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How to insert mathematical formulas on a website?

If you ever need to add one or two mathematical formulas to a web page, then the easiest way to do this is as described in the article: mathematical formulas are easily inserted onto the site in the form of pictures that are automatically generated by Wolfram Alpha. Besides simplicity, this universal method will help improve website visibility search engines. It has been working for a long time (and, I think, will work forever), but is already morally outdated.

If you regularly use mathematical formulas on your site, then I recommend you use MathJax - a special JavaScript library that displays mathematical notation in web browsers using MathML, LaTeX or ASCIIMathML markup.

There are two ways to start using MathJax: (1) using a simple code, you can quickly connect a MathJax script to your website, which will be automatically loaded from a remote server at the right time (list of servers); (2) download the MathJax script from a remote server to your server and connect it to all pages of your site. The second method - more complex and time-consuming - will speed up the loading of your site's pages, and if the parent MathJax server becomes temporarily unavailable for some reason, this will not affect your own site in any way. Despite these advantages, I chose the first method as it is simpler, faster and does not require technical skills. Follow my example, and in just 5 minutes you will be able to use all the features of MathJax on your site.

You can connect the MathJax library script from a remote server using two code options taken from the main MathJax website or on the documentation page:

One of these code options needs to be copied and pasted into the code of your web page, preferably between tags and or immediately after the tag. According to the first option, MathJax loads faster and slows down the page less. But the second option automatically monitors and loads the latest versions of MathJax. If you insert the first code, it will need to be updated periodically. If you insert the second code, the pages will load more slowly, but you will not need to constantly monitor MathJax updates.

The easiest way to connect MathJax is in Blogger or WordPress: in the site control panel, add a widget designed to insert third-party JavaScript code, copy the first or second version of the download code presented above into it, and place the widget closer to the beginning of the template (by the way, this is not at all necessary , since the MathJax script is loaded asynchronously). That's all. Now learn the markup syntax of MathML, LaTeX, and ASCIIMathML, and you are ready to insert mathematical formulas into your site's web pages.

Any fractal is constructed according to a certain rule, which is consistently applied an unlimited number of times. Each such time is called an iteration.

The iterative algorithm for constructing a Menger sponge is quite simple: the original cube with side 1 is divided by planes parallel to its faces into 27 equal cubes. One central cube and 6 cubes adjacent to it along the faces are removed from it. The result is a set consisting of the remaining 20 smaller cubes. Doing the same with each of these cubes, we get a set consisting of 400 smaller cubes. Continuing this process endlessly, we get a Menger sponge.

In this article you will learn how to find the area of ​​a figure bounded by lines using integral calculations. For the first time we encounter the formulation of such a problem in high school, when we have just completed the study of definite integrals and it is time to begin the geometric interpretation of the acquired knowledge in practice.

So, what is required to successfully solve the problem of finding the area of ​​a figure using integrals:

• Ability to make competent drawings;
• Ability to solve a definite integral using the well-known Newton-Leibniz formula;
• The ability to “see” more profitable option solutions - i.e. understand how it will be more convenient to carry out integration in one case or another? Along the x-axis (OX) or the y-axis (OY)?
• Well, where would we be without correct calculations?) This includes understanding how to solve that other type of integrals and correct numerical calculations.

Algorithm for solving the problem of calculating the area of ​​a figure bounded by lines:

1. We build a drawing. It is advisable to do this on a checkered piece of paper, on a large scale. We sign the name of this function with a pencil above each graph. Signing the graphs is done solely for the convenience of further calculations. Having received a graph of the desired figure, in most cases it will be immediately clear which limits of integration will be used. Thus, we solve the problem graphically. However, it happens that the values ​​of the limits are fractional or irrational. Therefore, you can make additional calculations, go to step two.

2. If the limits of integration are not explicitly specified, then we find the points of intersection of the graphs with each other and see whether our graphical solution coincides with the analytical one.

3. Next, you need to analyze the drawing. Depending on how the function graphs are arranged, there are different approaches to finding the area of ​​a figure. Let's consider different examples to find the area of ​​a figure using integrals.

3.1.

The most classic and simple version of the problem is when you need to find the area of ​​a curved trapezoid. What is a curved trapezoid? This is a flat figure limited by the x-axis (y = 0), straight lines x = a, x = b and any curve continuous in the interval from a to b. Moreover, this figure is non-negative and is located not below the x-axis. In this case, the area of ​​the curvilinear trapezoid is numerically equal to a certain integral, calculated using the Newton-Leibniz formula: Example 1

y = x2 – 3x + 3, x = 1, x = 3, y = 0.

