Let's consider a curved trapezoid bounded by the Ox axis, the curve y=f(x) and two straight lines: x=a and x=b (Fig. 85). Let's take an arbitrary value of x (just not a and not b). Let's give it an increment h = dx and consider a strip bounded by straight lines AB and CD, the Ox axis and the arc BD belonging to the curve under consideration. We will call this strip an elementary strip. The area of an elementary strip differs from the area of rectangle ACQB by curvilinear triangle BQD, and the area of the latter less area rectangle BQDM with sides BQ = =h=dx) QD=Ay and area equal to hAy = Ay dx. As side h decreases, side Du also decreases and simultaneously with h tends to zero. Therefore, the area of the BQDM is second-order infinitesimal. The area of an elementary strip is the increment of the area, and the area of the rectangle ACQB, equal to AB-AC ==/(x) dx> is the differential of the area. Consequently, we find the area itself by integrating its differential. Within the figure under consideration, the independent variable l: changes from a to b, so the required area 5 will be equal to 5= \f(x) dx. (I) Example 1. Let us calculate the area bounded by the parabola y - 1 -x*, straight lines X =--Fj-, x = 1 and the O* axis (Fig. 86). at Fig. 87. Fig. 86. 1 Here f(x) = 1 - l?, the limits of integration are a = - and £ = 1, therefore J [*-t]\- -fl -- Г -1-±Л_ 1V1 -l-l-Ii-^ 3) |_ 2 3V 2 / J 3 24 24* Example 2. Let's calculate the area limited by the sinusoid y = sinXy, the Ox axis and the straight line (Fig. 87). Applying formula (I), we obtain A 2 S= J sinxdx= [-cos x]Q =0 -(-1) = lf Example 3. Calculate the area limited by the arc of the sinusoid ^у = sin jc, enclosed between two adjacent intersection points with the Ox axis (for example, between the origin and the point with the abscissa i). Note that from geometric considerations it is clear that this area will be twice the area of the previous example. However, let's do the calculations: I 5= | s\nxdx= [ - cosх)* - - cos i-(-cos 0)= 1 + 1 = 2. o Indeed, our assumption turned out to be correct. Example 4. Calculate the area bounded by the sinusoid and the Ox axis at one period (Fig. 88). Preliminary calculations suggest that the area will be four times larger than in Example 2. However, after making calculations, we obtain “i Г,*i S - \ sin x dx = [ - cos x]0 = = - cos 2l -(-cos 0) = - 1 + 1 = 0. This result requires clarification. To clarify the essence of the matter, we also calculate the area limited by the same sinusoid y = sin l: and the Ox axis in the range from l to 2i. Applying formula (I), we obtain 2l $2l sin xdx=[ - cosх]l = -cos 2i~)-c05i=- 1-1 =-2. Thus, we see that this area turned out to be negative. Comparing it with the area calculated in exercise 3, we find that their absolute values are the same, but the signs are different. If we apply property V (see Chapter XI, § 4), we get 2l I 2l J sin xdx= J sin * dx [ sin x dx = 2 + (- 2) = 0What happened in this example is not an accident. Always the area located below the Ox axis, provided that the independent variable changes from left to right, is obtained when calculated using integrals. In this course we will always consider areas without signs. Therefore, the answer in the example just discussed will be: the required area is 2 + |-2| = 4. Example 5. Let's calculate the area of the BAB shown in Fig. 89. This area is limited by the Ox axis, the parabola y = - xr and the straight line y - = -x+\. Area of a curvilinear trapezoid The required area OAB consists of two parts: OAM and MAV. Since point A is the intersection point of a parabola and a straight line, we will find its coordinates by solving the system of equations 3 2 Y = mx. (we only need to find the abscissa of point A). Solving the system, we find l; = ~. Therefore, the area has to be calculated in parts, first square. OAM and then pl. MAV: .... G 3 2, 3 G xP 3 1/2 U 2. QAM-^x sq. units 2 = 2 sq. units
Example 5. Calculate the area of a figure bounded by lines: y 2 = x, yx = 1, x = 4
Here you need to calculate the area of a curvilinear trapezoid bounded by the upper branch of the parabola 2 = x, Ox axis and straight lines x = 1 and x = 4 (see figure)
According to formula (1), where f(x) = a = 1 and b = 4, we have = (= sq. units.
