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Splitting up an interval of integration

  1. Nov 28, 2014 #1
    How does one prove the following relation?

    [tex]\int_{a}^{b}f(x)dx= \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx [/tex]

    Initially, I attempted to do this by writing the definite integral as the limit of a Riemann sum, i.e.

    [tex] \int_{a}^{b}f(x)dx= \lim_{n\rightarrow\infty}\frac{(b-a)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k})[/tex]

    Where [itex] x^{\ast}_{k}\in\left[x_{k}, x_{k+1} \right] [/itex].


    [tex] \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx= \\ = \lim_{n\rightarrow\infty}\frac{(c-a)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k}) +\lim_{n\rightarrow\infty}\frac{(b-c)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k}) \\ = \lim_{n\rightarrow\infty}\frac{(c-a)+ (b-c)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k}) \\ = \lim_{n\rightarrow\infty}\frac{(b-a)}{n}\sum_{k=1}^{n}f(x^{\ast}_{k} = \int_{a}^{b}f(x)dx[/tex]

    But I have a feeling that this isn't correct?!
  2. jcsd
  3. Nov 28, 2014 #2


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    Gold Member

    Your feeling is correct!
    The two Riemann sums are not the same because they're taken in different intervals and so, in general, the terms aren't equal and so you can't factor them!
    Consider two functions h(x) and k(x) defined as:
    h(x)=\left\{ \begin{array}{cc} f(x) \ \ \ \ x \in (a,c) \\ 0 \ \ \ \ otherwise \end{array} \right.
    k(x)=\left\{ \begin{array}{cc} f(x) \ \ \ \ x \in [c,b) \\ 0 \ \ \ \ otherwise \end{array} \right.
    Its obvious that [itex] f(x)=h(x)+k(x) [/itex] when [itex] x \in (a,b) [/itex]. Use this for the proof!
    Last edited: Nov 28, 2014
  4. Nov 28, 2014 #3
    Ah, ok. thanks for your help.

    Would this be correct then?

    [tex] \int_{a}^{b}f(x)dx = \int_{a}^{b}\left(h(x)+k(x)\right)dx = \\ = \int_{a}^{b}h(x)dx +\int_{a}^{b}k(x)dx = \int_{a}^{c}h(x)dx +\int_{c}^{b}k(x)dx = \int_{a}^{c}f(x)dx +\int_{c}^{b}f(x)dx [/tex]

    As [itex] h(x)[/itex] is zero outside the interval [itex](a,c)[/itex] and so will provide no further contributions after this point in the integral, and similarly for [itex]k(x)[/itex].
  5. Nov 28, 2014 #4


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    Gold Member

    Yes, that's it!
    Last edited: Nov 28, 2014
  6. Nov 28, 2014 #5
    Ok, cool. Thanks very much for your help with this, much appreciated!
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