Does Swapping the Limits of Integration Change the Integral's Sign?

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    Identity Integral
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Discussion Overview

The discussion revolves around the mathematical property of integrals, specifically whether swapping the limits of integration changes the sign of the integral. Participants explore this concept through examples and seek clarification on the implications of negative integrals, particularly in relation to areas under curves.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant asserts that swapping the limits of integration results in a change of sign, stating that ∫ab f(x) dx = -∫ba f(x) dx.
  • Another participant expresses confusion regarding the negative integral when the function lies entirely above the x-axis, using the example of ∫0pi sin(x) dx = 2 and ∫pi0 sin(x) dx = -2.
  • A participant explains that a Riemann integral does not necessarily represent the area under a curve and emphasizes that the order of integration matters, referencing the definition in terms of Riemann sums.
  • One participant introduces the idea that negative areas can occur in other mathematical contexts, such as measuring flux through a surface, where the orientation of the area matters.
  • A participant questions the definition of the cross product, indicating a potential misunderstanding in the explanation provided earlier.

Areas of Agreement / Disagreement

Participants generally agree that swapping the limits of integration changes the sign of the integral, but there is confusion and debate regarding the interpretation of negative integrals, particularly in relation to areas under curves. The discussion remains unresolved regarding the intuitive understanding of these concepts.

Contextual Notes

Participants express uncertainty about the relationship between integrals and areas, particularly when functions are entirely above or below the x-axis. There are also unresolved questions about the definitions and implications of certain mathematical concepts, such as the cross product.

JustGaussing
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interval from a to b \int f(x) dx = interval from b to a (-)\int f(x) dx

Is this correct? Swapping the interval endpoints changes the sign of the integral? It seems like they should be equal. Thanks for the help.

By the way, I saw this property here: http://www.sosmath.com/calculus/integ/integ02/integ02.html.
 
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Welcome to PF!

Hi JustGaussing! Welcome to PF! :smile:

(have an integral : ∫ :wink:)

Yes, that's right … ∫ab f(x) dx = -∫ba f(x) dx.

Integrals add, in the sense that ∫ab f(x) dx + ∫bc f(x) dx = ∫ac f(x) dx …

now put c = a, and you get the result. :smile:
 
Thank you for your reply (and the integral!). I am confused by the concept of the negative integral where f(x) lies entirely above the x-axis. For example,

0pi sin(x) dx = 2

seems to make sense since this half of the sine wave is entirely above the x-axis and the area under the curve is all positive-y,

but

pi0 sin(x) dx = -2.

It's confusing since we are talking about the same curve. If you look at the graph, it's obviously still all above the x-axis.

Similarly, I would expect

2pi pi sin(x) dx

to be negative since the area of the curve is below the x-axis. But of course it's not.

Can anyone help explain this to me since it's so counter-intuitive (at least for me)? Thanks!
 
You have to understand that a Riemann integral, in general, does not give you the area under a curve. It might have originated from that idea, but, them mathematical abstraction takes over and generalizes things. By switching the limits of integration you change the sign of the integral as well, because order matters. Go to the definition of the integral in terms of Rieman sums (or Darboux sums) and you will see where it comes from.

if a<b and you are integrating from a to b, then when you partition the interval into n subintervals, then each length of the interval is (b-a)/n, wheras if you integrate from b to a, then it will be
(a-b)/n=-(b-a)/n<0.
 
Hi JustGaussing! :smile:

(I don't know whether this will make you happier, or even more confused :redface: …)

This isn't the only example in maths of area being negative …

when, for example, we measure the flux (of a force field) through a surface, we have to multiply by the area, but it is important to know which side of the surface is "positive" and which is "negative".

And the area of a parallelogram with sides a and b is a x b, which is abcosθ times a unit vector normal to the parallelogram … and obviously the area a x b is minus the area b x a.
 
Isn't the cross product defined as absinθ, not abcosθ?
 
oops! :redface:
 
Hey, no worries, although you probably know that already.
 

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