若f(x)在[0,1]上連續，對任意的x,y都有| f(x) - f(y) | < M*|x - y|，M為正常數
試證明 | ∫(0,1) f(x) dx - (1/n)Σ(k=1,n) f(k/n) | ≤ M/(2n)
含定積分的不等式證明
2014-05-13 10:54 pm
回答 (1)
2014-05-15 6:11 pm
✔ 最佳答案
| ∫[(k-1)/n,k/n] f(x) - (1/n)f(k/n)|=|∫[(k-1)/n, k/n] [f(x)-f(k/n)] dx |
< = ∫[(k-1)/n, k/n] |f(x)-f(k/n)| dx
< ∫[(k-1)/n, k/n] M(k/n - x) dx
= M/(2n^2)
故
|Σ(k=1,n) {∫[(k-1)/n, k/n] f(x) dx - 1/n f(k/n) }|
< = Σ(k=1,n) |[(k-1)/n, k/n] f(x) dx - (1/n) f(k/n) |
< Σ(k=1,n) M/(2n^2)
= M/(2n)
收錄日期: 2021-04-16 16:30:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140513000051KK00070