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I think the answer is in the negative. Here are two of the ways I know. Both of them use the Mean Value Theorem. The first one use in an indirect way, and the second uses it more forthrightly. The first proof goes something like this.

Prove that $F(x) = \int_a^x f(t) dt, a \le x \le b$ is a particular anti derivative and if $G$ is an anti derivative that is $G^\prime =(x) = f(x)$. Then apply the Mean Value Theorem to $F-G$ on any interval $(c,d) \subset (a,b)$ to conclude that $F(c) - G(d) = 0$. Now keep $c$ fixed and move $d$ in the interval $[a, b]$. Use $F(b) - G(b) = F(a)-G(a)$ to conclude $G(b) - G(a) = F(b) - F(a) = F(b) = \int_a^b f(x)dx$.

The second proof, which I prefer, goes something like this:

$G(b) - G(a) = \int dG = (G(b) - G(x_{n-1})) + (G(x_{n-1}-G(x_{n-2}) )+ \ldots (G(x_1) - G(a))$ for any partition $\{a, x_1, x_2,\ldots, x_{n-1}, b \}$ of $[a, b] $. Now you use the Mean Value Theorem to write replace each term $G(x_k) - G(x_{k-1})$ by $G^\prime(c_k)(x_k - x_{k-1})$ to get a Riemann sum which converges to $\int_a^b f(x)dx $.

I think one has to use some sort of theorem like mvt which gives you global information $f(b) - f(a)$ of $f$ using the derivative which can only provide local information on $f$.

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up vote 7 down vote accepted

As others have said, in the version of FTC in which we agree to integrate / antidifferentiate only continuous functions, the only place where a consequence of MVT occurs is when we want to assert the uniqueness of the antiderivative up to a constant, which -- as N.S. points out -- is equivalent to the Zero Velocity Theorem (ZVT):

Let $f: [a,b] \rightarrow \mathbb{R}$ be differentiable with $f'$ identically equal to zero. Then $f(x) = f(a)$ for all $x \in [a,b]$.

It is traditional to prove ZVT by noting that its contrapositive form is that a nonconstant differentiable function has a point at which the derivative is not equal to zero, and that this follows immediately from MVT. This is the way I proved ZVT in my recently completed Spivak Calculus class. Just for fun though, let me give an alternate proof using Real Induction (see e.g. page 14 of these course notes).

Note that it is no loss of generality to assume that $f(a) = 0$ and show that $f$ is identically zero on $[a,b]$. For $\epsilon > 0$, let $S(\epsilon)$ be the set of $x$ in $[a,b]$ such that for all $y \in [a,x]$, $|f(y)| \leq (x-a)\epsilon$. If we can show that $b \in S(\epsilon)$ for all $\epsilon > 0$, then we have that $|f(y)| \leq (b-a)\epsilon$ for all $y \in [a,b]$ for all $\epsilon > 0$, so $f \equiv 0$. Now:
(RI1) Since $f(a) = 0$, $a \in S(\epsilon)$.
(RI2) Suppose that for some $a \leq x < b$, $x \in S(\epsilon)$. Since $f'(x) = 0$, there exists $\delta > 0$ such that $|y-x| \leq \delta$ implies $|f(y)-f(x)| \leq \epsilon |y-x| \leq \epsilon \delta$. Thus for all $0 < \delta' \leq \delta$ and $y \in [x,x+\delta']$,

$|f(y)| \leq |f(x)| + |f(y) - f(x)| \leq |f(x)| + \epsilon \delta' \leq (x-a)\epsilon + \delta' \epsilon = (x+\delta'-a)\epsilon$,

so $[x,x+\delta] \subset S(\epsilon)$.

(RI3): Similarly, suppose $a < x \leq b$ and $[a,x) \in S(\epsilon)$. Choose $\delta > 0$ such that $|y-x| \leq \delta$ implies $|f(y) - f(x)| \leq \epsilon |y-x| \leq \ \delta \epsilon$. For $y \in [x-\delta,x]$, we have

$|f(y)| \leq |f(y) - f(x-\delta)| + |f(x-\delta)| \leq \delta \epsilon + (x-\delta-a)\epsilon = (x-a) \epsilon$.

I can't think of a situation in which I would rather give this proof than prove the Extreme Value Theorem and use that to to prove the Mean Value Theorem, but I find arguments by real induction to have a certain charming directness to them...

