设f(x)在区间【0,1】上连续,在(0,1)内可导,且∫_0^1▒〖f(x)dx〗=0.证明;存在一点ξ(0,1),使得2f(ξ)+ξf‘(ξ)=0,
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![设f(x)在区间【0,1】上连续,在(0,1)内可导,且∫_0^1▒〖f(x)dx〗=0.证明;存在一点ξ(0,1),使得2f(ξ)+ξf‘(ξ)=0,](/uploads/image/z/2624680-64-0.jpg?t=%E8%AE%BEf%EF%BC%88x%EF%BC%89%E5%9C%A8%E5%8C%BA%E9%97%B4%E3%80%900%2C1%E3%80%91%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%EF%BC%880%2C1%EF%BC%89%E5%86%85%E5%8F%AF%E5%AF%BC%2C%E4%B8%94%E2%88%AB_0%5E1%26%239618%3B%E3%80%96f%EF%BC%88x%EF%BC%89dx%E3%80%97%3D0.%E8%AF%81%E6%98%8E%EF%BC%9B%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9%CE%BE%EF%BC%880%2C1%EF%BC%89%2C%E4%BD%BF%E5%BE%972f%EF%BC%88%CE%BE%EF%BC%89%2B%CE%BEf%E2%80%98%EF%BC%88%CE%BE%EF%BC%89%3D0%2C)
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