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. In addition to simplicity, this universal method will help improve the visibility of the site in 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 It has been working for a long time (and, I think, will work forever), but is already morally outdated.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 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 your knowledge and skills in constructing drawings will be a much more pressing question. In this regard, it is useful to refresh your memory of the graphs of basic 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 the area of the corresponding curvilinear trapezoid.
Example 1
This is a typical assignment statement. The first and most important point in the decision is 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 build graphs of functions point by point.
In this problem, the solution might look like this.
Let's complete 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 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, the analytical method of finding limits still sometimes has to be used if, for example, the graph is large enough, 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:
Let's move on to consider applications of integral calculus. In this lesson we will look at the typical and most common problem of calculating the area of a plane figure using a definite integral. Finally, let all those who seek meaning in higher mathematics find it. You never know. In real life, you will have to approximate a dacha plot using elementary functions and find its area using a definite integral.
To successfully master the material, you must:
1) Understand the indefinite integral at least at an intermediate level. Thus, dummies should first familiarize themselves with the lesson of He.
2) Be able to apply the Newton-Leibniz formula and calculate the definite integral. You can establish warm friendly relations with definite integrals on the Definite Integral page. Examples of solutions. The task “calculate the area using a definite integral” always involves constructing a drawing, so your knowledge and drawing skills will also be an important issue. At a minimum, you need to be able to construct a straight line, parabola and hyperbola.
Let's start with a curved trapezoid. A curved trapezoid is a flat figure bounded by the graph of some function y = f(x), axis OX and lines x = a; x = b.
The area of a curvilinear trapezoid is numerically equal to a definite integral
Any definite integral (that exists) has a very good geometric meaning. In the lesson Definite Integral. Examples of solutions we said that a definite integral is a number. And now it’s time to state another useful fact. From the point of view of geometry, the definite integral is AREA. That is, a certain integral (if it exists) geometrically corresponds to the area of a certain figure. Consider the definite integral
Integrand
defines a curve on the plane (it can be drawn if desired), and the definite integral itself is numerically equal to the area of the corresponding curvilinear trapezoid.
Example 1
, , , .
This is a typical assignment statement. The most important point in the decision is the construction of the 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. The technique of pointwise construction can be found in the reference material Graphs and properties of elementary functions. There you can also find very useful material for our lesson - how to quickly build a parabola.
In this problem, the solution might look like this.
Let's do the drawing (note that the equation y= 0 specifies the axis OX):
We will not shade the curved trapezoid; here it is obvious what area we are talking about. The solution continues like this:
On the segment [-2; 1] function graph y = x 2 + 2 located above the axis OX, That's why:
Answer: .
Who has difficulties with calculating the definite integral and applying the Newton-Leibniz formula
,
Refer to the lecture Definite Integral. Examples of solutions. 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, we count the number of cells in the drawing “by eye” - well, there will be about 9, which 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 2
Calculate the area of a figure bounded by lines xy = 4, x = 2, x= 4 and axis OX.
This is an example for you to solve on your own. Full solution and answer at the end of the lesson.
What to do if a curved trapezoid is located under the axis OX?
Example 3
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 curved trapezoid is completely located under the axis OX, 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 lines y = 2x – x 2 , y = -x.
Solution: First you need to make a drawing. 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 y = 2x – x 2 and straight y = -x. This can be done in two ways. The first method is analytical. We solve the equation:
This means that the lower limit of integration a= 0, upper limit of integration b= 3. It is often more profitable and faster to construct lines point by point, and the limits of integration become clear “by themselves.” Nevertheless, the analytical method of finding limits still sometimes has to be used if, for example, the graph is large enough, or the detailed construction did not reveal the limits of integration (they can be fractional or irrational). 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:
Let us repeat that when constructing pointwise, the limits of integration are most often determined “automatically”.
And now the working formula:
If on the segment [ a; b] some continuous function f(x) is greater than or equal to some continuous function g(x), then the area of the corresponding figure 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, but what is important is 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 from 2 x – x 2 must be subtracted – x.
The completed solution might look like this:
The desired figure is limited by a parabola y = 2x – x 2 on top and straight y = -x below.
On segment 2 x – x 2 ≥ -x. According to the corresponding formula:
Answer: .
In fact, the school formula for the area of a curvilinear trapezoid in the lower half-plane (see example No. 3) is a special case of the formula
.
Because the axis OX given by the equation y= 0, and the graph of the function g(x) located below the axis OX, That
.
And now a couple of examples for your own solution
Example 5
Example 6
Find the area of a figure bounded by lines
When solving problems involving calculating area using a definite integral, a funny incident sometimes happens. The drawing was completed correctly, the calculations were correct, but due to carelessness... the area of the wrong figure was found.
