Using a definite integral, you can calculate the areas of plane figures, since this task always comes down to calculating the areas of curvilinear trapezoids.
The area of any figure in a rectangular coordinate system can be composed of the areas of curvilinear trapezoids adjacent to the axis Oh or to the axis OU.
It is convenient to solve problems on calculating the areas of plane figures using the following plan:
1. According to the conditions of the problem, make a schematic drawing
2. Present the required area as the sum or difference of the areas of curvilinear trapezoids. From the conditions of the problem and the drawing, the limits of integration are determined for each component of the curvilinear trapezoid.
3. Write each function in the form y = f(x).
4. Calculate the area of each curvilinear trapezoid and the area of the desired figure.
Let's consider several options for the arrangement of figures.
1). Let on the segment [ a; b] function f(x) takes non-negative values. Then the graph of the function y = f(x) located above the axis Oh.
S=
2). Let on the segment [ a; b] non-positive continuous function f(x). Then the graph of the function y = f(x) located under the axis Oh:
The area of such a figure is calculated by the formula: S = -
The area of such a figure is calculated by the formula: S= 
4). Let on the segment [ a; b] function f(x) takes both positive and negative values. Then the segment [ a; b] must be divided into parts in which the function does not change sign, then, using the above formulas, calculate the areas corresponding to these parts and add the found areas.
S 1 = S 2 = - S f = S 1 + S 2
Mathematics lesson for the first year of secondary vocational education institutions
Subject: “Calculating the areas of plane figures using a definite integral.”
Mathematics teacher S.B. Baranova
Educational objectives:
ensure repetition, generalization and systematization of material on this topic;
create conditions for control (self-control) of knowledge and skills.
Developmental tasks:
promote the formation of skills to apply techniques of comparison, generalization, and highlighting the main thing;
continue the development of mathematical horizons, thinking and speech, attention and memory.
Educational tasks:
promote interest in mathematics;
education of activity, mobility, communication skills.
Lesson type – a combined lesson with elements of problem-based learning.
Teaching methods and techniques – problematic, visual, independent work of students, self-test.
Equipment – lesson appendix, tables.
Lesson Plan
Organizing time. Preparing students for work in class.
Preparing students for active work (testing computational skills and tables of integrals by group).
Preparation for learning new material through repetition and updating of basic knowledge.
Working with new material.
Primary comprehension and application of the studied material, its consolidation.
Homework.
Application of knowledge.
Summarizing.
Reflection.
During the classes
1. Organizational moment.
The concept of a definite integral is one of the basic concepts of mathematics. By the end of the 17th century. Newton and Leibniz created the apparatus of differential and integral calculus, which forms the basis of mathematical analysis.
In previous lessons we learned to “take” indefinite integrals and calculate definite integrals. But much more important is the use of a definite integral. We know that it can be used to calculate the areas of curved trapezoids. Today we will answer the question: “How to do this?”
2. Preparing students for active work.
But first we need to test our computational skills and knowledge of the table of integrals. Before you is a task, the result of which will be a statement by the French mathematician S.D. Poisson (Life is enriched by two things: doing mathematics and teaching it).
The task is performed in pairs ().
3. Preparation for learning new material through repetition and updating of basic knowledge.
Let's move on to the topic of our lesson: “Calculating the areas of plane figures using a definite integral.” In addition to the ability to calculate a definite integral, we need to remember the properties of areas. What are they?
Equal figures have equal areas.
If a figure is divided into two parts, then its area is found as the sum of the areas of the individual parts.
We also need to repeat the sum integral rule and the Newton-Leibniz formula.
4. Working with the new integral
1. The definite integral is used to calculate the areas of curvilinear trapezoids. But in practice, there are more often figures that are not such, and we need to learn how to find the areas of just such figures.
Work according to the table “Basic cases of arrangement of a plane figure and the corresponding area formulas” ().
2. Let's test ourselves.
Work with the task () followed by verification (Table No. 3).
3. But the ability to choose the right formulas for area is not enough. In the following table () in each of the tasks there is an “external” reason that does not allow calculating the area of the figure. Let's find them.
a) formulas for graphs of functions are not indicated.
b) there are no limits of integration.
c) the names of the graphs are not indicated and there is no single limit.
d) the formula of one of the graphs is not indicated.
4. Taking into account the work done, we will formulate and write down an algorithm for solving problems on the topic of the lesson.
Construct graphs of these lines. Determine the desired figure.
Find the limits of integration.
Write down the area of the desired figure using a definite integral.
Calculate the resulting integral.
5. Primary comprehension and application of the studied material, its consolidation.
1. Taking into account the algorithm, let’s complete task No. 2 from the last table.
Picture 1
Solution:
For point A:
– does not satisfy the task conditions
For point B:
– does not satisfy the conditions of the problem.
