设f(x)在[a,b]上连续,f(a)=f(b)=0,定积分f^2(x)从b到a等于1,则定积分xf(x)f'(x)等于多少A.1/2 B.1 C.0 D.-1/2

问题描述:

设f(x)在[a,b]上连续,f(a)=f(b)=0,定积分f^2(x)从b到a等于1,则定积分xf(x)f'(x)等于多少
A.1/2 B.1 C.0 D.-1/2

刚回荅:
∫xf(x)f'(x)dx=(1/2)∫xdf(x)^2=(1/2)xf(x)^2-(1/2)∫f(x)^2dx,代入上下限后=-1/2.选D