Solve Log Problem: Integral of (1/x) from -b to -a with Formula ln(x)

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Homework Help Overview

The discussion revolves around evaluating the definite integral of the function \( \frac{1}{x} \) from \(-b\) to \(-a\), with a focus on the logarithmic function involved in the evaluation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the evaluation of the integral using the logarithmic function, questioning the conditions under which the logarithm is defined. There is also a discussion about the fundamental theorem of calculus and the correct form of the integral.

Discussion Status

Participants are actively engaging with the problem, with some providing insights on the correct interpretation of the logarithmic function involved. There is a recognition of the need for conditions on \(a\) and \(b\) for the logarithm to be defined, and a shift in understanding regarding the integral's expression.

Contextual Notes

There is an emphasis on the requirement that both \(a\) and \(b\) must be greater than zero for the logarithmic expressions to be valid. The discussion also highlights the distinction between \( \ln(x) \) and \( \ln|x| \) in the context of the integral.

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[tex]\int_{-b}^{-a} \frac{dx}{x} = ln(x)|_{-b}^{-a}[/tex]

What is this equal to? I don't know how to proceed. Please help.

Thank

Alan
 
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The right side evaluates to ln(-a) - ln(-b) = ln(-a/-b)

In order for ln(-a) and ln(-b) to be defined, both a and b have to be > 0.
 
How would you normally solve a definite integral? What does the fundamental theorem of calculus say?
 
Actually, the integral of 1/x isn't ln(x), but ln|x|. That makes everything possible.
 
Char. Limit said:
Actually, the integral of 1/x isn't ln(x), but ln|x|. That makes everything possible.
Good point...
 
So [tex]\int_{-b}^{-a}\frac {dx}{x} = ln|x|_{-b}^{-a}= ln\left (\frac a b\right )[/tex]

Thanks
 

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