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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 12D / b / i / Solution

No. Define

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.

Solved by gpt-5.6-sol high.

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