Example 6 . Calculate the area of the figure bounded by the lines: y = sinx, y = 0, x = 0, x= .
The required area is limited by the half-wave of the sinusoid and the Ox axis (see figure).
We have - cosx = - cos = 1 + 1 = 2 sq. units
Example 7. Calculate the area of the figure bounded by the lines: y = - 6x, y = 0 and x = 4.
The figure is located under the Ox axis (see figure).
Therefore, we find its area using formula (3)
= =
Example 8. Calculate the area of the figure bounded by the lines: y = and x = 2. Construct the y = curve from the points (see figure). Thus, we find the area of the figure using formula (4)
Example 9 .
X 2 + y 2 = r 2 .
Here you need to calculate the area enclosed by the circle x 2 + y 2 = r 2 , i.e. the area of a circle of radius r with the center at the origin. Let's find the fourth part of this area by taking the limits of integration from 0
before; we have: 1 = = [
Hence, 1 =
Example 10. Calculate the area of a figure bounded by lines: y= x 2 and y = 2x
This figure limited by the parabola y=x 2 and the straight line y = 2x (see figure) To determine the intersection points of the given lines, we solve the system of equations: x 2 – 2x = 0 x = 0 and x = 2
Using formula (5) to find the area, we obtain
= . That is, lines such as the cut of a mushroom are not taken into account, the stem of which fits well into this segment, and the cap is much wider.
Side segments can degenerate into points . If you see such a figure in the drawing, this should not confuse you, since this point always has its value on the “x” axis. This means that everything is in order with the limits of integration.
Now you can move on to formulas and calculations. So the area s curved trapezoid can be calculated using the formula
If f(x) ≤ 0 (the graph of the function is located below the axis Ox), That area of a curved trapezoid can be calculated using the formula
There are also cases when both the upper and lower boundaries of the figure are functions, respectively y = f(x) And y = φ (x) , then the area of such a figure is calculated by the formula
. (3)
Solving problems together
Let's start with cases where the area of a figure can be calculated using formula (1).
Example 1.Ox) and straight x = 1 , x = 3 .
Solution. Because y = 1/x> 0 on the segment , then the area of the curvilinear trapezoid is found using formula (1):
.
Example 2. Find the area of the figure bounded by the graph of the function, line x= 1 and x-axis ( Ox ).
Solution. The result of applying formula (1):
If then s= 1/2 ; if then s= 1/3, etc.
Example 3. Find the area of the figure bounded by the graph of the function, the abscissa axis ( Ox) and straight x = 4 .
Solution. The figure corresponding to the conditions of the problem is a curvilinear trapezoid in which the left segment has degenerated into a point. The limits of integration are 0 and 4. Since , using formula (1) we find the area of the curvilinear trapezoid:
.
Example 4. Find the area of the figure bounded by the lines , , and located in the 1st quarter.
Solution. To use formula (1), let’s imagine the area of the figure given by the conditions of the example as the sum of the areas of the triangle OAB and curved trapezoid ABC. When calculating the area of a triangle OAB the limits of integration are the abscissas of the points O And A, and for the figure ABC- abscissas of points A And C (A is the intersection point of the line O.A. and parabolas, and C- the point of intersection of the parabola with the axis Ox). Solving jointly (as a system) the equations of a straight line and a parabola, we obtain (the abscissa of the point A) and (the abscissa of another point of intersection of the line and the parabola, which is not needed for the solution). Similarly we obtain , (abscissas of points C And D). Now we have everything we need to find the area of a figure. We find:
Example 5. Find the area of a curved trapezoid ACDB, if the equation of the curve CD and abscissas A And B 1 and 2 respectively.
