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Fundamental theorem of calculus (something isn't right)

  1. Feb 12, 2010 #1
    The 2nd part of fundamental theorem of calculus says:
    Over what open interval does the formula


    represent antiderivative of f(x)=1/x ?

    By looking at the theorem I would say that f(x) is continuous only for [tex]x \neq 0[/tex]
    So I would say that F(x) is defined on [tex](-\infty, 0) U (0, +\infty)[/tex] that is F'(x)=f(x) for all numbers [tex]x \neq 0[/tex].

    But there is problem. F(x) actually equals Log(x) , because (Log(x))' = 1/x.

    Why this theorem does not work?

    Now there is problem with the fundamental theorem of calculus part 1:

    Here is the counterexample:


    But again the area is not -2 it is actually 5/2.

    One more thing:

    [tex](\frac{|2x-3|x(x-3)}{2x-3})' = |2x-3|[/tex] where [itex]x \neq 3/2[/itex] because F(x) is not defined at 3/2

    Which is not same as |2x-3| which is defined everywhere.

    Thanks in advance.
  2. jcsd
  3. Feb 12, 2010 #2
    Well your first definition of the theorem is quite wrong.

    A more accurate form will be:

    Let f(x) be integrable on some interval I, and let a be a real number in I. Then whenever
    the function


    Is continuous at a point x0, its derivative at this point will be equal to the original function: [tex]F'(x_{0})=f(x_{0})[/tex]

    Otherwise, if F is not continuous or defined at the point, the theorem fails.

    As for the second part, notice that the formula requires a function F which is continuous in [a,b], so it will be the real anti-derivative. Otherwise, again, the theorem fails.

    Of course your F is discontinuous at x=3/2: F(3/2-)=-3/2(3/2-3). F(3/2+)=3/2(3/2-3)
    And therefore your result should be wrong.
    But at any interval [0,3/2), (3/2,2] you may choose an anti-derivative up to a constant.
    Therefore there exist such constant that at x=3/2 the functions will be continuous (while not defined, it will have a removable discontinuity, which happens to work)

    Perhaps this will help you understand what was bothering you. I didn't quite catch it at the first part of the post.
  4. Feb 12, 2010 #3
    Thanks a lot for the reply.

    Now it make my things clearer by stating that F(x) must be continious so that it would be anti-derivative.

    Now I know that [tex]F(x)=\frac{|2x-3|x(x-3)}{2x-3}[/tex] is not the actual anti derivative of [tex]|2x-3|[/tex] and it is because the derivative of is not defined at 3/2, so that when I find F'(x) it would be equal to |2x-3| for [tex]x\neq 3/2[/tex]

    and |2x-3| is defined at all real numbers. Am I right?

    By writing the definite integral as:

    [tex] \int_{0}^{2}|2x-3|dx=\int_{0}^{3/2}(2x-3) + \int_{3/2}^{2}-(2x-3)[/tex]

    I somehow "skip" the discontinuity at 3/2.

    For the first part. I guess the theorem lacks of saying that F(x) must be defined on the interval I.

    For example. If I use the first theorem I would say that:


    F(x) is defined for [tex]x\neq 0[/tex] because f(x) is continuous for all reals except x=0, and the theorem says that if f(x) is continuous on interval I then also F(x) is continuous on interval I.

    But this is not true.

    Actually f(x)=1/x is continuous for all reals except x=0.

    BUT, F(x) = Log(x) and it is not continuous on all negative real numbers.

    I know that also F(x) could be Log|x| but it also could be Log(x).

    What is the problem? Is it with the book that I am using for learning?
  5. Feb 12, 2010 #4
    The problem is that you are not integrating 1/x correctly.

    The integral of 1/x is ln|x|, not ln x. It just happens that for positive x, ln|x|=ln x.
  6. Feb 12, 2010 #5
    But it can be ln(x). Why? Because (ln(x))' = 1/x. Because integration is anti-differentiation:

    [tex]\int 1/x = ln(x)[/tex]

    Am I right?

    Could you please possibly confirm my statement:

    Thank you.
  7. Feb 12, 2010 #6
    Your argument is flawed. You start with a function which by is defined on (0,∞). When you differentiate this fuction, it will only be valid on (0,∞), even if it is defined on (-∞,∞). Thus, you cannot expect the integral of 1/x to be ln(x) outside (0,∞).

    To account for the fact that 1/x is defined on (-∞,∞), its integral is ln|x|, not ln(x).
  8. Feb 12, 2010 #7
    Thank you, I understand now.

    In that case:
    is defined on real numbers except x=3/2

    So on F'(x)=|2x-3|=f(x) we cannot expect f(x) to be defined on x=3/2 since F'(3/2) is not defined because of the discontinuity of F(x) on x=3/2.

    But what we need to make to let |2x-3| be defined on (-∞,∞)?
  9. Feb 12, 2010 #8
    Any fuction with x'es in the denominator will have a discontinuity and thus can't be defined for all x.
  10. Feb 12, 2010 #9
    Does that mean that we need to set notice that the function f(x)=|2x-3| is valid on all real numbers except x=3/2, before stating the integral:


    so that F(x) would also not be defined on x=3/2?
  11. Feb 12, 2010 #10
    That fuction itself has no problem with being defined for all real x. why should it? |0| is just 0.
  12. Feb 12, 2010 #11

    F(x) is not defined on x=3/2 so that F'(x) would be not defined on x=3/2 so that F'(x)=f(x) would be only valid for [tex]x \neq 3/2[/tex], right?
  13. Feb 12, 2010 #12
    Yes, that looks right.
  14. Feb 12, 2010 #13
    Ok, that means for

    [tex]f(x)=|2x-3|[/tex] is valid only for [tex]x \neq 3/2[/tex] (even it is continuous on infinity), because of the discontinuity of its anti-derivative:


    And that is the problem with the theorem which says:
    "If f is continuous on an interval I, then f has an anti derivative on I."