Answer: (sq. units).
2. But when performing this task, the algorithm was not fully applied. To work it out, let's complete the following task:
Exercise. Find the area of a figure bounded by lines , .
Figure 2
Solution:
– parabola, vertex (m,n).
(0;2) – top
Let's find the limits of integration.
Answer: (sq. units).
6. Homework.
Calculate the area of a figure bounded by lines (disassemble the task).
7. Application of knowledge.
Independent work (Appendix No. 5))
8. Summing up.
learned to create formulas for finding the areas of plane figures;
find limits of integration;
calculate the area of figures.
9. Reflection.
Leaflets are distributed to students. They must evaluate their work by choosing one of the given answer options.
Assess the degree of difficulty of the lesson.
In class you had:
easily;
usually;
difficult.
I have fully grasped it and can apply it;
I have mastered it completely, but find it difficult to use;
learned partially;
didn't get it.
After reviewing the answers, draw a conclusion about the students’ preparedness for practical work.
Used Books:
Valutse I.I., Diligulin G.D. Mathematics for technical schools.
Kramer N.Sh., Putko B.A., Trishin I.M. Higher mathematics for economists.
Danko P.E., Popov A.G. Higher mathematics, part 1.
Zvanich L.I., Ryazanovsky A.R. M., New school.
Newspaper “Mathematics”. Publishing house “First of September”.
Appendix No. 1
Calculate definite integrals and you will recognize one of the statements of the French mathematician S.D. Poisson.
9
Life
Three
Two
Things
Occupation
Mathematics
Arithmetic
Teaching
Her
Decorated
By forgetting
Appendix No. 2
MAIN CASES OF LOCATION OF A FLAT FIGURE AND CORRESPONDING AREA FORMULAS
______________________________________
_
__________________________________
______
________________________________
______
___________________________________
A figure that is symmetrical about the ordinate axis or origin.
Appendix No. 3
Using a definite integral, write down formulas for calculating the areas of the shaded figures in the figure.
_________________________________________
__________________________________________
___________________________________________
___________________________________________
____________________________________________
Appendix No. 4
Find an “external” reason that does not allow you to calculate the area of the figure.
Picture 1
Figure 2
Figure 3
Figure 4
_____________________________
Appendix No. 5
Independent work
Option 1
Write using the integrals of the area of the figures and calculate them
Draw the shapes, plwhose areas are equal to the following integrals:
Independent work
Option 2
Determine whether the following statements are true:
Record withusing the area integrals of figures and calculate them
Draw figures whose areas are equal to the following integrals:
26. Calculating the areas of plane figures using a definite integral
Definition 1.Curvilinear trapezoid, generated by the graph of a non-negative function f on a segment, a figure bounded by a segment is called 
x-axis, line segments 
,
and the graph of the function 
on 
.
1. Let's split the segment 
points into partial segments.
2. In each segment 
(Where k=1,2,...,n) choose an arbitrary point 
.
3. Calculate the areas of rectangles whose bases are segments 
x-axes, and heights have lengths 
. Then the area of the stepped figure formed by these rectangles is equal to 
.
Note that the shorter the length of the partial segments, the more the stepped figure is close in location to the given curvilinear trapezoid. Therefore, it is natural to give the following definition.
Definition 2.Area of a curved trapezoid, generated by the graph of a non-negative function f on the segment 
, is called the limit (as the lengths of all partial segments tend to 0) of the areas of stepped figures if:
1) this limit exists and is finite;
2) does not depend on the way the segment is divided 
into partial sections;
3) does not depend on the choice of points 
.
Theorem 1.If the function
continuous and nonnegative on the interval 
, then the curvilinear trapezoidF,function generated by graphfon 
, has an area, which is calculated by the formula 
.
Using a definite integral, you can calculate the areas of plane figures and more complex ones.
If f And g- continuous and non-negative on the segment
functions for everyone x from the segment
inequality holds 
, then the area of the figure F, limited by straight lines 
,
and function graphs 
,
, calculated by the formula
.
Comment. If we discard the condition of non-negativity of functions f And g, the last formula remains true.
Topic 9. Differential equations
27. The concept of a differential equation. General and particular solution. Cauchy problem. The problem of constructing a mathematical model of the demographic process
The theory of differential equations arose at the end of the 17th century under the influence of the needs of mechanics and other natural science disciplines, essentially simultaneously with integral and differential calculus.
Definition 1.n-th order is an equation of the form in which 
- unknown function.
Definition 2. Function 
is called the solution of a differential equation on the interval I, if upon substitution of this function and its derivatives the differential equation becomes an identity.
Solve differential equation- is to find all its solutions.
