微積分:求救
2011-01-07 2:25 am
回答 (2)
2011-01-07 2:43 am
✔ 最佳答案
As follows:圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/ScreenHunter_06Jan061841.gif
2011-01-06 20:32:32 補充：
Thanks myisland!
2011-01-07 2:59 am
∫ 1/(9+4x^2) dx [from 0 to 3]
=∫ 1/(1+[(2/3)x]^2) dx [from 0 to 3]
=(3/2)(1/9) ∫ 1/(1+[(2/3)x]^2) d [(2/3)x] [from 0 to 3]
=(3/2)(1/9) ∫ 1/(1+u^2) du [from 0 to 2]
=(1/6) [arctan(2) - arctan(0)]
=(1/6) arctan(2)
=∫ 1/(1+[(2/3)x]^2) dx [from 0 to 3]
=(3/2)(1/9) ∫ 1/(1+[(2/3)x]^2) d [(2/3)x] [from 0 to 3]
=(3/2)(1/9) ∫ 1/(1+u^2) du [from 0 to 2]
=(1/6) [arctan(2) - arctan(0)]
=(1/6) arctan(2)
收錄日期: 2021-04-22 00:54:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110106000051KK00857