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Title: Another integral Post by ThudanBlunder on Oct 21st, 2008, 6:13pm 8 Evaluate http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifln(9 - x) / [ln(9 - x) + ln(x - 3)].dx 4 |
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Title: Re: Another integral Post by 1337b4k4 on Oct 22nd, 2008, 12:08am [hide]substitute u = 12 - x to get http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifln(u - 3) / [ln(9 - u) + ln(x - u)]dx So the original integral plus this one add up to the http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif1 from 4 to 8, which is 4. Thus the answer to the original is 2.[/hide] Also, my math input isn't working :( http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif I can see other people's math just fine. This was working before... am I not formatting it right? |
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Title: Re: Another integral Post by towr on Oct 22nd, 2008, 1:50am on 10/22/08 at 00:08:11, 1337b4k4 wrote:
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Title: Re: Another integral Post by 1337b4k4 on Oct 22nd, 2008, 11:31am There we go, thanks! |
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Title: Re: Another integral Post by adnanmat on Nov 17th, 2008, 11:08am Hello i am sorry but ı cant understood this problem Could you explain me? |
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Title: Re: Another integral Post by adnanmat on Nov 17th, 2008, 11:17am 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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Title: Re: Another integral Post by towr on Nov 17th, 2008, 11:24am 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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Title: Re: Another integral Post by adnanmat on Nov 17th, 2008, 1:28pm I saw him solution But i cant understand |
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Title: Re: Another integral Post by towr on Nov 17th, 2008, 2:02pm 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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Title: Re: Another integral Post by adnanmat on Nov 17th, 2008, 2:05pm Thanks a lot |
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