The second fundamental theorem of calculus states that if $f$ is continuous on $[a,b]$ and $F$ is an antiderivative of $f$ on the same interval, then: $$\int_a^b f(x) dx= F(b)-F(a).$$

The proof of this theorem in both my textbook and Wikipedia is pretty complex and long. It uses the mean value theorem of integration and the limit of an infinite Riemann summation. But I tried coming up with a proof and it was barely two lines. Here it goes:

Since $F$ is an antiderivative of $f$, we have $\frac{dF}{dx} = f(x)$. Multiplying both sides by $dx$, we obtain $dF = f(x)dx$. Now, $dF$ is just the small change in $F$ and $f(x)dx$ represents the infinitesimal area bounded by the curve and the $x$ axis. So integrating both sides, we arrive at the required result.

First, what is wrong with my proof? And if it is so simple, what is so fundamental about it?

Multiplying the equation by $dx$ should be an obvious step to find infinitesimal area, right? Why is the Wikipedia (and textbook) proof so long?

I have also read that the connection between differential and integral calculus is not obvious, making the fundamental theorem a surprising result. But to me, it seems trivial. So, what were the wrong assumptions I made in the proof and what am I taking for granted?

It should be noted that I have already learnt differential and integral calculus and I am being taught the "fundamental theorem" in the end and not as the first link between the two realms of calculus.

In response to the answers below: If expressing infinitesimals on their own is not "rigorous" enough to be used in a proof, then what more sense do they make when written along with an integral sign, or even in the notation for the derivative? The integral is just the continuous sum of infinitesimals, correct? And the derivative is just the quotient of two. How else should these be defined or intuitively explained? It seems to me that one needs to learn an entirely new part of mathematics before diving into differential or integral calculus. Plus we do this sort of thing in physics all the time.

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    $\begingroup$ Your intuition is good, and it really is an intuitively obvious theorem. So obvious. I like to say that the total change $F(b)-F(a)$ is the sum of all the little changes $F'(x) dx$. It does take a bit of work to turn the intuition into a rigorous proof. $\endgroup$
    – littleO
    Commented Oct 30, 2016 at 14:25
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    $\begingroup$ Regarding the "response": In mathematics, the "dx" in calculus books is not generally meant to represent an infinitesimal (apart from a few niche books such as Keisler's text). This is because the real line is usually taken to be Archimedian, which precludes the existence of infinitesimals. In many calculus books the "dx" is just formal notation, and it is not possible to "multiply both sides by dx". Most likely, nothing at all in your calculus book is intended to represent an infinitesimal. The integral is defined as a limit of Riemann sums, for example, not a sum of infinitesimals. $\endgroup$ Commented Oct 30, 2016 at 14:38
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    $\begingroup$ This might be rigorous in non-standard analysis. $\endgroup$ Commented Oct 30, 2016 at 14:59
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    $\begingroup$ In the last paragraph, you insist on infinitesimals. Yes, it is possible to develop rigourous notion of infinitesimal, but this is not how standard calculus works. Derivatives and Riemann integral are defined using limits, not infinitesimals. If you really want to know what $dx$ stands for, you should read about differential forms. $\endgroup$
    – Ennar
    Commented Oct 30, 2016 at 16:46
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    $\begingroup$ "If expressing infinitesimals on their own is not "rigorous" enough to be used in a proof, then what more sense do they make when written along with an integral sign, or even in the notation for the derivative?" It's just notation. We could use any symbols we want to notate it. But we choose $\int_a^b f(x)\,dx$ because of historical practice and because it reminds us of the process that led us to it. (Surely you know $dy/dx$ is not the quotient of two numbers. It's the limit of quotients of numbers. Same thing.) $\endgroup$
    – zhw.
    Commented Oct 30, 2016 at 17:03

10 Answers 10


The problem with your proof is the assertion

Now $dF$ is just the small change in $F$ and $f(x)dx$ represents the infinitesimal area bounded by the curve and the $x$ axis.

That is indeed intuitively clear, and is the essence of the idea behind the fundamental theorem of calculus. It's pretty much what Leibniz said. It may be obvious in retrospect, but it took Leibniz and Newton to realize it (though it was in the mathematical air at the time).

The problem calling that a "proof" is the use of the word "infinitesimal". Just what is an infinitesimal number? Without a formal definition, your proof isn't one.