Solution. Let's express given equation curve through the game: The area of a curvilinear trapezoid is found using formula (1):
.
Let's move on to cases where the area of a figure can be calculated using formula (2).
Example 6. Find the area of the figure bounded by the parabola and the x-axis ( Ox ).
Solution. This figure is located below the x-axis. Therefore, to calculate its area, we will use formula (2). The limits of integration are the abscissa and the points of intersection of the parabola with the axis Ox. Hence,
Example 7. Find the area enclosed between the abscissa axis ( Ox) and two adjacent sine waves.
Solution. The area of this figure can be found using formula (2):
.
Let's find each term separately:
.
.
Finally we find the area:
.
Example 8. Find the area of the figure enclosed between the parabola and the curve.
Solution. Let's express the equations of lines through the game:
The area according to formula (2) is obtained as
,
Where a And b- abscissas of points A And B. Let's find them by solving the equations together:
Finally we find the area:
And finally, cases when the area of a figure can be calculated using formula (3).
Example 9. Find the area of the figure enclosed between the parabolas 
And .
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:
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 solve for simple tasks. In reality, we will often have to work with more complex figures. In this regard, we will devote this section to an analysis of algorithms for calculating the area of figures that are limited by functions in explicit form, i.e. like y = f(x) or x = g(y).
TheoremLet the functions y = f 1 (x) and y = f 2 (x) be defined and continuous on the interval [ a ; b ] , and f 1 (x) ≤ f 2 (x) for any value x from [ a ; b ] . Then the formula for calculating the area of the figure G, bounded by the lines x = a, x = b, y = f 1 (x) and y = f 2 (x) will look like S (G) = ∫ a b f 2 (x) - f 1 (x) d x .
A similar formula will be applicable for the area of a figure bounded by the lines y = c, y = d, x = g 1 (y) and x = g 2 (y): S (G) = ∫ c d (g 2 (y) - g 1 (y) d y .
Proof
Let's look at three cases for which the formula will be valid.
In the first case, taking into account the property of additivity of area, the sum of the areas of the original figure G and the curvilinear trapezoid G 1 is equal to the area of the figure G 2. It means that
Therefore, S (G) = S (G 2) - S (G 1) = ∫ a b f 2 (x) d x - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) dx.
We can perform the last transition using the third property of the definite integral.
In the second case, the equality is true: S (G) = S (G 2) + S (G 1) = ∫ a b f 2 (x) d x + - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) d x
The graphic illustration will look like:
If both functions are non-positive, we get: S (G) = S (G 2) - S (G 1) = - ∫ a b f 2 (x) d x - - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) d x . The graphic illustration will look like:
Let's move on to consider the general case when y = f 1 (x) and y = f 2 (x) intersect the O x axis.
We denote the intersection points as x i, i = 1, 2, . . . , n - 1 . These points split the segment [a; b ] into n parts x i - 1 ; x i, i = 1, 2, . . . , n, where α = x 0< x 1 < x 2 < . . . < x n - 1 < x n = b . Фигуру G можно представить объединением фигур G i , i = 1 , 2 , . . . , n . Очевидно, что на своем интервале G i попадает под один из трех рассмотренных ранее случаев, поэтому их площади находятся как S (G i) = ∫ x i - 1 x i (f 2 (x) - f 1 (x)) d x , i = 1 , 2 , . . . , n
Hence,
S (G) = ∑ i = 1 n S (G i) = ∑ i = 1 n ∫ x i x i f 2 (x) - f 1 (x)) d x = = ∫ x 0 x n (f 2 (x) - f ( x)) d x = ∫ a b f 2 (x) - f 1 (x) d x
We can make the last transition using the fifth property of the definite integral.
Let us illustrate the general case on the graph.
The formula S (G) = ∫ a b f 2 (x) - f 1 (x) d x can be considered proven.
Now let's move on to analyzing examples of calculating the area of figures that are limited by the lines y = f (x) and x = g (y).
