How do I do this simple looking integral?

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Homework Help Overview

The discussion revolves around evaluating the integral of the function x/(x+1) from -1 to 0. Participants are exploring the nature of the integral and whether it converges or diverges.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial steps of rewriting the integral and consider the antiderivative of 1/(x+1). Questions arise regarding the necessity of substitution and the implications of limits of integration.

Discussion Status

The discussion is active, with participants offering guidance on substitution and the evaluation of limits. There is an exploration of different interpretations regarding the convergence of the integral, particularly focusing on the behavior of the logarithmic function at the limits.

Contextual Notes

Participants are considering the implications of the limits of integration and the nature of the function as it approaches certain values, particularly the undefined nature of ln(0). There is a mention of homework constraints regarding the evaluation of integrals.

frasifrasi
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How do I do this simple looking integral?

Ok, we are supposed to evaluate

int from -1 to 0 of x/(x+1) dx
---> my first step was to write (x+1) -1/(x+1) to form the two integrals: of 1 and 1/(x+1)


--> is this the right way? The answer is "it diverges."
 
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You are on the right path; what is the antiderivative of 1/(x+1)?
 
If I get the antiderivatives:

x] - [ln(u)] ? -- Am i supposed to substitute u for (x+1) before proceeding?
 
frasifrasi said:
If I get the antiderivatives:

x] - [ln(u)] ? -- Am i supposed to substitute u for (x+1) before proceeding?
you don't have to, this problem you can pretty much eyeball.

but if you had like 2x+1 in the denominator, then you would need worry about constants.
 
So, the lim n -->-1 of ln(n) is infinity?
 
If we let u=x+1 then du=dx

So then x=u-1 and dx=du

It's a way similar to what you did, and since ln(0) is undef. it diverges.
 
so, I do have to sub x+1 back I guess...
 
Either that or change the limits of integration as you go. If you make the substitution u= x+1 and your limits of integration, with respect to x, were x= -1 and x= 0, then the limits of integration become u= -1+1= 0 and u= 0+ 1= 1.
 
but then the integral wouldn't diverge, would it?
 
  • #10
ln(u), where u=0 will definitely diverge
 

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