What lines is the figure bounded by? We have a parabola y = x2 - 3x + 3, which is located above the OX axis, it is non-negative, because all points of this parabola have positive values. Next, the straight lines x = 1 and x = 3 are given, which run parallel to the axis of the op-amp and are the boundary lines of the figure on the left and right. Well, y = 0, which is also the x-axis, which limits the figure from below. The resulting figure is shaded, as can be seen from the figure on the left. In this case, you can immediately begin solving the problem. Before us is a simple example of a curved trapezoid, which we further solve using the Newton-Leibniz formula.

3.2.. Calculate the area of ​​the figure bounded by the lines y = x2 + 6x + 2, x = -4, x = -1, y = 0.

IN in this example we have a parabola y = x2 + 6x + 2, which originates from under the OX axis, straight lines x = -4, x = -1, y = 0. Here y = 0 limits the desired figure from above. The straight lines x = -4 and x = -1 are the boundaries within which the definite integral will be calculated. The principle of solving the problem of finding the area of ​​a figure almost completely coincides with example number 1. The only difference is that the given function is not positive, and is also continuous on the interval [-4; -1] . What do you mean not positive? As can be seen from the figure, the figure that lies within the given x's has exclusively “negative” coordinates, which is what we need to see and remember when solving the problem. We look for the area of ​​the figure using the Newton-Leibniz formula, only with a minus sign at the beginning.

The article is not completed.

Problem 1 (about calculating the area of ​​a curved trapezoid).

In the Cartesian rectangular coordinate system xOy, a figure is given (see figure) bounded by the x axis, straight lines x = a, x = b (a by a curvilinear trapezoid. It is required to calculate the area of ​​a curvilinear trapezoid.
Solution. Geometry gives us recipes for calculating the areas of polygons and some parts of a circle (sector, segment). Using geometric considerations, we can only find an approximate value of the required area, reasoning as follows.

Let's split the segment [a; b] (base of a curved trapezoid) into n equal parts; this partition is carried out using points x 1, x 2, ... x k, ... x n-1. Let us draw straight lines through these points parallel to the y-axis. Then the given curvilinear trapezoid will be divided into n parts, into n narrow columns. The area of ​​the entire trapezoid is equal to the sum of the areas of the columns.

Let us consider the k-th column separately, i.e. a curved trapezoid whose base is a segment. Let's replace it with a rectangle with the same base and height equal to f(x k) (see figure). The area of ​​the rectangle is equal to \(f(x_k) \cdot \Delta x_k \), where \(\Delta x_k \) is the length of the segment; It is natural to consider the resulting product as an approximate value of the area of ​​the kth column.

If we now do the same with all the other columns, we will come to the following result: the area S of a given curvilinear trapezoid is approximately equal to the area S n of a stepped figure made up of n rectangles (see figure):
\(S_n = f(x_0)\Delta x_0 + \dots + f(x_k)\Delta x_k + \dots + f(x_(n-1))\Delta x_(n-1) \)
Here, for the sake of uniformity of notation, we assume that a = x 0, b = x n; \(\Delta x_0 \) - length of the segment, \(\Delta x_1 \) - length of the segment, etc.; in this case, as we agreed above, \(\Delta x_0 = \dots = \Delta x_(n-1) \)

So, \(S \approx S_n \), and this approximate equality is more accurate, the larger n.
By definition, it is believed that the required area of ​​a curvilinear trapezoid is equal to the limit of the sequence (S n):
$$ S = \lim_(n \to \infty) S_n $$

Problem 2 (about moving a point)
Moves in a straight line material point. The dependence of speed on time is expressed by the formula v = v(t). Find the movement of a point over a period of time [a; b].
Solution. If the movement were uniform, then the problem would be solved very simply: s = vt, i.e. s = v(b-a). For uneven movement, you have to use the same ideas on which the solution to the previous problem was based.
1) Divide the time interval [a; b] into n equal parts.
2) Consider a period of time and assume that during this period of time the speed was constant, the same as at time t k. So we assume that v = v(t k).
3) Let’s find the approximate value of the point’s movement over a period of time; we’ll denote this approximate value as s k
\(s_k = v(t_k) \Delta t_k \)
4) Find the approximate value of displacement s:
\(s \approx S_n \) where
\(S_n = s_0 + \dots + s_(n-1) = v(t_0)\Delta t_0 + \dots + v(t_(n-1)) \Delta t_(n-1) \)
5) The required displacement is equal to the limit of the sequence (S n):
$$ s = \lim_(n \to \infty) S_n $$

Let's summarize. Solutions to various problems were reduced to the same mathematical model. Many problems from various fields of science and technology lead to the same model in the process of solution. So this mathematical model need to be specially studied.

