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This is my first post so bear with me, but something I've been thinking about lately is: Why didn't I ever question the relationship between the derivative and the integral when I was taking calculus?

Let me explain what I mean: In most courses, the derivative is introduced as the slope of a curve at a point, or the "instantaneous rate of change". Then the integral is introduced as the area under a curve. Then students are told that these two things "undo" each other (the fundamental theorem of calculus). So now I'm wondering, why didn't I ever question why these "undo" each other? It's not intuitive at all. For example, addition and subtraction, multiplication and division, logarithms and exponentials; all of these things I can intuitively understand why they "undo" each other. But how does something that represents the area under a curve "undo" something that represents an instantaneous rate of change? What is the connection between those two concepts? Is there a better way to explain these concepts that makes the fundamental theorem of calculus more intuitive?

Looking forward to hearing some of y'alls responses. Thanks!

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    $\begingroup$ $[\cdots]$ why didn't I ever question why these "undo" each other? --- Your calculus experience is probably in the minority, because it is fairly standard (at least in U.S. introductory calculus courses, both high school and college) to make a big deal about the connection, usually at the end of the first semester, this being especially the case since the rise of "reform calculus" from the late 1980s on. That said, see right brain explanation and left brain explanation. $\endgroup$ – Dave L Renfro Apr 2 at 5:26
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    $\begingroup$ Which resources have you perused before registering and asking this question? Voting to close. $\endgroup$ – Rusty Core Apr 2 at 21:01
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    $\begingroup$ In countries, where physics is a mandatory course in grade school, the traditional layman explanation is through kinematics. Average speed v, by definition is displacement s over time t. Start with the simplest case of uniform motion, draw a graph in (t, s), it would be a line climbing up. Tangent at any point is at the same angle (similarity of right triangles), this is the derivative by definition, which is speed. Draw speed in (t, v). Area below it is displacement. Continue with uniform acceleration and then with no-uniform motion, slicing up the area under speed into thin vertical bands. $\endgroup$ – Rusty Core Apr 2 at 21:14
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    $\begingroup$ @BenCrowell I think what makes this a math ed question is wanting better intuition to explain the relationship. Also, the fact that many of us didn't really get it until we were teaching speaks volumes to how much this is an education issue. $\endgroup$ – Sue VanHattum Apr 4 at 20:52
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    $\begingroup$ @SueVanHattum Exactly - that's why I posted it here. The funny thing is that when I learned how to prove the FTC in real analysis, I didn't find either of the proofs too difficult. But the intuition behind the overall concept still wasn't there. Then recently I realized that if a student came to me and asked "Why do these two concepts undo each other?", my best response would be to show them the proof. That's why I figured this was a math ed question, because I'm specifically asking what the best way is to explain this to a student and make it more intuitive. $\endgroup$ – Brain Gainz Apr 5 at 18:43
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You start by noticing that the Riemann sums (multiplication followed by addition) and the difference quotients (subtraction followed by division) undo each other. Their limits -- the integral and the derivative -- still undo each other.

Added: The first sentence is a part of what is called “Discrete Calculus” https://en.wikipedia.org/wiki/Discrete_calculus

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    $\begingroup$ Wow. I'm unsure whether this intuition can be made to correspond to some rigorous proof, but it's just solid enough to make me feel it gave me some insight into the mechanism of the theorem's truth and not mere word-soup hand-waving. And I think the maximal spacing in a partition (that goes to zero for the Riemann integral) even corresponds to the step size in the difference quotient in a suitable sense. $\endgroup$ – Vandermonde Apr 3 at 1:23
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    $\begingroup$ @Vandermonde You are correct. By the way, this will be in the third volume of my book Calculus Illustrated: amazon.com/dp/B082WKCYHY $\endgroup$ – Peter Saveliev Apr 3 at 1:44
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    $\begingroup$ Surprising how entirely abstract and formal mathematical thinking can be intuitive :-). Perhaps not for everybody, but even this engineer sees symmetries. $\endgroup$ – Peter - Reinstate Monica Apr 3 at 10:21
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    $\begingroup$ 1. Can you please exhibit the formula that you're hinting to? 2. Can you please clarify 1. which "Riemann sums" you mean? 3. Which "difference quotients"? 4. Which "limits"? $\endgroup$ – Accounting Apr 4 at 5:27
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    $\begingroup$ If one considers the differential $\text{d}$ and the integral $\int$ as the fundamental concepts of calculus (as opposed to $\frac{\text{d}(-)}{\text{d}x}$ and $\int(-)\text{d} x$), then the wording is even more simple: difference and sum undo each other. (Actually Leibniz called $\int$ the sum, before Bernoulli suggested integral) $\endgroup$ – Michael Bächtold Apr 6 at 15:58
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enter image description here

