设f(x+1)=xe-x，求∫02(x)dx．
令x=t+1，当x=0时，t=-1，x=2时，t=1， ∫02f(x)dx=∫-11f(t+1)dt=f(x+1)dx =∫-11xe-xdx=-xe-x|-11+∫-11e-xdx =-e-1-e-e-x|1-1=-(2/e)
设f(x+1)=xe-x，求∫02(x)dx．
令x=t+1，当x=0时，t=-1，x=2时，t=1， ∫02f(x)dx=∫-11f(t+1)dt=f(x+1)dx =∫-11xe-xdx=-xe-x|-11+∫-11e-xdx =-e-1-e-e-x|1-1=-(2/e)
