2. 考研
  3. 若f在[0，a]上连续可微，且f(0)=0，则 ∫0a｜f(x)f’(x)｜dx≤∫0a[f’(x)]2dx．

若f在[0，a]上连续可微，且f(0)=0，则 ∫0a｜f(x)f’(x)｜dx≤∫0a[f’(x)]2dx．

admin2022-11-23  33
问题 若f在[0，a]上连续可微，且f(0)=0，则
    ∫0a｜f(x)f’(x)｜dx≤∫0a[f’(x)]2dx．
选项
答案因为 ∫0a｜f(x)f’(x)｜dx=∫0a｜f(x)｜d｜f(x)｜=[*] 又f(0)=0，f(a)=∫0af’(x)dx，所以∫0a｜f(x)f’(x)｜dx=[*](∫0af’(x)dx)2． 由施瓦兹不等式可知， (∫0af’(x)dx)2=(∫0af’(x)·ldx)2≤(∫0a(f’(x))2dx)(∫0adx)≤a∫0a(f’(x))2dx． 因此，∫0a｜f(x)f’(x)｜dx≤[*][f’(x)]2dx．
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考研数学三

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