Calculating the areas of figures using integrals examples. Integral. Calculating areas using an integral. Find an “external” reason that does not allow you to calculate the area of ​​the figure

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

Class: 11

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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; exam preparation; continue work on developing attention, speech, logical thinking, and accuracy in writing; improve graphic culture; continue work on developing students’ creative abilities; increase interest in studying mathematics;

Equipment: multimedia projector, screen, presentation on the topic, developed in the Power Point environment.

During the classes

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)

1. Using the integral, express the areas of the figures shown in the figures:
2. 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 were solved by mathematicians of 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 identify the general content of integration techniques and concepts of the integral, much less 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 methods of calculation.

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 curvilinear 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 perform a parallel transfer of the figure 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. Summarizing.

• 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.

1. paragraph 4 pp. 228-230;
2. No. 1025 (c, d), No. 1037 (c, d), No. 1038 (c, d)

textbook: A. G. Mordkovich “Algebra and principles of analysis 10–11”

In fact, in order to find the area of ​​a figure, you don’t need that much knowledge of the indefinite and definite integral. The task “calculate the area using a definite integral” always involves constructing a drawing, so your knowledge and drawing skills will be a much more pressing issue. In this regard, it is useful to refresh your memory of the graphs of basic elementary functions, and, at a minimum, be able to construct a straight line and a hyperbola.

A curved trapezoid is a flat figure bounded by an axis, straight lines, and the graph of a function continuous on a segment that does not change sign on this interval. Let this figure be located not less x-axis:

Then the area of ​​a curvilinear trapezoid is numerically equal to a definite integral. Any definite integral (that exists) has a very good geometric meaning.

From the point of view of geometry, the definite integral is AREA.

That is, a certain integral (if it exists) geometrically corresponds to the area of ​​a certain figure. For example, consider the definite integral. The integrand defines a curve on the plane located above the axis (those who wish can make a drawing), and the definite integral itself is numerically equal to the area of ​​the corresponding curvilinear trapezoid.

Example 1

This is a typical assignment statement. The first and most important point of 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. It is more profitable to build graphs of functions point by point.

In this problem, the solution might look like this.
Let's draw the drawing (note that the equation defines the axis):

On the segment, the graph of the function is located above the axis, That's why:

Answer:

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, “by eye” we count the number of cells in the drawing - 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 3

Calculate the area of ​​the figure bounded by lines and coordinate axes.

Solution: Let's make a drawing:

If a curved trapezoid is located under the axle(or at least not higher given axis), then its area can be found using the formula:

In this case:

Attention! The two types of tasks should not be confused:

1) If you are asked to solve simply a definite integral without any geometric meaning, then it 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 the lines , .

Solution: First you need to complete the drawing. Generally speaking, when constructing a drawing in area problems, we are most interested in the points of intersection of lines. Let's find the intersection points of the parabola and the straight line. This can be done in two ways. The first method is analytical. We solve the equation:

This means that the lower limit of integration is , the upper limit of integration is .

If possible, it is better not to use this method..

It is much more profitable and faster to construct lines point by point, and the limits of integration become clear “by themselves.” Nevertheless, the analytical method of finding limits still sometimes has to be used if, for example, the graph is large enough, or the detailed construction did not reveal the limits of integration (they can be fractional or irrational). And we will also consider such an example.

Let's return to our task: it is more rational to first construct a straight line and only then a parabola. Let's make the drawing:

And now the working formula: If there is some continuous function on the segment greater than or equal to some continuous function , then the area of ​​the figure bounded by the graphs of these functions and the lines , , can be found using the formula:

Here you no longer need to think about where the figure is located - above the axis or below the axis, and, roughly speaking, it 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 it is necessary to subtract from

The completed solution might look like this:

The desired figure is limited by a parabola above and a straight line below.
On the segment, according to the corresponding formula:

Answer:

Example 4

Calculate the area of ​​the figure bounded by the lines , , , .

Solution: 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, a “glitch” often occurs that you need to find the area of ​​​​a 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 above the axis there is a graph of a straight line;

2) On the segment above the axis there is a graph of a hyperbola.

It is quite obvious that the areas can (and should) be added, therefore:

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).

A 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 curved 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 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 construct 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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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 ,CR . 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
.