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615

Part of 2010 Today's Calculation Of Integral

Problems(1)

Today's calculation of Integral 615

Source:

6/28/2010
For 0≤a≤20\leq a\leq 2, find the minimum value of ∫02∣11+ex−11+ea∣ dx.\int_0^2 \left|\frac{1}{1+e^x}-\frac{1}{1+e^a}\right|\ dx.
2010 Kyoto Institute of Technology entrance exam/Textile e.t.c.
calculusintegrationlogarithmsderivativecalculus computations