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: .
First, let's look at the problem in general form. 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 technique: iterated integrals can be calculated separately. First the inner integral, then the outer integral. I highly recommend this method to beginners in the subject.
1) Let's calculate the internal integral, and the integration is carried out over the variable “y”: 
The indefinite integral here is the simplest, and then the banal Newton-Leibniz formula is used, with the only difference that the limits of integration are not numbers, but functions. First, we substituted the upper limit into the “y” (antiderivative function), then the lower limit
2) The result obtained in the first paragraph must be substituted into the external integral: 
A more compact 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! Watch the lesson Calculating area using a definite integral, there she is at every step!
That is, 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 this 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 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 who read the second paragraph of the lesson How to calculate the volume of a body of revolution, he no longer experiences the slightest awkwardness with integration according to 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. This also happens.
The master class has come to an end, and it’s time to move on to the grandmaster level - How to calculate 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 on parsing geometric meaning 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. This 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 causes you difficulties, you can study the section on basic elementary functions, geometric transformation of graphs of functions, and also 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
Results
To 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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Calculating the area of a figure- This is perhaps one of the most difficult problems in area theory. In school geometry, they are taught to find the areas of basic geometric shapes such as, for example, a triangle, rhombus, rectangle, trapezoid, circle, etc. However, you often have to deal with calculating the areas of more complex figures. It is when solving such problems that it is very convenient to use integral calculus.
Definition.
Curvilinear trapezoid call some figure G bounded by the lines y = f(x), y = 0, x = a and x = b, and the function f(x) is continuous on the segment [a; b] and does not change its sign on it (Fig. 1). The area of a curved trapezoid can be denoted by S(G).
The definite integral ʃ a b f(x)dx for the function f(x), which is continuous and non-negative on the interval [a; b], and is the area of the corresponding curved trapezoid.
That is, to find the area of a figure G bounded by the lines y = f(x), y = 0, x = a and x = b, it is necessary to calculate the definite integral ʃ a b f(x)dx.
Thus, S(G) = ʃ a b f(x)dx.
If the function y = f(x) is not positive on [a; b], then the area of a curvilinear trapezoid can be found using the formula S(G) = -ʃ a b f(x)dx.
Example 1.
Calculate the area of the figure bounded by the lines y = x 3; y = 1; x = 2.
Solution.
The given lines form the figure ABC, which is shown by hatching in rice. 2.
The required area is equal to the difference between the areas of the curved trapezoid DACE and the square DABE.
Using the formula S = ʃ a b f(x)dx = S(b) – S(a), we find the limits of integration. To do this, we solve a system of two equations:
(y = x 3,
(y = 1.
Thus, we have x 1 = 1 – the lower limit and x = 2 – the upper limit.
So, S = S DACE – S DABE = ʃ 1 2 x 3 dx – 1 = x 4 /4| 1 2 – 1 = (16 – 1)/4 – 1 = 11/4 (sq. units).
Answer: 11/4 sq. units
Example 2.
Calculate the area of the figure bounded by the lines y = √x; y = 2; x = 9.
Solution.
The given lines form the ABC figure, which is limited above by the graph of the function
y = √x, and below is a graph of the function y = 2. The resulting figure is shown by hatching in rice. 3.
The required area is equal to S = ʃ a b (√x – 2). Let's find the limits of integration: b = 9, to find a, we solve a system of two equations:
(y = √x,
(y = 2.
Thus, we have that x = 4 = a - this is the lower limit.
So, S = ∫ 4 9 (√x – 2)dx = ∫ 4 9 √x dx –∫ 4 9 2dx = 2/3 x√x| 4 9 – 2х| 4 9 = (18 – 16/3) – (18 – 8) = 2 2/3 (sq. units).
Answer: S = 2 2/3 sq. units
Example 3.
Calculate the area of the figure bounded by the lines y = x 3 – 4x; y = 0; x ≥ 0.