It took mathematicians several centuries to straighten this out. One way to do that is the long proof with limits of Riemann sums you refer to. Another newer way is to make the idea of an infinitesimal number rigorous enough to justify your argument. That can be done, but it's not easy.

Edit in response to this new part of the question:

Plus we do this sort of thing in physics all the time.

Of course. We do it in mathematics too, because it can be turned into a rigorous argument if necessary. Knowing that, we don't have to write that argument every time, and can rely on our trained intuition. In fact you can safely use that intuition even if you don't personally know or understand how to formalize it.

Variations on your question come up a lot on this site. Here are some related questions and answers.

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    $\begingroup$ Indeed, the amount of effort required to make the infinitesimal notions rigorous (which can be done) is not going to be less than the amount of effort required to present the non-infinitesimal proof that is usually given in textbooks. $\endgroup$ Commented Oct 30, 2016 at 14:25
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    $\begingroup$ @CarlMummert, the point of the infinitesimal approach, as repeatedly argued by Terry Tao and others, is that once the basic framework is in place, you don't need to re-do the foundational material many times over, so if you have (a lot) more than one theorem to prove then you soon gain both in insight and in space. $\endgroup$ Commented Oct 30, 2016 at 14:29
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    $\begingroup$ As others have pointed out, if the proof in the book used infinitesimals, it would need to include a large amount of other material to verify that results using infinitesimals apply to the actual real line, so the overall proof in the book would be no shorter. This phenomenon is called "conservation of difficulty", and is an important thing for students to realize: a seemingly easier proof of a hard result usually incorporates equally or more difficult proofs somewhere else. $\endgroup$ Commented Oct 30, 2016 at 14:33
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    $\begingroup$ @Kalyan Any proof of a theorem about derivatives or integrals must deal somehow with limits, either the classical way (epsilon and delta) or with properly defined infinitesimals. To motivate the first fundamental theorem with infinitesimals, note that when $A$ is the area under $f$ the change $dA$ caused by the change $dx$ is product of the base $dx$ and the height $f(x)$ of the infinitesimal rectangle so the rate of change is $f(x)$. $\endgroup$ Commented Oct 30, 2016 at 16:23
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    $\begingroup$ @A---B: several reasons, I suspect. As I pointed out above, by the time you make it rigorous, the infinitesimal argument is no longer shorter than the standard argument. As several other answers point out, it is not a triviality to make the infinitesimal argument correct. Second, because the usual axiomatizations of the real line are Archimedean, and therefore do not have infinitesimals, only a tiny number of authors want to spend time developing a theory of infinitesimals. Students who learn the "standard" proof in calculus will see the same methods in more advanced analysis courses. $\endgroup$ Commented Oct 31, 2016 at 10:16

Many answers here seem to suggest that what your argument lacks is merely a rigorous theory of infinitesimals.

No. Your argument is simply wrong, regardless of whether there is a clear meaning of infinitesimals. Note that your argument doesn't make use of the condition that $f$ is continuous (hence integrable). However, there are examples of $F$ whose derivatives $f$ are not integrable (see this thread for instance).

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    $\begingroup$ This is a very very good point. I can give just one upvote, I would give more if possible. $\endgroup$ Commented Nov 2, 2016 at 15:00
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    $\begingroup$ @richard1941 I meant what I wrote. The OP tried to prove the second fundamental thm solely based on the assumption that $f=F'$ on $[a,b]$, but this is bound to fail (with or without a rigourous theory of infinitesimals), because the derivative $f$ of a differentiable function $F$ is not always Riemann integrable. In the counterexample where $F$ is Volterra's function, $F$ is differentiable on the real line but $f=F'$ is discontinuous on the Smith-Volterra-Cantor set, which has a positive measure. $\endgroup$
    – user1551
    Commented Nov 4, 2016 at 6:44
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    $\begingroup$ @VivaanDaga No. The problem isn't that the area is not differentiable, but that the area under a curve $y=f(x)$ may not exist even if $f$ has an antiderivative. Riemann integration and antidifferentiation are two different things. The former is a summing process and the latter is the solution of an inverse problem. There isn't any obvious reason why an "antidifferentiable" function must be integrable... $\endgroup$
    – user1551
    Commented Jul 6, 2020 at 18:28
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    $\begingroup$ ... The OP assumes that "$f(x)dx$ represents the infinitesimal area...", but why can we blindly assume that $f$ has a well-defined area around any small/infinitesimal interval? This is something that one needs to justify (under an appropriate setting) in mathematics. $\endgroup$
    – user1551
    Commented Jul 6, 2020 at 18:28
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    $\begingroup$ The problem of an $F$ whose derivative $f$ isn't integrable is due to the global limit taken in the Riemann sums (the limit is taken simultaneously over the whole range of the integral) whereas the limit in differentiation is local (taken at each point). There is the gauge integral which fixes this discrepancy. In other words, the intuition is not wrong per se, it just needs a slightly more flexible integral. $\endgroup$
    – tobi_s
    Commented Jul 21, 2021 at 2:44

