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

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．
选项
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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考研数学三

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