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Subdividing an integral into a sum of integrals over a given interval

  1. Jun 30, 2014 #1
    How does one prove the following:
    [tex] \int^{c}_{a} f\left(x\right)dx = \int^{b}_{a} f\left(x\right)dx +\int^{c}_{b} f\left(x\right)dx [/tex]
    where [itex] f\left(x\right)[/itex] is continuous in the interval [itex] x\in \left[a, b\right][/itex], and differentiable on [itex] x\in \left(a, b\right)[/itex].

    My approach was the following:

    Given that [itex] \int^{b}_{a} f\left(x\right)dx = \lim_{n\rightarrow \infty} \frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) [/itex], where [itex] x^{\ast}_{i} \in\left[x_{i},x_{i+1}\right] [/itex], we have that
    [tex] \int^{b}_{a} f\left(x\right)dx +\int^{c}_{b} f\left(x\right)dx = \lim_{n\rightarrow \infty} \frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) + \lim_{n\rightarrow \infty} \frac{\left(c-b\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) [/tex]
    [tex]\qquad\qquad\qquad\qquad\qquad\quad= \lim_{n\rightarrow \infty} \left[\frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)+\frac{\left(c-b\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)\right] = \lim_{n\rightarrow \infty} \frac{1}{n}\left[\left(b-a\right) +\left(c-b\right) \right]\sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)[/tex]
    [tex]\qquad\qquad\qquad\qquad\qquad\quad= \lim_{n\rightarrow \infty} \frac{\left(c-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) = \int^{c}_{a} f\left(x\right)dx[/tex]

    However, I have a feeling that this isn't quite correct?!
     
  2. jcsd
  3. Jun 30, 2014 #2

    micromass

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    You have made a notational error:

    You have said

    [tex]\int_a^b f(x)dx = \lim_{n\rightarrow \infty} \frac{b-a}{n} \sum_{i=1}^n f(x^*_i)[/tex]

    which is fine. But then you also do

    [tex]\int_b^c f(x)dx = \lim_{n\rightarrow \infty} \frac{c-b}{n} \sum_{i=1}^n f(x^*_i)[/tex]

    However, this is wrong. You cannot use the same notation ##x^*_i## in both equations. Indeed, the x_i^* of the integral ##\int_a^bf(x)dx## are not the same ##x_i^*## of the other integral. So you need to denote it as something else such as

    [tex]\int_b^c f(x)dx = \lim_{n\rightarrow \infty} \frac{c-b}{n} \sum_{i=1}^n f(y^*_i)[/tex]

    Also, your points ##x_i^*## should be depending on ##n##, since if ##n## changes, then so do these points. So you should probably call them ##x_{i,n}^*##.
     
  4. Jun 30, 2014 #3
    Ah ok, thanks for pointing that out. How should I proceed with the proof from that point then?
     
  5. Jul 1, 2014 #4

    HallsofIvy

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    What you can do is choose a refinement of every partition of [a, c] by adding the point "b". Then you can separate the sum from a to b and the sum from b to c.
     
  6. Jul 1, 2014 #5
    Ah ok, so I should start with [itex]\int_{a}^{c} f\left(x\right) dx = \lim_{n\rightarrow \infty} \frac{\left(c-a\right)}{n}\sum_{i=1}^{n} f\left(x_{i,n}^{\ast}\right) [/itex], and then redefine the point [itex] x_{i,n}^{\ast}[/itex] by adding the point "b", such that I can split the sum up on the RHS? (i.e. choose a set of points, [itex] x_{i,n}^{\ast} \in\left[ x_{i},x_{i+1} \right][/itex] in the interval [itex] x\in \left[a,b\right] [/itex], and then another one, [itex] y_{j,n}^{\ast} \in\left[ y_{j},y_{j+1} \right][/itex] in the interval [itex] x\in \left[b,c\right] [/itex])
     
    Last edited: Jul 1, 2014
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