Re: [微積] 反函數的定積分
※ 引述《Koguei (摳龜)》之銘言:
: 題目: http://ppt.cc/CkHS
: 想了許久總覺得卡卡的,
: 請各位幫忙高手解答,
: 題目我大概講一下 f(x)連續且為遞減函數
: g(x)為f(x)反函數,f(2)=1 ,f(4)=0 且f(x)從2積到4等於1
: 求g(x)從0積到1為?
: 麻煩各位高手了 謝謝 :)
1 4
S g(x) dx = [1 - 0]*[2 - 0] + S [f(x) - f(4)]dx
0 2
= 2 + 1
= 3
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 128.220.159.5
→
09/19 20:32, , 1F
09/19 20:32, 1F
→
09/19 20:33, , 2F
09/19 20:33, 2F
→
09/19 20:33, , 3F
09/19 20:33, 3F
→
09/19 20:46, , 4F
09/19 20:46, 4F
→
09/19 20:53, , 5F
09/19 20:53, 5F
→
09/19 21:07, , 6F
09/19 21:07, 6F
推
09/19 21:47, , 7F
09/19 21:47, 7F
推
09/19 22:00, , 8F
09/19 22:00, 8F
推
09/19 22:31, , 9F
09/19 22:31, 9F
推
09/19 23:44, , 10F
09/19 23:44, 10F
推
09/21 12:23, , 11F
09/21 12:23, 11F
→
01/02 15:32,
7年前
, 12F
01/02 15:32, 12F
→
07/07 11:26,
6年前
, 13F
07/07 11:26, 13F
討論串 (同標題文章)