 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-2/16/a/i/solution

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

and

F (r (t)) = (t^{6} - 2 t^{2}, 2 t^{5} - t^{3})^{T} .

The line integral is therefore

\int_{C} F \cdot d r = \int_{0}^{1} F (r (t)) \cdot r^{'} (t) d t = \int_{0}^{1} (8 t^{7} - 3 t^{5} - 4 t^{3}) d t = 1 - \frac{1}{2} - 1 = - \frac{1}{2} .

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