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FTC Question

  1. Aug 1, 2014 #1
    This is going to be a really dumb question, but somehow it's not really clicking.

    I read somewhere that: [itex]\int_{a}^{x} f(t)\,dt = F(x) - F(a)[/itex]

    But, this is also true: [itex]F(x) = \int_{a}^{x} f(t)\,dt[/itex].

    Does this imply that [itex]F(x) = F(x) - F(a)[/itex]?

    For example, at a random number, c,:

    [itex]F(c) = \int_{c}^{x} f(t)\,dt = F(c) - F(a)[/itex]
     
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  3. Aug 2, 2014 #2

    WWGD

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    A couple of problems, if you use the Riemann integral , F(x) with F'=f does not always exist. If it does, you use it in the top equation. But it is not the same F as the F in the second equation. And F(c) is the integral from a to c.
     
  4. Aug 2, 2014 #3

    HallsofIvy

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    Yes, if F'(x)= f(x) this is true.

    No, this is not true unless it happens that F(a)= 0. The difference is that F(x) is NOT 'well defined'. If F'= f then (F+ C)'= f for any constant C.

    What is true is
    1) [itex]\int_a^x f(t)dt= F(x)- F(a)[/itex] for any function, F, such that F'= f
    and
    2) There exist a function, F, such that F'= f and [tex]\int_a^x f(t)dt= F(x)[/tex]

    The specific function, F, in (2) is one of the infinite number of functions in (1).

    No. It implies that if [itex]F_1[/itex] and [itex]F_2[/itex] are any two of the infinite number of anti-derivatives of f, then [itex]F_1(x)= F_2(x)+ C[/itex] for some constant, C. (Clearly, [itex]C= F_1(a)- F_2(a)[/itex].)

    No, the "F" on the left has to be a different anti-derivative of f than the "F" on the right. You are using the same notation to indicate two different functions.
     
  5. Aug 2, 2014 #4
    Yes! I figured this out last night, as well! Thanks everyone!
     
  6. Aug 2, 2014 #5
    Actually, I thought I figured it out, but I don't quite understand 2. Can you elaborate on that? I understand that there are infinite number of antiderivatives of a function, and [itex]F1(x) = F2(x) + c[/itex].

    What I thought was that here [itex]F(x) = \int_{a}^{x} f(t)\,dt[/itex], I used [itex]F(x)[/itex] to represent the function as a function of x, where as here [itex]\int_{a}^{x} f(t)\,dt = F(x) - F(a)[/itex], [itex]F(x)[/itex] is used to denote the anti derivative. Because they are used to denote different things, they are not equal. That's what I thought was the answer last night.
     
  7. Aug 2, 2014 #6

    HallsofIvy

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    You are still using the phrase "the anti-derivative" though you say you understand "there are infinite number of antiderivatives of a function". When you write [tex]F(x)= \int_a^x f(t)dt[/tex] you are saying that "F" is specifically that anti-derivative such that F(a)= 0.
     
  8. Aug 2, 2014 #7

    WWGD

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    Besides, remember that even if the integral exists (when f is continuous outside of a set of measure zero), the integral may exist , i.e., the Riemann sums converge, but there may be no antiderivative F with F'=f. Notice that the contradiction in [tex]F(x)= \int_a^x f(t)dt=F(x)-F(a)[/tex] shows the two functions cannot be equal, i.e., F cannot be used for both equations.
     
  9. Aug 2, 2014 #8
    I realized that writing [itex]F(x)= \int_a^x f(t)dt[/itex] implies the case when [itex]F(a) = 0[/itex], since [itex]\int_a^x f(t)dt = F(x) - F(a)[/itex]. I think I was confused because of notation: sometimes people write [itex]F(x) = \int_a^x f(t)dt[/itex] to represent the integral as a function of x. i.e [itex]F(3) = \int_a^3 f(t)dt[/itex], and not using "[itex]F(x)[/itex]" to refer to the antiderivative. When people write [itex]\int_a^x f(t)dt = F(x) - F(a)[/itex], they are saying that the integral equals the antiderivative evaluated at x minus the antiderivative evaluated at a. So in this case, "F" is used to refer to the antiderivative. Am I right, or am I confused?
     
