Why are finding slope and finding area inverse processes?

[kartoshka]

I'm about one week into a Calc II course, and I realized that I have no idea why finding the area under the curve (integrating) would be the inverse operation of finding slope (differentiating). I get that they are opposites, but not why that should be the case... it's not at all intuitive. Is there some sort of intrinsic relationship between slope and area?

[Sorry if this question is poorly worded; I'm having a hard time putting my confusion into words.]

[SteamKing]

You might try this article:
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

[SteveL27]

I'm about one week into a Calc II course, and I realized that I have no idea why finding the area under the curve (integrating) would be the inverse operation of finding slope (differentiating). I get that they are opposites, but not why that should be the case... it's not at all intuitive. Is there some sort of intrinsic relationship between slope and area?

[Sorry if this question is poorly worded; I'm having a hard time putting my confusion into words.]

Are you looking for the intuition behind the Fundamental Theorem of Calculus? One informal approach is to think of it as a generalization of distance = rate x time.

If you are traveling at a certain rate of speed, your total distance covered is your rate of speed, times the length of time you travel.

Now if you are trying to integrate a function f over an interval [a,b], you can look at f over a tiny little interval, on which the rate of change of f is approximately the derivative f'. If the length of the little interval is dx, then the area under the curve (total distance travelled, amount of work done, etc.) is f'(x) dx. And if you add up all the little intervals, you are taking the integral of f'(x) dx over [a,b] to get the total distance represented by f over that interval.

Now that's a hand-wavy, imprecise argument, not a proof, and probably not what's in the book. But you can see that if you know your rate of speed over each small part of your trip, you can add up the rate times time for each part of the trip to get the total distance travelled.

Here's the same idea in a different form. You're driving your car from one point to another. Your passenger is assigned the task of writing down your speed, as indicated on your speedometer, at one-second intervals during the entire trip.

After you arrive, you look at the list of speeds and you see, well, I was going 50 mph for the first second, then 49mph in the second second, and then 51 mph in the third second, etc. Of course your speed during each second is not constant, you are only looking at one sample point taken during each interval. Yet, using this information, you can get a very good estimate of the total length of your trip. That's because your velocity is well-behaved in each one-second interval: if the sample is 50mph, your speed was reasonably close to that during the entire second. So if you add up all the d = rt calculations over each second, you can estimate the total length of the trip.

If that helps, good; if not, forget it. In your first week of Calc II you should just pay attention in class, do the homework, do some extra problems for practice, and get through the class. Intuition comes much later, if at all.

Or as John von Neumann once said: In mathematics you don't understand things. You just get used to them.

[Stephen Tashi]

Another way to build intuition about it is to look at material on "The Calculus Of Finite" differences. Look at the relation between summation and "anti-differencing". Notice how it resembles the approximation of derivatives and integrals by finite quantities.

The basic thought is that if we define
\triangle F(x) = \frac{ F(x+h) - F(x)}{h} then the Riemann-like sum:

h \sum_{k = 0} ^ n \triangle(F(a + kh)) = h (\frac{ F(a+h) - F(a)}{h} + ... \frac{ F(a + nh + h) - F(a+nh)}{h}) is a "telescoping sum", which collapses to

F(a + nh + h) - F(a) .

Think of F(x) as being the "anti-delta" of \triangle F(x) Think of letting n go to infinity while h goes to zero in such a manner that nh + h approaches the upper limit of integration b.

[Antiphon]

Minor point but important; the area and slope are not inverses; they're double inverses.

The integral computes the area. The inverse is the value of the function, not the slope.

The derivative computes the slope. The inverse is the function, not the area.

[kartoshka]

Thanks to everyone who answered, although I don't know how much better it made my gut intuition feel.

If that helps, good; if not, forget it. In your first week of Calc II you should just pay attention in class, do the homework, do some extra problems for practice, and get through the class. Intuition comes much later, if at all.

Or as John von Neumann once said: In mathematics you don't understand things. You just get used to them.


I think this actually makes me feel the most relieved. With pretty much everything in Calculus it's like, "why would THAT be the case?! This makes no sense!" But I took Calc in high school, so Calc I and II at college are basically the second time around for me, and stuff is almost starting to make sense to me, but not quite.

On a side note, I really wish they taught Calculus (concepts) in elementary school. I feel like just experiencing Calc a little bit when you're young would get rid of the whole "but this doesn't FEEL right" thing that Calculus has going on, at least for me.


And another question - what happens if you integrate twice? What does that tell you about the function (if anything). f''(x) is concavity, but the second integration of functions isn't really mentioned, as far as I can tell. I don't even know what the notation for it would be. And extra-capital f? Thanks again.