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FTC and integrals

  1. Mar 17, 2014 #1
    By FTC, every function f(x) can be expessed like: [tex]f(x) = \int_{x_0}^{x}f'(u)du + f(x_0)[/tex] Now, I ask: f(x) is a antiderivative of f'(x) or is a family of antiderivative of f'(x) ?
     
  2. jcsd
  3. Mar 17, 2014 #2

    Mark44

    Staff: Mentor

    f(x) is any antiderivative of f'.
     
  4. Mar 17, 2014 #3
    That's not what I asked.
     
  5. Mar 17, 2014 #4

    Mark44

    Staff: Mentor

    I actually did answer your question.
    f is one (pick any one that you like) antiderivative of f'. f is not a family of antiderivatives.
     
  6. Mar 17, 2014 #5
    Now yes!

    So, you are saying (implicitly) that f(x0) is not a arbitrary constant?
     
  7. Mar 17, 2014 #6

    Mark44

    Staff: Mentor

    Let's look at an example:
    $$\int_1^x t^2 dt$$

    Solution 1 -
    One antiderivative of t2 is (1/3)t3. Call this F1(t). By the FTC, the integral above is equal to F1(x) - F1(1) = (1/3)x3 - 1/3

    Solution 2
    Another antiderivative is t2 - 3. Call this F2(t). Again, by the FTC, the integral above is equal to F2(x) - F2(1) = [(1/3)x3 - 3] - [1/3 - 3] = (1/3)x3 - 1/3. This is the same value that was obtained in solution 1.

    What you get for F(t0) depends on which function you use for the antiderivative, but the resulting value of the definite integral doesn't depend on which antiderivative you pick. You can choose any function in the family of antiderivatives, without making any difference at all in the resulting value of the definite integral. For this reason, it's most convenient to work with the antiderivative for which the constant is 0.

    All of this seems very straightforward to me. Is there something I'm missing in what you're asking?
     
  8. Mar 17, 2014 #7
    Sorry my arrogance, but I still didn't understood if f(x0) is a arbitrary value...
     
  9. Mar 17, 2014 #8

    Mark44

    Staff: Mentor

    It's arbitrary in the sense that f is an unspecified antiderivative. In my two examples f(1) has two different values, depending on which antiderivative we're using. In my first example, f(1) (which I'm calling F1(1)) = 1/3. In the second example, F2(1) = 1/3 - 3 = -8/3.

    For a specified antiderivative F, F(t0) is a constant, but if the particular antiderivative is not known, then we don't know the value of F(t0).
     
  10. Mar 17, 2014 #9
    Based in your explanation, I can say that f(x0) isn't an arbitrary constant such as is an arbritrary constant C, that appears in integration*. Correct!?

    *∫ydx = Y + C
     
  11. Mar 17, 2014 #10

    Mark44

    Staff: Mentor

    Yeah, I think that's OK.
     
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