设f(x)在[a,b]上连续可导,且f(a)=f(b)=0.证明: |f(x)|≤∫ab|f’(x)|dx(a<x<b)

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问题 设f(x)在[a,b]上连续可导,且f(a)=f(b)=0.证明:
|f(x)|≤ab|f’(x)|dx(a<x<b)

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答案因为[*]且f(a)=f(b)=0,所以 [*]两式相加得|f(x)|≤[*]∫ab|f’(x)|dx.

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