  10. Aug 2, 2014 #9

    HallsofIvy

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    An anti-derivative, not "the" derivative! Other than that, yes, you are correct. Again, [tex]\int_a^x f(t)dt= F(x)- F(a)[/tex] for F any anti-derivative of f.
     
  11. Aug 2, 2014 #10
    Thought one: One thing, say we find the antiderivatives of [itex]y = x^2[/itex]. Obviously, [itex]\int x^2 dx = \frac{x^3}{3} + C[/itex], where [itex]\frac{x^3}{3} + C[/itex] represents all the possible antiderivatives. , i.e [itex]\frac{x^3}{3} + 4[/itex]. But we won't know which one it is until we get a definite integral, right?

    Thought two: Okay, if I'm getting it straight. Even if the definite integral is [itex]\int_{0}^{3} x^2\,dx = 9[/itex], there are still infinite antiderivatives of "F", right? For some odd reason, I used to think that the definite integral boils said infinite antiderivatives to just one.

    Which "thought" is correct, or are they both wrong?
     
    Last edited: Aug 2, 2014
  12. Aug 2, 2014 #11
    Wikipedia says (I think) that [itex]F(x) = \int_a^x f(t)dt[/itex] produces different antiderivatives depending on which value of "x" you put in there. But last time I checked, definite integrals yield a value, which is the area under the curve on that interval. How does said "number" or "value" represent an antiderivative? Antiderivatives are suppose to be represented by a function, no?
     
  13. Aug 3, 2014 #12

    HallsofIvy

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    Are you sure it doesn't say "depending on which value of "a" you put in there?

    That is not a "definite integral" in the sense that you are using. The "x" is a variable and so [itex]F(x)= \int_a^x f(t)dt[/itex] is a function of x not a number. For a specific value of x, say x= b, we have the number [itex]F(b)= \int_a^b f(t)dt[/itex].
     
  14. Aug 3, 2014 #13
    But [itex]F(x)= \int_a^x f(t)dt[/itex] doesn't represent antiderivatives right,? It just represents the area under the curve. If we put a value such as x = b like you said above: [itex]F(b)= \int_a^b f(t)dt[/itex] this yields a number, not a function.
     
  15. Aug 3, 2014 #14

    HallsofIvy

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    There is NO "area under the curve" unless you specify all boundaries of the region under the curve. [itex]\int_a^x f(t) dt[/itex] doesn't "represent the area under the curve" because "x" is not a specific number so does not define a boundary. Again, [itex]F(x)= \int_a^x f(t) dt[/itex] is one of the infinite number of antiderivatives of f.
     
  16. Aug 3, 2014 #15
    Alright, so the other antiderivatives are related to F(x) = G(x) + C, where G(x) could be [itex]\int_b^x f(t) dt[/itex], by changing the value of "a" at the bottom? I was saying that F(c), where c is a number = [itex]F(c)= \int_c^b f(t)dt[/itex] yields a number, because it has both boundaries.

    EDIT: A moment of epiphany, I do hope this is right:

    If we have [itex]F(x)= \int_a^x f(t) dt[/itex], we get all the different other antiderivatives if we change the value of "a"! Because [itex]\int_a^x f(t) dt = F(x) - F(a)[/itex], where F(a) is a constant term, and we have [itex]\int_b^x f(t) dt = F(x) - F(b)[/itex], where F(b) is a constant term. These all have the same derivative, namely, f(x) because the constant term disappears. Am I right now?
     
    Last edited: Aug 3, 2014
  17. Aug 3, 2014 #16

    HallsofIvy

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    Yes, that's pretty much what I have been trying to say!
     
  18. Aug 3, 2014 #17
    Alright, sorry if I been terribly slow. I finally understand all of this! Thanks HallsofIvy :D!
     
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