Allow me to translate your line "Multiplying both sides by $dx$, we obtain $dF=f(x)dx$." into what, interpreted strictly, you said:

"Pretending that the symbols $\mathrm{d}x$ and $\mathrm{d}F$ have existence outside of the symbol $\frac{\mathrm{d}F}{\mathrm{d}x}$, which is unjustified, we can multiply both sides by $\mathrm{d}x$, obtaining $$ 0 = \mathrm{d}F = f(x) \mathrm{d}x = 0 \text{,} $$ which, while true, has destroyed all information in our equation."

Why is this? Because $\frac{\mathrm{d}F}{\mathrm{d}x}$ is defined to be $$ \lim_{h \rightarrow 0} \frac{F(x+h) - F(x)}{(x+h) - x} \text{.} $$ Assuming this limit exists (which happily you have asserted), we could attempt to apply limit laws to obtain $$ \frac{\lim_{h \rightarrow 0} F(x+h) - F(x)}{\lim_{h \rightarrow 0} (x+h) - x} \text{.} $$ However, this gives a denominator of $0$, so is disallowed by the limit laws. (In fact, it gives $0/0$, suggesting that one should be more careful in explaining how one is sneaking up on this ratio.) Since you ignore this problem, you have multiplied both sides of your equation by $\mathrm{d}x = \lim_{h \rightarrow 0} (x+h) - x = 0$. Fortunately, your remaining left-hand-side is $\lim_{h \rightarrow 0} F(x+h) - F(x) = 0$. So you arrive at the true equation $0=0$, but this is completely uninformative. There are no infinitesimals (whatever those are) remaining.

In response to OP's general response:

An integral is a limit of sums of non-infinitesimal quantities. An integral cannot be the sum of infinitesimals because the sum of any number of zeroes, even infinitely many zeroes, is zero. This is quite easy to see by considering the (ordinal indexed) sequence of partial sums, which are always zero.

The derivative is an indeterminate form of type "$0/0$". The integral is an indeterminate form of the type "$\infty \cdot 0$". As I note above, we must be careful in how we sneak up on such forms to avoid absurdities.

Attempts to use infinitesimals rigorously failed. (From the "Continuity and Infinitesimals" article of the Stanford Encyclopedia of Philosophy)

However useful it may have been in practice, the concept of infinitesimal could scarcely withstand logical scrutiny. Derided by Berkeley in the 18th century as “ghosts of departed quantities”, in the 19th century execrated by Cantor as “cholera-bacilli” infecting mathematics, and in the 20th roundly condemned by Bertrand Russell as “unnecessary, erroneous, and self-contradictory”

You observe that it seems one must learn some other form of mathematics before attempting derivatives and integrals. I agree. To rigorously compute limits of difference quotients (derivatives) and limits of Riemann sums (integrals), one should first learn to find the limits of plain sequences. But there is a bootstrapping problem. As a consequence, in practice, we teach what one might call naive differentiation and integration in Calculus I/II/III and rigorous differentiation and integration in some class with a name like Advanced Calculus. The recipes for differentiating the common basket of functions (polynomials, trig function, exponentials, and logs) are simple enough to teach early on. But there is a full $\epsilon$-$\delta$ treatment of use to those who face functions not in that basket.

In the 20th century, there has been some progress in making infinitesimals rigorous. Useful articles are nonstandard analysis and dual numbers. (Aside: the first words of the nonstandard analysis article are

The history of calculus is fraught with philosophical debates about the meaning and logical validity of fluxions or infinitesimal numbers. The standard way to resolve these debates is to define the operations of calculus using epsilon–delta procedures rather than infinitesimals."

Since one wishes to perform mathematics starting from self-evident truths, one rejects objects with debatable meaning or questionable logical validity.) There are criticisms of nonstandard analysis. While I know that dual numbers can be used for automatic differentiation, I have never seen an attempt to use them as infinitesimals in a theory of integration.


