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Question about Indefinite integrals

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data

    I hope this is in the right forum, because this is a question on theory and not related to a specific problem.

    I was reading onlne about the Fundamental Theorem of Calculus. On one site the author wrote:

    [tex]F(x) = \int_{0}^{x} f(t) dt[/tex]

    Later, he wrote:

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



    However, I've been taught that

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

    equals the indefinite integral of f(x) evaluated at b minus the indefinite integral of f(x) evaluated at a.

    This leads me to ask:

    Is the indefinite integral of a function, lets call it [itex]G(x)[/itex], the same as the definite integral [tex]\int_{0}^{x} f(t) dt[/tex] ?

    I know this is a really basic question.

    ****
    Edited to fix what Mark44 pointed out
     
    Last edited: Sep 16, 2012
  2. jcsd
  3. Sep 15, 2012 #2

    Mark44

    Staff: Mentor

    I don't think you have copied this correctly. What you have above is a number, not a function. The value of the integral doesn't depend on x at all.

    It's more likely that the author wrote something like
    $$F(x) = \int_{0}^{x} f(t) dt $$

    This is now a function. Notice that the variable x appears as one of the limits of integration. The variable t (and dt) are called dummy variables. We could just as well have used r and dr or whatever, as long as we don't use x and dx.
    You can write
    $$ \int_{a}^{b} f(x) dx $$
    as
    $$ \int_{a}^{0} f(x) dx + \int_{0}^{b} f(x) dx $$
    Switch the limits of integration on the first integral to get
    $$ -\int_{0}^{a} f(x) dx + \int_{0}^{b} f(x) dx $$
    This is the same as -F(a) + F(b), using F as defined in the revised definition of F that I showed above.
    Better terminology would be "antiderivative of f" instead of "indefinite integral of f".
    Let me set things up a little better.
    Here's an indefinite integral:
    $$\int f(x) dx$$

    Let F(x) be defined as:
    $$ F(x) = \int_0^x f(t)dt$$

    Then, F'(x) = f(x) (by the Fund. Thm. of Calc), so F is an antiderivative of f.

    By the other part of the FTC, you can evaluate a definite integral by using an antiderivative.

    ## \int_a^b f(t)dt = \int_0^b f(t)dt - \int_0^a f(t)dt = F(a) - F(b)##
     
  4. Sep 15, 2012 #3
    Mark44 -

    You're right, I miss-copied. When I have access to a computer I'll try to explain what I meant.
    Thanks :)
     
  5. Sep 16, 2012 #4
    Ok, here's what I meant:

    I was looking at ## \int_a^b f(t)dt = \int_0^b f(t)dt - \int_0^a f(t)dt = F(a) - F(b)##

    and I've been taught that F(x), or F(a) or so on, in this context, is an indefinite integral. Like this:

    $$\int f(x) dx$$

    But that didn't make sense to me because I don't see how you could evaluate that up to a (in the case of F(a)). To evaluate up to a value you'd need to evaluate from some lower limit to an upper limit, which don't appear on the indefinite integral.

    So I was trying to ask what F(x) is. Your definition of it as

    [tex]F(x) = \int_0^x f(t) dt[/tex]

    makes much more sense.

    I'm a little confused about the difference between an antiderivative and an indefinite integral. Can you please explain?

    Thanks.
     
  6. Sep 16, 2012 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Now, this is wrong! The final term should be F(b)- F(a). Perhaps a typo?

    You may remember it that way but I doubt you were taught that. What is true is that [itex]F(x)= \int_0^x f(t)dt[/itex] is an indefinite integral but [itex]F(a)= \int_0^a f(t)dt[/itex] is a definite integral.

    The upper limit does appear on the indefinite integral- it is the variable x. To evaluate at, say, x= a, gives a number- the definite integral.

    They are "almost" the same thing. Though it would be better to say "an antiderivative" and "the indefinite integral". If, for example, f(x)= 2x, then the "indefinite integral" is [itex]\int 2xdx= x^2+ C[/itex]. Notice the "+C"? Choosing a specific "C" give one of the many possible antiderivatives, functions whose derivative is 2x.

     
    Last edited: Sep 17, 2012
  7. Sep 16, 2012 #6
    Absolutely a typo. :) sorry!


    You'd be surprised at the quality of the instruction here - for example, one of my teachers didn't know what this

    f(x+h)-f(x)
    ----------
    h

    lim h--> 0


    was.

    I understand that the upper limit is x, I just mean that that I would expect the definition of the indefinite integral to contain *some* reference to a lower bound of 0, which does in fact appear in the definition of the indefinite integral you guys gave (and does *not* appear in any definition I've been given). Otherwise the notation doesn't make sense. It would be like trying to evaluate

    [tex]\int_{}^x f(x) dx[/tex]

    That makes sense. It's like the indefinite integral describes a group of functions that are antiderivatives. Is that correct?

    Also, can you say,

    [tex]\int f(x) dx = F(x) + C[/tex]

    where F(x) is an antiderivative?

    Thanks :) :)
     
    Last edited: Sep 16, 2012
  8. Sep 16, 2012 #7

    SammyS

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The indefinite integral describes a family of functions each of which is an antiderivatives of the integrand, & these function differ from each other by at most a constant term (i.e. an additive constant).
    You say,
    [tex]\int f(x) dx = F(x) + C[/tex]
    where F(x) is an antiderivative of f(x) .
     
  9. Sep 17, 2012 #8
    Thanks a lot everyone!

    :D
     
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