设|f’(x)|≤M,x∈[0,1],且f(0)=f(1)=0,试证:|∫01f(x)dx|≤M.

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问题 设|f’(x)|≤M,x∈[0,1],且f(0)=f(1)=0,试证:|∫01f(x)dx|≤M.

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答案因为∫01f(x)dx=∫01f(x)d(x—c)=(x—c)f(x)|01一∫01(x—c)f’(x)dx =一∫01(x—c)f’(x)dx, 所以 |∫01f(x)dx|≤|∫01(x—c)f’(x)dx|≤∫01(x—c)||f’(x)dx ≤M∫01|x—c|dx=M[∫0c(c—x)dx+∫c1(x—c)dx] [*]

解析 要证结论是比较积分与被积函数的导函数值之大小,用分部积分法建立f(x)与f’(x)定积分的关系式,然后再放缩.由f(0)=f(1)=0可知,分部积分应注意应用小技巧dx=d(x—c),c∈[0,1].
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