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设f(x)在[a,b]上有连续二阶导函数,且f(a)=f(b
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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