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This is super-nice! You've used completeness of the reals in a different way. – Daniel Moskovich Dec 15 '11 at 16:36
@Daniel: Thanks. I have been thinking a lot about Real Induction recently. As you can see, I introduced it in my Spivak Calculus class. It turns out that you can prove almost all of the named theorems in elementary real analysis this way...but some of the proofs are nicer than others. Especially, I made a big(ger than Spivak does) deal about the role of uniform continuity, but the proof of the uniform continuity theorem using real induction was technical enough that I actually presented it incorrectly in class!... – Pete L. Clark Dec 15 '11 at 16:42
I was able to fix it, and the version I have in my notes amuses me in that it first proves a lemma about functions on two overlapping intervals which seems to be the simplest nontrivial case of the statement that every open cover of a compact space admits a Lebesgue number. But anyway I thought more and realized that I wanted UC only to prove the integrability of continuous functions, so why not try to prove that using Real Induction? That came out nicely -- the argument is actually quite similar to the one given here -- and I ended the entire course with it. – Pete L. Clark Dec 15 '11 at 16:45
I have one expository writeup about Real Induction, but now that I have thought more about it and, especially, actually used it in the classroom setting, I think it is out of date. I am contemplating writing an expository article called The Instructor's Guide to Real Induction. I'll let you know if/when this comes to fruition. – Pete L. Clark Dec 15 '11 at 16:47
____Yes please! – Daniel Moskovich Dec 15 '11 at 16:59

For an alternative approach to the FTC without the MVT, see this paper, which also contains other references:

Tucker, Thomas W.; Rethinking rigor in calculus: the role of the mean value theorem. Amer. Math. Monthly 104 (1997), no. 3, 231–240.; also available at .

This article has brought a reaction:

Swann, Howard; Commentary on rethinking rigor in calculus: the role of the mean value theorem. Amer. Math. Monthly 104 (1997), no. 3, 241–245.; also available at the author's site:

and also this

Scott E. Brodie; On "Rethinking Rigor in Calculus...," or Why We Don't Do Calculus on the Rational Numbers. The College Mathematics Journal , Vol. 30, No. 2 (Mar., 1999), pp. 135-138.

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i looked at the tucker paper you cite. he uses an MVT equivalent if $f^\prime >\ge 0$ then $f$ is not decreasing. other references i have not checked them out. – abel Dec 15 '11 at 17:41
@abel: I would call that a "variant" of MVT rather than an equivalent: it is not so easy to get from the increasing function theorem (IFT) to MVT (I believe some of the articles discuss this point). Moreover, I believe that the inductive argument I gave carries over easily to give a real induction proof of IFT, but I do not (yet, at least) see how to prove MVT using real induction (I mean directly: of course you can prove EVT using real induction and then apply that to prove MVT in the usual way). – Pete L. Clark Dec 15 '11 at 20:26

I think of the FTC as a sharpening of the MVT, so it would seem artificial to me to prove the FTC without the MVT.

Consider: The MVT states that, when $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in (a,b)$ such that the slope of the tangent line to the graph of $f(x)$ at $c$ is precisely equal to the slope of the line passing through $(a,f(a))$ and $(b,f(b))$. We don't know anything about the point c.

Wait a second! We do know something about c! We know that $f^{\prime}(c)$ is the average of the derivative between $a$ and $b$! How do we know this? Well, if the function were linear, the average would just be the slope itself; and now that it isn't linear, well, the average is still the slope of the aforementioned line. Think about it- the derivative is velocity. Sometimes the car goes faster and sometimes it goes slower, but its average velocity will be the velocity $v$ such that, if it were traveling at constant velocity $v$ all the time, it would get from $f(a)$ to $f(b)$ in time $b-a$.

Now we want a way to write down that $f^{\prime}(c)$ is this average. How do we do it? Well, the average is the total divided by the number of parts, each weighted by its length. Wait a second- the numerator of that is the Riemann integral! Taking limits, the average of $f^{\prime}$ on $(a,b)$ is $\frac{\int_{a}^b f^{\prime}(t)dt}{b-a}$. So we've rewritten the "average version" MVT as $\frac{f(b)-f(a)}{b-a}= \frac{\int_{a}^b f^{\prime}(t)dt}{b-a}$. Multiply both sizes by $b-a$ and you're finished.

To turn this into honest mathematics you use some kind of formal argument like you mentioned- but morally, there's nothing more to the FTC than the interpretation of $f^{\prime}(c)$ as not just $f^{\prime}$ evaluated at some strange point, but rather as an average.

