工程數學求解 - 考試
By Lily
at 2013-11-25T15:55
at 2013-11-25T15:55
Table of Contents
※ 引述《semmy214 (陳山河)》之銘言:
: [考題] 國考歷屆考題與考題觀念討論(書裡看到的選這個)請附上想法、出處
: Http://ppt.cc/-1HY
: 版上有大大說可以拆項
: 但我不知怎麼拆
: 我也不知這是工數那一章
: 請熱心的人說明一下 謝謝
∞
∫ (1/x^3+1) = lim 1/(x^3+1) - lim 1/(x^3+1)
-∞ x→∞ x→-∞
lim 1/(x^3+1) = (1/x^3)/(1+1/x^3) = 0/(1+0) = 0
x→∞
lim 1/(x^3+1) = (1/x^3)/(1+1/x^3) = 0/(1+0) = 0
x→-∞
∞
∫ (1/x^3+1) = lim 1/(x^3+1) - lim 1/(x^3+1) = 0 ........ANS
-∞ x→∞ x→-∞
有錯請指正,謝謝!
--
: [考題] 國考歷屆考題與考題觀念討論(書裡看到的選這個)請附上想法、出處
: Http://ppt.cc/-1HY
: 版上有大大說可以拆項
: 但我不知怎麼拆
: 我也不知這是工數那一章
: 請熱心的人說明一下 謝謝
∞
∫ (1/x^3+1) = lim 1/(x^3+1) - lim 1/(x^3+1)
-∞ x→∞ x→-∞
lim 1/(x^3+1) = (1/x^3)/(1+1/x^3) = 0/(1+0) = 0
x→∞
lim 1/(x^3+1) = (1/x^3)/(1+1/x^3) = 0/(1+0) = 0
x→-∞
∞
∫ (1/x^3+1) = lim 1/(x^3+1) - lim 1/(x^3+1) = 0 ........ANS
-∞ x→∞ x→-∞
有錯請指正,謝謝!
--
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考試
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