设f(x)=e-tcos tdt,求f(x)在[0,π]上的最大值与最小值.

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问题 设f(x)=e-tcos tdt,求f(x)在[0,π]上的最大值与最小值.

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答案由f(x)=[*]e-xcos tdt,令f(x)=e-xcos x=0,得x=π/2 又 f(x)=[*]e-tcos tdt=[*]e一d(sin t)=e-tsin t[*]e-tsin tdt =e-xsin x—[*]e-td(cos t)=e-xsin x一e-tcos t[*]cos tdt =e-x(sin x—cos x)+1-f(x), 所以 f(x)=[*]e-x=(sin x—cos x)+1/2 因此可知f(0)=0,f[*]+1),f(π)=1/2(e-x+1),所以 fmax(x)=f[*]+1),fmin(x)=f(0)=0.

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