Did I make a mistake in evaluating this integral?

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Discussion Overview

The discussion centers around the evaluation of the integral $$\int^4_0 \frac{6z + 5}{2z + 1} dz$$. Participants are reviewing the steps taken in the evaluation and identifying potential mistakes in the process.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents their evaluation of the integral, breaking it down into simpler components.
  • Another participant suggests that there is an error related to a factor of 2 in front of the logarithmic term in the final expression.
  • A participant questions whether the factor of 2 cancels out with the term 2z in the integral, indicating a realization of a potential mistake.
  • A later reply provides a general formula for integrating a function of the form $$\frac{1}{a\,x + b}$$, which may relate to the discussion but does not directly resolve the original question.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the original evaluation, as there are differing views on the presence and handling of the factor of 2.

Contextual Notes

There are unresolved aspects regarding the integration steps and the application of the logarithmic integration formula, which may depend on specific assumptions about the constants involved.

shamieh
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Can someone check my work?$$
\int^4_0 \frac{6z + 5}{2z + 1} dz$$

$$\int^4_0 3 + \frac{2}{2z + 1} dz$$
$$
[3z + 2\ln|2z + 1|]^4_0 = 12 + 2\ln|9| $$
 
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As far as I can see, you´ve done a great job - except for the factor 2 in front of ln|2z+1|
 
the 2 cancels out with the 2z doesn't it? i think i see where i made my mistake
 
shamieh said:
the 2 cancels out with the 2z doesn't it? i think i see where i made my mistake

$\displaystyle \begin{align*} \int{\frac{1}{a\,x + b} \, dx} = \frac{1}{a} \ln{ \left| a \, x + b \right| } + C \end{align*}$
 

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