Let's move on to consider applications of integral calculus. In this lesson we will analyze the typical and most common problem - how to calculate the area of a plane figure using a definite integral. Finally searching for meaning in higher mathematics - may they find him. 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 Not.
2) Be able to apply the Newton-Leibniz formula and calculate definite integral. You can establish warm friendly relations with definite integrals on the Definite Integral page. Examples of solutions.
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, to be able to construct a straight line, parabola and hyperbola. This can be done (for many, it is necessary) using methodological material and articles on geometric transformations of graphs.
Actually, everyone is familiar with the task of finding the area using a definite integral since school, and we will not go much further from school curriculum. This article might not have existed at all, but the fact is that the problem occurs in 99 cases out of 100, when a student suffers from a hated school and enthusiastically masters a course in higher mathematics.
The materials of this workshop are presented simply, in detail and with a minimum of theory.
Let's start with curved trapezoid.
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. In the lesson Definite Integral. Examples of solutions I 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. 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 - 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 pointwise; 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 draw the drawing (note that the equation defines the axis):
I will not hatch a curved trapezoid, it is obvious here what the area is we're talking about. The solution continues like this:
On the segment, the graph of the function is located above the axis, therefore:
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, 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 2
Calculate the area of a figure bounded by lines , , and axis
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?
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 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 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.” The technique of pointwise construction for various graphs is discussed in detail in the help Graphs and properties of elementary functions. 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: 
I repeat that when constructing pointwise, the limits of integration are most often found out “automatically”.
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 the 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:
In fact, the school formula for the area of a curvilinear trapezoid in the lower half-plane (see simple example No. 3) is a special case of the formula 
. Since the axis is specified by the equation, and the graph of the function is located not higher axes, then 
And now a couple of examples for your own solution
Example 5
Example 6
Find the area of the figure bounded by the lines , .
When solving problems involving calculating area using a definite integral, a funny incident sometimes happens. The drawing was done correctly, the calculations were correct, but due to carelessness... the area of the wrong figure was found, this is exactly how your humble servant went wrong several times. Here real case from life:
Example 7
Calculate the area of the figure bounded by the lines , , , .
Solution: First, let's make a drawing: 
...Eh, the drawing came out crap, but everything seems to be legible.
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 arises that you need to find the area of a figure that is shaded 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:
Answer:
Let's move on to another meaningful task.
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: 
From the drawing it is clear that our upper limit is “good”: .
But what is the lower limit?! It is clear that this is not an integer, but what is it? May be ? But where is the guarantee that the drawing is made with perfect accuracy, it may well turn out that... Or the root. 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 a straight line and a parabola.
To do this, we solve the equation: 
,
Really, .
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: 
Answer: 
Well, to conclude the lesson, let’s look at two more difficult tasks.
Example 9
Calculate the area of the figure bounded by the lines , ,
Solution: Let's depict this figure in the drawing. 
Damn, I forgot to sign the schedule, and, sorry, I didn’t want to redo the picture. Not a drawing day, in short, today is the day =)
For point-by-point construction you need to know appearance sinusoids (and 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 trigonometric table. In some cases (as 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 the segment, the graph of the function is located above the axis, therefore:
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 applied sequentially 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.
We begin to consider the actual process of calculating the double integral and get acquainted with its geometric meaning.
The double integral is numerically equal to the area of the plane figure (the region of integration). This is the simplest form of double integral, when the function of two variables is equal to one: .
Let's first consider the problem in general view. Now you will be quite surprised how simple everything really is! Let's calculate the area of a flat figure bounded by lines. For definiteness, we assume that on the segment . The area of this figure is numerically equal to:
Let's depict the area in the drawing: 
Let's choose the first way to traverse the area: 
Thus: 
And immediately an important technical trick: repeated integrals can be calculated separately. First the inner integral, then the outer integral. This method I highly recommend it to beginners in the subject.
1) Let's calculate the internal integral, and the integration is carried out over the variable “y”: 
Indefinite integral here is the simplest one, 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 representation of the entire solution looks like this:
The resulting formula 
is exactly the working formula for calculating the area of a plane figure using the “ordinary” definite integral! See the lesson Calculating area using a definite integral, there it is at every step!
That is, the problem of calculating area using double integral not much different from the problem of finding the area using a definite integral! In fact, it's the same thing!
Accordingly, no difficulties should arise! I won’t look at very many examples, since you, in fact, have repeatedly encountered this task.
Example 9
Solution: Let's depict the area in the drawing: 
Let us choose the following order of traversal of the area: 
Here and further I will not dwell on how to traverse the area, since very detailed explanations were given in the first paragraph.
Thus: 
As I already noted, it is better for beginners to calculate iterated integrals separately, and I will stick to the same method:
1) First, using the Newton-Leibniz formula, we deal with the internal integral: 
2) The result obtained in the first step is substituted into the external integral: 
Point 2 is actually finding the area of a plane figure using a definite integral.
Answer:
This is such a stupid and naive task.
An interesting example for an independent solution:
Example 10
Using a double integral, calculate the area of a plane figure bounded by the lines , ,
An approximate example of a final solution at the end of the lesson.
In Examples 9-10, it is much more profitable to use the first method of traversing the area; curious readers, by the way, can change the order of traversal and calculate the areas using the second method. If you do not make a mistake, then, naturally, you will get the same area values.
But in some cases, the second method of traversing the area is more effective, and at the end of the young nerd’s course, let’s look at a couple more examples on this topic:
Example 11
Using a double integral, calculate the area of a plane figure bounded by lines,
Solution: we are looking forward to two parabolas with a quirk that lie on their sides. There is no need to smile; similar things occur quite often in multiple integrals.
