设f(x)= S0=∫02f(x)e一xdx,S1=∫24f(x一2)e一xdx,…,Sn=∫2n2n+2+2f(x一2n)e一xdx,求

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问题 设f(x)=
S0=∫02f(x)e一xdx,S1=∫24f(x一2)e一xdx,…,Sn=∫2n2n+2+2f(x一2n)e一xdx,求

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答案S0=∫02f(x)e一xdx=∫01xe一xdx+∫12(2一x)e一xdx=[*] 令t=x一2,则S1=e一202f(t)e一tdt=e一2S0, 令t=x一2n,则Sn=e一2n02f(t)e一tdt=e一2nnS0, [*]

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