How to paste mathematical formulas to the 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 into the site in the form of pictures that Wolfram Alpha automatically generates. In addition to simplicity, this universal way will help improve the visibility of the site in search engines. It has been working for a long time (and I think it will work forever), but it is morally outdated.

If you are constantly using math formulas on your site, then I recommend you use MathJax, a special JavaScript library that displays math 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 site, which will be automatically loaded from a remote server at the right time (list of servers); (2) upload the MathJax script from a remote server to your server and connect it to all pages of your site. The second method is more complex and time consuming and will allow you to speed up the loading of the pages of your site, 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 within 5 minutes you will be able to use all the features of MathJax on your website.

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

One of these code options needs to be copied and pasted into the code of your web page, preferably between the tags and or right after the tag . According to the first option, MathJax loads faster and slows down the page less. But the second option automatically tracks and loads the latest versions of MathJax. If you insert the first code, then it will need to be updated periodically. If you paste the second code, then 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 load code above into it, and place the widget closer to the beginning of the template (by the way, this is not necessary at all , since the MathJax script is loaded asynchronously). That's all. Now learn the MathML, LaTeX, and ASCIIMathML markup syntax and you're ready to embed math formulas into your web pages.

Any fractal is built 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. It turns out a set consisting of 20 remaining smaller cubes. Doing the same with each of these cubes, we get a set consisting of 400 smaller cubes. Continuing this process indefinitely, we get the Menger sponge.

We begin to consider the actual process of calculating the double integral and get acquainted with its geometric meaning.

Double integral numerically equal to area plane figure (regions of integration). This is the simplest form of the double integral, when the function of two variables is equal to one: .

Let us first consider the problem in general terms. Now you will be surprised how simple it really is! Calculate the area of ​​a flat figure, bounded by lines. For definiteness, we assume that on the interval . The area of ​​this figure is numerically equal to:

Let's depict the area in the drawing:

Let's choose the first way to bypass the area:

In this way:

And immediately an important technical trick: iterated integrals can be considered separately. First the inner integral, then the outer integral. This method is highly recommended for beginners in the topic teapots.

1) Calculate the internal integral, while the integration is carried out over the variable "y":

The indefinite integral here is the simplest, and then the banal Newton-Leibniz formula is used, with the only difference that the limits of integration are not numbers, but functions. First, we substituted the upper limit into the “y” (antiderivative function), then the lower limit

2) The result obtained in the first paragraph must be substituted into the external integral:

A more compact notation for the whole solution looks like this:

The resulting formula - this is exactly the working formula for calculating the area of ​​\u200b\u200ba flat figure using the "ordinary" definite integral! See lesson Calculating area using a definite integral, there she is at every turn!

That is, the problem of calculating the area using a double integral little different from the problem of finding the area using a definite integral! In fact, they are one and the same!

Accordingly, no difficulties should arise! I will not consider very many examples, since you, in fact, have repeatedly encountered this problem.

Example 9

Solution: Let's depict the area in the drawing:

Let's choose the following order of traversal of the region:

Here and below, I won't go into how to traverse an area because the first paragraph was very detailed.

In this way:

As I already noted, it is better for beginners to calculate iterated integrals separately, I will adhere to the same method:

1) First, using the Newton-Leibniz formula, we deal with the internal integral:

2) The result obtained at the first step is substituted into the outer integral:

Point 2 is actually finding the area of ​​a flat figure using a definite integral.

Answer:

Here is such a stupid and naive task.

A curious example for an independent solution:

Example 10

Using the double integral, calculate the area of ​​a plane figure bounded by the lines , ,

An example of a final solution at the end of the lesson.

In Examples 9-10, it is much more profitable to use the first way to bypass the area, curious readers, by the way, can change the order of the bypass and calculate the areas in the second way. If you do not make a mistake, then, naturally, the same area values ​​\u200b\u200bare obtained.

But in some cases, the second way to bypass the area is more effective, and in conclusion of the young nerd's course, let's look at a couple more examples on this topic:

Example 11

Using the double integral, calculate the area of ​​a plane figure bounded by lines.

Solution: we are looking forward to two parabolas with a breeze that lie on their side. No need to smile, similar things in multiple integrals are often encountered.

What is the easiest way to make a drawing?

Let's represent the parabola as two functions:
- upper branch and - lower branch.

Similarly, imagine a parabola as an upper and lower branches.

Next, point-by-point plotting drives, resulting in such a bizarre figure:

The area of ​​the figure is calculated using the double integral according to the formula:

What happens if we choose the first way to bypass the area? First, this area will have to be divided into two parts. And secondly, we will observe this sad picture: . Integrals, of course, are not of a super-complex level, but ... there is an old mathematical saying: whoever is friendly with the roots does not need a set-off.

