设f(x)在[0,1]上连续,试证: ∫01dx∫0xdy∫0yf(x)f(y)dz=[∫01f(t)dt]3

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问题 设f(x)在[0,1]上连续,试证:
01dx∫0xdy∫0yf(x)f(y)dz=[∫01f(t)dt]3

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答案因为f(x)在[0,1]上连续,所以在[0,1]上存在原函数. 设F’(t)=f(t)(t∈[0,1]),则 ∫01dx∫0xdy∫0yf(x)f(y)f(z)dz=∫01f(x)dx∫0xf(y)[F(y)-F(0)]dy =∫01f(x)dx∫0x[F(y)-F(0)]d[F(y)-F(0)] [*]

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