Added: Because other answers are saying yes, I should stress that I'm saying no. There's an irreducible difficulty to the FTC, which is to translate between an epsilon-delta concept $\int_a^b f^{\prime}(t) dt$ and a number $f(b)-f(a)$. You have to cross that bridge somehow. And the only ways across are the IVT and the EVT, which in this context come in the guise of the MVT. I wish I had a logician to comment- but I think that, logically, I'm going to go out on a line here and assert that there can be no way to the FTC which doesn't pass through some form of the MVT.

Added 2: There are a number of different ways to formulate completeness of the real numbers. As other Pete L. Clark's and N.S.'s answers have shown, these, and techniques such as real induction, can be leveraged to yield genuinely different proofs of the FTC. So the answer turned out to be yes after all!

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$@$Daniel: I don't understand what "there can be no way to the FTC which doesn't pass through some form of the MVT" means as a formal mathematical statement: can you clarify? – Pete L. Clark Dec 14 '11 at 22:29
Two comments:1) The fact that one result is a stronger version of another, doesn't mean that one cannot be proven without using somehow the other. Riemann Hypothesis is stronger than the PNT, yet the PNT was proven without using the RH... And if RH will be proven, it is very improbable that the proof will rely on the PNT. 2)You can find a different bridge between the epsilon-delta concept $\int_a^b f′(t)dt$ and the number $f(b)-f(a)$ the following way: look to them not as numbers, but as functions in $b$. Sometimes it is much easier to prove that two functions are equal everywhere, than at a – N. S. Dec 15 '11 at 1:03
...cont: point. If you look over my comment to my answer, you'll see that the proof is based on the idea of prooving that $f(x)-f(a)=\int_a^x f'(t) dt$ as functions in $x$, not as numbers. This way I can use another bridge, namely differentiation.... – N. S. Dec 15 '11 at 1:08
@PeteL.Clark : I meant something between "FTC uses completeness of the real numbers in an essential way" and "FTC requires a theorem which relates the derivative of a function to that function, and I would consider any theorem which does this (which is not well-defined, I admit) to be some form of MVT in that it would use the same EVT-style argument to prove". – Daniel Moskovich Dec 15 '11 at 16:31
@N.S.: I agree! But isn't $f(x)-f(a)=\int_a^x f^{\prime}(t)dt$ just another form of the FTC? – Daniel Moskovich Dec 15 '11 at 16:32

If $f$ is defined on $[a,b)$ such that $F(x)=\int_a^x f(t)dt$ exists for all $x\in(a,b)$, and $f$ is continuous at $x_0\in (a,b)$, then, given $\epsilon>0$, find $\delta>0$ such that $I=(x_0-\delta,x_0+\delta)\subset (a,b)$ and $|f(x)-f(x_0)|<\epsilon$ when $x\in I$.

Then we can see that, for $0<h<\frac{\delta}2$, that $F(x_0+h)-F(x_0) = \int_{x_0}^{x_0+h} f(x)dx$ and

$$f(x_0)-\epsilon < f(x) < f(x_0)+\epsilon$$ on the interval $[x_0,x_0+h]$. So

$$hf(x_0) - h\epsilon \leq F(x_0+x) - F(x_0) = \int_{x_0}^{x_0+h} f(x) \leq hf(x_0)+h\epsilon$$.

Divide by $h$ and you get:

$$f(x_0)-\epsilon < \frac{F(x_0+h)-F(x_0)}h < f(x_0)+\epsilon$$

So $\frac{F(x_0+h)-F(x_0)}h$ has the limit of $f(x_0)$ as $h\rightarrow 0+$.

You can show the same for $h\rightarrow 0-$, so you are done.

Note, you need continuity at $x_0$ for this to work, since, if you define $f(x)=-1$ for $x<0$ and $f(x)=1$ for $x\geq 0$, then the integral of $f$ is not differentiable at $x=0$.

So, all you really need is that the following theorems:

If $u<f(x)<v$ for all $x\in[x_1,x_2]$ then $(x_2-x_1)u\leq \int_{x_1}^{x_2}f(x) dx \leq (x_2-x_1)v$.

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Are you sure that the proof of that theorem doesn't factor through the MVT? – Daniel Moskovich Dec 14 '11 at 21:29
Nope, it just uses the fact that the Riemann sum is limited to that range because the height of each rectangle is in that range and the sum of the widths of the rectangles is $x_2-x_1$. So any estimate of the integral with a Reimann sum is necessarily in this range, and so the limit has to be in this range. – Thomas Andrews Dec 14 '11 at 21:37
This is definitely correct. Note though that there is another half to the FTC: the one which says that we can evaluate $\int_a^b f$ using any antiderivative of $f$. This one does traditionally depend on MVT, but see N.S.'s answer and mine for alternate routes. – Pete L. Clark Dec 14 '11 at 22:25

You can prove that $F(x) =\int_a^x f(t) dt $ is an antiderivative of $f$ without Rolle or MVT.

