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Fundamental theorem of calculus - question & proof verifying

  1. Mar 4, 2015 #1
    I understand that the fundamental theorem of calculus is essentially the statement that the derivative of the anti-derivative [itex]F[/itex] evaluated at [itex]x\in (a,b)[/itex] is equal to the value of the primitive function (integrand) [itex]f[/itex] evaluated at [itex]x\in (a,b)[/itex], i.e. [itex]F'(x)=f(x)[/itex]. However, can one imply directly from this that [itex]F'=f\;\;\forall x\in (a,b)[/itex], such that [tex]F(b)-F(a)=\int_{a}^{b}F'(x)dx\;\;?[/tex]

    Also, I have attempted to prove the FTC and am interested to know whether my attempt is valid?!

    Let [itex]f[/itex] be a continuous function on [itex][a,b][/itex] and let [itex]F: \mathbb{R}\rightarrow\mathbb{R}[/itex] be a function that is continuous on [itex][a,b][/itex] and differentiable on [itex](a,b)[/itex], defined such that [tex]F(x)=\int_{a}^{x}f(t)dt[/tex] where [itex]x\in (a,b)[/itex].
    As [itex]f[/itex] is continuous on [itex][a,b][/itex] we therefore know that for any [itex]\varepsilon >0\;\;\exists\delta >0[/itex] such that [tex]0<\vert t-x\vert <\delta\;\;\Rightarrow\;\;\big\vert f(t)-f(x)\big\vert <\varepsilon\quad\forall x \in [a,b][/tex]
    Hence, consider the interval [itex](x,x+h)\subseteq (a,b)[/itex] and let [itex]0<h<\delta[/itex]. Now clearly [itex]x<t<x+h\;\;\Rightarrow\;\;0<t-x<h<\delta[/itex] and so [tex]\big\vert f(t)-f(x)\big\vert <\varepsilon\quad\forall t \in (x,x+h)[/tex] which implies that in this interval [tex]f(x)-\varepsilon<f(t)<f(x)+\varepsilon[/tex] Using that [itex]f(x)\leq g(x)\;\;\Rightarrow\;\;\int_{a}^{b}f(x)dx\leq \int_{a}^{b}g(x)dx[/itex] it follows that [tex]\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 \\ \Rightarrow\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.
    Next we consider the interval [itex](x-h,x)\subseteq (a,b)[/itex] and again let [itex]0<h<\delta[/itex]. We see that [itex]x-h<t<x\;\;\Rightarrow\;\;-h<t-x<0\;\;\Rightarrow\;\;0<-(t-x)<h[/itex] and so, following the same procedure as before [tex]\big\vert f(t)-f(x)\big\vert <\varepsilon\quad\forall t \in (x-h,x) \\ \Rightarrow\int_{x-h}^{x}\left[f(x)-\varepsilon\right]dt<\int_{x-h}^{x}f(t)dt<\int_{x-h}^{x}\left[f(x)+\varepsilon\right]dt \\ \Rightarrow \left[f(x)-\varepsilon\right]h<\int_{x-h}^{x}f(t)dt<\left[f(x)+\varepsilon\right]h\\ \Rightarrow f(x)-\varepsilon<\frac{1}{h}\int_{x-h}^{x}f(t)dt<f(x)+\varepsilon \\ \Rightarrow\Bigg\vert \frac{1}{h}\int_{x-h}^{x}f(t)dt - f(x)\Bigg\vert <\varepsilon [/tex] and as such the limit [tex]\lim_{h\rightarrow 0^{-}}\frac{1}{h}\int_{x-h}^{x}f(t)dt=f(x)[/tex] exists.
    Therefore, as the right-handed and left-handed limits exist and are equal, we have that for any [itex]\varepsilon >0\;\;\exists\delta >0[/itex] such that [tex]\left(0<\vert t-x\vert <h<\delta\;\;\Rightarrow\;\; 0<h<\delta\right)\;\;\Rightarrow\;\;\bigg\vert \frac{1}{h}\int_{x}^{x+h}f(t)dt -f(x)\bigg\vert <\varepsilon[/tex] i.e. the limit [tex]\lim_{h\rightarrow 0}\frac{1}{h}\int_{x}^{x+h}f(t)dt=f(x)[/tex] exists. As [tex]\frac{1}{h}\int_{x}^{x+h}f(t)dt= F(x+h)-F(x)[/tex] this implies that [tex]\lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h}=F'(x)=f(x)[/tex]
     
    Last edited: Mar 4, 2015
  2. jcsd
  3. Mar 4, 2015 #2
    The same holds if [itex]F\colon\mathbb{R}\to\mathbb{R}[/itex]. The demands apply only to the closed interval a,b such that F is differentiable for every x in the open interval a,b. It implies nothing on how F behaves in its entire domain. If you define F from [a,b] to R, then it somewhat implies that the Fundamental theorem is only applicable to functions that are continuous in their entire domain.
     
