∫[0→2] f(x-1)dx
令x-1=t,则dx=dt,t：-1→1
=∫[-1→1] f(t) dt
=∫[-1→0] f(t) dt+∫[0→1] f(t) dt
=∫[-1→0] (1+t²) dt+∫[0→1] te^(-t²) dt
=[t+(1/3)t³]+(1/2)∫[0→1] e^(-t²) d(t²) 前一项用[-1→0]代入
=4/3-(1/2)e^(-t²) |[0→1]
=4/3-(1/2)e^(-1)+1/2
=11/6-1/(2e)