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767

Part of 2011 Today's Calculation Of Integral

Problems(1)

Today's calculation of Integral 767

Source:

11/17/2011
For 0≤t≤10\leq t\leq 1, define f(t)=∫02π∣sin⁡x−t∣dx.f(t)=\int_0^{2\pi} |\sin x-t|dx. Evaluate ∫01f(t)dt.\int_0^1 f(t)dt.
calculusintegrationtrigonometrycalculus computations