2. 考研
  3. 设函数f(x)在[0，2π]上连续可微，f’(x)≥0，证明：对任意正整数n，有｜∫02πf(x)sinnxdx｜≤[f(2π)－f(0)]．

设函数f(x)在[0，2π]上连续可微，f’(x)≥0，证明：对任意正整数n，有｜∫02πf(x)sinnxdx｜≤[f(2π)－f(0)]．

admin2023-01-06  19
问题 设函数f(x)在[0，2π]上连续可微，f’(x)≥0，证明：对任意正整数n，有｜∫02πf(x)sinnxdx｜≤[f(2π)－f(0)]．
选项
答案因为f’(x)≥0，所以f(0)≤f(2π)，从而f(2π)－f(0)≥0． 由∫02xf(x)sinnxdx=－1/n∫02πf(x)d(cosnx) =－1/nf(x)cosnx｜02π+1/n∫02πf’(x)cosnxdx =－1/n[f(2π)－f(0)]+1/n∫02πf’(x)cosnxdx得 ｜∫02πf(x)sinnxdx｜≤1/n[f(2π)-f(0)]+1/n∫02π｜f’(x)cosnx｜dx ≤1/n[f(2π)－f(0)]+1/n∫02πf’(x)dx=2/n[f(2π)－f(0)]．
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考研数学二

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