Why is differential calculus often presented before integral calculus?

Question

Note: I'm still learning calculus at the moment.

It seems that many elementary calculus texts describe differential calculus before integral calculus. They start with an informal intuition into the concept of a limit and how to calculate various limits. They then go on to describe the derivative via physical applications and/or the tangent/secant line approach. To me it seems that the integral would be more intuitive to understand first, and then limits and derivatives. To the best of my knowledge, this is also how calculus was historically developed. Is there a particular reason for presenting the derivative before the integral in elementary calculus?

All of the first calculus textbooks I know treat differential calculus before integral (L'Hospital, Bernoulli, Euler). With Leibniz notation it also seems clear why: differentials appear in $\int y dx$. — Michael Bächtold, Jul 06 '21 at 05:57
See also https://math.stackexchange.com/questions/245047/differential-calculus-and-integral-calculus — J W, Jul 06 '21 at 06:37
I'm currently skimming an online text Calculus Made Easy: https://calculusmadeeasy.org/. This text seems to start with Leibniz dy/dx notation to begin with rather than the limit and tangent/secant line approach. It presents differential calculus before integral calculus. It seems to tie these concepts together well. Perhaps Leibniz notation and differentials are indeed why. — , Jul 06 '21 at 07:54
Also, as mentioned in the math SE post linked above, it seems that many authors prefer to present differential calculus first, whereas a few authors prefer to present integral calculus first. Perhaps there are merits to both approaches. Regardless, maybe this question should be marked closed. — , Jul 06 '21 at 08:05
Can you add some justification for the claim that this is also how calculus was historically developed? As far as I know, integration theory was not developed much before differentiation was introduced, but I certainly don't know all the facts, and I'm curious to know. — Michał Miśkiewicz, Jul 06 '21 at 17:09
@MichałMiśkiewicz I recall that Hersh and Davis asserted something like "the integral part of calculus was known to the ancients", in the sense that ancient Greek mathematics used limits of sums to compute areas and volumes, which is pretty close to the way we'd do it using integrals now. Of course the notation and conceptualisation were rather different, but as the results are the same, it seems a not unreasonable claim to make. Of course there was then a millennia-long gap before Newton and Leibniz related the (also ancient) work on derivatives to integrals. — tea-and-cake, Jul 07 '21 at 10:11
@RonJohn That's what I always assumed. Integration requires flexible out of the box thinking quite unlike any math you encounter before university. — DKNguyen, Jul 07 '21 at 17:42
@DKNguyen: Proofs (often taught in highschool geometry) require just as much out-of-the-box thinking if not more. — Ben Voigt, Jul 08 '21 at 15:00
@BenVoigt Proofs for sure, but I never ran into the concept of a proof until university, and I only ran into it because I took Math Honours in first year. — DKNguyen, Jul 08 '21 at 15:01
@Adam But you might equally say that anti-integration (which is what differentiation is) requires integrals. And it clearly doesn't. — Mike Scott, Jul 09 '21 at 11:20
@MikeScott I challenge you to prove the usual set of closed form expressions for antiderivatives without using derivatives. I'm not saying it is impossible. Linear and quadratic functions can be integrated using known results about arithmetic progressions and triangular numbers, respectively. But now try finding $\int \sin(x)dx$ without depending on derivatives at some point. — Adam, Jul 09 '21 at 19:18
@Adam In my (German) analysis textbook it's actually done: They calculate the integral of the cosine before introducing the FTC, using only trig identities. It's very cumbersome, though, which is probably the point they want to make. — Vercassivelaunos, Jul 14 '21 at 00:24
I'm kind of shy about this post. Anyway, perhaps there are merits to both approaches. I'm still learning Calculus at the moment. — , Jul 17 '21 at 11:17
Some might be surprised to learn that many integrals can be evaluated without finding antiderivatives. Consider the center of gravity of the northern hemisphere (assuming the earth is a sphere of uniform density). Archimedes showed that this is 5/8 of the way from the north pole to the center of the earth, without finding any antiderivatives. However, the way people think about calculus today is that the most elementary integrals are found by finding antiderivatives. (But in every high-school statistics course one encounters the "bell-shaped curve," whose equation is$\ldots\qquad$ — Michael Hardy, Jun 24 '22 at 04:43
$\ldots\qquad y = \text{constant}\times e^{-z^2/2}.$ Finding the area under this curve, and thus finding this constant, are not done by finding the antiderivative of that function. $\qquad4 — Michael Hardy, Jun 24 '22 at 04:44

score 37 · Answer 1 · answered Jul 06 '21 at 14:00

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While there are no theoretical difficulties with developing integration first ("from scratch" measure theory books demonstrate this), it presents some pedagogical challenges.

