设f(x)在[a,b]有连续的导数,求证: |∫abf(x)dx|+∫ab|f’(x)|dx.

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问题 设f(x)在[a,b]有连续的导数,求证:
|∫abf(x)dx|+∫ab|f’(x)|dx.

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答案可设[*]|f(x)|=|f(x0)|,即证 (b-a)|f(x0)|≤|∫abf(x)dx|+(b-a)∫ab|f’(x)|dx, 即证|∫abf(x0)dx|-|∫abf(x)dx|≤(b-a)∫ab|f’(x)|dx. 注意|∫abf(x0)dx|-|∫abf(x)dx|≤|∫ab[f(x0)-f(x)]-dx| =|∫ab[∫xx0f’(t)dt]dx|≤∫ab[∫ab|f’(t)|dt]dx=(b-a)∫ab|f’(x)|dx. 故得证.

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