 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/20/b/solution

The proposed parametric equation traces precisely the same oriented segment, with a different speed. Set

u = t^{n}

, which increases from zero to one, and

d u = n t^{n - 1} d t

. Direct substitution gives the line integrals

\int F \cdot d x \int G \cdot d x = \int_{0}^{1} (2 - 2 t^{n}) n t^{n - 1} d t = \int_{0}^{1} (2 - 2 u) d u = 1, = \int_{0}^{1} (2 - 4 t^{n} + 2 t^{2 n}) n t^{n - 1} d t = \int_{0}^{1} (2 - 4 u + 2 u^{2}) d u = \frac{2}{3} .

Thus both values are unchanged. The derivative of

u

vanishes at the initial endpoint for

n > 1

, but the monotone integration by substitution is still valid. This is invariance under reparametrization of one curve, which applies even to a nonconservative vector field; it is not evidence of general path independence.

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