F.6 math定積分

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]

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!!