 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-2/1c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

I'' = \int_{0}^{1} y^{4} e^{x y} d y

. Integrating the derivative of

e^{x y} (x y^{3} - 3 y^{2})

and then once more yields

x^{2} \int_{0}^{1} y^{4} e^{x y} d y - 12 \int_{0}^{1} y^{2} e^{x y} d y = e^{x} (x - 4)

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