Most problems in existing Calculus courses have "closed form" solutions. It is rare to just set up a problem and then numerically evaluate. The primary tool available for exact calculation of definite integrals is the Fundamental Theorem of Calculus. Finding $$\int_0^1 x^5 \textrm{d}x$$ can be done numerically or exactly using excruciating Riemann sum calculations. We like to know that $\frac{d}{dx} x^6 = 6x^5$ before covering integration so that this integral can be evaluated exactly using the fundamental theorem.

I personally think it would be a better story if we taught integration first, got practice setting up all kinds of integrals to solve practical problems, and used numerical approximation as our only evaluation tool. This would drive home what the definite integral IS. Then do differential calculus in all of its glory. Then the FTOC would actually be surprising and useful (Wow! Now we can calculate these integrals exactly instead of just approximating!). As it is, most students think the FTOC is the definition of the definite integral, which is unfortunate.

answered Jul 06 '21 at 14:00

Steven Gubkin

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Another issue is the ordering of corequisite courses. A student taking physics at the same time might need to understand velocity as the derivative of position before they get to integrals later. So making a change in the "standard curriculum" forces potentially unwelcome changes to courses in other departments. There is just a ton of institutional inertia to doing anything different with a calculus course. – Steven Gubkin Jul 06 '21 at 14:17
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You still have to do limits first though if you want real integral calculus and not just numerical math. Also limits that are much much harder than anything calc students normally cover. – DRF Jul 06 '21 at 18:37
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Also just the definition of a Riemann integral is pretty hard. I realize that calc students don't usually do proper proofs/definitions anyway, but you need a (proper) generalized definition of a limit if you want to define a Riemann integral. My experience that your normal calc student never even understands the basic limit definition. Which I think is another reason the order is what it is. You never actually learn differential calculus you learn a new kind of arithmetic. – DRF Jul 06 '21 at 18:41
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@DRF Regarding "Also just the definition of a Riemann integral is pretty hard", I grew up in Ontario, Canada back when they still had a grade $13$. In that year, we were taught about the basics of limits, derivatives and then integrals, mainly as anti-derivatives. However, at the University of Waterloo, in a first year advanced honors calculus course I took, the instructor started with the historical background of integrals, & then the theoretical aspects of Riemann integrals. Although this class consisted of many of the top math students from Canada, about half dropped out fairly early, ... – John Omielan Jul 06 '21 at 20:38
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@DRF (cont.) going from about $100$ to $50$. Also, the average mark on the first midterm was only about $20$%! The course continued to teach derivatives and then, near the end, showed with the FTC that, for most cases, integrals can, at least in theory, be calculated as anti-derivatives. Interestingly, I guess because they didn't want to bell-curve too much or fail too many students, the final exam had very few questions covering the first part of the course, with most of it being instead about derivatives & calculating standard integrals. – John Omielan Jul 06 '21 at 20:38
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@JohnOmielan Ah, yes. One of those classes. I took one of those once. Apparently I did astonishingly well, but I remember nothing from it and always felt I deserved 20%. – DKNguyen Jul 07 '21 at 15:37
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@user615 I agree this argument is not excruciating at all for us, but I think it would require at least a 50 minute lecture to reach 1/4 of the average freshman calculus population. You may also be interested in this: https://mathoverflow.net/questions/114738/integrating-powers-without-much-calculus – Steven Gubkin Jul 08 '21 at 13:26

score 19 · Answer 2 · edited Aug 26 '23 at 15:07

One issue is that for differentiation, you can find the derivatives of so-called "elementary functions" (e.g. powers, exponentials, logarithms, trig functions) directly from the definition of a limit, and there are then standard ways to find the derivatives of an expression knowing the derivatives of its component parts (e.g. the product and quotient rules, function-of-a-function, etc.)

Therefore it is straightforward to generate a large number of "textbook exercises" of graded difficulty.

On the other hand, most "simple-looking functions" do not have closed-form integrals, and the easiest way to find a set of basic functions that do have closed form integrals is to recognize that "the function you want to integrate looks like a derivative that you already know".

In fact many of the so-called "special functions" in mathematics are defined as the integral of a simple-looking function (e.g. the gamma function, which generalizes the idea of a factorial for non-integer values).

Of course as other answers have said, if you taught integration using only numerical methods as an initial way to evaluate integrals, that problem does not exist, but (at high school or university level) most students will not have much if any understanding of numerical methods, and therefore there are too many possibilities for "garbage-in garbage out" exercises where students have no way to check their work.

