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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答案方法一 令F(x)=∫0xf(t)dt,则F(x)=f(x),于是∫0x[∫0tf(μ)dμ]dt=∫0xF(t)dt, ∫0xf(t)(x-t)dt=x∫0xf(t)dt—∫0xtf(t)dt=xF(x)一∫0xtdF(t) =xF(x)一tF(t)|0x+∫0xF(t)dt=∫0xF(t)dt. 命题得证. 方法二 因为[*]∫0x[∫0tf(μ)dμ]dt=∫0xf(μ)dμ, [*][x∫0xf(t)dt-∫0xtf(t)]=∫0xf(t)dt, 所以∫0x[∫0tf(μ)dμ一∫0xf(t)(x—t)dt≡C0,取x=0得C0=0,故 ∫0x[∫0tf(μ)dμ]dt=∫0xf(t)(x-t)dt.

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