2. 考研
  3. 设f(x)在[0，+∞)上可导，f(0)=1，且f’(x)一f(x)+=0，求∫[f’’(x)－f’(x)]e－xdx．

设f(x)在[0，+∞)上可导，f(0)=1，且f’(x)一f(x)+=0，求∫[f’’(x)－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)+=0，求∫[f’’(x)－f’(x)]e－xdx．
选项
答案由题设知，f’(x)—f(x)+[*]∫0xf(t)dt=0，并且f’(0)=f(0)=1．于是，有(1+x)f’(z)一(1+x)f(z)+∫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．分离变量得[*]，即 [*] 由f’(0)=1，得c=1．代入上式，[*]故 ∫[f’’(x)一f’(x)]e－xdx=∫f’’(x)e－xdx—∫f’(x)e－xdx =f’(x)e－x+∫f’(x)e－xdx一∫f’(x)e－xdx =f’(x)e－x+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)．
含有抽象函数导数的积分，一般用分部积分法．
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考研数学一

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