The calculus I'm aiming to build in this series has its origins in antiquity. Archimedes was one of the first people to brush up on some concepts of this discipline with his ideas. This Greek philosopher from Sicily, whom we would now consider more of a mathematician and a physicist, has dealt with many problems. For example, he devised his own numerical system for counting large numbers, he designed the battle machines with which to sink enemy ships, he formulated the law of buoyancy, but what we'll be interested in is that he calculated the area under the segment of the parabola. Calculating the area under a curve is one of the fundamental functions of a differential count. Later, we will look at this more broadly and in more depth, showing what use we can find for this role. In the meantime, however, let's look at this role in a similar way to Archimedes -- purely as an interesting mathematical problem that has a different, more peculiar character than calculating the area of ordinary geometric formations as an example of $n$-angles.

Specifically, let's focus on the area segment below the parabola, which we'll take as a function of $y=x^2$, in the segment from $x=0$ to $x=1$. You can see the desktop in the image below. I will remind you that the $f(x)$ function is a so-called view, or assignment of elements from one set of elements from another. Our function assigns to each of the $x$ numbers belonging to the set of real numbers a different number $f(x)=y=x^2$ belonging to the same set. Sometimes the function is said to be like a box where we put a number in and it throws out another. Think about how these two definitions fit together. Below you can see a graph of said function with the indicated area counted.

I'm sure you've noticed how crooked this content is going to be, because the pattern isn't regular at all, and the curve is constantly changing, it's crooked at every point. But let's at least do an approximation, or an approximation, of that area. We use the familiar rule that the contents of a rectangle are equal to $S=a\cdot b$, where both $a$ and $b$ are lengths of sides. We divide the area under the parabola into two halves, and we approximate each one with one rectangle, like this:

Well, it's not a very good approach, but at least it's something. The first rectangle has zero height, so zero content, the second rectangle has a longitude of $a=1/2$, and a height of $b=f(1/2)=(1/2)^2=1/4$. So together we get for the content (index $2$, because we have $2$ rectangles): $$S_2=1/8\,.$$

Well, let's try another approximation, this time with three rectangles, intuition suggests we get a more accurate estimate.

In this approximation, rectangles have a base of $a=1/3$, height sequentially $b_0=(0/3)^2$, $b_1=(1/3)^2$, and $b_2=(2/3)^2$. Overall, we get: $$S_3 = \frac{1}{3}\left(\frac{1}{3}\right)^2 + \frac{1}{3}\left(\frac{2}{3}\right)^2 = \frac{1}{3^3}(1+4) \,.$$

We already have a slightly more accurate result, but let's try to go further and divide the content into quarters.

Now for the sum we get $$\begin{align*} S_4 &= \frac{1}{4}\left( \frac{1}{4}\right)^2 + \frac{1}{4}\left( \frac{2}{4}\right)^2 + \frac{1}{4}\left( \frac{3}{4}\right)^2\\ S_4 &= \frac{1}{4^3}\left( 1+2^2+3^2\right) \,. \end{align*}$$

One can certainly see from the diagrams that the finer the division we choose, the more accurate the result we receive. So let's try to figure out what the formula would be for the area of the rectangles if we were to divide the interval into the general $n$ parts. I recommend that you try to come to this result yourself, perhaps first constructing the $S_5$ case for dividing into five parts, so that you realize the principle properly, then generalizing for $n$. I'm enclosing a picture for dividing the interval into five parts (though it's better if you draw it yourself), and underneath it I'm describing, the formula for the $n$ part.

