Addition property of integration intervals proof

  • #1
First of all, apologies as I've asked this question before a while ago, but I never felt the issue got resolved on that thread.

Is it valid to prove that [tex]\int_{a}^{c}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx[/tex]
using the fundamental theorem of calculus (FTC)?! That is, would it be valid to do the following.

Let [itex]F[/itex] be an anti-derivative of [itex]f[/itex] in an interval [itex][a,c]=[a,b]\cup [b,c][/itex]. It follows then (from the FTC), that [tex]\int_{a}^{b}f(x)dx=F(b)-F(a)[/tex] and [tex]\int_{b}^{c}f(x)dx=F(c)-F(b)[/tex] As such, [tex]\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx= \left[F(b)-F(a)\right]+\left[F(c)-F(b)\right]=F(c)-F(a)=\int_{a}^{c}f(x)dx[/tex] where the last equality follows from the assumption that [itex]F[/itex] be an anti-derivative of [itex]f[/itex] in an interval [itex][a,c][/itex] and hence [tex]\int_{a}^{c}f(x)dx=F(c)-F(a)[/tex]

In a similar manner, is it valid to use the FTC to prove that [tex]\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx[/tex]
 

Answers and Replies

  • #3
Where do you think a problem lies?

No particular reason to be honest, more a lack of confidence in my own knowledge. I wasn't sure if it was rigorous enough or not?!

Thanks for the link by the way.
 
  • #4
34,935
6,698
No particular reason to be honest
Completely unrelated to mathematics, but "No particular reason to be honest, ..." and "No particular reason, to be honest, ..." mean entirely different things.
 
  • Like
Likes HallsofIvy
  • #5
22,089
3,297
Alright, this is valid. But it would be best not to give this proof for two reasons:

1) Using the fundamental theorem is really overkill for something that's supposed to be more simple
2) The fundamental theorem only has a quite limited range. That is: there are some functions for which integration and your results does make sense, but the fundamental theorem doesn't make sense (or isn't true).

Finally, saying that
[tex]\int_a^b f(t)dt = -\int_b^a f(t)dt[/tex]
is usually a definition, so it doesn't really require a proof.
 
  • #6
What would be a better way to prove it?
 
  • #7
22,089
3,297
Well, you can check any real analysis or rigorous calculus book. For example, see Spivak's calculus or Apostol's calculus. Basically, you first define it rigorously, usually with the help of Riemann sums. And then it's not so difficult using that definition.
 
  • #8
Ok, thanks for the advice. I'll have to take a look.
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,833
964
The best way to prove it is to use the "Riemann Sums" definition of the integral. Given a "partition" of the interval of integration, a to b, for any point, c, with a< c< b, we can always choose a "refinement" that has c as one of its endpoints. We can then break that into two Riemann sums, one from a to c, the other from c to b, and then get two different integrals.
 

Related Threads on Addition property of integration intervals proof

  • Last Post
Replies
3
Views
4K
Replies
7
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
7
Views
20K
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
18
Views
7K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
17
Views
4K
  • Last Post
Replies
1
Views
3K
Top