设f(x)在[a,b]上有连续的导数,且f(x)不恒等于0,f(a)=f(b)=0,证明∫(a,b)xf(x)f'(x)dx
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![设f(x)在[a,b]上有连续的导数,且f(x)不恒等于0,f(a)=f(b)=0,证明∫(a,b)xf(x)f'(x)dx](/uploads/image/z/2789572-4-2.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E6%9C%89%E8%BF%9E%E7%BB%AD%E7%9A%84%E5%AF%BC%E6%95%B0%2C%E4%B8%94f%28x%29%E4%B8%8D%E6%81%92%E7%AD%89%E4%BA%8E0%2Cf%28a%29%3Df%28b%29%3D0%2C%E8%AF%81%E6%98%8E%E2%88%AB%28a%2Cb%29xf%28x%29f%27%28x%29dx)
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设f(x)在[a,b]上有连续的导数,且f(x)不恒等于0,f(a)=f(b)=0,证明∫(a,b)xf(x)f'(x)dx
设f(x)在[a,b]上有连续的导数,且f(x)不恒等于0,f(a)=f(b)=0,证明∫(a,b)xf(x)f'(x)dx<0
设f(x)在[a,b]上有连续的导数,且f(x)不恒等于0,f(a)=f(b)=0,证明∫(a,b)xf(x)f'(x)dx
用分部积分就可以证明了,∫(a,b)xf(x)f'(x)dx=∫(a,b)xf(x)df(x)=1/2∫(a,b)xdf(x)^2=1/2x*f(x)^2|(a,b)-1/2∫(a,b)f(x)^2dx,因为f(a)=f(b)=0,所以有1/2x*f(x)^2|(a,b)=0,而∫(a,b)f(x)^2dx中被积函数是正数,所以积分大于零,从而得正,
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