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Fundamental theorem of calculus

  1. Jun 22, 2015 #1
    "If ##f## is continuous on ##[a,b]## and:
    $$g(x) = \int_a^x f(t) dt$$
    Then ##g## is continuous on ##[a,b]##, differentiable on ##(a,b)##, and ##g'(x) = f(x)##."

    This is the first fundamental theorem of calculus. I'm curious as to why ##g## is only differentiable on ##(a,b)##, but not ##[a,b]##.
     
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  3. Jun 22, 2015 #2

    micromass

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    Because you likely have not defined differentiability on closed intervals, only on open ones.
     
  4. Jun 22, 2015 #3

    Svein

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    Definition. Take point b. In order for the derivative to exist at b, the two limits [itex]\lim_{h\rightarrow 0}\frac{g(b-h)-g(b)}{h} [/itex] and [itex] \lim_{h\rightarrow 0}\frac{g(b+h)-g(b)}{h}[/itex] must both exist and be equal. But [itex]g(b+h) [/itex] is not defined and may not even exist!
    The same argument applies to point a (only here it is g(a-h) that is not defined).
     
  5. Jun 22, 2015 #4
    But can't we say the same about the continuity of ##f## and/or ##g##?
    In order for ##f## to be continuous at ##a## (implied by the statement "##f## is continuous on ##[a,b]##"), the following must hold true:
    $$\lim_{x \rightarrow a} f(x) = f(a)$$
    But ##\lim_{x \rightarrow a} f(x)## can only exist if ##\lim_{x \rightarrow a^+} f(x) = \lim_{x \rightarrow a^-} f(x)##.
     
  6. Jun 22, 2015 #5

    HallsofIvy

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    No. What is implied by the statement "f is continuous on [a, b]" is that f is continuous from the right so the only thing that is certain is that [itex]\lim_{x\to a^+} f(x)= f(a)[/itex].
     
  7. Jun 22, 2015 #6
    But again, why can't the same be said of the differentiability of a function on a closed interval ##[a,b]##?
     
  8. Jun 22, 2015 #7

    micromass

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    Because you haven't defined differentiability on closed intervals, only on open ones. You can define it on closed intervals and then the fundamental theorem holds on those closed intervals, but that is usually not done because it is annoying (and not very useful).
     
  9. Jun 23, 2015 #8

    Stephen Tashi

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    We need to straighten out whether this is an implication that follows simply from the ordinary definitions of the limit of a function and the definition of a function being continuous at a point or whether it comes from a special definition for the phrase "f(x) is continuous on [a,b]". For example, if a function f(x) does not exist outside the interval [a,b] then does the ordinary definition of "f(x) is continuous at x = c" imply that f(x) is NOT continuous at x = a? Or do the ordinary definitions imply [itex] lim_{x\rightarrow a^+} f(x) = f(a) [/itex] without any need for making a special definition for the phrase "f(x) is continuous on [a,b]" ?

    ( I think this is related to the controversies one sees in discussions about the meaning of the phrase "f(x) is everywhere continuous" when applied to functions whose domain is a proper subset of the real numbers. In a sophisticated treatment, the continuity of a function depends on the topology being used for the "spaces" of the domain and range. )
     
  10. Jun 24, 2015 #9
    What if the integrand, ##f##, were continuous over ##(a,b)## instead? Would it make any difference?
    And what if ##f## were defined for all ##x## in ##(a,b)##, but not at ##a## and/or ##b##?
     
  11. Jun 24, 2015 #10

    Svein

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    No. The integral has the same value (as long as we are talking about relatively standard integration).
    Same thing.
     
  12. Jun 24, 2015 #11
    But why is this not mentioned in the fundamental theorem of calculus? Or am I missing something?
    I know continuity on ##[a,b]## implies continuity on ##(a,b)## (not the other way around), but does that mean that based on the fundamental theorem (as stated above) I can deduce that continuity on ##(a,b)## doesn't make a difference?
     
  13. Jun 25, 2015 #12

    Svein

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    Observations: [itex] \int_{a}^{b}f(x)dx=\int_{a}^{a}f(x)dx+\int_{(a, b)}f(x)dx+\int_{b}^{b}f(x)dx[/itex]. Since f is continuous over [a, b], f(a) and f(b) exist and are finite. Therefore [itex] \int_{a}^{a}f(x)dx = f(a)\cdot (a - a)=0[/itex]. We also have [itex] \int_{b}^{b}f(x)dx = f(b)\cdot (b - b)=0[/itex]. Therefore [itex] \int_{a}^{b}f(x)dx=\int_{(a, b)}f(x)dx[/itex]
     
  14. Jun 26, 2015 #13

    Stephen Tashi

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    Try [itex] f(x) = 1 - x [/itex] if [itex] x \le 1 [/itex] and [itex] f(x) = x - 1 [/itex] otherwise and let [itex] a = 0,\ b = 1 [/itex]. Let [itex] g(x) = \int_0^x f(x) dx [/itex].
     
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