Typing formulas into Wolfram's integral evaluator isn't "learning math" - and it isn't even an interesting task, unless you are going to do something interesting with the output.

score 19 · Answer 3 · answered Jul 07 '21 at 14:39

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Because:

https://xkcd.com/2117/

And now I'm padding my answer with 21 characters.

answered Jul 07 '21 at 14:39

George Menoutis

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score 18 · Answer 4 · answered Jul 06 '21 at 14:16

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The text I know of that does integration first is: Apostol, Calculus, Vol. 1: One-Variable Calculus, with an Introduction to Linear Algebra.

An advantage of using this for top-rate incoming university students is that they do not think "I already know this" and tune out [as they may do when the course begins with differentiation].

answered Jul 06 '21 at 14:16

Gerald Edgar

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I just looked at the table of contents, and I find it interesting that continuity is treated after Riemann integrability. – Steven Gubkin Jul 06 '21 at 14:57
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Great point! I think teaching the easy error bound on the Riemann sum (with equal width rectangles) in this case is important. – Steven Gubkin Jul 07 '21 at 10:19
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I learned calculus from Apostol's book and found and still find its presentation of integration to be one of the cleanest and clearest available. – Dan Fox Jun 22 '22 at 16:25

score 8 · Answer 5 · answered Jul 06 '21 at 18:57

I think one of the main reasons for the ordering, limits-> derivatives -> integrals is that it follows in order of difficulty in some sense of the word.

In reality the situation is more complex, but if you start with the minimal (read $\epsilon$-$\delta$) definition of a limit it's fairly straightforward. It's also almost essential for even a semi rigorous treatment of both differential and integral calculus.

Differentiation comes pretty much next in complexity. In reality the basic version (assuming stuff you have is almost everywhere $C^\infty$) is actually quite a bit easier than limits, but since (it looks as if) you're using limits to calculate it, it's more complex. At the same time derivatives are important for a lot of other things you want the students to be learning, mainly basic physics, where you get differential equations at every corner.

Integration is by far the hardest both in terms of students understanding and actual computation. Unlike differentiation where (mostly) anything that can be written neatly has a neat (read made up of elementary functions) derivative, most integrals don't actually have a closed form solution. With "most" here meaning many of the ones students can think of.

All in all, this order is in many ways natural, progressing from easier to harder and building on previously learned material. In reality integration both 2d and 3d is a lot more complex than derivatives. Both in terms of the limits employed (you have to take a limit over partitions, so you need a new notion of limit) and the idea that's it's "finding the area under the curve" is actually surprisingly complex since you don't actually know what "area" is until you come up some version of integration.

score 8 · Answer 6 · answered Jul 06 '21 at 22:48

I'd like to comment on why, indeed, it would be reasonable to present the subject with integrals first... (which is not what is done, nearly universally, I understand).

In particular, I'd argue that the notion of "derivative" is more sophisticated than "area under a curve". This includes the subtlety of "instantaneous rate of change". Yes, people'd been thinking about this a long time (cf. Zeno's paradoxes).

But/and, yes, a fundamental reason for the success of "calculus" as a very basic mathematical gadget is the fundamental theorem of calculus, giving explicit and elementary (indefinite) integrals for many interesting functions. Explicit (elementary) expressions obviously allow many more manipulations and experimentation than subtler "estimates".

For that matter, the first really rigorous results about differential equations used conversion of differential equations to integral equations, and invocation of theorems about compact operators. (See Volterra, Hilbert, Schmidt.)

In my own experience, I did think it was charming that slopes to polynomial curves could be exactly determined, but it was really amazing that areas under such graphs could be precisely (not just numerically) determined, and in forms that were astonishingly simple.

score 7 · Accepted Answer · answered Jul 07 '21 at 21:36

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David Bressoud has a 2019 book that argues for teaching integration before derivatives, and only later teaching limits, partly because that matches the way that the ideas developed historically: "Calculus Reordered: A History of the Big Ideas".

answered Jul 07 '21 at 21:36

Andrew Ross

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I just purchased this book. I think the preface answers my question: "How did we wind up with a sequence that is close to the reverse of the historical order: limits first, then differentiation, integration, and finally series? The answer lies in the needs of the research mathematicians of the nineteenth century who uncovered apparent contradictions within calculus." I can't quote the entire preface here, but it describes this more in detail, and it also makes a good argument for why we might want to teach calculus via integration first, then differentiation, series, and finally limits. – Jul 12 '21 at 05:20
I would like to add that Bressoud's text refers to a "classic" calculus textbook that IS taught "integrals first", namely Tom Apostle's "Calculus Volume 1." As Apostle says, an integration-first approach is "historically correct and pedagogically sound. Moreover, it is the best way to make meaningful the true connection between the integral and the derivative."
I argue Apostle's books is more geared for students as a first view of calculus, while the Broussard book may be better described as a book for the teacher; or at least assumes you are already familiar with calculus.
– Michael G Sep 10 '23 at 20:21

score 4 · Answer 8 · answered Jul 07 '21 at 11:23

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As with a huge number of the "why do we do this order" questions, or the "why don't we teach real analysis before calculus" questions, they seem to tacitly assume that the order of teaching is based on philosophical explanation or even formal mathematical proof. WRONG!

