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First Fundamental Theorem of Calculus

  1. Jun 25, 2013 #1
    Hi,

    I just learned about the First Fundamental theorem of calculus. From my understanding, it talks specifically about definite integrals. I was wondering if there is any sort of theorem that proves that the derivative of the indefinite integral of a function is equal to the function itself.

    Thank you in advance!
     
  2. jcsd
  3. Jun 25, 2013 #2
    Me too,

    I have learned Differential Calculus recently, And then integral Calculus.

    I am not sure if there is a theorem because we derive other equations from calculus itself.

    I am not sure Peter, but i will get back to you after i figure it out.
     
  4. Jun 25, 2013 #3

    WannabeNewton

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    That is the very definition of the indefinite integral.
     
  5. Jun 25, 2013 #4

    lurflurf

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    The definition of indefinite integral is the derivative of the indefinite integral of a function is equal to the function itself. The fundamental theorem establishes that definite and indefinite integrals are almost the same thing. Which is why they have very similar notation. this causes confusion at times.
     
  6. Jun 25, 2013 #5
    Thank you very much guys! I hope this is useful to you too, utkarshraj!
     
  7. Jun 26, 2013 #6

    Bacle2

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    But there are some subtleties to consider too: Not every (differentiable) function can be recovered from its pointwise derivative . The Cantor function f is not the integral of its derivative, since f is a.e. constant , so you do not recover the integral from its pointwise derivative --since f' is a.e. 0, so that:

    ∫f' =∫0 * =0 ≠ f , but f is not constant.

    * f' is a.e. 0 but not actually 0 .

    http://en.wikipedia.org/wiki/Cantor_function
     
  8. Jun 26, 2013 #7

    micromass

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    OK, but the Cantor function is not even differentiable. It's only a.e. differentiable.
     
  9. Jun 26, 2013 #8

    Bacle2

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    Right, but the point is that it is Riemann-integrable but does not have an antiderivative, so you
    work a.e.
     
  10. Jun 26, 2013 #9

    micromass

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    Why wouldn't it have an antiderivative?
     
  11. Jun 27, 2013 #10

    Bacle2

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    By post #6 and the fact that integrals of a.e functions are equal, there is no F with F'=f , where f is the Cantor function.
     
  12. Jun 27, 2013 #11

    micromass

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    Since ##f## is continuous, it seems to me that ##\int_0^x f## would do the trick...
     
  13. Jun 27, 2013 #12

    Bacle2

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    ∫f'=∫0=0 , where the integral is over [0,1]
     
  14. Jun 27, 2013 #13

    micromass

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    That's true. So why does that imply that ##f## has no antiderivative?
     
  15. Jun 27, 2013 #14

    Bacle2

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  16. Jun 27, 2013 #15

    micromass

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    I absolutely agree that the cantor function isn't absolutely continuous.

    But why does that imply that it doesn't have an antiderivative. Since the Cantor function ##f## is continuous, doesn't the fundamental theorem of calculus imply that ##\int_0^x f## is an antiderviative?
     
  17. Jun 27, 2013 #16

    pwsnafu

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    Further, it's a standard exercise in measure theory to show that for any ##f \in L^1[0,1]##, the antiderivative ##\int_{[0,x]} f## is absolutely continuous. This of course is just a part of Fundamental Theorem of Lebesgue integration.
     
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