Integrating exponential functions

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Discussion Overview

The discussion revolves around the integration of the function xe^(x^2) using substitution methods, specifically the substitution u = x^2. Participants explore the implications of this substitution and the resulting integration process.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant asks how to integrate xe^(x^2) using the substitution u = x^2.
  • Another participant inquires about the differential du, identifying it as 2x.
  • A participant expresses confusion about how the substitution helps with the integration of e^u.
  • There is a mention of the integral of e^u, with one participant stating it is e^u + C, suggesting it as a solution.
  • Another participant reiterates the integral of e^u and expresses gratitude for the clarification provided.
  • A clarification is made that du is 2x(dx), emphasizing the importance of including dx in the expression.

Areas of Agreement / Disagreement

The discussion contains some agreement on the form of the integral of e^u, but participants express varying levels of understanding regarding the substitution process and its implications for the integration of the original function. Overall, the discussion remains somewhat unresolved as participants navigate their understanding of the integration steps.

Contextual Notes

There are limitations in the discussion regarding the clarity of the substitution process and the integration steps, as well as the potential confusion surrounding the role of dx in the expression for du.

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How would you integrate something like xe(x^2) by using a substitution like u=x^2
 
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What is du?
 
2x, but my problem is how will this help when you still have e raised to the power u
 
Do you know what's

\int e^{u}du

?
 
elibj123 said:
Do you know what's

\int e^{u}du

?

no this is what I am stuck on
 
Well, you know that integration is some sort of anti-derivative, and you know that a function of which derivative is e^x, is e^x it self. So it's safe to say that:

\int e^{u}du=e^{u}+C
 
elibj123 said:
Well, you know that integration is some sort of anti-derivative, and you know that a function of which derivative is e^x, is e^x it self. So it's safe to say that:

\int e^{u}du=e^{u}+C

Thanks for your help, that cleared things up.
 
just to clarify, du is 2x(dx), not just not 2x. I'm sure you already knew that though.
 

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