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=

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