Solution.
Let’s plot the function y = x 3 – 4x for x ≥ 0. To do this, find the derivative y’:
y’ = 3x 2 – 4, y’ = 0 at x = ±2/√3 ≈ 1.1 – critical points.
If we plot the critical points on the number line and arrange the signs of the derivative, we find that the function decreases from zero to 2/√3 and increases from 2/√3 to plus infinity. Then x = 2/√3 is the minimum point, the minimum value of the function y min = -16/(3√3) ≈ -3.
Let's determine the intersection points of the graph with the coordinate axes:
if x = 0, then y = 0, which means A(0; 0) is the point of intersection with the Oy axis;
if y = 0, then x 3 – 4x = 0 or x(x 2 – 4) = 0, or x(x – 2)(x + 2) = 0, whence x 1 = 0, x 2 = 2, x 3 = -2 (not suitable, because x ≥ 0).
Points A(0; 0) and B(2; 0) are the points of intersection of the graph with the Ox axis.
The given lines form the OAB figure, which is shown by hatching in rice. 4.
Since the function y = x 3 – 4x takes a negative value on (0; 2), then
S = |ʃ 0 2 (x 3 – 4x)dx|.
We have: ʃ 0 2 (x 3 – 4х)dx =(x 4 /4 – 4х 2 /2)| 0 2 = -4, whence S = 4 sq. units
Answer: S = 4 sq. units
Example 4.
Find the area of the figure bounded by the parabola y = 2x 2 – 2x + 1, the lines x = 0, y = 0 and the tangent to this parabola at the point with the abscissa x 0 = 2.
Solution.
First, let's create an equation for the tangent to the parabola y = 2x 2 – 2x + 1 at the point with the abscissa x₀ = 2.
Since the derivative y’ = 4x – 2, then for x 0 = 2 we get k = y’(2) = 6.
Let's find the ordinate of the tangent point: y 0 = 2 2 2 – 2 2 + 1 = 5.
Therefore, the tangent equation has the form: y – 5 = 6(x – 2) or y = 6x – 7.
Let's build a figure bounded by lines:
y = 2x 2 – 2x + 1, y = 0, x = 0, y = 6x – 7.
Г у = 2х 2 – 2х + 1 – parabola. Points of intersection with the coordinate axes: A(0; 1) – with the Oy axis; with the Ox axis - there are no points of intersection, because the equation 2x 2 – 2x + 1 = 0 has no solutions (D< 0). Найдем вершину параболы:
x b = 2/4 = 1/2;
y b = 1/2, that is, the vertex of the parabola point B has coordinates B(1/2; 1/2).
So, the figure whose area needs to be determined is shown by hatching on rice. 5.
We have: S O A B D = S OABC – S ADBC.
Let's find the coordinates of point D from the condition:
6x – 7 = 0, i.e. x = 7/6, which means DC = 2 – 7/6 = 5/6.
We find the area of triangle DBC using the formula S ADBC = 1/2 · DC · BC. Thus,
S ADBC = 1/2 · 5/6 · 5 = 25/12 sq. units
S OABC = ʃ 0 2 (2x 2 – 2x + 1)dx = (2x 3 /3 – 2x 2 /2 + x)| 0 2 = 10/3 (sq. units).
We finally get: S O A B D = S OABC – S ADBC = 10/3 – 25/12 = 5/4 = 1 1/4 (sq. units).
Answer: S = 1 1/4 sq. units
We've looked at examples finding the areas of figures bounded by given lines. To successfully solve such problems, you need to be able to draw lines and graphs of functions on a plane, find the points of intersection of lines, apply a formula to find the area, which implies the ability to calculate certain integrals.
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From the definition it follows that for a non-negative function f(x) the definite integral is equal to the area of a curvilinear trapezoid bounded by the curve y = f(x), straight lines x = a, x = b and the abscissa = 0 (Figure 4.1).