Definition 3. The graph of the solution to a differential equation is called integral curve differential equation.
Definition 4.Ordinary differential equation 1-th order called an equation of the form 
.
Definition 5. Equation of the form 
called differential equation 1-th order,resolved with respect to the derivative.
As a rule, any differential equation has infinitely many solutions. To select any one solution from the totality of all solutions, additional conditions must be imposed.
Definition 6. Type condition 
superimposed on the solution of a 1st order differential equation is called initial condition, or Cauchy condition.
Geometrically, this means that the corresponding integral curve passes through the point 
.
Definition 7.General solution 1st order differential equation 
on a flat area D is called a one-parameter family of functions 
, satisfying the conditions:
1) for anyone 
function 
is a solution to the equation;
2) for each point 
there is such a parameter value 
, that the corresponding function 
is a solution to the equation satisfying the initial condition 
.
Definition 8. The solution obtained from the general solution for a certain value of the parameter is called private solution differential equation.
Definition 9.By special decision A differential equation is any solution that cannot be obtained from the general solution for any value of the parameter.
Solving differential equations is a very difficult problem, and generally speaking, the higher the order of the equation, the more difficult it is to specify ways to solve the equation. Even for first-order differential equations, it is possible to indicate methods for finding a general solution only in a small number of special cases. Moreover, in these cases, the desired solution is not always an elementary function.
One of the main problems of the theory of differential equations, first studied by O. Cauchy, is to find a solution to a differential equation that satisfies given initial conditions.
For example, is there always a solution to the differential equation 
, satisfying the initial condition 
, and will it be the only one? Generally speaking, the answer is no. Indeed, the equation 
, the right side of which is continuous on the entire plane, has solutions y=0 and y=(x+C) 3 ,CR
. Therefore, through any point on the O axis X passes through two integral curves.
So the function must satisfy some requirements. The following theorem contains one of the variants of sufficient conditions for the existence and uniqueness of a solution to the differential equation 
, satisfying the initial condition 
.
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 quadrant.
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 downwards; 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).
Let's move on to consider applications of integral calculus. In this lesson we will analyze the typical and most common task calculating the area of a plane figure using a definite integral. Finally, let all those who seek meaning in higher mathematics find it. You never know. In real life, you will have to approximate a dacha plot using elementary functions and find its area using a definite integral.
To successfully master the material, you must:
1) Understand the indefinite integral at least at an intermediate level. Thus, dummies should first 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. Examples of solutions. The task “calculate the area using a definite integral” always involves constructing a drawing, so your knowledge and drawing skills will also be a relevant issue. At a minimum, you need to be able to construct a straight line, parabola and hyperbola.
Let's start with a curved trapezoid. A curved trapezoid is a flat figure bounded by the graph of some function y = f(x), axis OX and lines x = a; x = b.
The area of a curvilinear trapezoid is numerically equal to a definite integral
Any definite integral (that exists) has a very good geometric meaning. At the lesson Definite integral. Examples of solutions we said that a definite integral is a number. And now it’s time to state another useful fact. From the point of view of geometry, the definite integral is AREA. That is, the definite integral (if it exists) geometrically corresponds to the area of a certain figure. Consider the definite integral
Integrand
defines a curve on the plane (it can be drawn if desired), and the definite integral itself is numerically equal to the area of the corresponding curvilinear trapezoid.
Example 1
, , , .
This is a typical assignment statement. The most important point in the decision is the construction of the drawing. Moreover, the drawing must be constructed RIGHT.
When constructing a drawing, I recommend the following order: at first it is better to construct all straight lines (if they exist) and only Then– 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 very useful material for our lesson - how to quickly build a parabola.
In this problem, the solution might look like this.
Let's do the drawing (note that the equation y= 0 specifies the axis OX):
We will not shade the curved trapezoid; here it is obvious what area we are talking about. The solution continues like this:
On the segment [-2; 1] function graph y = x 2 + 2 located above the axisOX, That's why:
Answer: .
Who has difficulties with calculating the definite integral and applying the Newton-Leibniz formula
,
refer to lecture Definite integral. Examples of solutions. After the task is completed, it is always useful to look at the drawing and figure out whether the answer is real. In this case, we count the number of cells in the drawing “by eye” - well, there will be about 9, it seems to be true. It is absolutely clear that if we got, say, the answer: 20 square units, then it is obvious that a mistake was made somewhere - 20 cells obviously do not fit into the figure in question, at most a dozen. If the answer is negative, then the task was also solved incorrectly.
Example 2
Calculate the area of a figure bounded by lines xy = 4, x = 2, x= 4 and axis OX.
This is an example for you to solve on your own. Full solution and answer at the end of the lesson.
