What is the Indefinite Integral of 1/e^x?

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Discussion Overview

The discussion revolves around finding the indefinite integral of the function 1/e^x. Participants explore different approaches to solving the integral, including substitution methods and direct integration techniques.

Discussion Character

  • Mathematical reasoning, Homework-related

Main Points Raised

  • One participant asks for help with the indefinite integral of 1/e^x, indicating they are new to the forum.
  • Another participant provides a solution, stating that the integral is -e^{-x} + c, but also suggests the original poster may have posted in the wrong forum.
  • A different participant elaborates on the integration process, proposing a substitution method (-x=t) to arrive at the same result of -e^{-x} + c.
  • This participant also questions the necessity of substitution, suggesting that letting u = e^{-x} could simplify the process.
  • Another participant acknowledges the substitution method but explains their reasoning for using a more detailed approach, considering the original poster's potential lack of familiarity with integration techniques.

Areas of Agreement / Disagreement

There is no consensus on the best method to approach the integral, as participants present different techniques and reasoning. Some suggest substitution is unnecessary, while others advocate for it to aid understanding.

Contextual Notes

The discussion includes various assumptions about the original poster's familiarity with integration techniques, which may affect the clarity of the proposed solutions.

mathguyz
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Does anyone here know what the indefinite integal of 1/e^x is?

Im an new member and, this is my first post.
 
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mathguyz said:
Does anyone here know what the indefinite integal of 1/e^x is?

Im an new member and, this is my first post.

i guess u have posted on the wrong forum, but anyways!

[tex]\int e^{-x} dx= -e^{-x}+c[/tex]
 
here it is how you might want to think about it: i am assuming you know how to integrate this, since it is only a tabelar integral
[tex]\int e^{x}dx[/tex], for the other one[tex]\int e^{-x}dx[/tex], take the substitution -x=t, so now we have dx=-dt, now substitute back on the integral we get:
[tex]\int e^{t}(-dt)=-\int e^{t}dt=-e^t+c[/tex] substituting back for the original variable we get [tex]-e^{-x}+c[/tex]
 
sutupidmath said:
here it is how you might want to think about it: i am assuming you know how to integrate this, since it is only a tabelar integral
[tex]\int e^{x}dx[/tex], for the other one[tex]\int e^{-x}dx[/tex], take the substitution -x=t, so now we have dx=-dt, now substitute back on the integral we get:
[tex]\int e^{t}(-dt)=-\int e^{t}dt=-e^t+c[/tex] substituting back for the original variable we get [tex]-e^{-x}+c[/tex]

Why not just let u = e^-x?

No need for substitution here.
 
JasonRox said:
Why not just let u = e^-x?

No need for substitution here.

yeah, i know, but since the op was not able to integrate e^-x, i was assuming that approaching this problem like i did, it would make it easier for him/her to understand!
 

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