For a while I've been struggling with why the antiderivative should give the area under a curve, and a little more generally, how exactly integration and differentiation are inverses.

So recently, in learning more about the applications of integrals (specifically to do with volumes), the notion that what we're doing with the integral to find volume is adding up infinitely many 'slices' of the solid was explained to me. And seeing as taking a derivative is looking at a curve's slope point-by-point (broken up), my question is:

Is the intuition of going from 2D to 3D and vice versa a good way to think about the relationship between integrals and derivatives? Especially since the summing-up bit about integrals is what makes the solid, well solid. It also seems to reflect the loss of information in taking the derivative, in the way that when you only look at one dimension of a two dimensional shape, you don't get the whole idea.

Any corrections or thoughts would be helpful. Thanks!

EDIT It seems like the dimension bit isn't so clear, so I'll try to give more info. See, when my class was learning about volumes, the way my teacher described what the integral was in a way similar to this: since it's an infinite sum, it's taking all the slivers of area and adding them up to get a full solid. Thinking about integrals in that light, I focused on how we use 2D area of a slice, and integrate it to get the 3D volume of a solid. This put the idea in my brain that the process of integrating is similar to adding another dimension to a function. (in this case, integrating gave us a z-axis, kind of)

In that same thought process, differentiation can easily be thought of as the reverse: Taking a 2D curve and the process of homing in on one dimension, just a line. Also, I'll put a little more into my note on losing info while taking a derivative. As I said above, differentiation is like losing a dimension, or an axis in a way. So when we try to bring back another dimension through antidifferentiation, it's downright impossible to know where our line was in reference to the other axis without more information, since we dropped the axis and all the information it contained. That's why indefinite integrals require the constant added on: to make up for the lost y-axis information. I hope I cleared that up more, and thanks for any respones. :)

  • 1
    $\begingroup$ I don't understand where you are going with your 2D vs 3D and derivative vs integral comparison, but I highly recommend 3Blue1Brown's Essense of Calculus series. It covers this topic well. While the main discussion of the fundamental theorem of calculus occurs much later in the series, the very first video gives some good insight into why derivatives and integrals are opposites. $\endgroup$ Apr 7, 2018 at 16:43
  • $\begingroup$ His series is actually another place where I got this feeling. By the whole dimension thing, I mean that differentiation is like "closing in on a point", and that's a lot like how 3Blue1Brown explains it. And using this approach, integration is like starting from a point, and going through the process of expanding it over the function. Maybe I'm just not using terminology correctly, but that's I was trying to say. Sorry for the confusion. $\endgroup$
    – siren5474
    Apr 7, 2018 at 19:42

2 Answers 2


I don't quite understand the part about volumes, it seems like you understand about the "slicing up" of volumes.

I know this will be tedious notation (especially if you are unfamiliar calculus) but recall the standard (Riemann) definition of an integral

$$\int_{a}^{b} f(x) \cdot dx = \lim_{ || P || \to 0 } \sum_{k =1}^{n} f(x_{k}) \Delta x_{k} $$

All this is saying is as you take $ ||P|| \to 0$ where $||P||$ is the largest slice (called a partition) the area is a nice rectangle, or trapezoid depending on the value of $\Delta x_{k}$.

Now, look closely at the integral side of the equation. If you've been introduced to Newton's notation you know about $\frac{dy}{dx}$ Doesn't that $dx$ look familiar now? All we're doing with the right hand side is looking at the relationship between $x$ and $y$ also called - Area!

There's a much more serious looking post here

  • $\begingroup$ Nonono, I understand how we get area from an integral, and I'm familiar with the Riemann sum definition. I'll edit my question and try to expand on it. $\endgroup$
    – siren5474
    Apr 7, 2018 at 19:43
  • $\begingroup$ @siren5474 I understand what you're trying to say but don't think thats a good way of thinking of integrals/derivatives. It would get very confusing as you get into upper years especially with multivariable calculus that actually deals with a third dimension. $\endgroup$
    – Andre Fu
    Apr 7, 2018 at 20:53

You are right that integration is sort of combining slices of a lower dimensional object to get the full object, though 2D vs 3D isn't necessary. In fact, by definition, it is obtaining a 2D value (area) by adding 1D slices (heights). The 2D to 3D application that your teacher showed (summing areas of slices to get volume) comes from suppressing a dimension - treating the 2D areas as if they were 1D heights, so that we can use our existing 1D to 2D tool to also do 2D to 3D.

But with differentiation, I think you are misunderstanding the nature of it. A curve is not a 2D object. It is intrinsically 1-dimensional. In can exist in surroundings of any number of dimensions. For both convenience (you can't draw good curves in 3 dimensions on a 2D writing surface) and simplicity of examples, we mostly talk about curves on the plane, but in general, curves are not limited to 2D.

The derivative of a curve is not something intrinsic to the curve itself, but is about how the curve relates to its surroundings. When you look at the value of the function at just one point, what is tells you is exactly where the function is at that point. It says nothing about the behavior of the function anywhere else. but if you look at the derivative of the function at only that same point, it tells you something about how the function behaves near that point. There is no dimensional loss here. You may be comparing knowing the function everywhere to looking at the derivative at just that one point, and that of couse is going to less information. But if you look at know the original function everywhere (which is effectively a point in an infinite-dimensional space) to knowing the derivative everywhere, then the informational loss is trivial - one dimension out of infinitely many. This is why you can rebuild the original function except for a single constant of translation.

Derivatives are rates of change of one variable with respect to another, and the rate of change of the area under a curve is the height of curve at the moving border. This is why derivatives and integrals are inverses of each other.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.