设函数f(x)在[a,b]上连续,f’(x)在[a,b]上存在且可积,f(a)=f(b)=0,试证:|f(x)|≤∫abf(x)|dx (a<x<b)

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问题 设函数f(x)在[a,b]上连续,f’(x)在[a,b]上存在且可积,f(a)=f(b)=0,试证:|f(x)|≤abf(x)|dx (a<x<b)

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答案abf’(x)dx=∫axf’(t)dt+f’(t)dt

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