√2r ∫(0~2π) √(1-cosθ)dθ=?
列詳細過程。
積分題目√(1-cosθ)
2012-03-31 12:45 am
回答 (3)
2012-03-31 12:55 am
✔ 最佳答案
∫(θ = 0 → 2π) √(1 - cos θ) dθ= ∫(θ = 0 → 2π) √(2 sin2 θ/2) dθ
= √2 ∫(θ = 0 → 2π) sin θ/2 dθ
= 2√2 ∫(θ = 0 → 2π) sin θ/2 d(θ/2)
= 2√2 [- cos θ/2] (θ = 0 → 2π)
= 4√2
所以 √2r ∫(θ = 0 → 2π) √(1 - cos θ) dθ = 8r
參考: 原創答案
2012-03-31 3:36 am
OH, you are right , thanks !
2012-03-31 2:38 am
hung:
Isn't that odd f instead of even f?
For any even function of f(x)
∫f(x)dx [-a,a] = 2∫f(x)dx [0,a]
For any odd function of f(x)
∫f(x)dx [-a,a] = 0
Isn't that odd f instead of even f?
For any even function of f(x)
∫f(x)dx [-a,a] = 2∫f(x)dx [0,a]
For any odd function of f(x)
∫f(x)dx [-a,a] = 0
收錄日期: 2021-04-23 18:32:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120330000051KK00500