S5 Math M1 Definite Integral 2

YTC

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)

用戶9112

2014-11-29 8:06 pm

✔ 最佳答案

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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

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