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

[JustGaussing]

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.

 

[tiny-tim]

Welcome to PF!

Hi JustGaussing! Welcome to PF!

(have an integral : ∫ )

Yes, that's right … ∫_a^b f(x) dx = -∫_b^a f(x) dx.

Integrals add, in the sense that ∫_a^b f(x) dx + ∫_b^c f(x) dx = ∫_a^c f(x) dx …

now put c = a, and you get the result.

 

[JustGaussing]

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,

∫₀^pi 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!

 

[sutupidmath]

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.

 

[tiny-tim]

Hi JustGaussing!

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

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.

 

[Char. Limit]

Isn't the cross product defined as absinθ, not abcosθ?

 

[tiny-tim]

oops!

 

[Char. Limit]

Hey, no worries, although you probably know that already.