Swapping the limits of integration

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SUMMARY

Swapping the limits of integration in a one-dimensional integral results in a sign change, as established by the equation \(\int_a^b f(x) dx = -\int_b^a f(x) dx\). This principle is universally applicable, although it is sometimes presented as a special case in educational contexts. In electromagnetism, the potential is defined using the negative of the integral with swapped limits, which raises questions about the necessity of this convention. The practice likely stems from the convention of defining potential from a point at infinity to a local point.

PREREQUISITES
  • Understanding of one-dimensional integrals
  • Familiarity with the Fundamental Theorem of Calculus
  • Basic knowledge of electromagnetism concepts
  • Ability to manipulate mathematical expressions involving integrals
NEXT STEPS
  • Study the Fundamental Theorem of Calculus in detail
  • Explore the concept of potential in electromagnetism
  • Learn about the implications of sign changes in integrals
  • Investigate special cases of integration limits in various mathematical contexts
USEFUL FOR

Students of calculus, physicists working in electromagnetism, and educators seeking to clarify integral properties will benefit from this discussion.

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Can you always just swap the limits of integration and flip the sign of a one-dimensional integral or is there a time when you can't do this?
 
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Yes, \int_a^b f(x) dx= -\int_b^a f(x) dx. Let u= -x. I thought everyone knew this!
 
HallsofIvy said:
Yes, \int_a^b f(x) dx= -\int_b^a f(x) dx. Let u= -x. I thought everyone knew this!

I was taught that yes... but it's not too uncommon to be taught something that's a special case without being told it's a special case.

The question came from the fact that in electromagnetism, we define the potential by the negative of the integral with swapped limits. I'm not sure why you would put the extra step in there if there wasn't a case where the positive with the limits restored wouldn't be equivalent.

My assumption (given your response) is that they do it simply because we generally define potential from some point at infinity down to a local point.
 

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