 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-2/3b/a/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

x = t

y = 1 + t^{2}

d x = d t

, and

d y = 2 t d t

. Hence

\int_{0}^{1} [- 1 + 2 t ((1 + t^{2})^{2} + t)] d t = \int_{0}^{1} (- 1 + 2 t + 2 t^{2} + 4 t^{3} + 2 t^{5}) d t = 2 .

The difference from the straight-path value reflects that the one-form is not exact.

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