Question

设f(x)在[a,b]上有连续二阶导数，且f(a)=f(b)=0,试证f(x)在[a,b]上的积分=｛1/2(x-a)(x-b)*f(x)的二阶导数

设f(x)在[a,b]上有连续二阶导数，且f(a)=f(b)=0,试证f(x)在[a,b]上的积分=｛1/2(x-a)(x-b)*f(x)的二阶导数｝

Answer 1

∫[a,b]{1/2*(x-a)(x-b)f''(x)dx
=∫[a,b]1/2*(x-a)(x-b)df'(x)
=[1/2*(x-a)(x-b)f'(x)]|[a,b]-∫[a,b]1/2*(2x-a-b)f'(x)dx
=-∫[a,b]1/2*(2x-a-b)df(x)
=[-1/2*(2x-a-b)f(x)]|[a,b]+∫[a,b]f(x)dx
=∫[a,b]f(x)dx
所以原题应为：f(x)在[a,b]上的积分=｛1/2(x-a)(x-b)*f(x)的二阶导数｝在[a,b]上的积分