2. 考研
  3. 设f(x)在[0，+∞)上可导，f(0)=1，且f’(x)一f(x)+∫0xf(t)dt=0，求 ∫[f"(x)一f’(x)]e—xdx．

设f(x)在[0，+∞)上可导，f(0)=1，且f’(x)一f(x)+∫0xf(t)dt=0，求 ∫[f"(x)一f’(x)]e—xdx．

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问题 设f(x)在[0，+∞)上可导，f(0)=1，且f’(x)一f(x)+∫0xf(t)dt=0，求
    ∫[f"(x)一f’(x)]e—xdx．
选项
答案由题设知，f’(x)一f(x)+[*]∫0xf(t)dt=0，并且f’(0)=f(0)=1．于是，有 (1+x)f’(x)一(1+x)f(x)+∫0xf(t)dt=0． 两边对x求导得 f’(x)+(1+x)f"(x)一f(x)一(1+x)f’(x)+f(x)=0． 即(1+x)f"(x)一xf’(x)=0． 令f’(x)=p，则有(1+x)p’一xp=0． [*] ln｜p｜=x—ln(1+x)+ln｜c｜， [*] 由f’(0)=1，得c=1．代入上式得，p=f’(x)=[*]．故 ∫[f"(x)一f’(x)]e—xdx=∫f"(x)e—xdx一∫f’(x)e—xdx =f’(x)e—x+∫f’(x)e—xdx—∫f’(x)e—xdx =f’(x)e—x+c=[*]+c．
解析 ∫[f"(x)一f’(x)]e—xdx=∫f(x)e—xdx—∫f’(x)e—xdx
    =f’(x)e—x+∫f’(x)e—xdx—∫f’(x)e—xdx
    =f’(x)e—x+c．
  计算该积分的关键是求f’(x)．
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考研数学三

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