Therefore, from the misunderstanding that is given in the condition, we express the inverse functions:

Inverse functions in this example, they have the advantage that they immediately set the entire parabola without any leaves, acorns, branches and roots.

According to the second method, the area traversal will be as follows:

In this way:

As they say, feel the difference.

1) We deal with the internal integral:

We substitute the result into the outer integral:

Integration over the variable "y" should not be embarrassing, if there was a letter "zyu" - it would be great to integrate over it. Although who read the second paragraph of the lesson How to calculate the volume of a body of revolution, he no longer experiences the slightest embarrassment with integration over "y".

Also pay attention to the first step: the integrand is even, and the integration segment is symmetric about zero. Therefore, the segment can be halved, and the result can be doubled. This technique is commented on in detail in the lesson. Effective Methods calculation of a definite integral.

What to add…. Everything!

Answer:

To test your integration technique, you can try to calculate . The answer should be exactly the same.

Example 12

Using the double integral, calculate the area of ​​a plane figure bounded by lines

This is a do-it-yourself example. It is interesting to note that if you try to use the first way to bypass the area, then the figure will no longer be divided into two, but into three parts! And, accordingly, we get three pairs of iterated integrals. Sometimes it happens.

The master class has come to an end, and it's time to move on to the grandmaster level - How to calculate the double integral? Solution examples. I'll try not to be so manic in the second article =)

Wish you success!

Solutions and answers:

Example 2:Solution: Draw an area on the drawing:

Let's choose the following order of traversal of the region:

In this way:
Let's move on to inverse functions:


In this way:
Answer:

Example 4:Solution: Let's move on to direct functions:


Let's execute the drawing:

Let's change the order of traversal of the area:

Answer:

In fact, in order to find the area of ​​\u200b\u200ba figure, you do not need so much knowledge of the indefinite and definite integral. The task "calculate the area using a definite integral" always involves the construction of a drawing, so much more topical issue will be your knowledge and drawing skills. In this regard, it is useful to refresh the memory of the graphs of the main elementary functions, and, at a minimum, be able to build a straight line, and a hyperbola.

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

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

In terms of geometry, the definite integral is the AREA.

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

Example 1

This is a typical task statement. First and crucial point solutions - building a drawing. Moreover, the drawing must be built RIGHT.

When building a blueprint, I recommend the following order: first it is better to construct all lines (if any) and only after- parabolas, hyperbolas, graphs of other functions. Function graphs are more profitable to build pointwise.

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

On the segment, the graph of the function is located over axis, that's why:

Answer:

After the task is completed, it is always useful to look at the drawing and figure out if the answer is real. AT this case“By eye” we count the number of cells in the drawing - well, about 9 will be typed, it seems to be true. It is quite clear that if we had, say, the answer: 20 square units, then, obviously, a mistake was made somewhere - 20 cells clearly do not fit into the figure in question, at most a dozen. If the answer turned out to be 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 curvilinear trapezoid is located under axle(or at least not higher given axis), then its area can be found by the formula:

In this case:

Attention! Don't confuse the two types of tasks:

1) If you are asked to solve just a definite integral without any geometric meaning, then it can 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 considered.

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

Example 4

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

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

Hence, the lower limit of integration , the upper limit of integration .

It is best not to use this method if possible..

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

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

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

Here it is no longer necessary to think where the figure is located - above the axis or below the axis, and, roughly speaking, it matters which chart is ABOVE(relative to another graph), and which one 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 completion of the solution might look like this:

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

Answer:

Example 4

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

Solution: Let's make a drawing first:

The figure whose area we need to find is shaded in blue.(carefully look 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 ​​\u200b\u200bthe figure that is shaded in green!

This example is also useful in that in it the area of ​​\u200b\u200bthe figure is calculated using two definite integrals.

Really:

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

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

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

a)

Solution.

The first and most important moment of the decision is the construction of a drawing.

Let's make a drawing:

The equation y=0 sets the x-axis;

- x=-2 and x=1 - straight, parallel to the axis OU;

- y \u003d x 2 +2 - a parabola whose branches are directed upwards, with a vertex at the point (0;2).

Comment. To construct a parabola, it is enough to find the points of its intersection with the coordinate axes, i.e. putting x=0 find the intersection with the axis OU and deciding the appropriate quadratic equation, find the intersection with the axis Oh .

The vertex of a parabola can be found using the formulas:

You can draw lines and point by point.

On the interval [-2;1] the graph of the function y=x 2 +2 located over axis Ox , that's why:

Answer: S \u003d 9 square units

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

What to do if the curvilinear trapezoid is located under axle Oh?

b) Calculate the area of ​​a figure bounded by lines y=-e x , x=1 and coordinate axes.

