设f(x)在[a,b]上有连续的导数,证明 |∫abf(x)dx|+∫ab|f’(x)|dx.

admin2016-01-15  31

问题 设f(x)在[a,b]上有连续的导数,证明
    |∫abf(x)dx|+∫ab|f’(x)|dx.

选项

答案可设[*]|f(x)|=|f(x)|,即证 (b一a)|f(x0)|≤|∫abf(x)|+(b一a)∫ab|f’(x)|dx, 即|∫abf(x0)dx|—|∫abf(x)dx|≤(b—a)∫ab|f’(x)|dx. 事实上, |∫abf(x0)dx|—|∫abf(x)dx|≤|∫ab[f(x0)—f(x)]dx| =|∫ab[[*](t)dt]dx|≤∫ab[∫ab|f’(t)|dt]dx =(b一a)∫ab|f’(x)|dx. 故得证.

解析
转载请注明原文地址:https://jikaoti.com/ti/3IPRFFFM
0

随机试题
最新回复(0)