设f(x),g(x)在[0,1]上的导数连续,且f(0)=0,f’(x)≥0,g’(x)≥0.证明对任何a∈[0,1],有∫0ag(x)f’(x)dx+∫01f(x)g’(x)dx≥f(a)g(1).

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问题 设f(x),g(x)在[0,1]上的导数连续,且f(0)=0,f’(x)≥0,g’(x)≥0.证明对任何a∈[0,1],有∫0ag(x)f’(x)dx+∫01f(x)g’(x)dx≥f(a)g(1).

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答案设 F(x)=∫0xg(t)f’(t)dt+∫01f(t)g’(t)dt一f(x)g(1),则F(x)在[0,1]上的导数连续,并且 F’(x)=g(x)f’(x)-f’(x)g(1)=f’(x)[g(x)一g(1)]. 由于x∈[0,1]时,f’(x)≥0,g’(x)≥0,因此F’(x)≤0,即F(x)在[0,1]上单调递减. 注意到 F(1)=∫01g(t)f’(t)dt+∫01f(t)g’(t)dt一f(1)g(1), 而 ∫01g(t)f’(t)dt=∫01g(t)df(t)=g(t)f(t)|01一∫01f(t)g’(t)dt =f(1)g(1)一∫01f(t)g’(t)dt, 故F(1)=0. 因此x∈[0,1]时,F(x)≥F(1)=0,由此可得对任何a∈[0,1],有 ∫0ag(x)f’(x)dx+∫01f(x)g’(x)dx≥f(a)g(1).

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