设f(x)在[a,b]上有连续的导函数,且f(b)=0,当x∈[a,b]时|f’(x)|≤M,证明: |∫abf(x)dx|≤.

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问题 设f(x)在[a,b]上有连续的导函数,且f(b)=0,当x∈[a,b]时|f’(x)|≤M,证明:
    |∫abf(x)dx|≤

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答案对∫abf(x)d(x一a)利用分部积分公式.由于 ∫abf(x)dx=∫abf(x)d(x—a)=(x—a)f(x)|ab—∫ab(x—a)f’(x)dx =一∫ab(x一a)f’(x)dx, 因此 ∫abf(x)dx|=|∫ab(x一a)f’(x)dx|≤M∫ab(x一a)dx=[*].

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