∫√(x^2 - 4)/x dx
上限4 下限2
希望各位唔用設u,用其他方法.....
不過可以打出來比我睇..
唔該各位啦！！
積分法(定積分)急
2011-01-19 5:47 am
回答 (4)
2011-01-19 6:10 am
✔ 最佳答案
∫√(x^2 - 4)/x dx [2,4]Let x = 2secθ, dx = 2secθtanθdθx = 2, θ = (sec1)^(-1)x = 4, θ = (sec2)^(-1)√(x^2 - 4)= √[4(secθ)^2 - 4]= 2tanθ∫√(x^2 - 4)/x dx [2,4]= ∫ (2tanθ/2secθ)(2secθtanθ)dθ [(sec1)^(-1), (sec2)^(-1)]= 2 ∫ (tanθ)^2 dθ= 2 ∫ (secθ)^2 - 1 dθ= 2(tanθ - θ)= 2[(tan π/3 - π/3) - (tan 0 - 0)]= 2(√3 - π/3)
2011-01-19 8:36 am
你係味要好像大約金做....
∫ √(x^2-r^2) dx
=Area of the 1/4 circle (with the radius r)
= [(π)r^2]/4
∫ √(x^2-r^2) dx
=Area of the 1/4 circle (with the radius r)
= [(π)r^2]/4
2011-01-19 6:14 am
仲有無其他方法？如不用三角學跟設u...
2011-01-19 5:59 am
Technique of Trigo is needed to solve
收錄日期: 2021-04-26 14:43:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110118000051KK01357