So in the general case, we have to add together the contents of the total $n$ rectangles. Each has a sub-state of $a=1/n$. It may also be noted that the member $(1/n)^2$ can be touted from other members, so we get: $$\begin{align*}S_n &= \frac{1}{n^3} \left( 1+2^2+3^2+\dots + (n-1)^2\right)\\ S_n &= \frac{1}{n^3} \sum_{i=1}^{n-1} i^2\,.\end{align*}$$

But this expression contains a formula from the first chapter on sums! If we put him in, we get: $$S_n = \frac{1}{n^3}\frac{(n-1)(2(n-1)+1)((n-1)+1)}{6} \,.$$

But what now? That's not a formula that gives us directions about the content. But it's easy to see that as the $n$ increases, so does the accuracy of our estimate. Even for a very large $n$, we can get an accuracy, similar to the one on the bottom image, which shows the division into $n=100$ rectangles. One could mathematically say that whatever precision of approximation we require, with a sufficiently large $n$ we can achieve this precision. And if we go even $n$ to infinity, then even our accuracy is going to be close to infinite accuracy. So to figure out the content, we use the formula for the big $n$, which we also deduced in the chapter with Pascal. We're also going to say that for a big $n$, it's $n-1\approx n$, which for $10^{100}$, is such a small mistake that we can certainly afford it.

Together we get for infinitely accurate content $$S_{\infty} = \frac{1}{n^3}\frac{n^3}{3} = \frac{1}{3}\,.$$

Now we have finally got the right result! We received a simple number, which is correct, because it would seem strange if the final result depended on the number of pieces (which is supposed to be infinite).

You may be somewhat surprised at the way in which we arrived at the result, or even doubt its veracity. Indeed, it seems counterintuitive that mathematics, which is supposed to be a precise science, first calculates content with error, second uses a formula for large numbers that we have received by some vague simplification. I described this simplification a little more clearly in the chapter with Pascal, but still not enough. You may still find the procedure suspicious.

But it turns out that my ideas are correct. Later mathematicians such as Isaac Newton were able to build a theory in which they credibly work with infinitesimal error and simplifying formulas for large numbers. Nevertheless, it is still a conceptual leap, one must simply trust it at first, and only later come a certain intuition.

The adding of the content under the curve that we just did is called integration. We might also want to do this integration on a different scale from maybe $0 to $2. This content, the integral, let's mark it $I_0^2$. We'll split the area from $0 to $2 into $n$ parts, so one of them will have a sub $2/n$. Next, the parts will be multiples of $2/n$. So gradually we get for the $n$-th sum: $$S_n=\frac{2}{n} \left( (2/n\cdot 1)^2 +(2/n\cdot 2)+ \dots +(2/n \cdot (n-1) \right)\,.$$

Now we're going to raise $2/n$: $$S_n = 2^3 \frac{1}{n^3} \left( 1 + 2^2 + \dots + (n-1)^2 \right) \,.$$

So the total is $2 ^3 $ times larger. We will $$I_0^2 = \frac{8}{3}\,.$$

If we reconsider this course of action, it becomes obvious that $$I_0^x = \frac{x^3}{3}\,.$$

If we want to calculate the integral backwards from the $y$position, we simply get the result by adding the areas (draw): $$I_y^x = I_0^x - I_0^y\,.$$

But we can also integrate other functions. We can do a similar method of dividing into rectangles. Nevertheless, for a certain class of functions, we can do integration very simply. I mean functions in the form of $f(x)=x^m$. If we use the same division into rectangles, we get $$S_n= \frac{1}{n^{m+1}} \left( 1^m + 2^m + \dots + (n-1)^m \right) \,.$$

When we think of the sum formulas for the big $n$ chapter with Pascal: $$\sum_{i=1}^{n} i^m \approx \frac{n^{m+1}}{m+1} \,, $$

we can easily figure out what the integral of $f(x)=x^m$ is from $0$ to $x$. This I will mark $I_0^x(f(x))$ and it is: $$I_0^x(x^m) = \frac{x^{m+1}}{m+1} \,. $$

In this chapter, we have shown ourselves a powerful tool — integration that we can use to add areas under functions. For $f(x)=x^m$, we have shown how it can be used effectively, for other functions, it is more difficult to use. However, we did not derive integration completely rigorously, it may be that you still do not believe it applies. In the next chapter we will take a completely different subject, but we will return to integration later and show how it can still be generalised.