The concept of a definite integral

Let us give a mathematical description of the model that was built in the three considered problems for the function y = f(x), continuous (but not necessarily non-negative, as was assumed in the considered problems) on the interval [a; b]:
1) split the segment [a; b] into n equal parts;
2) make up the sum $$ S_n = f(x_0)\Delta x_0 + f(x_1)\Delta x_1 + \dots + f(x_(n-1))\Delta x_(n-1) $$
3) calculate $$ \lim_(n \to \infty) S_n $$

I know mathematical analysis it has been proven that this limit exists in the case of a continuous (or piecewise continuous) function. It is called the definite integral of the function y = f(x) over the segment [a; b] and denoted as follows:
\(\int\limits_a^b f(x) dx \)
The numbers a and b are called the limits of integration (lower and upper, respectively).

Let's return to the tasks discussed above. The definition of area given in Problem 1 can now be rewritten as follows:
\(S = \int\limits_a^b f(x) dx \)
here S is the area of ​​the curved trapezoid shown in the figure above. This is the geometric meaning of the definite integral.

The definition of the displacement s of a point moving in a straight line with a speed v = v(t) over the period of time from t = a to t = b, given in Problem 2, can be rewritten as follows:

Newton - Leibniz formula

First, let's answer the question: what is the connection between the definite integral and the antiderivative?

The answer can be found in Problem 2. On the one hand, the displacement s of a point moving in a straight line with a speed v = v(t) over the period of time from t = a to t = b is calculated by the formula
\(S = \int\limits_a^b v(t) dt \)

On the other hand, the coordinate of a moving point is an antiderivative for speed - let's denote it s(t); this means that the displacement s is expressed by the formula s = s(b) - s(a). As a result we get:
\(S = \int\limits_a^b v(t) dt = s(b)-s(a) \)
where s(t) is the antiderivative of v(t).

The following theorem was proven in the course of mathematical analysis.
Theorem. If the function y = f(x) is continuous on the interval [a; b], then the formula is valid
\(S = \int\limits_a^b f(x) dx = F(b)-F(a) \)
where F(x) is the antiderivative of f(x).

The above formula is usually called the Newton-Leibniz formula in honor of the English physicist Isaac Newton (1643-1727) and the German philosopher Gottfried Leibniz (1646-1716), who obtained it independently of each other and almost simultaneously.

In practice, instead of writing F(b) - F(a), they use the notation \(\left. F(x)\right|_a^b \) (sometimes called double substitution) and, accordingly, rewrite the Newton-Leibniz formula in this way form:
\(S = \int\limits_a^b f(x) dx = \left. F(x)\right|_a^b \)

When calculating a definite integral, first find the antiderivative, and then carry out a double substitution.

Based on the Newton-Leibniz formula, we can obtain two properties of the definite integral.

Property 1. The integral of the sum of functions is equal to the sum of the integrals:
\(\int\limits_a^b (f(x) + g(x))dx = \int\limits_a^b f(x)dx + \int\limits_a^b g(x)dx \)

Property 2. The constant factor can be taken out of the integral sign:
\(\int\limits_a^b kf(x)dx = k \int\limits_a^b f(x)dx \)

Calculating the areas of plane figures using a definite integral

Using the integral, you can calculate the areas not only of curvilinear trapezoids, but also of flat figures more complex type, for example the one shown in the figure. The figure P is limited by straight lines x = a, x = b and graphs of continuous functions y = f(x), y = g(x), and on the segment [a; b] the inequality \(g(x) \leq f(x) \) holds. To calculate the area S of such a figure, we will proceed as follows:
\(S = S_(ABCD) = S_(aDCb) - S_(aABb) = \int\limits_a^b f(x) dx - \int\limits_a^b g(x) dx = \)
\(= \int\limits_a^b (f(x)-g(x))dx \)

So, the area S of a figure bounded by straight lines x = a, x = b and graphs of functions y = f(x), y = g(x), continuous on the segment and such that for any x from the segment [a; b] the inequality \(g(x) \leq f(x) \) is satisfied, calculated by the formula
\(S = \int\limits_a^b (f(x)-g(x))dx \)

Table of indefinite integrals (antiderivatives) of some functions $$ \int 0 \cdot dx = C $$ $$ \int 1 \cdot dx = x+C $$ $$ \int x^n dx = \frac(x^(n +1))(n+1) +C \;\; (n \neq -1) $$ $$ \int \frac(1)(x) dx = \ln |x| +C $$ $$ \int e^x dx = e^x +C $$ $$ \int a^x dx = \frac(a^x)(\ln a) +C \;\; (a>0, \;\; a \neq 1) $$ $$ \int \cos x dx = \sin x +C $$ $$ \int \sin x dx = -\cos x +C $$ $ $ \int \frac(dx)(\cos^2 x) = \text(tg) x +C $$ $$ \int \frac(dx)(\sin^2 x) = -\text(ctg) x +C $$ $$ \int \frac(dx)(\sqrt(1-x^2)) = \text(arcsin) x +C $$ $$ \int \frac(dx)(1+x^2 ) = \text(arctg) x +C $$ $$ \int \text(ch) x dx = \text(sh) x +C $$ $$ \int \text(sh) x dx = \text(ch ) x +C $$

In fact, in order to find the area of ​​a figure, you don’t need that much knowledge of the indefinite and definite integral. The task “calculate the area using a definite integral” always involves constructing a drawing, so it is much more topical issue will be your knowledge and skills in drawing. In this regard, it is useful to refresh your memory of the graphs of the main elementary functions, and, at a minimum, be able to construct a straight line and a hyperbola.

