1/[(2 - x)[(1 - x)^(1/2)]]
http://farm8.staticflickr.com/7116/7084046545_95d89f0fcd.jpg
How to do? Thanks.
更新1:
How about if (1) Integrate from x = 0 to x = 1? (2) I only evaluate the indefinite integral 1/[(2 - x)[(1 - x)^(1/2)]]? Thanks.
Integrate from x = 0 to x = 2,
2012-04-17 6:57 am
Integrate from x = 0 to x = 2,
1/[(2 - x)[(1 - x)^(1/2)]]
http://farm8.staticflickr.com/7116/7084046545_95d89f0fcd.jpg
How to do? Thanks.
1/[(2 - x)[(1 - x)^(1/2)]]
http://farm8.staticflickr.com/7116/7084046545_95d89f0fcd.jpg
How to do? Thanks.
回答 (3)
2012-04-17 5:09 pm
✔ 最佳答案
As follows:圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Apr12/Crazyint4.jpg
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參考: 原創答案
2012-04-17 3:58 pm
when x = 2, 1/(2 - x) is infinity.
When x > 1, sqrt ( 1 - x) is unreal.
So there is no solution for x = 0 to x = 2.
When x > 1, sqrt ( 1 - x) is unreal.
So there is no solution for x = 0 to x = 2.
2012-04-17 3:47 pm
此積分發散，無法求出
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120416000051KK00918