Example 7
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, people often decide that they need to find the area of the figure that is shaded in green!
This example is also useful because it calculates the area of a figure using two definite integrals. Really:
1) On the segment [-1; 1] above the axis OX the graph is located straight y = x+1;
2) On a segment above the axis OX the graph of a hyperbola is located y = (2/x).
It is quite obvious that the areas can (and should) be added, therefore:
Answer:
Example 8
Calculate the area of a figure bounded by lines
Let’s present the equations in “school” form
and make a point-by-point drawing:
It is clear from the drawing that our upper limit is “good”: b = 1.
But what is the lower limit?! It is clear that this is not an integer, but what is it?
May be, a=(-1/3)? But where is the guarantee that the drawing is made with perfect accuracy, it may well turn out that a=(-1/4). What if we built the graph incorrectly?
In such cases, you have to spend additional time and clarify the limits of integration analytically.
Let's find the intersection points of the graphs
To do this, we solve the equation:
.
Hence, a=(-1/3).
The further solution is trivial. The main thing is not to get confused in substitutions and signs. The calculations here are not the simplest. On the segment
, 
,
according to the corresponding formula:
To conclude the lesson, let's look at two more difficult tasks.
Example 9
Calculate the area of a figure bounded by lines
Solution: Let's depict this figure in the drawing.
To construct a point-by-point drawing, you need to know the appearance of a sinusoid. In general, it is useful to know the graphs of all elementary functions, as well as some sine values. They can be found in the table of values of trigonometric functions. In some cases (for example, in this case), it is possible to construct a schematic drawing, on which the graphs and limits of integration should be fundamentally correctly displayed.
There are no problems with the limits of integration here; they follow directly from the condition:
– “x” changes from zero to “pi”. Let's make a further decision:
On a segment, the graph of a function y= sin 3 x located above the axis OX, That's why:
(1) You can see how sines and cosines are integrated in odd powers in the lesson Integrals of trigonometric functions. We pinch off one sinus.
(2) We use the main trigonometric identity in the form
(3) Let's change the variable t=cos x, then: is located above the axis, therefore:
.
.
Note: note how the integral of the tangent cubed is taken; a corollary of the basic trigonometric identity is used here
.
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” a more profitable solution option - i.e. understand how in one case or another it will be more convenient to carry out integration? 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 look at different 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 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. Further, 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. In the previous paragraph 3.1 we examined the case when a curved 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. We will consider how to solve such a problem below.
Example 2. Calculate the area of the figure bounded by the 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 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.
In the previous section, devoted to the analysis of the geometric meaning of a definite integral, we received a number of formulas for calculating the area of a curvilinear trapezoid:
Yandex.RTB R-A-339285-1
S (G) = ∫ a b f (x) d x for a continuous and non-negative function y = f (x) on the interval [ a ; b ] ,
S (G) = - ∫ a b f (x) d x for a continuous and non-positive function y = f (x) on the interval [ a ; b ] .
These formulas are applicable to solving relatively simple problems. In reality, we will often have to work with more complex figures. In this regard, we will devote this section to an analysis of algorithms for calculating the area of figures that are limited by functions in explicit form, i.e. like y = f(x) or x = g(y).
TheoremLet the functions y = f 1 (x) and y = f 2 (x) be defined and continuous on the interval [ a ; b ] , and f 1 (x) ≤ f 2 (x) for any value x from [ a ; b ] . Then the formula for calculating the area of the figure G, bounded by the lines x = a, x = b, y = f 1 (x) and y = f 2 (x) will look like S (G) = ∫ a b f 2 (x) - f 1 (x) d x .
A similar formula will be applicable for the area of a figure bounded by the lines y = c, y = d, x = g 1 (y) and x = g 2 (y): S (G) = ∫ c d (g 2 (y) - g 1 (y) d y .
Proof
Let's look at three cases for which the formula will be valid.
In the first case, taking into account the property of additivity of area, the sum of the areas of the original figure G and the curvilinear trapezoid G 1 is equal to the area of the figure G 2. It means that
Therefore, S (G) = S (G 2) - S (G 1) = ∫ a b f 2 (x) d x - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) dx.
We can perform the last transition using the third property of the definite integral.
In the second case, the equality is true: S (G) = S (G 2) + S (G 1) = ∫ a b f 2 (x) d x + - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) d x
The graphic illustration will look like:
If both functions are non-positive, we get: S (G) = S (G 2) - S (G 1) = - ∫ a b f 2 (x) d x - - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) d x . The graphic illustration will look like:
Let's move on to consider the general case when y = f 1 (x) and y = f 2 (x) intersect the O x axis.
