设f’(x)在[0,1]上连续,且f(1)=f(0)=1.证明:∫01f’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:∫01f’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)2≤∫0112dxf’2(x)dx=∫01f’2(x)dx,即∫01f’2(x)dx≥1.

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