设f(x)在[a,b]上连续,证明:∫abf(x)dx∫xbf(y)dy=[∫abf(x)dx]2.

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问题 设f(x)在[a,b]上连续,证明:∫abf(x)dx∫xbf(y)dy=[∫abf(x)dx]2

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答案令F(x)=∫axf(t)dt, 则 ∫f(x)dx∫f(y)dy =∫f(x)[F(b)-F(x)]dx =F(b)∫abf(x)dx-∫abf(x)F(x) =F(b)-∫abF(x)dF(x) =F2(b)-[*]F2(x)|ab =[*]F2(b) =[*][∫abf(x)dx]2.

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