设,求∫01x2f(x)dx.

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问题,求∫01x2f(x)dx.

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答案01x2f(x)dx=1/3∫01f(x)d(x3)=1/3x3f(x)|01-1/3∫01x3f’(x)dx=-1/3∫01x3e-xdx=-1/6∫01te-tdt=1/6(te-t01-∫01e-tdt)=1/6(2e-1-1).

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