How to insert mathematical formulas on the site?
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 method 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 It has been working for a long time (and I think it will work forever), but it is morally outdated.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.
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.
We now turn to the consideration of applications of the integral calculus. In this lesson, we will analyze a typical and most common task. calculating the area of a flat figure using a definite integral. Finally, all those who seek meaning in higher mathematics - may they find it. You never know. In real life, you will have to approximate a summer cottage with elementary functions and find its area using a certain integral.
To successfully master the material, you must:
1) Understand the indefinite integral at least at an intermediate level. Thus, dummies should first read the lesson Not.
2) Be able to apply the Newton-Leibniz formula and calculate the definite integral. You can establish warm friendly relations with certain integrals on the page Definite integral. Solution examples. The task "calculate the area using a definite integral" always involves the construction of a drawing, therefore, your knowledge and drawing skills will also be an urgent issue. At a minimum, one must be able to build a straight line, a parabola and a hyperbola.
Let's start with a curvilinear trapezoid. A curvilinear 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 certain integral
Any definite integral (that exists) has a very good geometric meaning. On the lesson Definite integral. Solution examples 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 the AREA. That is, the definite integral (if it exists) geometrically corresponds to the area of some 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 task statement. The most important point of the decision is the construction of 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. The point-by-point construction technique can be found in the reference material Graphs and properties of elementary functions. There you can also find material that is very useful in relation to our lesson - how to quickly build a parabola.
In this problem, the solution might look like this.
Let's make a drawing (note that the equation y= 0 specifies the axis OX):
We will not hatch the curvilinear trapezoid, it is obvious what area we are talking about here. The solution continues like this:
On the interval [-2; 1] function graph y = x 2 + 2 located over axisOX, that's why:
Answer: .
Who has difficulty calculating the definite integral and applying the Newton-Leibniz formula
,
refer to the lecture Definite integral. Solution examples. 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 obviously 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 2
Calculate the area of a figure bounded by lines xy = 4, x = 2, x= 4 and axis OX.
This is a do-it-yourself example. Full solution and answer at the end of the lesson.
What to do if the curvilinear trapezoid is located under axleOX?
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 curvilinear trapezoid completely under the axle OX , then its area can be found by the formula:
In this case:
.
Attention! The two types of tasks should not be confused:
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 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 intersection points of lines. Find the intersection points of the parabola y = 2x – x 2 and straight y = -x. This can be done in two ways. The first way is analytical. We solve the equation:
So 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, 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). 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:
We repeat that in pointwise construction, the limits of integration are most often found out “automatically”.
And now the working formula:
If on the segment [ a; b] some continuous function f(x) greater than or equal some continuous function g(x), then the area of the corresponding figure 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, but 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 from 2 x – x 2 must be subtracted - x.
The completion of the solution might look like this:
The desired figure is limited by a parabola y = 2x – x 2 top and straight y = -x from 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
.
Since the axis OX is given by the equation y= 0, and the graph of the function g(x) is located below the axis OX, then
.
And now a couple of examples for an independent solution
Example 5
Example 6
Find the area of a figure bounded by lines
In the course of solving problems for calculating the area using a certain integral, a funny incident sometimes happens. The drawing was made correctly, the calculations were correct, but, due to inattention, ... found the area of the wrong figure.
Example 7
Let's draw 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, they often decide that they 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 [-1; 1] above axle OX the graph is straight y = x+1;
2) On the segment above the axis OX the graph of the 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 the "school" form
and do the line drawing:
It can be seen 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?
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 didn't get the graph right at all?
In such cases, one has to spend additional time and refine the limits of integration analytically.
Find the intersection points of the graphs
To do this, we solve the equation:
.
Consequently, 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 easiest. On the segment
, 
,
according to the corresponding formula:
Answer: 
In conclusion of the lesson, we will consider two tasks more difficult.
Example 9
Calculate the area of a figure bounded by lines
Solution: Draw this figure in the drawing.
To draw a drawing point by point, you need to know the appearance of the sinusoid. In general, it is useful to know the graphs of all elementary functions, as well as some values of the sine. They can be found in the table of values trigonometric functions. In some cases (for example, in this case), it is allowed to construct a schematic drawing, on which graphs and integration limits must be displayed in principle correctly.
There are no problems with the integration limits here, they follow directly from the condition:
- "x" changes from zero to "pi". We make a further decision:
On the segment, the graph of the 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 sine.
(2) We use the basic trigonometric identity in the form
(3) Let us change the variable t= cos x, then: located above the axis , so:
.
.
Note: note how the integral of the tangent in the cube is taken, here the consequence of the basic trigonometric identity is used
.
In the previous section, devoted to the analysis of the geometric meaning of a definite integral, we obtained 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 segment [ a ; b] ,
S (G) = - ∫ a b f (x) d x for a continuous and non-positive function y = f (x) on the segment [ a ; b] .
These formulas are applicable for solving relatively simple problems. In fact, we often have to work with more complex shapes. In this regard, we will devote this section to the analysis of algorithms for calculating the area of figures, which are limited by functions in an 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 segment [ a ; b ] , and f 1 (x) ≤ f 2 (x) for any value x from [ a ; b] . Then the formula for calculating the area of \u200b\u200ba figure Gbounded by lines x \u003d a, x \u003d b, y \u003d f 1 (x) and y \u003d f 2 (x) will look like S (G) \u003d ∫ a b f 2 (x) - f 1 (x) d x .
