设f(x)在【0,1】上连续且∫(0,1)f(x)dx=A,证明∫(0,1)dx∫(x,1)f(x)f(y)dy=A∧2/2,谢谢!
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![设f(x)在【0,1】上连续且∫(0,1)f(x)dx=A,证明∫(0,1)dx∫(x,1)f(x)f(y)dy=A∧2/2,谢谢!](/uploads/image/z/2651170-58-0.jpg?t=%E8%AE%BEf%EF%BC%88x%EF%BC%89%E5%9C%A8%E3%80%900%2C1%E3%80%91%E4%B8%8A%E8%BF%9E%E7%BB%AD%E4%B8%94%E2%88%AB%280%2C1%29f%28x%29dx%3DA%2C%E8%AF%81%E6%98%8E%E2%88%AB%280%2C1%29dx%E2%88%AB%28x%2C1%29f%28x%29f%28y%29dy%3DA%E2%88%A72%2F2%2C%E8%B0%A2%E8%B0%A2%21)
设f(x)在【0,1】上连续且∫(0,1)f(x)dx=A,证明∫(0,1)dx∫(x,1)f(x)f(y)dy=A∧2/2,谢谢!
设f(x)在【0,1】上连续且∫(0,1)f(x)dx=A,证明∫(0,1)dx∫(x,1)f(x)f(y)dy=A∧2/2,谢谢!
设f(x)在【0,1】上连续且∫(0,1)f(x)dx=A,证明∫(0,1)dx∫(x,1)f(x)f(y)dy=A∧2/2,谢谢!
设∫f(x)dx=F(x),则F(0)=0,F(1)=A,
∫[∫f(x)f(y)dy]dx
=∫f(x)[∫dF(y)] dx
=∫[A-F(x)]dF(x )
=A∫f(x)dx -(A^2)/2
=(A^2)/2