We will begin our consideration of any of the examples by constructing a graph. The image will allow us to represent complex shapes as unions of simpler shapes. If constructing graphs and figures on them is difficult for you, you can study the section on basic elementary functions, geometric transformation of graphs of functions, as well as constructing graphs while studying a function.
Example 1
It is necessary to determine the area of the figure, which is limited by the parabola y = - x 2 + 6 x - 5 and straight lines y = - 1 3 x - 1 2, x = 1, x = 4.
Solution
Let's draw the lines on the graph in the Cartesian coordinate system.
On the segment [ 1 ; 4 ] the graph of the parabola y = - x 2 + 6 x - 5 is located above the straight line y = - 1 3 x - 1 2. In this regard, to obtain the answer we use the formula obtained earlier, as well as the method of calculating the definite integral using the Newton-Leibniz formula:
S (G) = ∫ 1 4 - x 2 + 6 x - 5 - - 1 3 x - 1 2 d x = = ∫ 1 4 - x 2 + 19 3 x - 9 2 d x = - 1 3 x 3 + 19 6 x 2 - 9 2 x 1 4 = = - 1 3 4 3 + 19 6 4 2 - 9 2 4 - - 1 3 1 3 + 19 6 1 2 - 9 2 1 = = - 64 3 + 152 3 - 18 + 1 3 - 19 6 + 9 2 = 13
Answer: S(G) = 13
Let's look at a more complex example.
Example 2
It is necessary to calculate the area of the figure, which is limited by the lines y = x + 2, y = x, x = 7.
Solution
In this case, we have only one straight line located parallel to the x-axis. This is x = 7. This requires us to find the second limit of integration ourselves.
Let's build a graph and plot on it the lines given in the problem statement.
Having the graph in front of our eyes, we can easily determine that the lower limit of integration will be the abscissa of the point of intersection of the graph of the straight line y = x and the semi-parabola y = x + 2. To find the abscissa we use the equalities:
y = x + 2 O DZ: x ≥ - 2 x 2 = x + 2 2 x 2 - x - 2 = 0 D = (- 1) 2 - 4 1 (- 2) = 9 x 1 = 1 + 9 2 = 2 ∈ O DZ x 2 = 1 - 9 2 = - 1 ∉ O DZ
It turns out that the abscissa of the intersection point is x = 2.
We draw your attention to the fact that in the general example in the drawing, the lines y = x + 2, y = x intersect at the point (2; 2), so such detailed calculations may seem unnecessary. We have provided such a detailed solution here only because in more complex cases the solution may not be so obvious. This means that it is always better to calculate the coordinates of the intersection of lines analytically.
On the interval [ 2 ; 7] the graph of the function y = x is located above the graph of the function y = x + 2. Let's apply the formula to calculate the area:
S (G) = ∫ 2 7 (x - x + 2) d x = x 2 2 - 2 3 · (x + 2) 3 2 2 7 = = 7 2 2 - 2 3 · (7 + 2) 3 2 - 2 2 2 - 2 3 2 + 2 3 2 = = 49 2 - 18 - 2 + 16 3 = 59 6
Answer: S (G) = 59 6
Example 3
It is necessary to calculate the area of the figure, which is limited by the graphs of the functions y = 1 x and y = - x 2 + 4 x - 2.
Solution
Let's plot the lines on the graph.
Let's define the limits of integration. To do this, we determine the coordinates of the points of intersection of the lines by equating the expressions 1 x and - x 2 + 4 x - 2. Provided that x is not zero, the equality 1 x = - x 2 + 4 x - 2 becomes equivalent to the third degree equation - x 3 + 4 x 2 - 2 x - 1 = 0 with integer coefficients. To refresh your memory of the algorithm for solving such equations, we can refer to the section “Solving cubic equations.”
The root of this equation is x = 1: - 1 3 + 4 1 2 - 2 1 - 1 = 0.