    Last edited: Mar 4, 2015
  4. Mar 4, 2015 #3
    You're right, I think I have edited my original post correctly?
    Is the proof (that I gave above) itself ok though?
     
  5. Mar 4, 2015 #4
    You define [itex]F(x) = \int_a^x f(t)dt[/itex] The way I understand it is that F(a) = 0. So does the fundamental theorem only work if (for this case) F(a) = 0? The constant plays no role there once you dive into the limit definition, but it works for every constant : F(a) does not have to be 0 or have I mis-understood that part?

    The limit part is handled nicely and concisely. The only thing I would be curious of would be if [itex]F(a)\neq 0[/itex]
     
  6. Mar 4, 2015 #5
    I'm not sure I understand why [itex] F(a)=0[/itex]? As far as I understand [itex]a[/itex] is taken to be arbitrary, and similarly, there is no reason a priori for [itex] F(a)=0[/itex]. The definition of [itex]F[/itex] is just that, a way of defining a function that we wish to consider in our proof. We assume this, and then show that if this is the case then [itex]F'(x)=f(x)[/itex], however, this does not imply that [itex] F(a)=0[/itex] as any two anti-derivatives that differ by a constant give the same result, so if we had [itex]F(a)\neq 0[/itex], then [itex]\left(F(x)-F(a)\right)'=F'(x)=f(x)[/itex]. The proof does not mean that we can immediately conclude that [tex]F(x)=\int_{a}^{x}F'(t)dt[/tex] however, if I've understood it correctly it does mean that we can conclude that [tex]F(x)=\int_{a}^{x}F'(t)dt +c[/tex] where [itex]c[/itex] is some constant. Using the other part of the FTC (that I didn't state in my OP) we can then conclude that [tex]F(x)=\int_{a}^{x}F'(t)dt +c=F(x)-F(a)+c\;\;\Rightarrow\;\; c=F(a)[/tex] and therefore [tex]F(x)=F(a)+\int_{a}^{x}F'(t)dt[/tex]
     
  7. Mar 4, 2015 #6

    HallsofIvy

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    You said, originally, that you were defining F(x) by [itex]F(x)= \int_a^x F'(t)dt[/itex]. From that it follows immediately that [itex]F(a)= \int_a^a F'(t)dt= 0[/itex] no matter what F' was. That was what nuuskar was complaining about. Now you have changed to [itex]F(x)= F(a)+ \int_a^x F'(t)dt[/itex] which fixes that problem. Of course you would have to have another equation defining F(a) separately.
     
  8. Mar 4, 2015 #7
    Apologies, hadn't meant to cause confusion or come across agressive. It was two separate questions in my first post, the first was that can one imply from [itex]F'(x)=f(x)[/itex] that [itex]F(b)-F(a)=\int_{a}^{b}F'(x)dx[/itex]? And then whether my proof is correct or not?
     
  9. Mar 4, 2015 #8
    That implication is a bit tough for me too. I tend to think that [itex]\int_a^b F'(x)dx = F(b) - F(a)[/itex] implies Riemann integrability on [a,b] specifically which does not come from [itex]F'(x) = f(x)[/itex] , not directly
    With the defined F that I growled at, the argument is conclusive for one specific case and not for every case. You would need to redefine your F such that it accounts for every case possible.
     
    Last edited: Mar 4, 2015
  10. Mar 4, 2015 #9
    Yeah, it has been bothering me, as I would've thought that from [itex]F'(x)=f(x)[/itex] that one can imply that [itex]F'=f\;\;\forall x\in (a,b)[/itex] as [itex]x[/itex] was chosen arbitrarily from the interval [itex](a,b)[/itex]?!
     
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