设f(x)连续,且=∫0xf(x—t)costdt,求∫01f(x)dx.

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问题 设f(x)连续,且=∫0xf(x—t)costdt,求∫01f(x)dx.

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答案[*] =cosx∫0xf(u)cosudu+sinx∫0xf(u)sinudu, 且 2e2x=一sinx∫0xf(u)cosudu+f(x)cos2x+cosx∫0xf(u)sinudu+f(x)sin2x, 4e2x=f’(x)一cosx∫0xf(u)cosudu一f(x)sinxcosx—sinx∫0xf(u)sinudu +f(x)sinxcosx =f’(x)一e2x, 从而f’(x)=5e2x.于是,有 ∫01f(x)dx=xf(x)|01—5∫01xedx [*]

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