A curved trapezoid is a flat figure bounded by an axis, straight lines, and the graph of a function continuous on a segment that does not change sign on this interval. Let this figure be located not less x-axis:

Then the area of ​​the curvilinear trapezoid is numerically equal to the definite integral. Any definite integral (that exists) has a very good geometric meaning.

From a geometry point of view, the definite integral is AREA.

That is, a certain integral (if it exists) geometrically corresponds to the area of ​​a certain figure. For example, consider the definite integral. The integrand defines a curve on the plane located above the axis (those who wish can make a drawing), and the definite integral itself is numerically equal to area corresponding curved trapezoid.

Example 1

This is a typical assignment statement. First and the most important moment solutions - construction of a drawing. Moreover, the drawing must be constructed CORRECTLY.

When constructing a drawing, I recommend the following order: first, it is better to construct all the straight lines (if any) and only then - parabolas, hyperbolas, and graphs of other functions. It is more profitable to construct graphs of functions point by point.

In this problem, the solution might look like this.
Let's draw the drawing (note that the equation defines the axis):

On the segment, the graph of the function is located above the axis, therefore:

Answer:

After the task is completed, it is always useful to look at the drawing and figure out whether the answer is real. In this case, “by eye” we count the number of cells in the drawing - well, there will be about 9, it seems to be true. It is absolutely clear that if we got, say, the answer: 20 square units, then it is obvious that a mistake was made somewhere - 20 cells obviously do not fit into the figure in question, at most a dozen. If the answer is negative, then the task was also solved incorrectly.

Example 3

Calculate the area of ​​the figure bounded by lines and coordinate axes.

Solution: Let's make a drawing:

If the curved trapezoid is located under the axis (or at least not higher given axis), then its area can be found using the formula:

In this case:

Attention! The two types of tasks should not be confused:

1) If you are asked to solve simply a definite integral without any geometric meaning, then it may be negative.

2) If you are asked to find the area of ​​a figure using a definite integral, then the area is always positive! That is why the minus appears in the formula just discussed.

In practice, most often the figure is located in both the upper and lower half-plane, and therefore, from the simplest school problems we move on to more meaningful examples.

Example 4

Find the area of ​​a plane figure bounded by the lines , .

Solution: First you need to make a drawing. Generally speaking, when constructing a drawing in area problems, we are most interested in the points of intersection of lines. Let's find the intersection points of the parabola and the straight line. This can be done in two ways. The first method is analytical. We solve the equation:

This means that the lower limit of integration is , the upper limit of integration is .

It is better, if possible, not to use this method.

It is much more profitable and faster to construct lines point by point, and the limits of integration become clear “by themselves.” Nevertheless, analytical method finding limits still sometimes has to be used if, for example, the graph is quite large, or the detailed construction did not reveal the limits of integration (they can be fractional or irrational). And we will also consider such an example.

Let's return to our task: it is more rational to first construct a straight line and only then a parabola. Let's make the drawing:

And now the working formula: If on a segment some continuous function is greater than or equal to some continuous function, then the area of ​​the figure limited by the graphs of these functions and straight lines can be found using the formula:

Here you no longer need to think about where the figure is located - above the axis or below the axis, and, roughly speaking, it is important which graph is HIGHER (relative to another graph) and which is BELOW.

In the example under consideration, it is obvious that on the segment the parabola is located above the straight line, and therefore it is necessary to subtract from

The completed solution might look like this:

The desired figure is limited by a parabola above and a straight line below.
On the segment, according to the corresponding formula:

Answer:

Example 4

Calculate the area of ​​the figure bounded by the lines , , , .

Solution: First, let's make a drawing:

The figure whose area we need to find is shaded in blue (look carefully at the condition - how the figure is limited!). But in practice, due to inattention, a “glitch” often occurs that you need to find the area of ​​​​a figure that is shaded in green!

This example is also useful in that it calculates the area of ​​a figure using two definite integrals.

Really :

1) On the segment above the axis there is a graph of a straight line;

2) On the segment above the axis there is a graph of a hyperbola.

It is quite obvious that the areas can (and should) be added, therefore: