设f(x)为连续函数,则定积分上限是1,下限是0,f(x/2)的导数,的定积分等于()A.f(1/2)-f(0) B.2[f(1/2)-f(0)] C.1/2[f(1/2)-f(0)] D.f(2)-f(0)
问题描述:
设f(x)为连续函数,则定积分上限是1,下限是0,f(x/2)的导数,的定积分等于()
A.f(1/2)-f(0) B.2[f(1/2)-f(0)] C.1/2[f(1/2)-f(0)] D.f(2)-f(0)
答
∫f'(x/2)dx=2∫f'(x/2)d(x/2)=2f(x/2)|(0,1) =2[f(1/2)-f(0)] 积分上下限打不上去……