We denote the intersection points as x i, i = 1, 2, . . . , n - 1 . These points split the segment [a; b ] into n parts x i - 1 ; x i, i = 1, 2, . . . , n, where α = x 0< x 1 < x 2 < . . . < x n - 1 < x n = b . Фигуру G можно представить объединением фигур G i , i = 1 , 2 , . . . , n . Очевидно, что на своем интервале G i попадает под один из трех рассмотренных ранее случаев, поэтому их площади находятся как S (G i) = ∫ x i - 1 x i (f 2 (x) - f 1 (x)) d x , i = 1 , 2 , . . . , n
Hence,
S (G) = ∑ i = 1 n S (G i) = ∑ i = 1 n ∫ x i x i f 2 (x) - f 1 (x)) d x = = ∫ x 0 x n (f 2 (x) - f ( x)) d x = ∫ a b f 2 (x) - f 1 (x) d x
We can make the last transition using the fifth property of the definite integral.
Let us illustrate the general case on the graph.
The formula S (G) = ∫ a b f 2 (x) - f 1 (x) d x can be considered proven.
Now let's move on to analyzing examples of calculating the area of figures that are limited by the lines y = f (x) and x = g (y).
We will begin our consideration of any of the examples by constructing a graph. The image will allow us to represent complex shapes as unions of simpler shapes. If constructing graphs and figures on them is difficult for you, you can study the section on basic elementary functions, geometric transformation of graphs of functions, as well as constructing graphs while studying a function.
Example 1
It is necessary to determine the area of the figure, which is limited by the parabola y = - x 2 + 6 x - 5 and straight lines y = - 1 3 x - 1 2, x = 1, x = 4.
Solution
Let's draw the lines on the graph in the Cartesian coordinate system.
On the segment [ 1 ; 4 ] the graph of the parabola y = - x 2 + 6 x - 5 is located above the straight line y = - 1 3 x - 1 2. In this regard, to obtain the answer we use the formula obtained earlier, as well as the method of calculating the definite integral using the Newton-Leibniz formula:
S (G) = ∫ 1 4 - x 2 + 6 x - 5 - - 1 3 x - 1 2 d x = = ∫ 1 4 - x 2 + 19 3 x - 9 2 d x = - 1 3 x 3 + 19 6 x 2 - 9 2 x 1 4 = = - 1 3 4 3 + 19 6 4 2 - 9 2 4 - - 1 3 1 3 + 19 6 1 2 - 9 2 1 = = - 64 3 + 152 3 - 18 + 1 3 - 19 6 + 9 2 = 13
Answer: S(G) = 13
Let's look at a more complex example.
Example 2
It is necessary to calculate the area of the figure, which is limited by the lines y = x + 2, y = x, x = 7.
Solution
In this case, we have only one straight line located parallel to the x-axis. This is x = 7. This requires us to find the second limit of integration ourselves.
Let's build a graph and plot on it the lines given in the problem statement.
Having the graph in front of our eyes, we can easily determine that the lower limit of integration will be the abscissa of the point of intersection of the graph of the straight line y = x and the semi-parabola y = x + 2. To find the abscissa we use the equalities:
y = x + 2 O DZ: x ≥ - 2 x 2 = x + 2 2 x 2 - x - 2 = 0 D = (- 1) 2 - 4 1 (- 2) = 9 x 1 = 1 + 9 2 = 2 ∈ O DZ x 2 = 1 - 9 2 = - 1 ∉ O DZ
It turns out that the abscissa of the intersection point is x = 2.
We draw your attention to the fact that in the general example in the drawing, the lines y = x + 2, y = x intersect at the point (2; 2), so such detailed calculations may seem unnecessary. We have provided such a detailed solution here only because in more complex cases the solution may not be so obvious. This means that it is always better to calculate the coordinates of the intersection of lines analytically.
On the interval [ 2 ; 7] the graph of the function y = x is located above the graph of the function y = x + 2. Let's apply the formula to calculate the area:
S (G) = ∫ 2 7 (x - x + 2) d x = x 2 2 - 2 3 · (x + 2) 3 2 2 7 = = 7 2 2 - 2 3 · (7 + 2) 3 2 - 2 2 2 - 2 3 2 + 2 3 2 = = 49 2 - 18 - 2 + 16 3 = 59 6
Answer: S (G) = 59 6
Example 3
It is necessary to calculate the area of the figure, which is limited by the graphs of the functions y = 1 x and y = - x 2 + 4 x - 2.