Solution.

Let's make a drawing.

If a curvilinear trapezoid completely under the axle Oh , then its area can be found by the formula:

Answer: S=(e-1) sq. unit" 1.72 sq. unit

Attention! Don't confuse the two types of tasks:

1) If you are asked to solve just a definite integral without any geometric meaning, then it can 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 considered.

In practice, most often the figure is located in both the upper and lower half-planes.

With) Find the area of ​​a plane figure bounded by lines y \u003d 2x-x 2, y \u003d -x.

Solution.

First you need to make a drawing. Generally speaking, when constructing a drawing in area problems, we are most interested in the intersection points of lines. Find the intersection points of the parabola and direct This can be done in two ways. The first way is analytical.

We solve the equation:

So the lower limit of integration a=0 , the upper limit of integration b=3 .

We build the given lines: 1. Parabola - vertex at the point (1;1); axis intersection Oh - points(0;0) and (0;2). 2. Straight line - the bisector of the 2nd and 4th coordinate angles. And now Attention! If on the interval [ a;b] some continuous function f(x) greater than or equal to some continuous function g(x), then the area of ​​the corresponding figure can be found by the formula: . And it does not matter where the figure is located - above the axis or below the axis, but it is important which chart is HIGHER (relative to another chart), and which one 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

It is possible to construct lines point by point, while the limits of integration are found out as if "by themselves". Nevertheless, the analytical method of finding the limits still sometimes has to be used if, for example, the graph is large enough, or the threaded construction did not reveal the limits of integration (they can be fractional or irrational).

The desired figure is limited by a parabola from above and a straight line from below.

On the segment , according to the corresponding formula:

Answer: S \u003d 4.5 sq. units

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 the study of certain integrals has just been completed and it is time to start the geometric interpretation of the knowledge gained in practice.

So, what is required to successfully solve the problem of finding the area of ​​\u200b\u200ba figure using integrals:

• Ability to correctly draw drawings;
• Ability to solve a definite integral using the well-known Newton-Leibniz formula;
• The ability to "see" a more profitable solution - i.e. to understand how in this or that case it will be more convenient to carry out the integration? Along the x-axis (OX) or y-axis (OY)?
• Well, where 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 piece of paper in a cage, on a large scale. We sign with a pencil above each graph the name of this function. The signature of the graphs is done solely for the convenience of further calculations. Having received the graph of the desired figure, in most cases it will be immediately clear which integration limits 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 integration limits are not explicitly set, then we find the intersection points of the graphs with each other, and see if our graphical solution matches the analytical one.

3. Next, you need to analyze the drawing. Depending on how the graphs of functions are located, there are different approaches to finding the area of ​​\u200b\u200bthe figure. Consider various examples of finding the area of ​​​​a figure using integrals.

3.1. The most classic and simplest version of the problem is when you need to find the area of ​​a curvilinear trapezoid. What is a curvilinear trapezoid? This is a flat figure bounded by the x-axis (y=0), straight x = a, x = b and any curve continuous on the interval from a before b. At the same time, this figure is non-negative and is located not lower than the x-axis. In this case, the area of ​​the curvilinear trapezoid is numerically equal to the definite integral calculated using the Newton-Leibniz formula:

Example 1 y = x2 - 3x + 3, x = 1, x = 3, y = 0.

What lines define the figure? We have a parabola y = x2 - 3x + 3, which is located above the axis OH, it is non-negative, because all points of this parabola are positive. Next, given straight lines x = 1 and x = 3 that run parallel to the axis OU, are the bounding lines of the figure on the left and right. Well y = 0, she is the x-axis, which limits the figure from below. The resulting figure is shaded, as seen in the figure on the left. In this case, you can immediately begin to solve the problem. Before us is a simple example of a curvilinear trapezoid, which we then solve using the Newton-Leibniz formula.

3.2. In the previous paragraph 3.1, the case was analyzed when the curvilinear trapezoid is located above the x-axis. Now consider the case when the conditions of the problem are the same, except that the function lies under the x-axis. A minus is added to the standard Newton-Leibniz formula. How to solve such a problem, we will consider further.

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

In this example, we have a parabola y=x2+6x+2, which originates from under the axis OH, straight x=-4, x=-1, y=0. Here y = 0 limits the desired figure from above. Direct x = -4 and x = -1 these are the boundaries within which the definite integral will be calculated. The principle of solving the problem of finding the area of ​​\u200b\u200ba 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 does not positive mean? As can be seen from the figure, the figure that lies within the given x has exclusively "negative" coordinates, which is what we need to see and remember when solving the problem. We are looking for the area of ​​\u200b\u200bthe figure using the Newton-Leibniz formula, only with a minus sign at the beginning.

The article is not completed.