}{n} + d^{3} \frac{( n + 1 ) ( 2 n + 1 )}{6 n ^{2}} .

An equivalent expression separates its limit and finite-

n

corrections:

S_{n} = \frac{b ^{3} - a ^{3}}{3} + \frac{( b - a ) ( b ^{2} - a ^{2} )}{2 n} + \frac{( b - a ) ^{3}}{6 n ^{2}} .

Consequently

n \to \infty lim S_{n} = \frac{b ^{3} - a ^{3}}{3} = \int_{a}^{b} x^{2} d x .

The function is continuous, so the Riemann sums converge to its definite integral, the desired area under the parabola. Both corrections are positive when

0 < a < b

, agreeing with the geometric fact that these right-endpoint rectangles overestimate an increasing curve.

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