设f(x)在区间[0,1]上可积,当0≤x

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问题 设f(x)在区间[0,1]上可积,当0≤x01f(x)dx|≤ln2.

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答案由|f(x)|=|f(x)-f(1)|≤|arctanx-arctan1|=|arctanx-π/4|得 |∫01f(x)dx|≤∫01|f(x)|dx≤∫01 |arctanx-π/4|dx=∫01(π/4-arctanx)dx =π/4-∫01arctanrdx=π/4-xarctanx|01+∫01x/(1+x2)dx=(1/2)∫ln(1+x2)|01=ln2/2.

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