设f(x)在[0,1]上连续,求∫01xnf(x)dx.

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问题 设f(x)在[0,1]上连续,求01xnf(x)dx.

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答案因为∫01xndx=[*],且连续函数|f(x)|在[0,1]存在最大值记为M,于是 |∫01xnf(x)dx|≤∫01xn|f(x)|dx≤M∫01xndx=[*] 又[*]∫01xnf(x)dx=0.

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