 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-1/12d/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

f (0) = 0, f (x) = \frac{1}{x} (0 < x \leq 1) .

It is continuous at every point of

(0, 1)

and is unbounded. Its truncation is Riemann integrable, but

\int_{0}^{1} min (\frac{1}{x}, r) d x = \int_{0}^{1/ r} r d x + \int_{1/ r}^{1} \frac{d x}{x} = 1 + lo g r ⟶ \infty.

Thus it is not improperly integrable with a finite integral.

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