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Help understanding the First Fundamental Theorem of Calculus

  1. Apr 13, 2012 #1
    The first fundamental theorem of calculus begins by defining a function like this:


    (sorry was not sure how to write this legibly in this post so I just uploaded on imgur)

    I kind of have a hard time wrapping my mind aruond this. How do you chose a? I feel like in the proof for the fundamental theorem, a ends up canceling out and doesn't matter, but it's hard for me to look at this and see and understand that. If F is an antiderivative of f, then you would end up with F(x)-F(a) isn't it? By the second fundamental theorem? I'm so confused every time I look at this. I just don't understand how a is or is not relevant and how this function is defined this way.
  2. jcsd
  3. Apr 13, 2012 #2
    That "a" is any point in the definition domain of f(x), and so is x, and we have thus that f(x) is defined

    and Riemann integrable in [a,x], or in [x,a]...what's so strange to you in this?

    For example, [itex]F(x)=\int^x_a\frac{1}{x}\, dx\,,\,\,a\,,\,x>0\,,\,\, [/itex] gives us [itex]F(x)=\log x-\log a[/itex] , and choosing a different positive

    a changes F only by an additive constant...

  4. Apr 13, 2012 #3
    So does the "a part" end up being the constant of integration?
  5. Apr 13, 2012 #4
    The first fundamental theorem only states the derivative of F is f.
    And when you differentiate, the constants don't matter, so the fact that F is an antiderivative of f is actually independent from which a you choose. More concretely, and without any mathematical rigor :

    [tex]\dfrac{d}{dx}\int_b^x f(t)dt = \dfrac{d}{dx}\left(\int_b^a f(t)dt + \int_a^x f(t)dt \right) = \dfrac{d}{dx} \int_a^x f(t)dt = f(x) [/tex]
  6. Apr 13, 2012 #5

    The is no constant of integration in definite integration.

  7. Apr 13, 2012 #6
    ooooooooh sachav thank you, this clears up my main misunderstanding about what the first fundamental theorem is saying. Though I do have one further question. While it makes sense that the derivative of F is f (because the constant goes away during the differentiation), I still have a difficult time undersatnding that way the function defined (in the imgur link in my first post).

    DonAntonion I think you answered this well, but I think I messed up in my reply when I mentioned constant of integration. What I mean is, can you chose any "a" value (defined in the domain of course) because you end up with F(x)-F(a) and depending then on which a you chose you will get this different associated constant, which would still be an antiderivative of f because that constant will go away in the integration? So for example, if you chose constant a vs. constant b you would end up with two sepearate functions which are both antiderivatives which are different only based on an additive constant. Is that why the choice of a doesn't matter?
  8. Apr 13, 2012 #7
    A possible example of why I am confused:

    this is a screenshot of the calc book I am reviewing http://i.imgur.com/d8KHi.jpg

    I guess I have a difficult time for example understanding natural log as defined this way, because I don't understand why 1 has been chosen as a. I guess it makes sense because they are trying to equate the integral to lnx and ln(1)=0. EDIT: Oh wait nevermind I think I understand. Ok. So it makes sense why 1 is the lower limit on integration there. Because say you are looking at x = 2, you end up evaluating ln(2)-ln(1) = ln(2).

    sorry if these questions are stupid and basic. I just want to make sure I am understanding all of this thoroughly rather than just memorizing definitions
  9. Apr 14, 2012 #8
    Actually, having 1 as the lower bound of integration makes it more coherent, since it makes it so that ln is the inverse function of the exponential (with another lower bound, its inverse would be the exponential multiplied by the lower bound constant).
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