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Topic: Another integral (Read 2279 times) |
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ThudnBlunder
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Another integral
« on: Oct 21st, 2008, 6:13pm » |
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8
Evaluate  ln(9 - x) / [ln(9 - x) + ln(x - 3)].dx
4
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teekyman
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Re: Another integral
« Reply #1 on: Oct 22nd, 2008, 12:08am » |
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substitute u = 12 - x to get ln(u - 3) / [ln(9 - u) + ln(x - u)]dx So the original integral plus this one add up to the 1 from 4 to 8, which is 4. Thus the answer to the original is 2.
Also, my math input isn't working 
I can see other people's math just fine. This was working before... am I not formatting it right?
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| « Last Edit: Oct 22nd, 2008, 11:30am by teekyman » |
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towr
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Re: Another integral
« Reply #2 on: Oct 22nd, 2008, 1:50am » |
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on Oct 22nd, 2008, 12:08am, 1337b4k4 wrote:| I can see other people's math just fine. This was working before... am I not formatting it right? |
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Have you ticked the "Insert Symbols" box? If you don't use the "Add symbols" menu, and you don't quote a post that uses symbols, then you have to tick that box yourself (otherwise it's automatic).
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teekyman
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Re: Another integral
« Reply #3 on: Oct 22nd, 2008, 11:31am » |
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There we go, thanks!
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adnanmat
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Re: Another integral
« Reply #4 on: Nov 17th, 2008, 11:08am » |
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Hello
i am sorry but ı cant understood this problem
Could you explain me?
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adnanmat
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Re: Another integral
« Reply #5 on: Nov 17th, 2008, 11:17am » |
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Hello
My name is Adnan Donmez from Turkey
I have problem with this integral problem
Can you explain me?
8
Evaluate ln(9 - x) / [ln(9 - x) + ln(x - 3)].dx
4
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towr
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Re: Another integral
« Reply #6 on: Nov 17th, 2008, 11:24am » |
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If you select the orange area in 1337b4k4's post, you should be able to see his solution.
He makes a clever substitution, adds the two integrals, and then the solution is suddenly very simply.
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adnanmat
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Re: Another integral
« Reply #7 on: Nov 17th, 2008, 1:28pm » |
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I saw him solution
But i cant understand
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towr
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Re: Another integral
« Reply #8 on: Nov 17th, 2008, 2:02pm » |
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Make a substitution
ln(9 - x) / [ln(9 - x) + ln(x - 3)]
=> x=v
ln(9 - v) / [ln(9 - v) + ln(v - 3)]
=> v= 12-x
ln(9 - (12-x)) / [ln(9 - (12-x)) + ln((12-x) - 3)]
=>
ln(x-3) / [ln(x - 3) + ln(9-x)]
=>
ln(x-3) / [ln(9-x) + ln(x - 3) ]
(As for the rest of the integral, the boundaries are switched, but so is the direction of integration 4->8, 8->4, dx -> d(12-x)= -dx)
Add the original and new expressions together
ln(9 - x) / [ln(9 - x) + ln(x - 3)] + ln(x-3) / [ln(9-x) + ln(x - 3) ]
=>
[ln(9 - x) + ln(x-3)] / [ln(9 - x) + ln(x - 3)]
=>
1
Integrating over a constant is fairly simple, and will get you twice the value you're looking for.
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adnanmat
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Re: Another integral
« Reply #9 on: Nov 17th, 2008, 2:05pm » |
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Thanks a lot
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