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

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．
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)－[*][∫abf(x)dx]2．
解析
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考研数学三

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