What is the easiest way to make a drawing?
Let's imagine a parabola in the form of two functions:
– the upper branch and – the lower branch.
Similarly, imagine a parabola in the form of upper and lower 
branches.
Next, point-wise plotting of graphs rules, resulting in such a bizarre figure: 
We calculate the area of the figure using the double integral according to the formula:
What happens if we choose the first method of traversing the area? Firstly, 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-complicated level, but... there is an old mathematical saying: those who are close to their roots do not need a test.
Therefore, from the misunderstanding given in the condition, we express the inverse functions: 
Inverse functions in in this example have the advantage that they specify the entire parabola at once without any leaves, acorns, branches and roots.
According to the second method, the area traversal will be as follows: 
Thus: 
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 confusing; if there were a letter “zy”, it would be great to integrate over it. Although anyone who has read the second paragraph of the lesson How to calculate the volume of a body of rotation no longer experiences the slightest awkwardness with integration using the “Y” method.
Also pay attention to the first step: the integrand is even, and the interval of integration is symmetrical about zero. Therefore, the segment can be halved, and the result can be doubled. This technique is commented in detail in the lesson. Effective methods calculation of a definite integral.
What to add…. All!
Answer:
To test your integration technique, you can try to calculate . The answer should be exactly the same.
Example 12
Using a double integral, calculate the area of a plane figure bounded by lines 
This is an example for you to solve on your own. It is interesting to note that if you try to use the first method of traversing the area, the figure will no longer have to be divided into two, but into three parts! And, accordingly, we get three pairs of repeated 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 a double integral? Examples of solutions. I’ll try not to be so maniacal in the second article =)
I wish you success!
Solutions and answers:
Example 2:Solution:
Let's depict the area on the drawing:
Let us choose the following order of traversal of the area:
Thus: 
Let's move on to inverse functions:
Thus: 
Answer:
Example 4:Solution:
Let's move on to direct functions:
Let's make the drawing:
Let's change the order of traversing the area:
Answer:
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 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 causes you difficulties, you can study the section on basic elementary functions, geometric transformation of function graphs, as well as the construction of graphs during the study of 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 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 brought this here detailed solution only because there are more difficult 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
We will draw a line on the graph with a red line, given 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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Task No. 3. Make a drawing and calculate the area of the figure bounded by the linesApplication of the integral to the solution of applied problems
Area calculation
The definite integral of a continuous non-negative function f(x) is numerically equal to the area of a curvilinear trapezoid bounded by the curve y = f(x), the O x axis and the straight lines x = a and x = b. In accordance with this, the area formula is written as follows:
Let's look at some examples of calculating the areas of plane figures.
Task No. 1. Calculate the area bounded by the lines y = x 2 +1, y = 0, x = 0, x = 2.
Solution. Let's construct a figure whose area we will have to calculate.
y = x 2 + 1 is a parabola whose branches are directed upward, and the parabola is shifted upward by one unit relative to the O y axis (Figure 1).
Figure 1. Graph of the function y = x 2 + 1
Task No. 2. Calculate the area bounded by the lines y = x 2 – 1, y = 0 in the range from 0 to 1.
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Solution. The graph of this function is a parabola of branches that are directed upward, and the parabola is shifted relative to the O y axis downward by one unit (Figure 2).
Figure 2. Graph of the function y = x 2 – 1
Task No. 3. Make a drawing and calculate the area of the figure bounded by the lines
y = 8 + 2x – x 2 and y = 2x – 4.
Solution. The first of these two lines is a parabola with its branches directed downward, since the coefficient of x 2 is negative, and the second line is a straight line intersecting both coordinate axes.
To construct a parabola, we find the coordinates of its vertex: y’=2 – 2x; 2 – 2x = 0, x = 1 – abscissa of the vertex; y(1) = 8 + 2∙1 – 1 2 = 9 is its ordinate, N(1;9) is the vertex.
Now let’s find the intersection points of the parabola and the straight line by solving the system of equations:
Equating the right sides of an equation whose left sides are equal.
We get 8 + 2x – x 2 = 2x – 4 or x 2 – 12 = 0, whence 
.
So, the points are the intersection points of a parabola and a straight line (Figure 1).
Figure 3 Graphs of functions y = 8 + 2x – x 2 and y = 2x – 4
Let's construct a straight line y = 2x – 4. It passes through the points (0;-4), (2;0) on the coordinate axes.
To construct a parabola, you can also use its intersection points with the 0x axis, that is, the roots of the equation 8 + 2x – x 2 = 0 or x 2 – 2x – 8 = 0. Using Vieta’s theorem, it is easy to find its roots: x 1 = 2, x 2 = 4.
Figure 3 shows a figure (parabolic segment M 1 N M 2) bounded by these lines.
The second part of the problem is to find the area of this figure. Its area can be found using a definite integral according to the formula 
.
In relation to this condition, we obtain the integral: 
2 Calculation of the volume of a body of rotation
The volume of the body obtained from the rotation of the curve y = f(x) around the O x axis is calculated by the formula:
When rotating around the O y axis, the formula looks like:
Task No. 4. Determine the volume of the body obtained from the rotation of a curved trapezoid bounded by straight lines x = 0 x = 3 and curve y = around the O x axis.
Solution. Let's draw a picture (Figure 4).
Figure 4. Graph of the function y =
The required volume is 
Task No. 5. Calculate the volume of the body obtained from the rotation of a curved trapezoid bounded by the curve y = x 2 and straight lines y = 0 and y = 4 around the O y axis.
Solution. We have:
Review questions