Then, you need MVT to prove the following result: if $H'=0$ on an interval, then $H$ is constant.

I think you can prove this result without MVT. I will prove it for any close interval $I$, if your interval is open, you can get it by looking to any closed subinterval.

Fix $\epsilon >0$. Then for each $x \in I$, there exists some $\delta_x$ so that

$$|x-y|< \delta_x $$ implies $|H(x)-H(y)| < \epsilon |x-y| $

Now $\cup (x-\delta_x, x+\delta_x)$ is an open cover for $I$, thus you can find a finite subcover.

Let $a < b$ be in $I$. Pick $x_0=a, x_1,..,x_n=b$ some increasing sequence so that $x_i, x_{i+1}$ are in the same intervals. This can be done since the subcover is finite... (See the details at the end of the proof)


$$|H(a)-H(b)| = |\sum H(x_{i+1})-H(x_i)| < \epsilon(b-a) \,.$$

The rest is simple.

Edited Here is how you can prove the existence of $x_n$.

Let $I_1,.., I_k$ be the finite open cover, with $I_k= (c_k,d_k )$.

$x_0=a$. This is in some interval $I_{k_0}=(c_{k_0}, d_{k_0})$.

If $b \in I_{k_0}$ then we are done. Otherwise, $d_{k_0} \leq b$.

$d_{k_0} \in I$, means $d_{k_0} \in some I_{k_1}$. Since $I_{k_1}$ is open and contains the end point of $I_{k_0}$, we have $I_{k_0} \cap I_{k_1} \neq \emptyset$.

Pick some $x_1 \in I_{k_0} \cap I_{k_1}$.

Repeat the process, picking at each step some $x_i \in I_{k_i} \cap I_{k_{i-1}}$ so that $x_i$ is at the right at $I_{k_{i-2}}$.

This can be done since $d_{k_{i-2}} < d_{k_{i-1}} $ and $ d_{k_{i-1}}$ is an interiour point in $I_{k_i}$. Thus, $\max( d_{k_{i-2}}, c_{k_i}, c_{k_{i-1}} ) < d_{k_{i-1}}$ and all you need is to pick some $x_i$ in between.

Once you pick a point in an interval, tyou'll leave that interval in at most two steps and never get back... And the process can only end when $b$ is in some $I_{k_n}$, but needs to end in finitely many steps....

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I'm guessing that no you can't. To relate properties of the derivative to properties of the function, you're going to need the MVT every time, because what other bridge between the two have you got? Epsilon-delta won't help unless you leverage the EVT to shift the discussion to points (rather than just neighbourhoods), and basically you'll be reproving some version of Rolle's Theorem. – Daniel Moskovich Dec 14 '11 at 21:41
@DanielMoskovich I will add the details. Basically the proof is based on the following idea: If $H'=0$ on an interval, the local rate of change is smaller than any $\epsilon$. Thus the global change cannot be possitive... – N. S. Dec 14 '11 at 22:21
If you means that $F$ is an antiderivative of $f$ that's easy: $F(x)-F(x_0)=$ area below the graph of $f$ between $x-x_0$. So if $m$ is the min of $f$ on $[x_0,x]$ or $[x,x_0]$ and $M$ is the max, $m \leq \frac{F(x)-F(x_0)}{x-x_0} \leq M$. Use now the fact that $f$ is continuous at $x_0$ to get some interval around $x_0$ on which $f(x_0)-\epsilon < m < M <f(x_0)+\epsilon$ and done. – N. S. Dec 14 '11 at 22:41
I see how this works... you're using the nested interval property instead of the least upper bound property, so completeness of the reals is being used in a different way. Nice! – Daniel Moskovich Dec 15 '11 at 16:51

"Can it be proved..."

Yes, strictly speaking. One could go to the extremes - one can prove the Lebesgue Differentiation Theorem, which I would not at all call an immediate corollary of the MVT. Nor does one need the MVT to prove it at all. For example, see here.

Really though, all we need is a bound. We use the MVT because it's pretty good, but just knowing that the value of a function is bounded above and below on an interval, then it's integral is bounded above and below by the max and min times the length of the integral.

Any sort of average will work, really. But these are all more or less equivalent, so it's not so exciting.

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To get from the Lebesgue FTC to the Riemann FTC we need to know that every $C^1$ function is absolutely continuous. Isn't this (traditionally, at least) shown using MVT? – Pete L. Clark Dec 15 '11 at 0:19

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