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#1 2006-02-25 06:53:18

mikau
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infinite integrals

Use y as the variable of integration to find the area between the graph of y = 1/x and the y axis from y = 1 to y = infinity

Ok. y = 1/x so x = 1/y. The length of each "rectangle" should be x and the width should be dy.

So it should be the limit of  ∫1/y dy from y = 1 to y = b as b approaches infinity, = lim (ln b - ln 1 ) as b approaches infinity, = ∞

Rewrite the integral found in the preceding problem using x as the variable of integration.

Ok first I tried using the change of variable system, then rewriting from scratch. Both produced the same answer. Rewriting from scratch:

The area should be  ∫  x dy from y = 1 to y = infinity.   dy = -1/x^2 dx so we can replace this with:

∫ -1/x dx   from y = 1 to y = infintiy 

Lets reverse the evaluation limits to make it positive:

∫ 1/x dx  from y = infinity to y = 1

Ok. From y = infinity to y = 1. y = 1/x so x = 1/y   when y = infinity, x aproaches zero to the right.
When y = 1, x = 1 so we should be able to rewrite this as:

limit of ∫ 1/x dx  from x = b to x = 1 as b approaches zero from the right.  (This should be the answer to the problem.)

= limit of [ ln 1 - ln b ]  as b approaches zero from the right.

= 0 - (-∞) = ∞

In both cases, we ended up evaluating ln ∞ + ln 1. Well technically we ended up with - ln 1 in the first, but thats the same as ln 1/1 = ln 1 = 0. So it ended up being the same thing.

The bizzare answer my book gave is as follows:

The limit of ∫ (1/x - 1) dx from x = 0 to x = 1 as b approaches zero from the right.


Whhhaaaat??? As b approaches zero from the right? I don't see a b in that expression at all! And where did that -1 come from?

Either its a misprint or I am doing something COMPLETELY wrong.

Last edited by mikau (2006-02-25 06:56:51)


A logarithm is just a misspelled algorithm.

#2 2006-02-25 06:55:30

espeon
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Re: infinite integrals

Sorry can't help you.I don't think they'll teach us this in primary school.Sorry


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#3 2006-02-25 08:31:26

irspow
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Re: infinite integrals

As y varies from 1 to ∞ x varies from 1 to 0.  But we never integrate in the negative direction, so your translated function now one of f(x) would need to be integrated from 0 to 1.

  Okay, I too have no idea how they came up with that function, so uh er, I guess that I can't help you either.  Sorry.

#4 2006-02-25 09:32:04

mikau
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Re: infinite integrals

I'm at least glad to see usually when I get stuck on a problem, no one else can solve it either. lol...


A logarithm is just a misspelled algorithm.

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