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)]dt=∫0xF(t)dt, ∫0xf(t)(x-t)dt=x∫0xf(t)dt-∫0xtf(t)dt=xF(x)-∫0xtdF(t)dt=xF(x)-tF(t)|0x+∫0xF(t)dt=∫0xF(t)dt 命颢得证.

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