A similar formula will be applicable for the area of \u200b\u200bthe figure bounded by the lines y \u003d c, y \u003d d, x \u003d g 1 (y) and x \u003d g 2 (y): S (G) \u003d ∫ c d (g 2 (y) - g 1 (y) d y .
Proof
We will analyze three cases for which the formula will be valid.
In the first case, taking into account the additivity property of the 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)) d x .
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 the consideration of the general case when y = f 1 (x) and y = f 2 (x) intersect the axis O x .
We will denote the intersection points as x i , i = 1 , 2 , . . . , n - 1 . These points break 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
Consequently,
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.
And now let's move on to the analysis of examples of calculating the area of \u200b\u200bfigures that are limited by the lines y \u003d f (x) and x \u003d g (y) .
Considering any of the examples, we will begin with the construction of a graph. The image will allow us to represent complex shapes as combinations of simpler shapes. If plotting graphs and shapes on them is difficult for you, you can study the section on basic elementary functions, geometric transformation of graphs of functions, as well as plotting during the study of a function.
Example 1
It is necessary to determine the area of \u200b\u200bthe figure, which is limited by the parabola y \u003d - x 2 + 6 x - 5 and straight lines y \u003d - 1 3 x - 1 2, x \u003d 1, x \u003d 4.
Solution
Let's plot the lines on the graph in the Cartesian coordinate system.
On the interval [ 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 an answer, we use the formula obtained earlier, as well as the method for calculating a 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 parallel to the x-axis. This is x = 7 . This requires us to find the second integration limit ourselves.
Let's build a graph and put on it the lines given in the condition of the problem.
Having a graph in front of our eyes, we can easily determine that the lower limit of integration will be the abscissa of the intersection point of the graph with a straight line y \u003d x and a semi-parabola y \u003d 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 D G x 2 = 1 - 9 2 = - 1 ∉ O D G
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 redundant. 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 better to always 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 . 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 \u200b\u200bthe figure, which is limited by the graphs of the functions y \u003d 1 x and y \u003d - x 2 + 4 x - 2.
Solution
Let's draw 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 equal to zero, the equality 1 x \u003d - x 2 + 4 x - 2 becomes equivalent to the equation of the third degree - x 3 + 4 x 2 - 2 x - 1 \u003d 0 with integer coefficients. You can refresh the memory of the algorithm for solving such equations by referring to the section “Solution of 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 \u003d 3 - 13 2 ≈ - 0. 3
We have found an interval x ∈ 1; 3 + 13 2 , where G is enclosed above the blue line and below the red line. This helps us determine the area of the shape:
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) \u003d 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 \u003d x 3, y \u003d - log 2 x + 1 and the x-axis.
Solution
Let's put 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 place it symmetrically about the x-axis and move it up one unit. The equation of the x-axis y \u003d 0.
Let's denote the points of intersection of the lines.
As can be seen from the figure, the graphs of the functions y \u003d x 3 and y \u003d 0 intersect at the point (0; 0) . This is because x \u003d 0 is the only real root of the equation x 3 \u003d 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 \u003d x 3 and y \u003d - log 2 x + 1 intersect at the point (1; 1) . The last statement may not be obvious, but the equation x 3 \u003d - log 2 x + 1 cannot have more than one root, since the function y \u003d x 3 is strictly increasing, and the function y \u003d - log 2 x + 1 is strictly decreasing.
The next step involves several options.
Option number 1
We can represent the figure G as the sum of two curvilinear trapezoids located above the abscissa axis, the first of which is located below the midline on the segment x ∈ 0; 1 , and the second one 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 number 2
The 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 one is between the red and blue lines on the segment x ∈ 1 ; 2. This allows us to find the area like this:
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) \u003d ∫ c d (g 2 (y) - g 1 (y)) d y. In fact, the lines that bound the shape can be represented as functions of the y argument.
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 \u003d x, y \u003d 2 3 x - 3, y \u003d - 1 2 x + 4.
Solution
Draw a line on the chart with a red line, given by the function y = x . Draw the line y = - 1 2 x + 4 in blue, and mark the line y = 2 3 x - 3 in black.
Note the intersection points.
Find the intersection points of the graphs of 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 \u003d 144 x 1 \u003d 20 + 144 2 \u003d 16; x 2 = 20 - 144 2 = 4 i 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
Find the intersection point of the graphs of 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 \u003d 2 3 9 - 3 \u003d 3 ⇒ x 1 \u003d 9 is the solution to the equation ⇒ (9; 3) point and intersection 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 is not a solution to the equation
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 number 1
We represent 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 number 2
The area of the original figure can be represented as the sum of the other two figures.
Then we solve the line equation for x, and only after that we apply the formula for calculating the area of \u200b\u200bthe 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 i l i n i i
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 match.
Answer: S (G) = 11 3
Results
To find the area of a figure that is bounded by given lines, we need to draw lines on a plane, find their intersection points, and apply the formula for finding the area. In this section, we have reviewed the most common options for tasks.
If you notice a mistake in the text, please highlight it and press Ctrl+Enter
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 solving the corresponding 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 .
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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 segment [ 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
