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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/20/b/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 / 2021 / Paper 1 / 20 / b / Solution by

Codex  0 2026-10-18

Apply the Leibniz integral rule: differentiate both moving endpoints and the integrand's explicit

x

dependence. The partial derivative is

\partial_{x} sin (x t^{2}) = t^{2} cos (x t^{2})

. Hence

f^{'} (x) = β e^{β x} sin (x e^{2 β x}) - α e^{αx} sin (x e^{2 αx}) + \int_{e^{αx}}^{e^{β x}} t^{2} cos (x t^{2}) d t .

(1)

The signs distinguish upper and lower endpoint contributions. The formula also holds when the limits occur in reversed order, by the oriented-integral convention. If

α = β

, the integral defining

f

is identically zero and all displayed terms cancel, so

f^{'} (x) = 0 when α = β .

(2)

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