求∫(b&a) [v(t) / u(t) - 2v(t)] dt
∫(b&a)..means F(b) - F(a)
更新1:
∫(b&a) [u(x) / u(x) - 2v(x)] dx = A ∫(b&a) [v(t) / u(t) - 2v(t)] dt ∫(b&a)..means F(b) - F(a)
更新2:
求過程,Ans. is (1/2)(A-b+a)
F.6 math定積分
2009-09-25 4:48 am
已知∫(b&a) [u(x) / u(x) - 2v(x)] dx = A
求∫(b&a) [v(t) / u(t) - 2v(t)] dt
∫(b&a)..means F(b) - F(a)
求∫(b&a) [v(t) / u(t) - 2v(t)] dt
∫(b&a)..means F(b) - F(a)
回答 (2)
2009-09-25 4:57 am
✔ 最佳答案
朋友﹐這是一個concept問題∫(b&a) [u(x) / u(x) - 2v(x)] dx = A
但您知道對定積分而言﹐可以任意進行換元
因此 ∫(b&a) [u(x) / u(x) - 2v(x)] dx
= ∫(b&a) [v(t) / u(t) - 2v(t)] dt
= A
2009-09-26 12:40:41 補充：
∫(b&a) [v(t) / u(t) - 2v(t)] dt
= (1/2)∫(b&a) [2v(t)-u(t)+u(t)] / [u(t) - 2v(t)] dt
= (1/2)∫(b&a) [2v(t)-u(t)] / [u(t) - 2v(t)] dt + (1/2)∫(b&a) u(t) / [u(t) - 2v(t)] dt
=(1/2)[A-b+a]
2009-09-26 8:22 am
朋友..你concept問題都係認識太淺,比我淺
Ans. is (1/2)(A-b+a)
2009-09-26 23:52:39 補充：
感謝你的幫忙!^^thank you!!
Ans. is (1/2)(A-b+a)
2009-09-26 23:52:39 補充：
感謝你的幫忙!^^thank you!!
收錄日期: 2021-04-26 13:59:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090924000051KK01551