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Today's calculation of Integral 800

Source: 2012 Kumamoto University entrance exam/Medicine.

March 6, 2012
calculusintegrationtrigonometryfunctioncalculus computations

Problem Statement

For a positive constant aa, find the minimum value of f(x)=∫0π2∣sin⁡t−axcos⁡t∣dt.f(x)=\int_0^{\frac{\pi}{2}} |\sin t-ax\cos t|dt.
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