2. 考研
  3. 设f(x)在[a,b]上连续可导，且f(a)=0,证明：∫abf2(x)dx≤∫ab[f’(x)]2dx.

设f(x)在[a,b]上连续可导，且f(a)=0,证明：∫abf2(x)dx≤∫ab[f’(x)]2dx.

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问题 设f(x)在[a,b]上连续可导，且f(a)=0,证明：∫abf2(x)dx≤∫ab[f’(x)]2dx.
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答案由f(a)=0得f(x)－f(a)=f(x)=∫axf’(t)dt,由柯西不等式得 f2(x)=(∫axf’(t)dt)2≤∫axl2dt∫axf’2(t)dt≤(x－a)∫abf’2(x)dx 积分得∫abf2(x)dx≤∫ab(x－a)dx·∫abf’2(x)dx=[*]∫abf’2(x)dx.
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考研数学二

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