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)－[*]F2(x)｜ab=[*]F2(b)=[*][∫abf(x)dx]2
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考研数学二

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