    In this case f is continuous on infinity, but the anti derivative is not defined on infinity.

    elibj123 gave the correct theorem, but how is possible that book for Calculus from Irl Bivens, Steven Davis and Howard Anton made such a mistake?

    Also in the solutions of the text book they say that:


    is defined on open interval (0, +∞) when it is defined on (-∞,0) U (0,+∞)?

    It is the same for:


    except for the interval (3,+∞) whether it should be (-∞,3) U (3,+∞).

    They wrote the theorem and disregard it:

    "If f is continuous on an interval I, then f has an anti derivative on I."

    even it is incorrect.
  15. Feb 12, 2010 #14


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    No - the function used does not have a removable discontinuity at x = 3/2. the function is

    If x < 3/2 this simplifies to

    F(x) = -x(x-3)

    For x > 3/2 it simplifies to [tex] F(x) = x(x-3) [/tex]. You cannot provide any value at x = 3/2 to turn this into a continuous function (try graphing it to see this).
  16. Feb 12, 2010 #15


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    The function

    F(x) = \int_1^x \frac 1 t \, dt

    is not defined on [tex] (-\infty, 0) [/tex] since, if [tex] x < 0 [/tex], the integral

    \int_1^x \frac 1 t \, dt

    does not exist (it doesn't converge). [tex] F [/tex] IS continuous on [tex] (0,\infty) [/tex]
  17. Feb 12, 2010 #16


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    I have no idea what you mean by ``defined on infinity''. Your problem comes from the way you are dealing with [tex] |2x-3| [/tex].

    Think this way. You are saying that the derivative of

    F(x) = \frac{|2x-3|x(x-3)}{2x-3}, \quad x \ne 3/2


    F'(x) = |2x-3|, \quad x \ne 3/2

    and that is true - but you have to stick with the condition [tex] x \ne 3/2 [/tex]. In this case, you can say this: since |2x-3| is defined and continuous on the interval [tex] (-\infty, 3/2) \cup (3/2, \infty)[/tex], it has a derivative there (one that works is your function F).

    What you do in the second part is to consider the function

    g(x) = |2x - 3|

    as a starting point. This function is continuous for all real numbers, so it has an antiderivative. This is not the contradiction you claim: in the case above the absolute value function wasn't defined at 3/2; its antiderivative was F. The current function is defined at 3/2, so its antiderivative is defined at 3/2 and so the antiderivative IS NOT THE FUNCTION F. Here

    g(x) = \begin{cases}
    -(2x-3) &\text{ if } x < 3/2 \\
    \hphantom{-}(2x-3) &\text{ if } x \ge 3/2

    An antiderivative can be formed from this.

    In summary: in the first situation you arrived at |2x-3| by considering it to be derivative function not defined at 3/2, so in this context it is not appropriate to treat |2x -3| as though it is defined at 3/2.

    In the second part you start with |2x-3| as its own function, so it is defined and continuous everywhere; that means it has an antiderivative everywhere, but that antiderivative cannot be F. There is no contradiction with the theorem.
  18. Feb 13, 2010 #17
    Thanks a lot for the posts.

    First you're right by saying that the anti-derivative is not F(x).

    But the theorem says:

    F(x) is an anti-derivative of f on I, that is F'(x) = f(x) for each x in I.

    Also there is theorem which says:

    This is not true.

    f(t)=|2t-3| is continuous on [0,2] but F(x) is not continuous on [0,2] it has gap in x=3/2.

    I am just following the theorem. That's the problem with the theorem.

    The theorem must state that F(x) must be continuous so that F'(x)=f(x) will work for all x in the interval. Do you agree?

    Or maybe I found such F(x) which is continuous on any interval [a,b] :confused:

    x(x-3), x > \frac{3}{2}\\
    \frac{-9}{4}, x=3/2 \\
    \end{matrix}\right. [/tex]
    Last edited: Feb 13, 2010
  19. Feb 13, 2010 #18
    if f(x) is continuous from x=a to x=b, then f(x) is integrable from a to b.
    IF F(x) (the antiderivative of f(x)) is not continuous from a to b, then it will not be integrable from a to b
    but im very sure that not only is 2t-3 integrable from 0 to 2,
    its antiderivative t^2 - 3t + c is also
  20. Feb 13, 2010 #19
    You missread my post and perhaps didn't understand the subject yourself.

    The antiderivative of a function is not unique, a function has infinitely many anti-derivatives differing by a constant. That constant doesn't have to be "constant" on the entire interval.

    For example

    F(x)=2x+3 for x>0
    F(x)=2x+100 for x<0

    Is still an antiderivative of f(x)=2, but it's not of the form F(x)=2x+c (c changes between the intervals).

    What I said is that while


    is discontinuous, you can choose such c's, that the function F(x) (while not defined with a single formula) will be now continuous on the entire interval (neglecting the minor removable discontinuity at x=3/2) and then the definite integral identity applies.
  21. Feb 13, 2010 #20


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    Apparently I understand this issue better than you - so let's stop this back and forth.
    No, you cannot make F(x) continuous by defining its value at 3/2. Perhaps you should graph it to see this. This is not a case where there is a hole in the graph but the two one-sided limits agree: the one-sided limits at 3/2 are different.
    Last edited: Feb 13, 2010
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