∵对任意的x，f(0)＝f(x)+f'(x)(0－x)f(1)＝f(x)+f'(x)(1－x)两式相加得∴2f(x)＝（2x-1)f'(x)即f(x)=(x-1/2)f'(x)且0≤x≤1∴l∫ f(x)dx l= l∫ (x-1/2)f'(x)dx l≤ ∫ |(x-1/2)f'(x)| dx= 1/2 ∫ |f’(x) |dx
