 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/18/c/ii/solution

The square is even, and direct integration gives

\int_{- 1}^{1} f (x)^{2} d x = 2 \int_{0}^{1} x^{2} (1 - x)^{2} d x = 2 (\frac{1}{3} - \frac{1}{2} + \frac{1}{5}) = \frac{1}{15} .

\frac{1}{30} = \frac{1}{2} \sum_{m = 0}^{\infty} \frac{64}{π ^{6} ( 2 m + 1 ) ^{6}} .

Hence

m = 0 \sum \infty \frac{1}{( 2 m + 1 ) ^{6}} = \frac{π ^{6}}{960}, A = \frac{1}{960} .

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