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Integral on domain equals minus integral on the reverse domain?

  1. Aug 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Why integral from a to b of f(x) equals "minus (-)" integral from b to a of f(x)??? when a<b or a>b

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 17, 2013 #2
    Is this a homework question or your own question?
     
  4. Aug 17, 2013 #3
    I am reading Apostol calculus and I couldn't figure out how the above relation got derived!
     
  5. Aug 17, 2013 #4

    Zondrina

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    You want this I believe :

    ##\int_{a}^{b} f(x) dx = - \int_{b}^{a} f(x) dx##

    Is it that you're not understanding how to prove it or is it conceptually confusing?
     
  6. Aug 17, 2013 #5
    Exactly that I meant! It is conceptually confusing ( even though I thought I am good at that) and I don't know where it came from!
    Many thanks for help
     
  7. Aug 17, 2013 #6

    Zondrina

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    Think about this integral :

    ##\int_{a}^{b} f(x) dx## where you assume ##a≤b##.

    In words, it represents the area under the curve when we integrate f(x) from a to b. Right?

    Now think about this one :

    ##- \int_{b}^{a} f(x) dx## now you assume ##a≥b##.

    This one will represent the area under the curve when we integrate f(x) from b to a. The negative sign 'flips' the answer around so you will get the same answer as if you integrated from a to b.

    I hope that clears that up a bit.

    Do you know about Riemann sums at all? The mesh of a partition? Any definitions you have would be great.
     
  8. Aug 17, 2013 #7
    Thanks for your time! But I still don't understand the geometric representation for this. And, both integrals are related to each other only when a<b for both the positive and the negative.
     
  9. Aug 17, 2013 #8

    HallsofIvy

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    If I remember correctly, Apostol defines the integral [itex]\int_a^b f(x)dx[/itex] through "Riemann sums" only for a< b, the defines [itex]\int_b^a f(x)dx= -\int_a^b f(x)dx[/itex].
     
  10. Aug 17, 2013 #9
    Here is a simple way to look at this. When [itex]\frac{d}{dx}F(x)=f(x)[/itex], [itex]\int_{a}^{b}f(x)dx=F(b)-F(a)[/itex]. Also, [itex]\int_{b}^{a}f(x)dx=F(a)-F(b)=-(F(b)-F(a))[/itex]. Therefore, [itex]\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx[/itex].
     
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