设f(x)在[a,b]上连续,证明:∫abf(x)dx∫xbf(x)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(x)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)∫abf(x)dx-∫abf(x)F(x)dx=F2(b)-∫abF(x)dF(x)=F2(b)-1/2F2(x)|ab=1/2F2(b)=1/2[∫abf(x)dx]2

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