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Integral proof

  1. Feb 7, 2005 #1

    quasar987

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    Consider f:[a,b] --> R, an integrable and bounded (ain't that implied by "integrable"?!) function and consider {a_n} a sequence that converges towards a and such that a < a_n < b (for all n). Show and rigorously justify that

    [tex]\lim_{n \rightarrow \infty} \int_{a_n}^{b} f(x)dx = \int_{a}^{b} f(x)dx[/tex]

    All we found is the, imo, not very rigorous and seemingly too easy,

    [tex]\int_{a}^{b} f(x)dx = \int_{a}^{a_n} f(x)dx + \int_{a_n}^{b} f(x)dx \Rightarrow \lim_{n \rightarrow \infty} \int_{a}^{b} f(x)dx = \lim_{n \rightarrow \infty} \int_{a}^{a_n} f(x)dx + \lim_{n \rightarrow \infty} \int_{a_n}^{b} f(x)dx[/tex]
    [tex]\Rightarrow \int_{a}^{b} f(x)dx = 0 + \lim_{n \rightarrow \infty}\int_{a_n}^{b} f(x)dx[/tex] qed

    Does anyone with more insight see how to do this more rigorously or is this the way?
     
  2. jcsd
  3. Feb 7, 2005 #2

    learningphysics

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    Yes. If it's Riemann integrable, then it must be bounded. I don't know anything about Lebesgue integrals.

    I'd say, use the second fundamental theorem of calculus to define an anti-derivative of f... call it F(x).

    Then use the first fundamental theorem of calculus, to write your integral of f(x) from [tex]a_n[/tex] to b, as F(b)-F(an). Then the rest should be easy using the properties of limits.
     
    Last edited: Feb 7, 2005
  4. Feb 7, 2005 #3

    quasar987

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    I think you're assuming that there exists a primitive. "Integrable" does not imply "there exist a primitive to f(x) such that the integral is F(b) - F(a)". Maybe f is discontinuous.
     
  5. Feb 7, 2005 #4

    learningphysics

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    Yes, you're right. Forget about the first fundamental theorem of calculus.

    Your proof is right... except you need to show that:

    [tex]lim_{n \rightarrow \infty} \int_{a}^{a_n} f(x)dx =0[/tex]

    According to the second fundamental theorem of calculus, we can define:

    [tex]F(z)=\int_{a}^{z} f(x)dx [/tex]

    We don't need f(x) to be continuous to do this. So:

    [tex]F(a_n)=\int_{a}^{a_n} f(x)dx [/tex]

    We can put this in your limit and the limit becomes:

    [tex]lim_{n \rightarrow \infty} F(a_n)[/tex]

    But according to the second fundamental theorem of calculus F(z) is continuous. So:

    [tex]lim_{n \rightarrow \infty} F(a_n)= F(lim_{n \rightarrow \infty}a_n)[/tex]

    which equals:

    [tex]F(a)[/tex]

    and [tex]F(a)=\int_{a}^{a} f(x)dx =0[/tex]

    So that should do it.
     
    Last edited: Feb 8, 2005
  6. Feb 7, 2005 #5

    quasar987

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    Very nice! Thank you for that.
     
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