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

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

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

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