f(x)=1/2+x x>0 1+e的x次方 x

∫ [0-->2] f(x-1) dx
令x-1=t,dx=dt,t：-1-->1
=∫ [-1-->1] f(t) dt
=∫ [-1-->0] 1/(2+t) dt+∫ [0-->1] (1+e^t) dt
=ln|2+t| | [-1-->0] +(t+e^t) | [0-->1]
=ln2+1+e-1
=ln2+e

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