设f’(x)在[0,1]上连续,且f(1)-f(0)=1.证明:∫0af’2(x)dx≥1.

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问题 设f’(x)在[0,1]上连续,且f(1)-f(0)=1.证明:∫0af’2(x)dx≥1.

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答案由1=f(1)-f(0)=∫01f’(x)dx, 得12=1=(∫01f’(x)dx)≤∫0112dx∫01f’2(x)dx= ∫01f’2(x)dx,即∫01f’2(x)dx≥1.

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