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

设f(x)在[a，b]上连续，证明：∫abf(x)dx∫abf(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∫abf(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)一F(x)]dx =F(b)∫abf(x)dx—∫abf(x)F(x)dx=F2(b)一∫abF(x)dF(x) =F2(b)一[*][∫abf(x)dx]2．
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