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Sunday, October 9th 2016



  • STATE OF EXISTENCE: pretty exhausted

I really had no intention of writing yet another piece of math for the blog. As I was going through my e-mail around noon, I came across some notification from Quora.com, which eventually led me to the question:

“What does the integral in math mean?”

If you should look it up, you may agree with me that the answers currently in place pretty much assume that the person who asked it is already familiar with calculus and that comments get pretty esoteric pretty fast. So, I thought I would write a quick answer that would be intelligible to someone who didn’t already know calculus. Well, seven pages later, I hope that I have possibly helped someone understand the basic nature of calculus a little better without totally scaring them off. However, my answer is quite a bit longer than anticipated (as always), so I will post it on my blog instead in the hope that someone may see it and read it. I am trading possible “up-votes” for potential intelligibility.

There are two parts to calculus. One of them is “integration,” and the other one is called “differentiation.” I will mention differentiation in passing in order to highlight what integration is, but I’ll make sure that I won’t stray too far on a tangent. (Wow! Sorry, that pun was not intended.) The question is one about the meaning of a term in mathematics, so my explanation is not going to include proofs, lengthy calculations, or application of complex formulas, though I may reproduce a few just as visual aids. I’m going to use two illustrations spread over three sections. Section 1 consists of a basic explanation of integration; section 2 brings up a badly needed reality check; and the third section is something totally different. I would suggest that if you don’t want to follow the whole exposition here, but you want to get something out of it, you might want to go straight to section 3.

(1) I can’t imagine why I would want to buy a flock of sheep, but let’s pretend that I do, and I visit a man who has kept sheep for a several decades, but wants to retire now. As I look over the entire collection of wool bearers, I realize that there are too many of them for me to count, not even if I resort to one of the standard tricks, such as counting all the legs and dividing by four. So, I ask the shepherd how many there are. He replies, “There are about a thousand sheep now. I started out twenty years ago with just ten, but the flock has grown enormously.” (The story is getting less and less probable; a shepherd would surely know exactly how many sheep he has.)

Having received that number, I want to mine it for some more information, and so I set up an equation that relates the number of sheep to the 20 years. The number of 10 for the original sheep does not change, and I will simply add it as needed after basic calculations. I’m thinking of dividing 1,000 sheep by 20 years, and come up with a rate of 50 sheep per year). I write out an equation:

number of sheep = number of years × 50 + 10

s = 50y + 10,

where s stands for the number of sheep and y for the years.

So, after 5 years, there would have been 260 (5×50+10) sheep and 1010 (20×50+10) after 20 years. Five years from now I’m expecting to own around 1260 (50 × 25 +10) sheep, a manageable number. [If you are catching a bit of a misunderstanding on my part, please follow me here and wait for the next section.]

I’d like to visualize those numbers, and I make myself a graph that plots the number of sheep against the years. Since I don’t intend to reverse the flow of time, and since I don’t have any idea what a “negative sheep” would be, I can stick to positive numbers. 

straightline graph 1

This a nice straight line containing no surprises. I extended the line to 25 years since I want to see what to expect in the future, but I dropped a vertical line down to the 20-year point because that's our present location in time.

Even though we already know the rate of change, allow me to add a little bit of terminology. We can understand the equation as a function.


The rate of change is called the “derivative” of a function. There are various ways of writing this out. Because differentiation is not the point here, I’ll just choose the simplest one. The derivative of function f is f', and thereby


Since the line is straight and we already knew that information anyway, we’ll just leave it at that and move into the other direction, which is integration.

I know how many sheep I have and at what rate the flock grows, but I’m a little afraid of how much work will be involved. For example, I’m wondering about how often I will have to shear those sheep, not just in one year but over many years. I know that I should do so halfway through every year. In order to make this this calculation easy for now, let me make the somewhat bizarre assumption that the growth of the flock occurs continuously over any given year. There will be a new sheep in my enclosure every 7.3 days, which translates into 7 days, 7 hours, and 12 minutes. We know it doesn’t work that way, but let’s just keep it simple. The point is not the number; the idea is to avoid unneeded complexity and think of the equation as an unbroken continuous line. We’ll have to get back to this assumption soon.

Given these two considerations: (1) the continuity of the function and (2) the need to shear the sheep halfway throughout the year, let’s figure out how many sheep our friendly shepherd and his helpers must have sheared so far. Since we have just finished the twentieth year, the last shearing was ½ year ago, at 19.5 on the year axis. Now, it might be tempting simply to plug in 19.5 as y and calculate

50 × 19.5 + 10, which comes out to 985.

But then we’ve merely calculated the number of sheep that were sheared on the last occasion. I want to know the total amount over all of the years.

