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Proving the fundamental theorem of calculus using limits

  1. Feb 24, 2015 #1
    Would it be a legitimate (valid) proof to use an [itex]\epsilon[/itex]-[itex]\delta[/itex] limit approach to prove the fundamental theorem of calculus?
    i.e. as the FTC states that if [itex]f[/itex] is a continuous function on [itex][a,b][/itex], then we can define a function [itex]F: [a,b]\rightarrow\mathbb{R}[/itex] such that [tex]F(x)=\int_{a}^{x}f(t)dt[/tex]
    Then [itex]F[/itex] is differentiable on [itex](a,b)[/itex], and [itex]\forall\;\; x\in (a,b)[/itex],
    [tex]F'(x)=f(x)[/tex] Is it possible to prove that this is true by showing that for any given [itex]\varepsilon >0[/itex] we can find a [itex]\delta >0[/itex] such that [tex]\Bigg\vert\frac{F(x+h)-F(x)}{h}-f(x)\Bigg\vert <\varepsilon[/tex] whenever [itex]0<\delta < h[/itex]?

    (I've found these set of notes https://math.berkeley.edu/~ogus/Math_1A/lectures/fundamental.pdf that uses this approach and it seems legitimate and easy to follow, but I'm unsure whether it's valid or not)?!

    (Also, is it valid to say that [itex]F'(x)=f(x)\Rightarrow F'=f[/itex], i.e. the derivative of the function [itex]F[/itex] is equal to the original function [itex]f[/itex], such that [itex]F(x)=F(a)+\int_{a}^{x}\frac{dF}{dt}dt[/itex]?)
     
    Last edited: Feb 24, 2015
  2. jcsd
  3. Feb 24, 2015 #2

    lavinia

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    You can definitely use the Newton quotient method to prove the fundamental theorem of calculus.
    I would try approximating the integral from x to x+h by hf(x) and thus the Newton quotient by f(x).
    I would then use the continuity of f to show that this approximation differs from the actual integral by epsilon as h goes to zero.
     
    Last edited: Feb 24, 2015
  4. Feb 24, 2015 #3

    PeroK

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    Why wouldn't that be valid? (Although, you've got ##h## and ##\delta## the wrong way round in the final inequality.)
     
  5. Feb 24, 2015 #4
    So the proof given in the link I gave would be a valid one then?! (Sorry for going on a bit, just wanted to double check)

    Whoops, yes thanks for pointing out the mistake. Yes, I couldn't see anything wrong with it myself, but I'm not particularly confident in myself, so wanted to clarify it.
     
  6. Feb 24, 2015 #5

    lavinia

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    Try doing the proof as I sketched it for yourself.
     
  7. Feb 24, 2015 #6
    Would the following be correct?

    Let [itex]f[/itex] be a continuous function on [itex][a,b][/itex] and define a function [itex]F: [a,b]\rightarrow\mathbb{R}[/itex] such that [tex]F(x)=\int_{a}^{x}f(t)dt[/tex]
    Now, for small [itex]h\in\mathbb{R}[/itex] we can approximate the difference between the values of [itex]F[/itex] at [itex]x[/itex] and at [itex]x+h[/itex] by [itex]hf(x)[/itex], i.e. [tex]F(x+h)-F(x)\approx hf(x)[/tex]
    As [itex]f[/itex] is continuous on [itex][a,b][/itex] we know that, for a given number [itex]t\in [x,x+h][/itex] the following limit exists [tex]\lim_{h\rightarrow 0}f(t)=f(x)[/tex]
    In other words, there exists a [itex]\delta >0[/itex] such that for any [itex]\varepsilon >0[/itex], [tex]0<h<\delta\;\;\Rightarrow\;\; \bigg\vert f(t)-f(x)\bigg\vert <\varepsilon[/tex] This implies that [tex]f(x)-\varepsilon <f(t)< f(x)+\varepsilon \\ \Rightarrow\;\;\int_{x}^{x+h}\left(f(x)-\varepsilon\right)dt < \int_{x}^{x+h}f(t)dt < \int_{x}^{x+h}\left(f(x)+\varepsilon\right)dt \\ \Rightarrow\;\; \left(f(x)-\varepsilon\right)h < \int_{x}^{x+h}f(t)dt < \left(f(x)+\varepsilon\right)h \\ \Rightarrow\;\; f(x)-\varepsilon < \frac{1}{h}\int_{x}^{x+h}f(t)dt < f(x)+\varepsilon [/tex] which further implies that [tex]\bigg\vert \frac{1}{h}\int_{x}^{x+h}f(t)dt -f(x)\bigg\vert < \varepsilon[/tex] and as such the limit [tex]\lim_{h\rightarrow 0} \frac{1}{h}\int_{x}^{x+h}f(t)dt = f(x)[/tex] exists.
    Finally we note that [tex]F(x+h)-F(x)=\int_{a}^{x+h}f(t)dt - \int_{a}^{x}f(t)dt \\ \qquad\qquad\qquad\quad\;= \left(\int_{a}^{x}f(t)dt +\int_{x}^{x+h}f(t)dt\right) - \int_{a}^{x}f(t)dt \\ \qquad\qquad\qquad\quad\;= \int_{x}^{x+h}f(t)dt[/tex]
    and so [tex] \lim_{h\rightarrow 0} \frac{1}{h}\int_{x}^{x+h}f(t)dt= \lim_{h\rightarrow 0} \frac{F(x+h)-F(x)}{h} = F'(x) = f(x) [/tex]
     
  8. Feb 24, 2015 #7

    PeroK

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    Some comments:

    I'm not sure why you included this, as you don't use it. That's the "informal" proof, just to use that approximation.


