 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/20/c/iii/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 / 2024 / Paper 1 / 20 / c / iii / Solution by

Codex  0 2026-10-18

At

x = 0

, the integrand vanishes for

0 < t < 1

, so

I (0) = 0

and

g (0) = 1.

Solving the first-order linear differential equation from part (i) on

(- 1, \infty)

gives

g = C (x + 1)

. The initial condition fixes

C = 1

, so

g (x) = x + 1, x > - 1.

Equivalently, the defining exponent is the logarithmic integral representation of the natural logarithm,

I (x) = ln (x + 1)

.

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