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≤∫0112dx∫01f′2(x)dx=∫01f′2(x)dx,即∫01f′2(x)dx≥1.

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