设f(x)在[a,b]上连续,证明:∫abf(x)dx∫xbf(y)dy=1/2[∫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=1/2[∫abf(x)dx]2.

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答案令F(x)=∫axf(t)dt, 则∫abf(x)dx∫xbf(y)dy=∫abf(x)[F(b)-F(x)]dx =F(b)|f(x)dx-∫abf(x)F(x)dx=F2(b)-∫abF(x)dF(x) =F2(b)-(1/2)F2(x)|ab=1/2F2(b)=1/2[∫abf(x)dx]2.

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