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881

Part of 2013 Today's Calculation Of Integral

Problems(1)

Today's calculation of Integral 881

Source: 2013 Kagoshima University entrance exam/Medicine

6/29/2013
Evaluate ∫−ππ(∑k=12013sin⁡kx)2dx\int_{-\pi}^{\pi} \left(\sum_{k=1}^{2013} \sin kx\right)^2dx.
calculusintegrationtrigonometrycalculus computationsDefinite integral