Dividing the expression - x 3 + 4 x 2 - 2 x - 1 by the binomial x - 1, we get: - x 3 + 4 x 2 - 2 x - 1 ⇔ - (x - 1) (x 2 - 3 x - 1) = 0
We can find the remaining roots from the equation x 2 - 3 x - 1 = 0:
x 2 - 3 x - 1 = 0 D = (- 3) 2 - 4 · 1 · (- 1) = 13 x 1 = 3 + 13 2 ≈ 3 . 3; x 2 = 3 - 13 2 ≈ - 0 . 3
We found the interval x ∈ 1; 3 + 13 2, in which the figure G is contained above the blue and below the red line. This helps us determine the area of the figure:
S (G) = ∫ 1 3 + 13 2 - x 2 + 4 x - 2 - 1 x d x = - x 3 3 + 2 x 2 - 2 x - ln x 1 3 + 13 2 = = - 3 + 13 2 3 3 + 2 3 + 13 2 2 - 2 3 + 13 2 - ln 3 + 13 2 - - - 1 3 3 + 2 1 2 - 2 1 - ln 1 = 7 + 13 3 - ln 3 + 13 2
Answer: S (G) = 7 + 13 3 - ln 3 + 13 2
Example 4
It is necessary to calculate the area of the figure, which is limited by the curves y = x 3, y = - log 2 x + 1 and the abscissa axis.
Solution
Let's plot all the lines on the graph. We can get the graph of the function y = - log 2 x + 1 from the graph y = log 2 x if we position it symmetrically about the x-axis and move it up one unit. The equation of the x-axis is y = 0.
Let us mark the points of intersection of the lines.
As can be seen from the figure, the graphs of the functions y = x 3 and y = 0 intersect at the point (0; 0). This happens because x = 0 is the only real root of the equation x 3 = 0.
x = 2 is the only root of the equation - log 2 x + 1 = 0, so the graphs of the functions y = - log 2 x + 1 and y = 0 intersect at the point (2; 0).
x = 1 is the only root of the equation x 3 = - log 2 x + 1 . In this regard, the graphs of the functions y = x 3 and y = - log 2 x + 1 intersect at the point (1; 1). The last statement may not be obvious, but the equation x 3 = - log 2 x + 1 cannot have more than one root, since the function y = x 3 is strictly increasing, and the function y = - log 2 x + 1 is strictly decreasing.
The further solution involves several options.
Option #1
We can imagine the figure G as the sum of two curvilinear trapezoids located above the x-axis, the first of which is located below the midline on the segment x ∈ 0; 1, and the second is below the red line on the segment x ∈ 1; 2. This means that the area will be equal to S (G) = ∫ 0 1 x 3 d x + ∫ 1 2 (- log 2 x + 1) d x .
Option No. 2
Figure G can be represented as the difference of two figures, the first of which is located above the x-axis and below the blue line on the segment x ∈ 0; 2, and the second between the red and blue lines on the segment x ∈ 1; 2. This allows us to find the area as follows:
S (G) = ∫ 0 2 x 3 d x - ∫ 1 2 x 3 - (- log 2 x + 1) d x
In this case, to find the area you will have to use a formula of the form S (G) = ∫ c d (g 2 (y) - g 1 (y)) d y. In fact, the lines that bound the figure can be represented as functions of the argument y.
Let's solve the equations y = x 3 and - log 2 x + 1 with respect to x:
y = x 3 ⇒ x = y 3 y = - log 2 x + 1 ⇒ log 2 x = 1 - y ⇒ x = 2 1 - y
We get the required area:
S (G) = ∫ 0 1 (2 1 - y - y 3) d y = - 2 1 - y ln 2 - y 4 4 0 1 = = - 2 1 - 1 ln 2 - 1 4 4 - - 2 1 - 0 ln 2 - 0 4 4 = - 1 ln 2 - 1 4 + 2 ln 2 = 1 ln 2 - 1 4
Answer: S (G) = 1 ln 2 - 1 4
Example 5
It is necessary to calculate the area of the figure, which is limited by the lines y = x, y = 2 3 x - 3, y = - 1 2 x + 4.
Solution
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
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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