This is NOT the reason why instruction is structured the way it is. It is structured based on imperfect beings that learn harder subjects after easier ones, because that works better pedagogically. Instruction is a maximization problem of time, brains, etc. To get the most sense pounded into skulls within the practical limits that pertain in the practical world. And in the case of integration, it is harder than differentiation (going backwards) and makes more sense afterwards. So. Practical pedagogy uber theoretical explication.

answered Jul 07 '21 at 11:23

guest troll

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I do not see any pedagogical obstacles to an integration first presentation. In some ways (mentioned in other answers) integration is conceptually easier. I think the barriers are mostly institutional inertia rather than based in the relative difficulty of the topics.
Your claim that integration is the inverse of differentiation is the same mistaken belief that many students have: that the definite integral is defined by the Fundamental Theorem rather than by Riemann sums. You do not need the derivative to study integration.
– Steven Gubkin Jul 07 '21 at 15:42
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@StevenGubkin: Explaining the geometric meaning of integration is easier, yes. But doing integration is not easier, conceptually or practically. Integration is not defined as the inverse of differentiation, but performing the inverse of differentiation is a useful method for performing integration. – Ben Voigt Jul 07 '21 at 20:12
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@BenVoigt Numerical integration is just multiplying and adding repeatedly. Almost all integrands do not have "closed form" antiderivatives, so the illusion that the antiderivative is useful for computation is ONLY an illusion. Better to be honest, teach the numerics, and bound the error. – Steven Gubkin Jul 07 '21 at 20:40
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Or even just mention the issue of bounding error, but don't do it. Promise to learn it in a higher course. Trust the numerical integration button on your calculator for now. We do this all the time. We teach students ALL ABOUT the natural logarithm, but the only calculation tool is the ln button on the calculator. We can do the same for the integral. – Steven Gubkin Jul 07 '21 at 20:41
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This situation is even better than the logarithm, because at least we would be teaching how the calculator gets the answer: it just does a Riemann sum with a lot of rectangles. – Steven Gubkin Jul 07 '21 at 20:42
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You seem to be operating under the illusion that there's any kind of consistency between integration buttons on calculators. In fact, the only calculators I'm aware of with any numerical integration features also have computer algebra systems (how else do you input the expression to be integrated?) and those will prefer an algebraic integration and only perform a sum if algebraic methods fail. – Ben Voigt Jul 07 '21 at 20:47
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@BenVoigt Calculus students at my institution are actually required to have a calculator with numerical integration ability, but no symbolic algebra capacity at all. – Steven Gubkin Jul 07 '21 at 22:37
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@StevenGubkin numerical integration is only simple at the conceptual level, with simple integrands. Too often do I come across practical quadrature problems that require a change of variables, Taylor expansion, or some other transformation in order to avoid numerical instability or overflow (e.g. evaluation of the integrand near removable singularities; computing improper integrals). And this already involves derivatives. – Ruslan Jul 08 '21 at 10:46
@Ruslan Sure. Most of the integrands which we deal with in a freshman calculus course are continuous and piecewise monotone. Such integrands have simple error bounds on the error of a Riemann sum, and cannot behave too badly. Of course, improper integrals provide challenges. Numerical instability is a challenge related to the real world implementation of real number arithmetic, rather than a theoretical problem. In other words, 0.0000000...01/0.0000000...02 provides no conceptual challenge to a student, but they might be confused about why a computer might call this 0 instead of 0.5. – Steven Gubkin Jul 08 '21 at 13:24
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@StevenGubkin: So you're teaching one particular model of calculator, not calculus. I fail to see how that's better than teaching derivatives first. And note that even integral-as-sums requires teaching limits first. – Ben Voigt Jul 08 '21 at 14:58
@BenVoigt No more than teaching logarithms without teaching how to calculate them is teaching "one particular calculator". The students understand that they could approximate the integrals by hand (it is just repeated multiplication and addition). We could have them write their own numerical integrator in an excel spreadsheet, but that would detract from the main concepts of the course: it is not a programming class. – Steven Gubkin Jul 08 '21 at 15:01
Consider the integral $\int_2^7 \sqrt{x^5 - x - 1}$. When you ask student A what this means they draw a picture of the integrand, and indicate that it is the area under the curve from $x=2$ to $x=7$. When you ask them how to calculate it, they say "I cannot do it exactly, but I can get close by using many rectangles. I can calculate 5 rectangles by hand to show you what I mean, but the calculator can do thousands of rectangles when I push these buttons". – Steven Gubkin Jul 08 '21 at 15:05
Compare to student B who says "I do the reverse of differentiating and then plug in the bounds". You prod them a little further: is that all there is to it? How do you do the "reverse of differentiating" in this case? They get stuck. So does this integral exist? "Maybe I just can't figure it out, or maybe the integral actually doesn't exist after all". – Steven Gubkin Jul 08 '21 at 15:06
Who has the better understanding of the definite integral at this point? Who is in a better position to learn the fundamental theorem of calculus? – Steven Gubkin Jul 08 '21 at 15:07
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@StevenGubkin: The natural logarithm is of course a function with a single input, and all calculators treat it like other standard functions with a single input, such as the trig functions. Numeric integration features have neither similar simplicity nor consistency. Also, I certainly learned that the logarithm was the anti-function of exponentiation long before I started using the calculator button. – Ben Voigt Jul 08 '21 at 15:08
@StevenGubkin: You are attacking a strawman, the student who doesn't know that the integral represents an area. I refer you to my very first comment. – Ben Voigt Jul 08 '21 at 15:10
Let us continue this discussion in chat. – Steven Gubkin Jul 08 '21 at 15:16

