作业帮设f(x)在[a,b]上有连续二阶导函数,且f(a)=f(b)=0,证明∫[a,b][2f(x)-(x-a)(x-b)f''(x)]dx=0
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![设f(x)在[a,b]上有连续二阶导函数,且f(a)=f(b)=0,证明∫[a,b][2f(x)-(x-a)(x-b)f''(x)]dx=0](/uploads/image/z/5350006-46-6.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%E4%BA%8C%E9%98%B6%E5%AF%BC%E5%87%BD%E6%95%B0%2C%E4%B8%94f%28a%29%3Df%28b%29%3D0%2C%E8%AF%81%E6%98%8E%E2%88%AB%5Ba%2Cb%5D%5B2f%28x%29-%28x-a%29%28x-b%29f%27%27%28x%29%5Ddx%3D0)
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设f(x)在[a,b]上有连续二阶导函数,且f(a)=f(b)=0,证明∫[a,b][2f(x)-(x-a)(x-b)f''(x)]dx=0
设f(x)在[a,b]上有连续二阶导函数,且f(a)=f(b)=0,证明∫[a,b][2f(x)-(x-a)(x-b)f''(x)]dx=0
设f(x)在[a,b]上有连续二阶导函数,且f(a)=f(b)=0,证明∫[a,b][2f(x)-(x-a)(x-b)f''(x)]dx=0
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