设f(x)在[0,1]上可导,且f(0)=0,0<f’(x)<1,证明:[∫01f(x)dx]2>∫01f3(x)dx.

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问题 设f(x)在[0,1]上可导,且f(0)=0,0<f’(x)<1,证明:[∫01f(x)dx]2>∫01f3(x)dx.

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答案令φ(x)=[∫0xf(t)dt]2-∫0xf3(t)dt,φ(0)=0, φ’(x)=2f(x)∫0xf(t)dt-f3(x)=f(x)[2∫0xf(t)dt-f2(x)]. 再令h(x)=2∫0xf(t)dt-f2(x),h(0)=0,h’(x)=2f(x)[1-f’(x)]. 由f(0)=0,0<f’(x)<1得f(x)>0(0<x≤1), 则h’(x)=2f(x)[1-f’(x)]>0(0<x≤1), [*] 于是φ(1)>0,即[∫01f(x)dx]2>∫01f3(x)dx.

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