score 3 · Answer 9 · answered Jun 22 '22 at 14:51

I have come to realize another very practical reason for teaching differentiation before integration.

In most applications of integration we are splitting something (area, volume, arclength, work, etc) into lots of tiny pieces, compute an approximation of each piece, and then sum. Taking the limit as the size of the pieces tends to $0$ yields an integral.

What is underappreciated, I think, is that often the "approximate each piece" step will involve differentiation.

Take arc length as an example.

We have a function $f: [a,b] \to \mathbb{R}$ and we want to calculate the arc length of the graph of $f$.

It is natural to first subdivide the interval $[a,b]$ into $N$ equal sized subintervals $[x_k, x_{k+1}]$ with $x_k = a + \frac{b-a}{N}(k-1)$ for $k=0,1,2, 3 \dots, N, N+1$.

Since arclength is additive, the length of the whole curve is the sum of the length over each subinterval.

We then approximate each small arc by the length of the secant line connecting the two endpoints $\sqrt{(f(x_{k+1}) - f(x_k))^2 + (x_{k+1}-x_k)^2}$.

The issue is that

$$\sum_0^{N-1} \sqrt{(f(x_{k+1}) - f(x_k))^2 + (x_{k+1}-x_k)^2}$$

is not in the form of a Riemann sum for any function.

So we need to approximate each of these summands again using the derivative:

$$ \sqrt{(f(x_{k+1}) - f(x_k))^2 + (x_{k+1}-x_k)^2} \approx \sqrt{(f'(x_k)(x_{k+1}-x_k))^2 + (x_{k+1}-x_k)^2} $$

So we obtain

$$\sum_0^{N-1} \sqrt{(f'(x_k))^2 + 1} \Delta x$$

as our approximation of the arc length.

This is an extremely common phenomenon: forming an integral to compute a quantity of interest relies on decomposition into small pieces, but approximating these small pieces requires differentiation. This is a good mathematical reason for studying differentiation first.

You are describing applications of integrals rather than integration per se. The definition of the integral requires no notion of derivative. You are using derivatives to motivate the defining geometric quantities such as arclength and surface via poligonal approximation in a way that requires derivatives. It would be quite reasonable to teach integration, then teach differentiation, and only then teach such applications. — Dan Fox, Jun 22 '22 at 16:29
I actually see this as a justification for transitioning from integration to differentiation. As I see it, the narrative starts with the classical Greek question "How do you compute area?", which leads to the Riemann integral. Then you start trying to use this integral to do other things (e.g. compute arc length), and you start running into problems which are solved by "smoothing things out" via differentiation. — Xander Henderson, Jun 22 '22 at 22:28
@DanFox I agree that there is no formal dependence between the two topics: one can structure a logically consistent path through the material in either order. I would prefer to teach applications hand in hand with the theory, however, and for this it would make sense to teach differentiation first. — Steven Gubkin, Jun 23 '22 at 11:07

Why is differential calculus often presented before integral calculus?

9 Answers9