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Today's calculation of Integral 881

Source: 2013 Kagoshima University entrance exam/Medicine

June 29, 2013
calculusintegrationtrigonometrycalculus computationsDefinite integral

Problem Statement

Evaluate ∫−ππ(∑k=12013sin⁡kx)2dx\int_{-\pi}^{\pi} \left(\sum_{k=1}^{2013} \sin kx\right)^2dx.
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