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Aug 17 2014 11:14pm
consider we want to integrate -1/(y-10)

what I would personally do is pull the (-1) out of the integral sign and integrate 1/(y-10), giving me an answer of -ln(y-10)

the book i'm using//wolfram alpha instead multiplies top and bottom by (-1), changing it to 1/(10-y) and yielding an answer of -ln(10-y)

Why is this? I don't think we can say -ln(10-y) = -ln(y-10)

but at the same time, whichever you choose to derive will get you an answer of 1/(10-y)

what is going on here
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Aug 17 2014 11:20pm
thats cause the integral isnt -ln(10-y) its ln(10-y).
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Aug 17 2014 11:22pm
you also cannot pull the -1 out without distributing it to the denominator either. when you distribute it, itll give the answer you have when you multiply top and bottom by -1
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Aug 17 2014 11:23pm
Quote (CrackerJacker @ Aug 18 2014 12:14am)
consider we want to integrate -1/(y-10)

what I would personally do is pull the (-1) out of the integral sign and integrate 1/(y-10), giving me an answer of -ln(y-10)
Because this.....................................^............is (1)


see what im sayin

This post was edited by Genky on Aug 17 2014 11:23pm
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Aug 17 2014 11:24pm
Quote (Yawn @ Aug 17 2014 09:20pm)
thats cause the integral isnt -ln(10-y) its ln(10-y).


wolfram alpha disagrees with you

edit: link wont work

This post was edited by CrackerJacker on Aug 17 2014 11:26pm
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Aug 17 2014 11:25pm
damn thats pretty hard, im good in math and physics but this is nuts
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Aug 17 2014 11:28pm
Quote (Yawn @ Aug 17 2014 09:22pm)
you also cannot pull the -1 out without distributing it to the denominator either. when you distribute it, itll give the answer you have when you multiply top and bottom by -1



I can rewrite it as (-1) * 1/(y-10)

with (-1) acting as a constant theres no reason I shouldn't be able to take it outside of the integral
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Aug 17 2014 11:42pm
Quote (CrackerJacker @ Aug 18 2014 12:28am)
I can rewrite it as (-1) * 1/(y-10)

with (-1) acting as a constant theres no reason I shouldn't be able to take it outside of the integral


This is true. Your book is not saying that -ln(10-y) = -ln(y-10) because this is not equivalent.

However what it is saying is that -1/(y-10) = 1/(10-y).

Your issue lies when you take the integral of -1/(y-10). The answer isnt so much -ln(y-10) as it is -ln( |y-10| )

The same applies if simplify as your book does to 1/(10-y) Then your integral evaluation is -ln( |10-y| )

Which in turn are equivalent.

Let y = 5

Case 1: -ln( | y-10 | )

-ln ( | 5 - 10 | ) = -ln ( | -5 | ) = - ln (5)

Case 2: -ln ( | 10-y | )

-ln ( | 10-5 | ) = -ln (5)

This post was edited by trojan24 on Aug 17 2014 11:43pm
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Aug 17 2014 11:54pm
Quote (trojan24 @ Aug 17 2014 09:42pm)
This is true. Your book is not saying that -ln(10-y) = -ln(y-10) because this is not equivalent.

However what it is saying is that -1/(y-10) = 1/(10-y).


but this is the issue i am having, depending how I decide to factor my integral I get a completely different answer.. i understand that you can rewite -1/(y-10) as 1/(10-y) but that doesn't explain why they yield completely different integrals =/

it seems that puling the constant -1 out of the integral is an illegal move here and i don't understand why..



This post was edited by CrackerJacker on Aug 17 2014 11:54pm
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Aug 18 2014 12:01am
Quote (CrackerJacker @ Aug 18 2014 12:54am)
but this is the issue i am having, depending how I decide to factor my integral I get a completely different answer.. i understand that you can rewite -1/(y-10) as 1/(10-y) but that doesn't explain why they yield completely different integrals =/

it seems that puling the constant -1 out of the integral is an illegal move here and i don't understand why..

http://i.imgur.com/RVd5Hrd.png


pmd
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