Plus we do this sort of thing in physics all the time.

This is worth a short, slightly philosophical answer of its own.

It's worth understanding how physics and mathematics relate at handwavey boundaries like these. In physics, you do steps like this knowing you might be wrong. Then, you simultaneously search for experiments to back up the calculation, and for mathematical proof. And in the cases where you find experimental justification but lack mathematical proof, mathematical physicists use your experiment as a starting point from which to search for mathematical proof.

The use of the word "infinitesimal" is a unique point where "slop" meets "rigor" and has tremendous history behind it. The very concise history is that while intuition leads to correct results in two line proofs a lot of the time, it leads to overtly or subtly wrong proofs some of the time. Mathematicians circa Leibniz resolved this conflict by going full rigor.

In your case, it's really just the fact that the mathematical theory is well-understood enough that the physicist can be sloppy and retire none the wiser. But physicists also use experiment to justify their findings, and their findings are well-supported by mathematicians interested in rigorous argument. It's best not to be too arrogant about shortcuts like these, when they work because of a combination of experimental evidence, the work of other rigor-driven scientists past and present, and some tolerance for the chance of being wrong.

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    $\begingroup$ And in the cases where you find mathematical proof but lack experimental justification, experimental physicists use your proof as a funding application! $\endgroup$
    – Shai
    Commented Nov 3, 2016 at 17:52

And if it is so simple, what is so fundamental about it?

One reason this theorem can be said to be "fundamental" precisely because it is the basic tool that lets us turn informal arguments such as yours into precisely stated facts.

(incidentally, $\int_a^b f$ and $\int_a^b f(x) \, \mathrm{d}x$ are both reasonable notations, but $\int_a^b f(x)$ is very much not)

Since $F$ is an antiderivative of $f$, we have $\frac{dF(x)}{dx} = f(x)$. Multiplying both sides by $dx$, we obtain $dF(x) = f(x)dx$. Now $dF(x)$ is just the small change in $F(x)$ and $f(x)dx$ represents the infinitesimal area bounded by the curve and the $x$ axis. So integrating both sides, we arrive at the required result.

(note: I've made grammatical corrections to the math in this quote

Sure, but you're begging the question. — you're using the fundamental theorem of calculus to say "integrating $\mathrm{d}F(x)$ over an interval gives you the change in $F(x)$", so it doesn't make for a very good proof of the theorem.

The way a Calc II student would translate this into a rigorous argument would be

  • Substituting $u = F(x)$ gives $\int_{F(a)}^{F(b)} \mathrm{d}u = \int_a^b f(x) \mathrm{d}x $
  • Applying the fundamental theorem of calculus tells us $\int_{F(a)}^{F(b)} \mathrm{d}u = F(b) - F(a)$

Your argument has the further complication of working in terms of differentials — which, while a great thing, at this point in your education you probably don't really know what those are even though you've seen them used enough to be able to mimic the arguments people make with them. The "infinitesimal change in $x$" heuristic is an analogy, and doesn't really hold up when stressed.

Don't get me wrong — in my opinion differentials are great things and more of calculus should be formulated in terms of them.

However, that approach usually generally isn't taught, presumably since it has the added complication of actually having to learn what differentials are, and working with the various derivative and integral rules makes is a perfectly good substitute for most purposes.


If you manage to give a precise mathematical meaning to $\Bbb d x$ and to the multiplication of a function by it, then indeed, your proof is correct. But what meaning do you give to these? In fact, the whole theory necessary to do this takes tens of pages, and since your proof would rely on them, this means that it wouldn't be just a one-liner.

The next part would be to formally justify the assertion that $f(x) \Bbb d x$ is just an "infinitesimal area" (what could this mean?).

  • $\begingroup$ I saw a documentary where the presenter read directly from Leibniz' own notes: $dx$ is the difference between two $x$'s. $\endgroup$
    – user301988
    Commented Dec 9, 2016 at 6:29
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    $\begingroup$ @selfawareuser: If Leibniz (and Newton, for that matter) were ressurected today, he wouldn't recognize his ideas anymore. Don't forget that he died in 1716, which means the prehistory of mathematical analysis. $\endgroup$
    – Alex M.
    Commented Dec 9, 2016 at 12:25

There are several places in your proof where you make approximations.