If the function – f(x) is non-positive, then the definite integral 
equal to the area of the corresponding curvilinear trapezoid, taken with a minus sign (Figure 4.7).
Figure 4.7 – Geometric meaning of a definite integral for a non-positive function
For an arbitrary continuous function f(x), the definite integral 
is equal to the sum of the areas of curvilinear trapezoids lying under the graph of the function f(x) and above the abscissa axis, minus the sum of the areas of curvilinear trapezoids lying above the graph of the function f(x) and below the abscissa axis (Figure 4.8).
Figure 4.8 – Geometric meaning of a definite integral for an arbitrary continuous function f(x) (the plus sign marks the area that is added, and the minus sign marks the area that is subtracted).
When calculating the areas of curvilinear figures in practice, the following formula is often used: 
, where S is the area of the figure enclosed between the curves y = f 1 (x) and y = f 2 (x) on the segment [a,b], and f 1 (x) and f 2 (x) are continuous functions defined on this segment, such that f 1 (x) ≥ f 2 (x) (see Figures 4.9, 4.10).
When studying the economic meaning of the derivative, it was found that the derivative acts as the rate of change of some economic object or process over time or relative to another factor under study. To establish the economic meaning of a certain integral, it is necessary to consider this speed itself as a function of time or another factor. Then, since a definite integral represents a change in the antiderivative, we get that in economics it evaluates the change in this object (process) over a certain period of time (or with a certain change in another factor).
For example, if the function q=q(t) describes labor productivity depending on time, then the definite integral of this function 
represents the volume of output Q for the period of time from t 0 to t 1.
Methods for calculating definite integrals are based on the integration methods discussed earlier (we will not carry out proofs).
When finding the indefinite integral, we used the variable change method based on the formula: f(x)dx= =f((t))`(t)dt, where x =(t) is a function differentiable on the considered in between. For a definite integral, the variable change formula takes the form 
, Where 
and for everyone.
Example 1. Find
Let t= 2 –x 2. Then dt= -2xdx and xdx= - ½dt.
At x = 0 t= 2 – 0 2 = 2. At x = 1t= 2 – 1 2 = 1. Then
Example 2. Find
Example 3. Find
The integration by parts formula for a definite integral takes the form: 
, Where 
.
Example 1. Find
Let u=ln(1 +x),dv=dx. Then
Example 2. Find
Calculating the areas of plane figures using a definite integral
Example 1. Find the area of the figure bounded by the lines y = x 2 – 2 and y = x.
The graph of the function y= x 2 – 2 is a parabola with a minimum point at x= 0, y= -2; The abscissa axis intersects at the points 
. The graph of the function y = x is a straight line, the bisector of a non-negative coordinate quarter.
Let’s find the coordinates of the intersection points of the parabola y = x 2 – 2 and the straight line y = x by solving the system of these equations:
x 2 – x - 2 = 0
x = 2; y= 2 or x = -1;y= -1
Thus, the figure whose area needs to be found can be represented in Figure 4.9.
Figure 4.9 – Figure bounded by the lines y = x 2 – 2 and y = x
On the segment [-1, 2] x ≥ x 2 – 2.
Let's use the formula 
, putting f 1 (x) = x; f 2 (x) = x 2 – 2;a= -1;b= 2.
Example 2. Find the area of the figure bounded by the lines y = 4 - x 2 and y = x 2 – 2x.
The graph of the function y = 4 - x 2 is a parabola with a maximum point at x = 0, y = 4; The x-axis intersects at points 2 and -2. The graph of the function y = x 2 – 2x is a parabola with a minimum point at 2x- 2 = 0, x = 1; y = -1; The x-axis intersects at points 0 and 2.
Let's find the coordinates of the intersection points of the curves:
4 - x 2 = x 2 – 2x
2x 2 – 2x - 4 = 0
x 2 – x - 2 = 0
x = 2; y= 0 or x = -1;y= 3
Thus, the figure whose area needs to be found can be represented in Figure 4.10.
