S5 Math M1 Definite Integral 2

2014-11-29 7:44 pm
1. If ∫(upper limit= 2, lower limit= 1) a^x dx=6/ln a, where a>0 and a≠1, find the
value of a.

The question is given that a≠1, but the lower limit is equal to 1, then how can I
solve the question?


2. Proved that
∫(upper limit= 0, lower limit= -a)f(x)dx= ∫(upper limit= a, lower limit= 0)f(-x)dx
Evaluate ∫(upper limit=1, lower limit=-1) [(e^x)-(e^-x)]/(1+x^4) dx

I try to let f(x)=[(e^x)-(e^-x)]/(1+x^4) and a=1, but i cannot solve the question, Why?
Please help! Thank you very much!~

回答 (1)

2014-11-29 8:06 pm
✔ 最佳答案
===============================

圖片參考:https://s.yimg.com/rk/HA05107138/o/1149110608.png


2014-11-29 12:09:16 補充:
For (2b), you can also split the integral into 2 parts, one from -1 to 0 and the other 0 to 1
For the part from -1 to 0, using (a), you will get integral from 0 to 1 and f(-x)
and f(-x) = -f(x)
Hence the sum of the 2 parts = 0


收錄日期: 2021-04-15 17:24:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20141129000051KK00028

檢視 Wayback Machine 備份