Evaluate Integral

Is there any condition on m? Positive integer or any real number?

2009-11-02 21:03:58 補充：
Riemann sum definition既意思，係指你想計Riemann Integral？
因為Henstock-Kurzweil Integral同埋McShane Integral都係用Riemann Sum黎定義……

2009-11-03 12:42:43 補充：
oh，明你既意思，不過呢個唔係definition of integral。

因為Riemann Integral既定義係對任何partition of interval，
但佢呢度係將個interval分成好多相等長度既subinterval，
所以其實佢係已經given左個function係integrable，
然後想你用計infinite sum既方法去計呢個integral……

2009-11-03 13:13:29 補充：
How come I cannot post my answer???

答完之後佢竟然話「請詳述你的回答內容」……想點？

2009-11-07 22:43:16 補充：
The solution is as follows:

Let 0<r<1, with the following partition:
0<...<r^n<...<r^3<r^2<r<1, which is NOT equal length as usual.
(I use this partition to avoid the evaluation of the sum Σk^m in equal partition)

Note that it is NOT a finite partition, it is infinite,
and when r→1, the sub-intervals would become smaller
and the situation becomes more likely as the usual sense.

Also, take the left end-point of sub-intervals as the tags,
Then we have

S(r) = (1-r)r^m + (r-r^2)(r^2)^m + (r^2-r^3)(r^3)^m + ...
= (1 - r) r^m (1 + r^{m+1} + r^{2m+2} + ...)
= r^m (1 - r) / (1 - r^{m+1})
= r^m / [1 + r + r^2 + ... + r^m] [note that it has m+1 terms]

Finally, we have
∫{0~1} x^m dx = lim{r→1} S(r) = 1/(m+1)

PS. It is the Fermat's Method of Dynamic Exhaustion

2009-11-07 22:44:22 補充：
唉，真係好艱難先研究到點可以post答案成功……
其實都唔知點解之前一直都唔成功，今次又點解得返……=.="

2009-11-07 22:48:59 補充：
之前意見欄講錯左D野。

睇返Riemann Integral既資料，一個partition已經可以ensure個integral存在，
唔需要先證佢存在先用呢個方法計，而可以計完之後就話佢存在。

當m係正整數~

2009-11-02 23:03:14 補充：
我諗好似http://archives.math.utk.edu/visual.calculus/4/definite.1/riemann.pdf咁計掛~

2009-11-08 18:41:02 補充：
多謝哂EMK兄你的回答...等我慢慢參詳下先~