Figure 4.10 - Figure bounded by the lines y = 4 - x 2 and y = x 2 – 2x
On the segment [-1, 2] 4 - x 2 ≥ x 2 – 2x.
Let's use the formula 
, putting f 1 (x) = 4 - - x 2; f 2 (x) = x 2 – 2x;a= -1;b= 2.
Example 3. Find the area of the figure bounded by the lines y = 1/x; y= x 2 and y= 4 in a non-negative coordinate quadrant.
The graph of the function y = 1/x is a hyperbola; for positive x it is convex downward; the coordinate axes are asymptotes. The graph of the function y = x 2 in a non-negative coordinate quadrant is a branch of a parabola with a minimum point at the origin. These graphs intersect at 1/x = x 2 ; x 3 = 1; x = 1; y = 1.
The graph of the function y = 1/x intersects the straight line y = 4 at x = 1/4, and the graph of the function y = x 2 at x = 2 (or -2).
Thus, the figure whose area needs to be found can be represented in Figure 4.11.
Figure 4.11 - Figure bounded by lines y = 1/x; y= x 2 and y= 4 in non-negative coordinate quadrant
The required area of the figure ABC is equal to the difference between the area of the rectangle ABHE, which is equal to 4 * (2 - ¼) = 7, and the sum of the areas of two curvilinear trapezoids ACFE and CBHF. Let's calculate the area ACFE:
Let's calculate the area SVНF:
.
So, the required area is 7 – (ln4 + 7/3) = 14/3 –ln43.28 (unit 2).
Class: 11
Presentation for the lesson
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Lesson objectives: derive a formula for calculating the areas of plane figures using a definite integral; develop the skill of calculating the areas of plane figures using a definite integral; repeat known and provide new information from the history of integral calculus; preparation for the exam; continue work on developing attention, speech, logical thinking, and accuracy in writing; improve graphic culture; continue development work creativity students; increase interest in studying mathematics;
Equipment: multimedia projector, screen, presentation on the topic, developed in the Power Point environment.
Lesson progress
I. Organizational moment, message of the topic and purpose of the lesson.
II. Checking homework.
Checking additional homework (the teacher shows the solution on a previously prepared drawing, the solution is on the back of the board):
Calculate the area of the figure bounded by the graphs of the functions y = 1+ 3cos(x/2), x = -π/2, x = 3π/2, y = 0
III. Updating basic knowledge.
1. Oral work(Slides 3-4)
- Using the integral, express the areas of the figures shown in the figures:
- Calculate the integrals:
2. A little history. ( Slides 5-9)
A fragment of a students’ computer project on the topic “From the history of integral calculus.”
1 student
Integral- one of the most important concepts of mathematics, which arose in connection with the need, on the one hand, to find functions by their derivatives, and on the other, to measure areas, volumes, lengths of arcs, the work of forces over a certain period of time, etc.
The word integral itself was invented by J. Bernoulli(1690). It comes from Latin integero, translated as bring to the previous state, restore.
Other terms related to integral calculus that you may know appeared much later. Current name antiderivative function replaced an earlier one "primitive function", which was introduced by Joseph Louis Lagrange(1797). Latin word primitivus translated as "initial".
The emergence of problems of integral calculus is associated with finding areas and volumes. A number of problems of this kind have been solved by mathematicians ancient Greece. The first known method for calculating integrals is the Eudoxus exhaustion method ( approximately 370 BC BC), who tried to find areas and volumes by breaking them into an infinite number of parts for which the area or volume was already known. This method was taken up and developed by Archimedes, and was used to calculate the areas of parabolas and approximate the area of a circle.
However, Archimedes did not single out general content integration techniques and concepts of the integral, and even more so did not create an algorithm for integral calculus.
The works of Archimedes, first written in 1544, were one of the most important starting points for the development of integral calculus.
