设f(x)在[a,b]上连续且严格单调增加,证明: (a+b)∫abf(x)dx<2∫abxf(x)dx.

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问题 设f(x)在[a,b]上连续且严格单调增加,证明:
    (a+b)∫abf(x)dx<2∫abxf(x)dx.

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答案令F(t)=(a+t)∫atf(x)dx一2∫atxf(x)dx,则 F’(t)=∫atf(x)dx+(a+t)f(t)一2tf(t) =∫atf(x)dx一(t一a)f(t)=∫atf(x)dx一∫atf(t)dx =∫at[f(x)—f(t)]dx. 因为a≤x≤t,且f(x)在[a,b]上严格单调增加,所以f(x)一f(t)≤0,于是有 F’(t)=∫at[f(x)一f(t)]dx≤0, 即F(t)单调递减,又F(a)=0,所以F(b)<0,即 (a+b)∫abf(x)dx一2∫abxf(x)dx<0, 即(a+b)∫abf(x)dx<xf(x)dx.

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