证明f(x)在[a,b]上可导,导函数f‘(x)可积,并且f(b)-f(a)=1证明∫a到b[f’(x)]^2dx>=1/(b-a)
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![证明f(x)在[a,b]上可导,导函数f‘(x)可积,并且f(b)-f(a)=1证明∫a到b[f’(x)]^2dx>=1/(b-a)](/uploads/image/z/9855714-66-4.jpg?t=%E8%AF%81%E6%98%8Ef%EF%BC%88x%EF%BC%89%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E5%8F%AF%E5%AF%BC%2C%E5%AF%BC%E5%87%BD%E6%95%B0f%E2%80%98%EF%BC%88x%EF%BC%89%E5%8F%AF%E7%A7%AF%2C%E5%B9%B6%E4%B8%94f%EF%BC%88b%EF%BC%89-f%EF%BC%88a%EF%BC%89%3D1%E8%AF%81%E6%98%8E%E2%88%ABa%E5%88%B0b%5Bf%E2%80%99%EF%BC%88x%EF%BC%89%5D%5E2dx%3E%3D1%2F%28b-a%29)
证明f(x)在[a,b]上可导,导函数f‘(x)可积,并且f(b)-f(a)=1证明∫a到b[f’(x)]^2dx>=1/(b-a)
证明f(x)在[a,b]上可导,导函数f‘(x)可积,并且f(b)-f(a)=1证明∫a到b[f’(x)]^2dx>=1/(b-a)
证明f(x)在[a,b]上可导,导函数f‘(x)可积,并且f(b)-f(a)=1证明∫a到b[f’(x)]^2dx>=1/(b-a)
∫a到b[f’(x)]^2dx*∫a到b dx≥[∫a到bf'(x)dx]^2,这步是由许尔瓦兹不等式来的,左边等于(b-a)*∫a到b[f’(x)]^2dx,右边等于1,结论就出来了.
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