What to do if the curved trapezoid is located under the 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 curved trapezoid completely located under the axis OX , then its area can be found using the formula:
In this case:
.
Attention! The two types of tasks should not be confused:
1) If you are asked to solve simply a definite integral without any geometric meaning, then it may be negative.
2) If you are asked to find the area of a figure using a definite integral, then the area is always positive! That is why the minus appears in the formula just discussed.
In practice, most often the figure is located in both the upper and lower half-plane, and therefore, from the simplest school problems we move on to more meaningful examples.
Example 4
Find the area of a plane figure bounded by lines y = 2x – x 2 , y = -x.
Solution: First you need to make a drawing. When constructing a drawing in area problems, we are most interested in the points of intersection of lines. Let's find the intersection points of the parabola y = 2x – x 2 and straight y = -x. This can be done in two ways. The first method is analytical. We solve the equation:
This means that the lower limit of integration a= 0, upper limit of integration b= 3. It is often more profitable and faster to construct lines point by point, and the limits of integration become clear “by themselves.” Nevertheless, the analytical method of finding limits still sometimes has to be used if, for example, the graph is large enough, or the detailed construction did not reveal the limits of integration (they can be fractional or irrational). Let's return to our task: it is more rational to first construct a straight line and only then a parabola. Let's make the drawing:
Let us repeat that when constructing pointwise, the limits of integration are most often determined “automatically”.
And now the working formula:
If on the segment [ a; b] some continuous function f(x) greater than or equal to some continuous function g(x), then the area of the corresponding figure can be found using the formula:
Here you no longer need to think about where the figure is located - above the axis or below the axis, but it matters which graph is HIGHER(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 completed solution might look like this:
The desired figure is limited by a parabola y = 2x – x 2 on top and straight y = -x below.
On segment 2 x – x 2 ≥ -x. According to the corresponding formula:
Answer: .
In fact, the school formula for the area of a curvilinear trapezoid in the lower half-plane (see example No. 3) is a special case of the formula
.
Because the axis OX given by the equation y= 0, and the graph of the function g(x) located below the axis OX, That
.
And now a couple of examples for your own solution
Example 5
Example 6
Find the area of a figure bounded by lines
When solving problems involving calculating area using a definite integral, a funny incident sometimes happens. The drawing was done correctly, the calculations were correct, but due to carelessness... The area of the wrong figure was found.
Example 7
First let's make a drawing:
The figure whose area we need to find is shaded blue(look carefully at the condition - how the figure is limited!). But in practice, due to inattention, people often decide that they need to find the area of the figure that is shaded in green!
This example is also useful in that it calculates the area of a figure using two definite integrals. Really:
1) On the segment [-1; 1] above the axis OX the graph is located straight y = x+1;
2) On a segment above the axis OX the graph of a hyperbola is located y = (2/x).
It is quite obvious that the areas can (and should) be added, therefore:
Answer:
Example 8
Calculate the area of a figure bounded by lines
Let’s present the equations in “school” form
and make a point-by-point drawing:
From the drawing it is clear that our upper limit is “good”: b = 1.
But what is the lower limit?! It is clear that this is not an integer, but what is it?
May be, a=(-1/3)? But where is the guarantee that the drawing is made with perfect accuracy, it may well turn out that a=(-1/4). What if we built the graph incorrectly?
In such cases, you have to spend additional time and clarify the limits of integration analytically.
Let's find the intersection points of the graphs
To do this, we solve the equation:
.
Hence, a=(-1/3).
The further solution is trivial. The main thing is not to get confused in substitutions and signs. The calculations here are not the simplest. On the segment
, 
,
according to the corresponding formula:
Answer: 
To conclude the lesson, let's look at two more difficult tasks.
Example 9
Calculate the area of a figure bounded by lines
Solution: Let's depict this figure in the drawing.
To construct a point-by-point drawing, you need to know the appearance of a sinusoid. In general, it is useful to know the graphs of all elementary functions, as well as some sine values. They can be found in the table of values trigonometric functions. In some cases (for example, in this case), it is possible to construct a schematic drawing, on which the graphs and limits of integration should be fundamentally correctly displayed.
There are no problems with the limits of integration here; they follow directly from the condition:
– “x” changes from zero to “pi”. Let's make a further decision:
On a segment, the graph of a function y= sin 3 x located above the axis OX, That's why:
(1) You can see how sines and cosines are integrated in odd powers in the lesson Integrals of trigonometric functions. We pinch off one sinus.
(2) We use the main trigonometric identity in the form
(3) Let's change the variable t=cos x, then: is located above the axis, therefore:
.
.
Note: note how the integral of the tangent cubed is taken; a corollary of the basic trigonometric identity is used here
.