(this is from my calculus notes, see page 233 of: http://www.supermath.info/OldschoolCalculusII.pdf)

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    $\begingroup$ So in words: the change in area is equal to the height of the function times the change in x? That actually makes sense. $\endgroup$ – Brain Gainz Apr 3 at 0:31
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    $\begingroup$ @BrainGainz thanks! Incidentally, I had exactly your feeling after taking years of calculus and an Advanced Calculus course where I suspect we proved the FTC. Your feeling is not unusual in my estimation. $\endgroup$ – James S. Cook Apr 3 at 2:52
  • $\begingroup$ I understand the pedagogical need, but I cringe when I read (or say myself when teaching calculus) "arbitrary function" and the picture is that of a differentiable function. $\endgroup$ – Martin Argerami Apr 3 at 10:36
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    $\begingroup$ @BrainGainz yep, I have a few Topology lectures posted. I am that James Cook. $\endgroup$ – James S. Cook Apr 4 at 21:57
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    $\begingroup$ @StevenGubkin cool, if I ever get to teach Calculus I again, I should use that. So many options to play with... $\endgroup$ – James S. Cook Apr 6 at 3:49
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Derivative of integral is the original function: Let $F$ be the integral function of $f$, so that $F(x)$ is the area under the graph from zero to $x$. For small $h>0$ the difference $F(x+h)-F(x)$ is the area of a narrow vertical strip. The width is $h$ and height approximately $f(x)$. As $h\to0$, this means $F'(x)=f(x)$.

If you like thinking in terms of infinitesimals, write $dF=F(x+dx)-F(x)=f(x)dx$ and divide by the differential. This can be a cleaner way to think as the limit process is left implicit in a way.

Integral of derivative is the original function: The integral is roughly $$ \int_0^x f'(x)dx = \sum_{k=1}^n f'(x_k)(x_k-x_{k-1}) $$ and the derivative is roughly $$ f'(x_k) = \frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}. $$ Combine these and you get a telescoping sum and the desired result.

This is the way it makes intuitive and graphical sense to me. The link is not immediate; that's why it's a great theorem.

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My answer also comes from physics.

Say p(t) is the position of an object in the time t. For concreteness, suppose you are traking a truck and the truck is going forward on a road from A to B. (i.e., all derivatives are positive)

It is very natural to graph p(t), and derive it, arriving at v(t), the speed of the truck for each time. After all, the derivative is just taking a small time increment $dt$, calculating the corresponding $ds$ (a space increment) and dividing

However, when I look at the graph of v(t), I can also obtain (almost everything) from p(t). Say I am at the position 100 km in the time t=0. After a $dt$, I shall be at $p(t+dt) = 100+dt\cdot v(t)$. And I can do it again and again $p(t+dt+dt+dt) = 100+dt\cdot v(t)+dt\cdot v(t+dt)+dt\cdot v(t+2\cdot dt)+dt\cdot v(t+3\cdot dt)$ (notice that after some $dt$s, I use the updated speed)

But that (taking a limit) is an integral!

The only weird thing is my assuption that I am at position 100km at time t=0. The graph of v(t) cannot tell me that. Intuitively, it does tell me how fast I am going, and therefore can tell me how much how much I walked, but it cannot tell me at which specific part of the road I started. That is that weird constant +c that appears on the indefinite integral.

But then again, we dont use the indefinite integral much, do we? We are more interested in summing $dt$*v(t) for ranges. Say, from t=3 to t=10. That gives me the amount I walked between those two times. So, mercifully, the +c disappears in most use cases: to know how much I walked, there is no need to know where I started.

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    $\begingroup$ I love how velocity and position helps to cement the relationship. In one class, my students asked why area below the axis should count as negative. We agreed that negative velocity meant going backwards, and therefore decreased your distance from start. (Distance being measured by area under velocity curve, a negative velocity curve has to have "negative area" to decrease the total distance.) $\endgroup$ – Sue VanHattum Apr 4 at 20:47
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When I first began teaching Calculus, I realized that I really didn't understand the Fundamental Theorem. So I looked for something that would help me have a deep understanding, that would also help me help students to see it.

