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Antiderivatives and the fundamental theorem

  1. Jun 30, 2015 #1
    I know that according to the first fundamental theorem of calculus:
    $$\frac{d}{dx} \int_a^x f(t) dt = f(x)$$
    I also know that if ##F## is an antiderivative of ##f##, then the most general antiderivative is obtained by adding a constant.
    My question is, can every single antiderivative of ##f## be expressed as:
    $$\int_{a_n}^x f(t) dt = F_n (x)$$
    where ##a_n## is some constant (every ##a_n## generates a different antiderivative)? Or is it not possible in some cases?
    In other words, can every antiderivative of a function be expressed as a definite integral (one term only)?
     
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  3. Jun 30, 2015 #2

    RUber

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    The antiderivative is a function F(x) such that F'(x) = f(x). The most basic form might be given with ##a_{null} ## in the null space of F(x), such that ##F(a_{null}) = 0##, and ## F(x) = \int_{a_{null}}^x f(t) dt## so the constant c = 0.
    Any constant at the bottom of the integral will correspond to a constant in the evaluation ##F(x)-F(a) = F(x) + c## and the derivative of such a function will always be f(x).
    What sort of exceptions are you looking for? There are some requirements on f so that you can even begin to find an antiderivative...are you assuming that those basic conditions are met?
     
  4. Jun 30, 2015 #3
    ##c## can be any real number, however, ##F(a)## (as you described it) might belong in some restricted interval. So this means that not every antiderivative can be expressed as a single definite integral, right?
     
  5. Jun 30, 2015 #4

    verty

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    Hint: Integrate the zero function. What are the antiderivatives?
     
  6. Jun 30, 2015 #5

    mathwonk

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    in the first place, the fundamental theorem of calculus does not say what you wrote. that statement has no modifiers, i.e. no hypotheses. rather it says that IF f is a continuous function on the interval [a,x]. it may seem pedantic, but theorems have two parts, and the hypothesis is the most important part in many ways.
     
  7. Jun 30, 2015 #6

    pwsnafu

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    The answer is no. mathwonk brought up the point
    The key is that the domain needs to be connected for this to work. If you look at domains which are disconnected then it fails.
    Consider ##f(x) = -\frac{1}{x^2}##. Clearly, ##F(x) = \frac{1}{x}+c## right? But what about
    ##g(x) = \frac{1}{x} + 1## when ##x >0## and ##g(x) = \frac{1}{x}-1## when ##x<0##.
    We have ##g'(x) = f(x)## but now what is c? The "arbitrary constant" changes as x crosses over zero. You can't write this as a single definite integral. Your best effort would a piecewise function where each is a definite integral, or the "constant" contains the Heaviside function.
     
    Last edited: Jun 30, 2015
  8. Jul 1, 2015 #7

    lavinia

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    If your function, f(x), is continuous on an open interval then
    $$\int_{a}^x f(t) dt = F (x)$$ is an anti-derivative of f(x).

    If you change a then the anti-derivative changes by a constant.
    But not all anti-derivatives can be obtained in this way. For instance take the function,f(x), that is identically equal to zero on the interval (0,1). Then $$\int_{a}^x f(t) dt = F(x)$$ is equal to zero. But any constant function is also an anti-derivative.

    What is true is that if F(x) is an anti-derivative of f(x) then $$\int_{a}^x f(t) dt = F(x) - F(a)$$
     
    Last edited: Jul 1, 2015
  9. Jul 10, 2015 #8

    mathwonk

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    actually the correct hypotheses on the function f are quite interesting. I.e. there are two facts you want to assume: 1) the riemann integral of f exists on [a,b], 2) f has an "antiderivative" F. and then you want to conclude 3) that the integral of f equal F(b) - F(a).

    Continuity of f suffices for all 3 of these to hold, but is not necessary. E.g. if f is a step function then it has a finite set of discontinuities and its indefinite integral is differentiable elsewhere with derivative equal to f. If however we define an "antiderivative" of f to be a continuous function with derivative equal to f except at the finite set of discontimnuities, then again the indefinite integral satisfies both 2) and 3).

    There are however more complicated functions f whose riemann integral exists on [a,b] and such that an antiderivative F in this sense, i.e. F continuous everywhere and derivative equal to f where f is continuous, does not satisfy 3). an example is the characteristic function f of the cantor set, which is zero off the cantor set, hence has integral zero. the cantor function F which is locally constant off the cantor set, hence has derivative zero there, is continuous everywhere, but climbs from 0 to 1 over [a,b], hence does not compute the integral of f. We can however strengthen the requirement on an "antiderivative" to be Lipschitz continuous.

    Notice the key point that failed was that with the weaker definition of antiderivative we did not get that any two of them differ by a constant, i.e. a continuous function (even on a connected interval), and with derivative zero almost everywhere, does not have to be constant, but we do get that with the addition of the Lipschitz property.
     
    Last edited: Jul 10, 2015
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