设f(x)在[0,1]上可微,且f(1)=2∫0~1/2 xf(x)dx,证明存在ξ属于(0,1),使f(ξ)+ξf'(ξ)=1
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![设f(x)在[0,1]上可微,且f(1)=2∫0~1/2 xf(x)dx,证明存在ξ属于(0,1),使f(ξ)+ξf'(ξ)=1](/uploads/image/z/3858906-66-6.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%5B0%2C1%5D%E4%B8%8A%E5%8F%AF%E5%BE%AE%2C%E4%B8%94f%281%29%3D2%E2%88%AB0%7E1%2F2+xf%28x%29dx%2C%E8%AF%81%E6%98%8E%E5%AD%98%E5%9C%A8%CE%BE%E5%B1%9E%E4%BA%8E%280%2C1%29%2C%E4%BD%BFf%28%CE%BE%29%2B%CE%BEf%27%28%CE%BE%29%3D1)
设f(x)在[0,1]上可微,且f(1)=2∫0~1/2 xf(x)dx,证明存在ξ属于(0,1),使f(ξ)+ξf'(ξ)=1
设f(x)在[0,1]上可微,且f(1)=2∫0~1/2 xf(x)dx,证明存在ξ属于(0,1),使f(ξ)+ξf'(ξ)=1
设f(x)在[0,1]上可微,且f(1)=2∫0~1/2 xf(x)dx,证明存在ξ属于(0,1),使f(ξ)+ξf'(ξ)=1
证明:由积分中值定理,存在η∈(0,1/2)使
2∫[0→1/2] xf(x) dx=2*ηf(η)*(1/2)=ηf(η)=f(1)
令g(x)=xf(x),则g(η)=ηf(η)=f(1),g(1)=f(1)
因此g(x)在[η,1]内满足罗尔中值定理条件,
即存在ξ∈(η,1),使g'(ξ)=0,且g'(x)=f(x)+xf '(x)
因此:g'(ξ)=0即:f(ξ)+ξf '(ξ)=0.证毕
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是证明f(ξ)+ξf'(ξ)=1 ?
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