    You need to sort out the order of ##\epsilon-\delta## here. And also, you need to think about how ##t## relates to this.

    Hint: it should be ##\epsilon## then ##\delta## then ##h## then ##t##

    I think the rest is fine, although you might consider why you can take ##h > 0##. This seems to be a flaw in the link as well.

     
  9. Feb 24, 2015 #8
    Thanks for your feedback.

    So would it be the following?

    As [itex]f[/itex] is continuous at [itex]x\in [a,b][/itex], we have that for any [itex]\varepsilon >0[/itex] there exists a [itex]\delta >0[/itex] such that [itex]0<h< \delta [/itex] (where [itex]h>0[/itex] is some small positive number). Then, for all [itex]t\in [x,x+h][/itex] that satisfy [itex] 0<(t+h)-t=h< \delta[/itex], we have that the following inequality holds [itex]\bigg\vert f(t)-f(x)\bigg\vert <\varepsilon[/itex], and as such the following limit exists [tex]\lim_{h\rightarrow 0}f(t)=f(x)[/tex] where [itex]f(x)\in\mathbb{R}[/itex]
     
    Last edited: Feb 24, 2015
  10. Feb 24, 2015 #9

    PeroK

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    That's essentially correct but if you break it down a bit more I think it's more logical. For example:

    Let ##\epsilon > 0##

    As f is continuous ##\exists \delta > 0## such that ##|t - x| < \delta \Rightarrow |f(t) - f(x)| < \epsilon##

    Let ##0 < h < \delta## ...
     
  11. Feb 25, 2015 #10
    Is what you've written the case because of the following:

    [itex]f[/itex] is continuous [itex]\forall\;\; t\in [a,b][/itex] and in particular at [itex]t=x[/itex] and so [itex]\exists\delta >0[/itex] such that [itex]\vert t-x\vert <\delta\Rightarrow\big\vert f(t)-f(x)\big\vert <\varepsilon[/itex], i.e. the following limit exists: [tex]\lim_{t\rightarrow x}f(t)=f(x)[/tex] Given this we can consider the interval [itex][x,x+h][/itex] (where [itex]h>0[/itex] is some small number) such that [tex]x\leq t\leq x+h \;\;\Rightarrow\;\; 0\leq t-x \leq h<\delta[/tex] which motivates us to assume the same [itex]\delta >0[/itex] such that we have [itex]0<h<\delta[/itex]. We can then use that, from the above criteria, when [itex]t\in [x,x+h][/itex] we have that [itex]\vert t-x\vert <\delta[/itex] and so we know that the following inequality holds, [itex] f(x)-\varepsilon <f(t)< f(x)+\varepsilon[/itex]. From this, we can use that [tex]f(x)-\varepsilon <f(t)< f(x)+\varepsilon\;\;\Rightarrow\;\;\int_{x}^{x+h}\left[f(x)-\varepsilon \right]dt < \int_{x}^{x+h}f(t)dt < \int_{x}^{x+h}\left[f(x)+\varepsilon \right]dt \\ \Rightarrow \left[f(x)-\varepsilon \right]h <\int_{x}^{x+h}f(t)dt < \left[f(x)+\varepsilon \right]h \;\;\Rightarrow\;\; f(x)-\varepsilon <\frac{1}{h}\int_{x}^{x+h}f(t)dt < f(x)+\varepsilon [/tex] and this implies that [tex]\Bigg\vert \frac{1}{h}\int_{x}^{x+h}f(t)dt -f(x)\Bigg\vert <\varepsilon[/tex] whenever [itex]0<h<\delta[/itex] and the rest follows...?
     
  12. Feb 25, 2015 #11

    PeroK

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    You are over-complicating things. What I wrote is based on the clear and simple definition of the limits we are considering. To extend this a bit further:

    Let ##x \in (a, b)##

    Let ##\epsilon > 0##

    As ##f## is continuous at ##x## ##\exists \delta > 0## such that ##|t - x| < \delta \Rightarrow |f(t) - f(x)| < \epsilon##

    Let ##0 < h < \delta##

    ##t \in [x, x+h] \Rightarrow |t-x| < \delta \Rightarrow |f(t)-f(x)| < \epsilon##

    ##\Rightarrow \int_{x}^{x+h}|f(t)-f(x)|dt < h\epsilon \Rightarrow |\int_{x}^{x+h}f(t)-f(x)dt| < h\epsilon \Rightarrow |\int_{x}^{x+h}f(t)dt -hf(x)| < h\epsilon##

    ##\Rightarrow |\frac{1}{h}\int_{x}^{x+h}f(t)dt -f(x)| < \epsilon##

    ##\therefore \lim_{h \rightarrow 0^{+}}\frac{1}{h}\int_{x}^{x+h}f(t)dt = f(x)##

    Which, hopefully, is clear and logical.
     
    Last edited: Feb 25, 2015
  13. Feb 25, 2015 #12
    Thanks for your input, I admit I do have a tendency to over-complicate things at times, apologies for that!

    Does this follow because if [itex]t\in [x,x+h][/itex] then [itex]x\leq x\leq x+h[/itex] which implies that [itex]0\leq t-x \leq h < \delta[/itex] (in the limit definition doesn't it have to be [itex] 0<\vert t-x\vert<\delta[/itex] though?)

    Also, can we get away with just considering [itex]h>0[/itex] by noting that [itex] f[/itex] is continous in this interval and hence the right-hand limit must equal the left-hand limit and so the derivative [itex]F'(x)[/itex] exists?
     
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