I found a lovely project, which I have modified over the years. Here are links to the pdf and to a .doc version (in which the formulas are messed up, because google docs couldn't read them). If you use this, you'll want to change the part near the end where I reference the textbook.

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    $\begingroup$ Wow, this is really great. I'll definitely save this. Thanks! $\endgroup$ – Brain Gainz Apr 3 at 0:29
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I gave a similar post to this one touching on this on Math.StackExchange.

Basically, the way I would go about it is to say that there is a very easy way by which one can think of at least Riemann integration (the usual definition given in a "most courses" calculus course) as an inverse of differentiation by construction: that is, the relationship between the two is not an "accident", but design, so that given one, you could fairly easily be led to the other.

And to do that, I'd first suggest getting rid of the whole "slope" vs "area" business altogether - if anything, those are best given as theorems to be proved, after you have other, free-standing definitions of a tangent line and an area which, by the way, can be done, but just too-often aren't, in favor of various hand-waving arguments.

Instead, the relevant idea is change: the derivative of a function $f$, i.e. $f'$, stands for a kind of "sensitivity" measure. Suppose that $f$ were like a kind of meter or instrument, with a knob attached to it, and a readout of some kind. The knob attached is the function's input argument, typically denoted $x$. The indicator on the readout is the return value, $f(x)$. If $x$ is set at some value of interest $x_0$, then likewise the readout will be at $f(x_0)$. Now suppose you "wiggle" the knob $x$ back and forth a little bit, and you see how the needle $f(x)$ responds to that small impulse. For a continuous function, the size of the output's wiggle will be smaller in absolute terms the smaller you make the input, but the proportionate size, i.e. how much it wiggles relative to how much you wiggle the input value, may not be. When we say $f$ is differentiable, what that means is that there exists at each point $x_0$ a proportionality factor $f'(x_0)$ such that

$$f(x \pm \underbrace{dx}_\mbox{"wiggle" in $x$}) \approx f(x_0) \pm \underbrace{[f'(x_0)\ dx]}_\mbox{"wiggle" in $f(x)$}$$

so that

$$\mbox{proportional "wiggle"} = \frac{\mbox{"wiggle" in $f(x)$}}{\mbox{"wiggle" in $x$}}$$

so long as the change $dx$ is suitably small

Integration, then, goes the other way. Suppose that I am given now, not the function $f$, but only its derivative, $f'$, and want to find $f$. First off, one should observe that since $f'$ only deals with changes in the input, to start, we actually need one more piece of information, and that is some sort of initial value, i.e. $f(0)$. Suppose this to be given as well.

Starting at $f(0)$, suppose we apply a small, but nonzero, change $\Delta x$ to the input, i.e. we ask, "given $f(0)$, what is $f(0 + \Delta x)$, to the best we can do?" Well, since we know $f(0)$ and $f'$, then since $\Delta x$ is a small change or "wiggle", we can say that approximately,

$$f(0 + \Delta x) \approx f(0) + [f'(0)\ \Delta x]$$

which is just what you see above.

Now, suppose we take another step of $\Delta x$. We're now going from $x = 0 + \Delta x$ to $x = (0 + \Delta x) + \Delta x$ (or $2\ \Delta x$, but I find writing it this way makes it clearer what is going on - "simpler" isn't necessarily "better"). At this second step, likewise, treating $0 + \Delta x$ as a prior input in and of itself, we have

$$f([0 + \Delta x] + \Delta x) \approx f([0 + \Delta x]) + [f'([0 + \Delta x])\ \Delta x]$$

which, combining with the previous expression, becomes

$$f([0 + \Delta x] + \Delta x) \approx f(0) + [f'(0)\ \Delta x] + [f'([0 + \Delta x])\ \Delta x]$$

and it is not hard to then continue this process so that we see for $N$ steps,

$$f(0 + N[\Delta x]) \approx f(0) + \sum_{i=0}^{N-1} f'(0 + i[\Delta x])\ \Delta x$$

