求下列积分: (Ⅰ)设f(x)=,求∫01x2f(x)dx; (Ⅱ)设函数f(x)在[0,1]连续且∫01f(x)dx=A,求∫01dxf(x)f(y)dy.

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问题 求下列积分:
(Ⅰ)设f(x)=,求∫01x2f(x)dx;
(Ⅱ)设函数f(x)在[0,1]连续且∫01f(x)dx=A,求∫01dxf(x)f(y)dy.

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答案(Ⅰ) [*] (Ⅱ)令Ф(x)=∫x1f(y)dy,则Ф’(x)=-f(x),于是 ∫01dx∫x1f(x)f(y)dy=∫01[∫x1f(y)dy]f(x)dx =-∫01Ф(x)dФ(x)=[*]

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