 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/14/a/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

0 \leq x \leq 2

0 \leq y \leq x^{3}

. Reversing the order in the double integral and then putting

u = x^{4} + 1

gives

\int_{0}^{2} \int_{0}^{x^{3}} x^{4} + 1 d y d x = \int_{0}^{2} x^{3} x^{4} + 1 d x = [\frac{( x ^{4} + 1 ) ^{3/2}}{6}]_{0}^{2} .

Therefore

I = \frac{17 17 - 1}{6} .

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