or, if we set $x_i := 0 + i[\Delta x]$, we can say more neatly as

$$f(0 + N[\Delta x]) \approx f(0) + \sum_{i=0}^{N-1} f'(x_i)\ \Delta x$$

and to be more general, if we take $N$ steps of suitable, perhaps different, sizes $\Delta x_i$ to reach from $0$ some fixed point $x_0$,

$$f(x_0) \approx f(0) + \sum_{i=0}^{N-1} f'(x_i)\ \Delta x_i$$

and then we consider what happens as the steps become arbitrarily fine, at which point we hope - and need to prove - that

$$f(x_0) = f(0) + \left[\lim_{||\Delta|| \rightarrow 0}\ \sum_{i=0}^{N-1} f'(x_i)\ \Delta x_i\right]$$

which leads us to define this new operation, given by the limit on the right...

$$\int_{0}^{x_0} f'(x)\ dx := \lim_{||\Delta|| \rightarrow 0}\ \sum_{i=0}^{N-1} f'(x_i)\ \Delta x_i$$

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  • $\begingroup$ This does a great job of explaining how to think of $\int_a^b f'(x)\mathrm dx = f(b) - f(a)$, but, in suggesting that we think of it as a definition, I'm left to wonder how to define $\int_a^b F(x)\mathrm dx$ without proving some separate theorems about existence of anti-derivatives. $\endgroup$ – LSpice Apr 6 at 0:04
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Intuitively, the fundamental theorem of calculus states that "the total change is the sum of all the little changes". $f'(x) dx$ is a tiny change in the value of $f$. We sum up all these little changes to get the total change $f(b) - f(a)$.

I elaborated on this explanation here: https://math.stackexchange.com/a/1537836/40119.

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    $\begingroup$ Very nice! The idea that $\mathrm dx$ is not punctuation, but (conceptually) an actual infinitesimal measure of $x$-length, so that $f'(x)\mathrm dx$ has units of (slope)($x$-length) = $y$-length, I think is probably the most important pre-proof part of understanding the connection. That is, I think that most people think of integration as "something you do to $f'(x)$", not "something you do to $f'(x)\mathrm dx$"—hence later difficulties with differential forms, and earlier difficulties with changes of variable. $\endgroup$ – LSpice Apr 6 at 0:06
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Well, for me, the nicest intuition comes from physics. If $F=F(x)$ is some force applied to an object $O$ casuing $O$ to move, where $x$ is its displacement, when the work $W(x)$ produced by that force from $x_0=0$ to $x$ metres is given by:

$$W(x):=\int_0^xF(s)ds.$$

Now, what if we ask what is the rate of change of that work with respect to displacement? At some certain point $x$, the work provided to or substracted by that object $O$ is determined by the force $F$ applied on it. So, one can intuitively expect that:

$$W'(x)=\left(\int_0^xF(s)ds\right)'=F'(x).$$

That is, the larger the force, the faster the energy flows from the one applying the force towoards the object. The less the force, the slower that energy flows.

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If you want more clearly "undoing" operations, define "stacking" as the following:

Take some $\Delta x$. Chop the curve into rectangles of width $\Delta x$ and height $f(x)$. (There's some leeway as to take $f(x)$ at the left side, right side, middle, minimum, maximum, etc. The rest of my description will be right side.) Now take each rectangle and put the bottom of each rectangle at the top of the previous rectangle. For instance, if you have $f(x) = x^2$ and $\Delta x = 0.1$, you'd have a rectangle with lower left corner at the origin, and top left at $(0.1, 0.01)$. Then the next rectangle would have lower left corner at $(0.1, 0.01)$, and upper right at $(0.2, 0.05)$, and so on.

Define "unstacking" as follows:

Take some $\Delta x$. Chop the curve into rectangles of width $\Delta x$ and height $f(x)$. Now, take each rectangle and shift it vertically down the height of the previous rectangle. For instance, if $f(x) = x^2$, the first rectangle will have height $0.01$. The second one will have height $0.04$, so for the unstacked version, we move it down $0.01$, leaving its new height as $0.03$. Another way of phrasing it is for each $x$, take the rectangle whose bottom left corner is at $(x, f(x))$ and top right corner is $(x+\Delta x, f(x+\Delta x))$. Then create a chart out of all of those rectangle, with all of them moved down so their bottom is on the x-axis.

For "well-behaved" functions (I believe that uniform continuity is a sufficient condition), as $\Delta x$ goes to zero, stacking becomes integration, and unstacking become differentiation (with the proper scaling).

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