2. 考研
  3. 设f(x)连续，证明：∫0x[∫0tf(u)du]dt=∫0xf(x)(x—t)dt．

设f(x)连续，证明：∫0x[∫0tf(u)du]dt=∫0xf(x)(x—t)dt．

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问题 设f(x)连续，证明：∫0x[∫0tf(u)du]dt=∫0xf(x)(x—t)dt．
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答案令F(x)=∫0πf(t)dt，则F’(x)=f(x)，于是∫0x[∫0tf(u)du]dt=∫0xF(t)dt， ∫0xf(t)(x一t)dt=x∫0xxf(t)dt一∫0xtf(t)dt=xF(x)一∫0xtdF(t) =xF(x)一tF(x)|0x+∫0xF(t)dt=∫0xF(t)dt．命题得证．
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考研数学三

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