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Addition property of integration intervals proof

  1. Feb 26, 2015 #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]
  2. jcsd
  3. Feb 26, 2015 #2


    Staff: Mentor

  4. Feb 26, 2015 #3
    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.
  5. Feb 27, 2015 #4


    Staff: Mentor

    Completely unrelated to mathematics, but "No particular reason to be honest, ..." and "No particular reason, to be honest, ..." mean entirely different things.
  6. Feb 27, 2015 #5
    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.
  7. Feb 27, 2015 #6
    What would be a better way to prove it?
  8. Feb 27, 2015 #7
    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.
  9. Feb 27, 2015 #8
    Ok, thanks for the advice. I'll have to take a look.
  10. Feb 28, 2015 #9


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    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.
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