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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/6/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 6 / b / Solution by

Codex  0 2026-10-31

Use the substitution

u = x^{2}

, so

d u = 2 x d x

. The antiderivative is

\int 4 x e^{x^{2}} d x = 2 \int e^{u} d u = 2 e^{x^{2}} + C .

Differentiation by the chain rule gives

4 x e^{x^{2}}

, verifying the result.

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