Multiplying both sides by $dx$, (…)

The rules of multiplications apply to numbers, but $dx$ isn't a number. If $dx$ was a nonzero number then $\dfrac{\cdot}{dx} dx$ would cancel out, but does that work for $dx$?

$f(x)dx$ represents the infinitesimal area bounded by the curve and the $x$ axis.

$f(x) dx$ is the area of a rectangle of height $f(x)$ and width $dx$. The area between the $x$ axis and the curve is not a rectangle (unless $f$ is constant around $x$). Why would summing approximate areas give the right result in the end, rather than an approximation which may or may not be good?

So integrating both sides, (…)

This is an infinite sum. Do the rules of finite sums work for infinite sums?

All of these approximations work out providing that the function is reasonably regular. Well, that's why the theorem has some hypotheses — “$f$ is continuous and is a derivative of $F$” is a sufficient condition for “reasonably regular”.

The classic proof of the theorem which you've read in your book and on Wikipedia and elsewhere follows the same route as yours, but it takes the time to justify all the approximations:

  • Instead of reasoning about “infinitesimal” $dx$, it reasons about real numbers. The intuition of infinitesimals arise from numbers that tend to zero.
  • It verifies that there is a way to equate the area under the curve to the area of a rectangle of width $dx$.
  • It gives a precise notion of summing those very small numbers, and justifies the idea of an infinite sum by showing that it doesn't matter how the interval is broken up into small slices.

(It's also possible to reason formally on infinitesimal, and then less work is needed to make each of those steps work, but more work is needed to prepare the grounds at the beginning.)

Physicists make approximations all the time, but they need to justify these approximations, either by mathematical arguments (“this is the first-order effect so it's valid for small amounts”) or by experimental arguments (do the approximate computation, measure the real thing and check that they agree). In order to get a feeling for when approximations are justified, you need to have some physical intuition about the phenomenon that's modeled by the equations. In particular, physicists know that all functions are infinitely regular — except when they aren't, and that's called a singularity.

Singularities are precisely where the fundamental theorem of calculus breaks down! Intuitively speaking, the regularity hypothesis is “no singularity”. (It's also possible to make it work with singularities, but then $f$ is no longer a function but a distribution.)

For example, consider the Dirac delta function. That's $F(x) = 0$ for $x \lt 0$, $F(x) = 1$ for $x \gt 0$, $f(x) = 0$ for $x \lt 0$ and for $x \gt 0$. It's not clear how to define $F(0)$ and $f(0)$, but it doesn't really matter, since it's just one point, it has zero width… right?

Well then, $f(x) dx = 0$ everywhere since $f(x) = 0$, so if you sum them all up you get $F(x) = 0$ everywhere. Oops, where did we go wrong?

Something has to give. It turns out that it does matter how $F$ and $f$ are defined at $0$. You can say that $F$ has no derivative at $0$ and so the theorem does not apply: in the first step, there's no equation where you can multiply by $dx$. Or you can say that the derivative of $F$ is not a function (there's no function that fits) but some object that sometimes behaves like a function and sometimes not; that's why distributions were invented. Then in the second step $f(x)dx$ is not infinitesimal at $x=0$: that's where all the area is located. Whatever approach you take, there's a singularity at $0$ and the approximations permitted by regularity break down.


Your proof can indeed be made rigorous in Robinson's framework for calculus with infinitesimals; see e.g., Keisler's textbook Elementary Calculus.

Your final comment just added to the question indicates that you are a physicist. If so then you can safely ignore most of the other answers here.

At the Physics SE you may get some answers that address your concerns more directly; see for example this answer.

It should be noted that Keisler's development of the calculus using infinitesimals is fully rigorous. Some technical aspects are addressed in the companion volume Foundations of Infinitesimal Calculus.

That some foundational details should be assumed is natural in a freshman calculus course. For example, the typical calculus course does not construct the real number field, either via Cantor's approach or via Dedekind's approach. This material is appropriately left for a more advanced course.

