 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-2/5/a/ii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

For the curved path

r (t) = (t, t^{2}, t^{3})

r^{'} (t) = (1, 2 t, 3 t^{2})

v (r (t)) = (t^{2} e^{t}, e^{t} + 2 t^{11}, 3 t^{10}) .

(t^{2} + 2 t) e^{t} + 13 t^{12}

. Therefore the direct line integral is

\int_{C} v \cdot d r = \int_{0}^{1} [(t^{2} + 2 t) e^{t} + 13 t^{12}] d t = [t^{2} e^{t} + t^{13}]_{0}^{1} = e + 1.

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