Now we are moving into the area of integration. To put it briefly we want to calculate the area underneath the line of the function.

straightline graph 2

Given the simplicity of our example, it’s not too hard to calculate the number the numbers for each time (e.g., at 0.5, 1.5, etc.) and add them all up. There is no great need for calculus and integration here, but that’s because I’m using a simple example to illustrate the method. There are many far more complex problems for which a simple count-and-add procedure is not possible. We can get the information we’re after by dropping a vertical line at 19.5 and then calculating the area of that triangle. There are several methods with which we could do so.

One would be to try to figure out the area of the triangle, given whatever values we have and plugging them into the formula. A more generally useful method in the long run is to apply a formula for “indefinite integration,” which runs in its most basic form (for exhibition purposes only for now):

integration formula

                                       The integral of the equation above  =  the formula on the right                                                                   

 where C stand for an undetermined constant. I'll plug in one set of values so you can at least make a bit of sense out of it. Let's say you have the formula


Then its integral will be


As promised, I’m not going to apply any really complex formulas or do any long calculations. Besides, I have a huge problem on my hands all of a sudden. The shepherd is laughing his head off.

(2) The shepherd has glanced at my graph and, after just a short look, seems to think that it’s the funniest thing he’s ever seen. “It doesn’t work that way, Win. We’re talking about living animals. A flock doesn’t grow along a straight-line graph.”  I get it. Some sheep get sick and die; some may be sold; others may be stolen. One can add to a flock by buying more sheep as well as having them make lambs. If they all would reproduce according to the proper mathematical pattern, their growth would geometric. In the real world, that would not the case for many practical reasons. However, if we looked at their growth pattern in an ideal setting, the sheep would multiply by an exponential curve. In plotting this graph, I had to settle for a relatively slow rate of growth in order to squeeze the growth from 10 to 1,000 into 20 years, but it’s still going to make the difference between the two approaches very clear.

Here is what a graph for what this equation looks like:

parabola graph 1

The geometric form is that of a parabola; not only does the number of sheep grow every year, but the rate at which it does so gets faster. The longer it takes, the steeper the curve gets. What starts out as almost a horizontal line eventually will be almost vertical. Because of the example I chose, I had to make this graph come out at 1,000 just like the previous one, and If you remember the straight-line graph, you can see that up until that point the parabolic line actually runs underneath the linear one. However, given the nature of parabolic lines, there must be a point where it exceeds its linear counterpart, and it’s right about here. It cannot remain below the straight line forever. In another 5 years, I will not have merely 1,260 sheep, but 1,598. And looking at ten years from now, the straight-line curve would only give me 1,510 sheep, but under the exponential growth curve, I would have 2,250. The rate of growth is slow, but getting faster and faster. I’m not entirely sure about the numbers that Wolfram-Alpha has given me after my somewhat quirky request that the first 20 points on the x-axis had to stay below 1,000, but they make the point without my having to fit in a graph that wants to zoom toward the sky so fast that the lower section becomes unreadable.

If we want to find the integral in order to figure out how often the shepherds have sheared the sheep so far, we have to look underneath the line again. This time we’re not getting a nice right triangle. We need to use the integration formula, and I’m keeping my promise not to make you slog through line after line of equations, but obviously these numbers will become much higher than those of the straight-line function as well.

parabola graph 2


(3) Okay, everybody (almost?) knows the two celebrated formulas concerning a circle. The area of a circle is calculated by the equation

area of a circle

and we get the circumference of a circle with


Here’s what I would like you to see because here we have both integration and differentiation in action. π is a constant that we can ignore here for our demonstration purposes. If we want to get the integral for a circle, that’s kind of a tricky business. There is no genuine function under which we can calculate the area. However, we can calculate the area of the inside of the circle. That procedure is analogous to finding the integral of true functions, and it's easy to remember as long as you know the two circle formulas. 

In this case, take the formula for the circumference, 2πr, make 2 the exponent of r, and divide the formula by 2. You get: πr2, which we immediately recognize as the formula for the circle’s area. It is also the analog for an integral.

Conversely, the pseudo-derivative of a circle measures the length of the curve surrounding the area. We get it by taking the exponent 2 from πr2 and making it a coefficient, 2πr. Voila! There’s the formula for the circumference again.

Since, as I said, the graph for a circle is not truly a function in the technical sense, these are not actually instances of differentiation and integration. However, they are close, probably easier to remember, and a good introduction to these two sides of calculus.

I think I have successfully answered the question on What is integration in math? My answer is way too long to fit into that forum, but I really wanted it to be of help to someone. Not wanting it to go to waste, I made it a blog entry. Hopefully, people who are interested in such a question will see it and find it helpful. 
If you read this entry and followed my line of thought, and you’ve never studied calculus before: Congratulations! You are now an initiate on the very first rung of that mysterious part of mathematics. 

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