Solution
Let's plot the lines on the graph.
Let's define the limits of integration. To do this, we determine the coordinates of the points of intersection of the lines by equating the expressions 1 x and - x 2 + 4 x - 2. Provided that x is not zero, the equality 1 x = - x 2 + 4 x - 2 becomes equivalent to the third degree equation - x 3 + 4 x 2 - 2 x - 1 = 0 with integer coefficients. To refresh your memory of the algorithm for solving such equations, we can refer to the section “Solving cubic equations.”
The root of this equation is x = 1: - 1 3 + 4 1 2 - 2 1 - 1 = 0.
Dividing the expression - x 3 + 4 x 2 - 2 x - 1 by the binomial x - 1, we get: - x 3 + 4 x 2 - 2 x - 1 ⇔ - (x - 1) (x 2 - 3 x - 1) = 0
We can find the remaining roots from the equation x 2 - 3 x - 1 = 0:
x 2 - 3 x - 1 = 0 D = (- 3) 2 - 4 · 1 · (- 1) = 13 x 1 = 3 + 13 2 ≈ 3 . 3; x 2 = 3 - 13 2 ≈ - 0 . 3
We found the interval x ∈ 1; 3 + 13 2, in which the figure G is contained above the blue and below the red line. This helps us determine the area of the figure:
S (G) = ∫ 1 3 + 13 2 - x 2 + 4 x - 2 - 1 x d x = - x 3 3 + 2 x 2 - 2 x - ln x 1 3 + 13 2 = = - 3 + 13 2 3 3 + 2 3 + 13 2 2 - 2 3 + 13 2 - ln 3 + 13 2 - - - 1 3 3 + 2 1 2 - 2 1 - ln 1 = 7 + 13 3 - ln 3 + 13 2
Answer: S (G) = 7 + 13 3 - ln 3 + 13 2
Example 4
It is necessary to calculate the area of the figure, which is limited by the curves y = x 3, y = - log 2 x + 1 and the abscissa axis.
Solution
Let's plot all the lines on the graph. We can get the graph of the function y = - log 2 x + 1 from the graph y = log 2 x if we position it symmetrically about the x-axis and move it up one unit. The equation of the x-axis is y = 0.
Let us mark the points of intersection of the lines.
As can be seen from the figure, the graphs of the functions y = x 3 and y = 0 intersect at the point (0; 0). This happens because x = 0 is the only real root of the equation x 3 = 0.
x = 2 is the only root of the equation - log 2 x + 1 = 0, so the graphs of the functions y = - log 2 x + 1 and y = 0 intersect at the point (2; 0).
x = 1 is the only root of the equation x 3 = - log 2 x + 1 . In this regard, the graphs of the functions y = x 3 and y = - log 2 x + 1 intersect at the point (1; 1). The last statement may not be obvious, but the equation x 3 = - log 2 x + 1 cannot have more than one root, since the function y = x 3 is strictly increasing, and the function y = - log 2 x + 1 is strictly decreasing.
The further solution involves several options.
Option #1
We can imagine the figure G as the sum of two curvilinear trapezoids located above the x-axis, the first of which is located below the midline on the segment x ∈ 0; 1, and the second is below the red line on the segment x ∈ 1; 2. This means that the area will be equal to S (G) = ∫ 0 1 x 3 d x + ∫ 1 2 (- log 2 x + 1) d x .
Option No. 2
Figure G can be represented as the difference of two figures, the first of which is located above the x-axis and below the blue line on the segment x ∈ 0; 2, and the second between the red and blue lines on the segment x ∈ 1; 2. This allows us to find the area as follows:
S (G) = ∫ 0 2 x 3 d x - ∫ 1 2 x 3 - (- log 2 x + 1) d x
In this case, to find the area you will have to use a formula of the form S (G) = ∫ c d (g 2 (y) - g 1 (y)) d y. In fact, the lines that bound the figure can be represented as functions of the argument y.
Let's solve the equations y = x 3 and - log 2 x + 1 with respect to x:
y = x 3 ⇒ x = y 3 y = - log 2 x + 1 ⇒ log 2 x = 1 - y ⇒ x = 2 1 - y
We get the required area:
S (G) = ∫ 0 1 (2 1 - y - y 3) d y = - 2 1 - y ln 2 - y 4 4 0 1 = = - 2 1 - 1 ln 2 - 1 4 4 - - 2 1 - 0 ln 2 - 0 4 4 = - 1 ln 2 - 1 4 + 2 ln 2 = 1 ln 2 - 1 4
Answer: S (G) = 1 ln 2 - 1 4
Example 5
It is necessary to calculate the area of the figure, which is limited by the lines y = x, y = 2 3 x - 3, y = - 1 2 x + 4.
