设f(x)在[a,b]上连续可导,且f(a)=0.证明: ∫ab(x)dx≤∫ab[f’(x)]2dx.

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问题 设f(x)在[a,b]上连续可导,且f(a)=0.证明:
ab(x)dx≤ab[f’(x)]2dx.

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答案由f(a)=0,得f(x)-f(a)=f(x)=∫axf’(t)dt,由柯西不等式得 f2(x)=(∫axf’(t)dt)≤∫ax12dt∫axf’2(t)dt≤(x-a)∫abf’2(x)dx 积分得∫abf2(x)dx≤∫ab(x-a)dx.∫abf’2(x)dx=[*]∫abf’2(x)dx.

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