 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/12/a/iii/solution

With

y

held fixed, the inner antiderivative gives

\int_{2 y}^{2} x^{2} y^{2} d x = \frac{y ^{2}}{3} [x^{3}]_{2 y}^{2} = \frac{8}{3} (y^{2} - y^{5}) .

The remaining integral is therefore

\frac{8}{3} \int_{0}^{1} (y^{2} - y^{5}) d y = \frac{8}{3} (\frac{1}{3} - \frac{1}{6}) = \frac{4}{9} .

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