2 student
The concept of integral is directly related to integral calculus, a branch of mathematics that deals with the study of integrals, their properties and calculation methods.
We came closer and more accurately to the concept of integral Isaac Newton. He was the first to construct differential and integral calculus and called it the “Method of Fluxions...” (1670-1671, published 1736). Newton named the variables fluents(current values, from lat. fluo – flow). Rate of change fluent Newton – fluxions, and the infinitesimal changes in fluxions necessary to calculate fluxions are " moments"(Leibniz called them differentials). Thus, Newton based the concepts of fluxions (derivative) and fluents (antiderivative, or indefinite integral).
This immediately made it possible to solve a wide variety of mathematical and physical problems.
Simultaneously with Newton, another outstanding scientist came to similar ideas - Gottfried Wilhelm Leibniz.
Reflecting on philosophical and mathematical issues, Leibniz became convinced that mathematics could be the most reliable means of seeking and finding truth in science. The integral sign (∫) was first used by Leibniz at the end of the 17th century. This symbol is formed from the letter S - an abbreviation of the Latin word. summa(sum).
Newton and Leibniz developed two interpretations of the concept of an ordinary definite integral.
Newton interpreted the definite integral as the difference between the corresponding values of the antiderivative function:
,
Where F`(x)=f(x).
For Leibniz, the definite integral was the sum of all infinitesimal differentials.
The formula that Newton and Leibniz discovered independently of each other was called Newton–Leibniz formula.
Thus, the concept of integral was associated with the names of famous scientists: Newton, Leibniz, Bernoulli, who laid the foundation for modern mathematical analysis.
IV. Explanation of new material.
Using the integral, you can calculate the areas not only of curved trapezoids, but also of plane figures of a more complex type.
Let the figure P limited to straight lines X = a, x = b and function graphs y = f(x) And y = g(x), and on the segment [ a;b] the inequality holds g(x)f(x).
To calculate the area of a figure, we will reason as follows. Let's do it parallel transfer figures P on m units up so that the figure P turned out to be located in the coordinate plane above the abscissa axis.
Now it is limited above and below by function graphs y = f(x)+m And
y = g(x)+m, and both functions are continuous and non-negative on the interval [ a;b].
We denote the resulting figure ABCD. Its area can be found as the difference between the areas of the figures:
S ABCD = S aDCb – S aABb =
=
=
Thus, the area of the figure S bounded by straight lines X = a, x = b and function graphs y = f(x) And y = g(x), continuous on the interval [ a;b] and those that are for everyone X from the segment [ a;b] the inequality holds g(x)f(x), calculated by the formula
Example.(Slide 11) Calculate the area of the figure bounded by the lines y = x, y = 5 – x, x = 1, x = 2.
From these formulas for calculating the area of a figure, select the one that fits one of the six drawings. (Slide 14)
Task 3.(Slide 15) Calculate the area of the figure bounded by the graph of the function y = 0,5x 2+ 2, tangent to this graph at the abscissa point X= -2 and straight X = 0.
1. Let’s create an equation for the tangent to the graph of the function y = 0,5x 2+ 2 at the abscissa X = -2:
y = f(x 0) + f"(x 0)(x – x 0)
f(-2) = 0,5∙(-2) 2 + 2 = 4
f"(x) = (0,5x 2 + 2)"= x
f"(-2) = -2
y = 4 – 2(x + 2)
y = -2x
2. Let's build graphs of functions.
3. Find the area of the figure ABC.
VI. Summing up.
- formula for calculating the areas of plane figures;
- writing formulas for the areas of plane figures using a definite integral;
- repeating the equation of a tangent to the graph of a function and solving the equation with a modulus;
- grading students.
VII. Homework.
- paragraph 4 pp. 228-230;
- No. 1025 (c, d), No. 1037 (c, d), No. 1038 (c, d)
textbook: A. G. Mordkovich “Algebra and principles of analysis 10–11”
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