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    $\begingroup$ It should be noted, however, that Keisler's development of the definite integral is not fully rigorous, because he fails to show that when you approximate the area under a curve by the sum of the areas of infinitesimal rectangles, each introducing a small error, that the error accumulated by summing over an infinite number of rectangles remains negligible. Rigorous nonstandard developments are a bit more complicated. $\endgroup$
    – pash
    Commented Oct 30, 2016 at 22:59
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    $\begingroup$ @pash Do you know a better (and accessible) alternative to Keisler, one in which the problem you point out is considered? $\endgroup$
    – Red Banana
    Commented Oct 31, 2016 at 21:08
  • $\begingroup$ @pash, yes, see the Foundations of Infinitesimal Calculus linked in my answer. $\endgroup$ Commented Nov 1, 2016 at 12:15
  • $\begingroup$ @OppaHilbertStyle, there are some books on the market that attempt to use intuitive infinitesimals without bothering about rigor, but Keisler's book is the best that I know that combines intuition with rigor. We have been using it for several years to teach freshman calculus to hundreds of students with excellent outcome; and yes the students are satisfied also. $\endgroup$ Commented Nov 1, 2016 at 12:49
  • $\begingroup$ @MikhailKatz Thanks. Does the foundations book address the problem shown by pash? I took a quick skimming at the book but I'm not sure if I actually saw it. Also, what do you think about Henle/Kleinenberg's Infinitesimal Calculus? $\endgroup$
    – Red Banana
    Commented Nov 11, 2016 at 11:33

When you see infinitesimals ($dx, dy$) in an expression, it is helpful to think of them as small positive numbers ($\Delta x, \Delta y$), together with the understanding that you are not finished until you take the limit (i.e. where $\Delta x$ goes to zero).

This is basically what we do in calculus proofs--we work with deltas and then take the limit of the resulting expression. Before taking the limit, we are just working with numeric quantities. So, in some cases, there may be common delta factors in the numerator and denominator that are both going to zero at the same rate and can be cancelled out. If you can get the expression reduced to one where setting the delta values to zero will not lead to a singularity or an indeterminate expression, then you can safely replace them with zero to take the limit.


$$\frac{d}{dx}x^2 = \frac{d(x^2)}{dx}$$

$$= \lim_{\Delta x\to 0}\frac{(x+\Delta x)^2 - x^2}{\Delta x}$$

$$= \lim_{\Delta x\to 0}\frac{(x^2 + 2x\Delta x + \Delta x^2) - x^2}{\Delta x}$$

$$= \lim_{\Delta x\to 0}\frac{x^2 + (2x\Delta x + \Delta x^2) - x^2}{\Delta x}$$

$$= \lim_{\Delta x\to 0}\frac{2x\Delta x + \Delta x^2}{\Delta x}$$

$$= \lim_{\Delta x\to 0} ( 2x + \Delta x )$$

$$= 2x$$

As long as $\Delta x$ is not zero, you can divide by $\Delta x$, which allows you to factor the common $\Delta x$ from numerator and denominator.

In the remaining expression, $\Delta x$ is just one term of the sum, and now, if it goes to zero, it can simply be dropped.

This may help explain why "multiplying by $dx$" seems to work, since, before you actually take the limit, it is valid to multiply by $\Delta x$. But at some point, you need to take the limit, and the pivotal question is whether you can do that without having to perform an invalid operation such as dividing by zero.

Note that you can always turn a false equation, such as $3=5$, into a true one by multiplying both sides by zero, but it doesn't prove anything about the original expression to do that. So "multiplying both sides by $dx$" does not necessarily accomplish anything meaningful.


Here is a proof I think you'll like: Take g(x) = $ \int_{a}^{x} f(t) dt $ then by Part 1 of FTC we know that g' = f(X). Now assume F(x) is another antiderivative of f, then we know that

F(x) = g(x) + C

Now notice if we put x = a in the formula for g(x) we get:

g(a) = $ \int_{a}^{a} f(t) dt $ = 0

And finally

F(b)-F(a) = [g(b) + C] - [(a) + C] = g(b) - g(a) = g(b) - 0 = $ \int_{a}^{b} f(t) dt $

I hope this helps

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    $\begingroup$ "Now assume F(x) is another antiderivative of f..." is mean value theorem. $\endgroup$
    – djechlin
    Commented Oct 31, 2016 at 3:12
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    $\begingroup$ @djechlin: Do you mean that the existence of multiple antiderivatives for a given function requires the mean value theorem? How exactly? $\endgroup$
    – Alex M.
    Commented Aug 24, 2017 at 10:07
  • $\begingroup$ @AlexM. the "then we know that F(x) = g(x) + C" part, I meant to quote the whole sentence. $\endgroup$
    – djechlin
    Commented Sep 2, 2017 at 5:11

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