设函数f(x)在[0,1]上连续且非负,证:存在ζ∈(0,1)使ζf(ζ)=∫(1,ζ)f(x)dx
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![设函数f(x)在[0,1]上连续且非负,证:存在ζ∈(0,1)使ζf(ζ)=∫(1,ζ)f(x)dx](/uploads/image/z/5392636-52-6.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%5B0%2C1%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%E4%B8%94%E9%9D%9E%E8%B4%9F%2C%E8%AF%81%3A%E5%AD%98%E5%9C%A8%CE%B6%E2%88%88%280%2C1%29%E4%BD%BF%CE%B6f%28%CE%B6%29%3D%E2%88%AB%281%2C%CE%B6%29f%28x%29dx)
设函数f(x)在[0,1]上连续且非负,证:存在ζ∈(0,1)使ζf(ζ)=∫(1,ζ)f(x)dx
设函数f(x)在[0,1]上连续且非负,证:存在ζ∈(0,1)使ζf(ζ)=∫(1,ζ)f(x)dx
设函数f(x)在[0,1]上连续且非负,证:存在ζ∈(0,1)使ζf(ζ)=∫(1,ζ)f(x)dx
证明:令F(x)=x*积分(从x到1)f(t)dt,0
令ζ=0和ζ=1,就可以证明了
F(ζ)=ζf(ζ)-∫(1,ζ)f(x)dx
设函数f(x)在[0,1]上连续且非负,证:存在ζ∈(0,1)使ζf(ζ)=∫(1,ζ)f(x)dx
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