Solution
With a red line we plot the line defined by the function y = x. We draw the line y = - 1 2 x + 4 in blue, and the line y = 2 3 x - 3 in black.
Let's mark the intersection points.
Let's find the intersection points of the graphs of the functions y = x and y = - 1 2 x + 4:
x = - 1 2 x + 4 O DZ: x ≥ 0 x = - 1 2 x + 4 2 ⇒ x = 1 4 x 2 - 4 x + 16 ⇔ x 2 - 20 x + 64 = 0 D = (- 20) 2 - 4 1 64 = 144 x 1 = 20 + 144 2 = 16 ; x 2 = 20 - 144 2 = 4 Check: x 1 = 16 = 4, - 1 2 x 1 + 4 = - 1 2 16 + 4 = - 4 ⇒ x 1 = 16 not Is the solution to the equation x 2 = 4 = 2, - 1 2 x 2 + 4 = - 1 2 4 + 4 = 2 ⇒ x 2 = 4 is the solution to the equation ⇒ (4; 2) point of intersection i y = x and y = - 1 2 x + 4
Let's find the intersection point of the graphs of the functions y = x and y = 2 3 x - 3:
x = 2 3 x - 3 O DZ: x ≥ 0 x = 2 3 x - 3 2 ⇔ x = 4 9 x 2 - 4 x + 9 ⇔ 4 x 2 - 45 x + 81 = 0 D = (- 45 ) 2 - 4 4 81 = 729 x 1 = 45 + 729 8 = 9, x 2 45 - 729 8 = 9 4 Check: x 1 = 9 = 3, 2 3 x 1 - 3 = 2 3 9 - 3 = 3 ⇒ x 1 = 9 is the solution to the equation ⇒ (9 ; 3) point a s y = x and y = 2 3 x - 3 x 2 = 9 4 = 3 2, 2 3 x 1 - 3 = 2 3 9 4 - 3 = - 3 2 ⇒ x 2 = 9 4 There is no solution to the equation
Let's find the point of intersection of the lines y = - 1 2 x + 4 and y = 2 3 x - 3:
1 2 x + 4 = 2 3 x - 3 ⇔ - 3 x + 24 = 4 x - 18 ⇔ 7 x = 42 ⇔ x = 6 - 1 2 6 + 4 = 2 3 6 - 3 = 1 ⇒ (6 ; 1) point of intersection y = - 1 2 x + 4 and y = 2 3 x - 3
Method No. 1
Let us imagine the area of the desired figure as the sum of the areas of individual figures.
Then the area of the figure is:
S (G) = ∫ 4 6 x - - 1 2 x + 4 d x + ∫ 6 9 x - 2 3 x - 3 d x = = 2 3 x 3 2 + x 2 4 - 4 x 4 6 + 2 3 x 3 2 - x 2 3 + 3 x 6 9 = = 2 3 6 3 2 + 6 2 4 - 4 6 - 2 3 4 3 2 + 4 2 4 - 4 4 + + 2 3 9 3 2 - 9 2 3 + 3 9 - 2 3 6 3 2 - 6 2 3 + 3 6 = = - 25 3 + 4 6 + - 4 6 + 12 = 11 3
Method No. 2
The area of the original figure can be represented as the sum of two other figures.
Then we solve the equation of the line relative to x, and only after that we apply the formula for calculating the area of the figure.
y = x ⇒ x = y 2 red line y = 2 3 x - 3 ⇒ x = 3 2 y + 9 2 black line y = - 1 2 x + 4 ⇒ x = - 2 y + 8 s i n i a l i n e
So the area is:
S (G) = ∫ 1 2 3 2 y + 9 2 - - 2 y + 8 d y + ∫ 2 3 3 2 y + 9 2 - y 2 d y = = ∫ 1 2 7 2 y - 7 2 d y + ∫ 2 3 3 2 y + 9 2 - y 2 d y = = 7 4 y 2 - 7 4 y 1 2 + - y 3 3 + 3 y 2 4 + 9 2 y 2 3 = 7 4 2 2 - 7 4 2 - 7 4 1 2 - 7 4 1 + + - 3 3 3 + 3 3 2 4 + 9 2 3 - - 2 3 3 + 3 2 2 4 + 9 2 2 = = 7 4 + 23 12 = 11 3
As you can see, the values are the same.
Answer: S (G) = 11 3
ResultsTo find the area of a figure that is limited by given lines, we need to construct lines on a plane, find their intersection points, and apply the formula to find the area. In this section, we examined the most common variants of tasks.
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