已知f''(x)在[0,1]上连续,f'(1)=0,且f(1)-f(0)=2,则∫(0,1)xf''(x)dx=

∫(0,1)xf''(x)dx
=∫(0,1)xdf'(x)
=xf'(x)|(0,1)-∫(0,1)f'(x)dx
=f'(1)-0-f(x)|(0,1)
=0-[f(1)-f(0)]
=-2