设f''(x)在[0,1]上连续,f'(1)=0,且f(1)-f(2)=2,则∫(0,1)xf''(x)dx=
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![设f''(x)在[0,1]上连续,f'(1)=0,且f(1)-f(2)=2,则∫(0,1)xf''(x)dx=](/uploads/image/z/8756655-15-5.jpg?t=%E8%AE%BEf%27%27%28x%29%E5%9C%A8%5B0%2C1%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2Cf%27%281%29%3D0%2C%E4%B8%94f%281%29-f%282%29%3D2%2C%E5%88%99%E2%88%AB%280%2C1%29xf%27%27%28x%29dx%3D)
设f''(x)在[0,1]上连续,f'(1)=0,且f(1)-f(2)=2,则∫(0,1)xf''(x)dx=
设f''(x)在[0,1]上连续,f'(1)=0,且f(1)-f(2)=2,则∫(0,1)xf''(x)dx=
设f''(x)在[0,1]上连续,f'(1)=0,且f(1)-f(2)=2,则∫(0,1)xf''(x)dx=
原式=∫(0,1)xdf'(x)
=xf'(x)-∫(0,1)f'(x)dx
=[xf'(x)-f(x)](0,1)
=[1*f'(1)-f(1)]-[0*f'(0)-f(0)]
=f'(1)+f(0)-f(1)
因为不知道f(0)-f(1)
所以没法求
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