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109

Part of 2006 Today's Calculation Of Integral

Problems(1)

Today's calculation of integral 109

Source: created by kunny

6/7/2006
Let In=∫20062006+1nxcos⁡2(x−2006) dx (n=1,2,⋯ ).I_n=\int_{2006}^{2006+\frac{1}{n}} x\cos ^ 2 (x-2006)\ dx\ (n=1,2,\cdots). Find lim⁡n→∞nIn.\lim_{n\to\infty} nI_n.
calculusintegrationtrigonometrylimitinequalitiesfunctioncalculus computations