How would one teach to young pupils derivative, integral and when the are inverse "operations". Namely, if one has an intuition that $D(\int f(x)dx)=f(x)$ so then of course $\int(D(f(x))dx)=f(x)$. But hey, isn't $\int(D(f(x))dx=f(x)+C$. Or if $\int f(x)dx$ is a set of functions $\{F(x)+C|C\in\mathbb{R}\}$ then isn't $D(\int f(x)dx)$ a derivative of a set?
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- 5$\begingroup$ For teaching young pupils, get a good textbook and follow it. I think "$D(\int f(x)dx)=f(x)$" is not appropriate for young pupils. $\endgroup$Gerald Edgar– Gerald Edgar2025-02-12 15:53:17 +00:00Commented Feb 12 at 15:53
- $\begingroup$ I agree with Gerald! but also when your young kids know derivatives they also Know that all functions f(x)+c have the same derivatives for any c. $\endgroup$trula– trula2025-02-12 18:06:42 +00:00Commented Feb 12 at 18:06
- 9$\begingroup$ I'm voting to close this question as written because it feels more like a query about how to understand this fact, not about how to teach this result. I suggest expanding the post to clarify what is meant by "young pupils" and to specify what prior knowledge is assumed of the audience. $\endgroup$Brendan W. Sullivan– Brendan W. Sullivan2025-02-12 19:39:04 +00:00Commented Feb 12 at 19:39
- 1$\begingroup$ As a prerequisite, the student needs to be able to treat functions as objects and think of differentiation and integration as operators on those objects. Is meeting this prerequisite part of the problem? $\endgroup$Alexander Woo– Alexander Woo2025-02-14 16:01:31 +00:00Commented Feb 14 at 16:01
- $\begingroup$ From a certain point of view it is wrong to say that integration and derivation are inverse operations. Integration is a pairing between a functional object (e..g. a function) and geometric object (e.g. an oriented interval). Differentiation is a linear operator sending a functional object to another functional object. Its dual with respect to the integration pairing is the boundary operator on geometric objects. This is what the fundamental theorem of calculus, which is the one-dimensional Stokes theorem, says. From this point of view, differentiation is not an inverse of integration. $\endgroup$Dan Fox– Dan Fox2025-04-04 07:47:15 +00:00Commented Apr 4 at 7:47
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2 Answers
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Just a simple remark: imagine $F(x)$ and $f(x)$ and $F'(x)=f(x)$.
You can say that $F(x)$ is achieved by closing in and summing $(f(x_1)-f(x_2)) \cdot (x_1 - x_2)$. $(1)$
You can say that $f(x)$ is achieved by closing in $\frac{F(x_1)-F(x_2)}{x_1 - x_2}$. $(2)$
The relation between $(1)$ and $(2)$ is obvious.
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I always find it odd when math educators ask for intuitive strategies regarding analytical statements when my experience has been that most students intuit using their visual capacities. Consider epsilon-delta. It is far more likely to be understood as a shrinking rectangle imposed on a curve then it is as a series of algebraic statements. As the epsilon diminishes, so too does the rectangular neighborhood. This also makes it easier to understand that points can be viewed as locii and why the differential is actually non-zero. I believe that those of us with a predisposition to mathematics tend to miss the important point of self-reflection that our intuitions are fundamentally different than of many of our students, particularly if they are not STEM-oriented in their educations.
But to answer your question, the inverse nature to me of the derivative and anti-derivative is best conveyed, perhaps using Geogebra, as a geometric representation. For instance, the derivative is the slope of the line tangent to an approximated point on a curve, and an integral is the area under a curve. By showing this relationship as metrics of a geometric representation, I believe this is far more effective in demonstrating the relationship. I was 16 when I first saw this, and more than 30 years later, it's how I still fundamentally understand the relationship. Something like this lecture are, IMHO, more meaningful to students than an analytic approach.
My final recommendation, and again, I saw this several decades ago, is to consider the relationship in physics between position, velocity, and acceleration. I was actually shown the basics of calculus junior year in high school by an excellent physics teacher so we could understand a calculus-based approach to mechanics. Having graphs that represent constant velocity, for instance, and then some that represent variable velocity in comparison to position and acceleration make it obvious how derivatives and integrals are interrelated. Having a descriptive application (and a series of practical equations related by x, v, and a) was far more meaningful than just presenting mathematical statements devoid of context. After all, Newton's discovery of the fluxion was very much an exercise in physics so that it's rather odd to think that many math teachers approach the topic as if it's category theory: a mere abstraction.
