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 ​​the rectangle ACQB by the curvilinear triangle BQD, and the area of ​​the latter is less than the area of ​​the 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's 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 more area 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+\. Square curved trapezoid The required area of ​​the 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 = [replacement:

] =

This means that the improper integral converges and its value is equal to .

Back forward

Attention! Slide previews are for informational purposes only and may not represent all the features of the presentation. If you are interested this work, please download the full version.

Keywords: integral, curvilinear trapezoid, area of ​​figures bounded by lilies

Equipment: marker board, computer, multimedia projector

Lesson type: lesson-lecture

Lesson Objectives:

• educational: to create a culture of mental work, create a situation of success for each student, and create positive motivation for learning; develop the ability to speak and listen to others.
• developing: formation of independent thinking of the student in applying knowledge in different situations, the ability to analyze and draw conclusions, the development of logic, the development of the ability to correctly pose questions and find answers to them. Improving the formation of computational skills, developing students’ thinking in the course of completing proposed tasks, developing an algorithmic culture.
• educational: to form concepts about a curvilinear trapezoid, about an integral, to master the skills of calculating the areas of plane figures

Teaching method: explanatory and illustrative.

During the classes

In previous classes we learned to calculate the areas of figures whose boundaries are broken lines. In mathematics, there are methods that allow you to calculate the areas of figures bounded by curves. Such figures are called curvilinear trapezoids, and their area is calculated using antiderivatives.

Curvilinear trapezoid ( slide 1)

A curved trapezoid is a figure bounded by the graph of a function, ( sh.m.), straight x = a And x = b and x-axis

Various types of curved trapezoids ( slide 2)

We are considering different kinds curvilinear trapezoids and notice: one of the lines degenerates into a point, the role of the limiting function is played by the line

Area of ​​a curved trapezoid (slide 3)

Let us fix the left end of the interval A, and the right one X we will change, i.e., we move the right wall of the curvilinear trapezoid and get a changing figure. The area of ​​a variable curvilinear trapezoid bounded by the graph of the function is an antiderivative F for function f

And on the segment [ a; b] area of ​​a curvilinear trapezoid formed by the function f, is equal to the increment of the antiderivative of this function:

Exercise 1:

Find the area of ​​a curvilinear trapezoid bounded by the graph of the function: f(x) = x 2 and straight y = 0, x = 1, x = 2.

Solution: ( according to the algorithm slide 3)

Let's draw a graph of the function and lines

Let's find one of antiderivative functions f(x) = x 2 :

Self-test on slide

Integral

Consider a curvilinear trapezoid defined by the function f on the segment [ a; b]. Let's break this segment into several parts. The area of ​​the entire trapezoid will be divided into the sum of the areas of smaller curved trapezoids. ( slide 5). Each such trapezoid can be approximately considered a rectangle. The sum of the areas of these rectangles gives an approximate idea of ​​the entire area of ​​the curved trapezoid. The smaller we divide the segment [ a; b], the more accurately we calculate the area.

Let us write these arguments in the form of formulas.

Divide the segment [ a; b] into n parts by dots x 0 = a, x1,…, xn = b. Length k- th denote by xk = xk – xk-1. Let's make a sum

Geometrically, this sum represents the area of ​​the figure shaded in the figure ( sh.m.)

Sums of the form are called integral sums for the function f. (sh.m.)

Integral sums give an approximate value of the area. Exact value is obtained by passing to the limit. Let's imagine that we are refining the partition of the segment [ a; b] so that the lengths of all small segments tend to zero. Then the area of ​​the composed figure will approach the area of ​​the curved trapezoid. We can say that the area of ​​a curved trapezoid is equal to the limit of integral sums, Sc.t. (sh.m.) or integral, i.e.,

Definition:

Integral of a function f(x) from a before b called the limit of integral sums

= (sh.m.)

Newton-Leibniz formula.

We remember that the limit of integral sums is equal to the area of ​​a curvilinear trapezoid, which means we can write:

Sc.t. = (sh.m.)

On the other hand, the area of ​​a curved trapezoid is calculated by the formula

S k.t. (sh.m.)

Comparing these formulas, we get:

= (sh.m.)

This equality is called the Newton-Leibniz formula.

For ease of calculation, the formula is written as:

= = (sh.m.)

Tasks: (sh.m.)

1. Calculate the integral using the Newton-Leibniz formula: ( check on slide 5)

2. Compose integrals according to the drawing ( check on slide 6)

3. Find the area of ​​the figure bounded by the lines: y = x 3, y = 0, x = 1, x = 2. ( Slide 7)

Finding the areas of plane figures ( slide 8)

How to find the area of ​​figures that are not curved trapezoids?

Let two functions be given, the graphs of which you see on the slide . (sh.m.) Find the area of ​​the shaded figure . (sh.m.). Is the figure in question a curved trapezoid? How can you find its area using the property of additivity of area? Consider two curved trapezoids and subtract the area of ​​the other from the area of ​​one of them ( sh.m.)

Let's create an algorithm for finding the area using animation on a slide:

1. Graph functions
2. Project the intersection points of the graphs onto the x-axis
3. Shade the figure obtained when the graphs intersect
4. Find curvilinear trapezoids whose intersection or union is the given figure.
5. Calculate the area of ​​each of them
6. Find the difference or sum of areas

Oral task: How to obtain the area of ​​a shaded figure (tell using animation, slide 8 and 9)

Homework: Work through the notes, No. 353 (a), No. 364 (a).

Bibliography

1. Algebra and the beginnings of analysis: a textbook for grades 9-11 of evening (shift) school / ed. G.D. Glaser. - M: Enlightenment, 1983.
2. Bashmakov M.I. Algebra and the beginnings of analysis: a textbook for 10-11 grades of secondary school / Bashmakov M.I. - M: Enlightenment, 1991.
3. Bashmakov M.I. Mathematics: textbook for institutions beginning. and Wednesday prof. education / M.I. Bashmakov. - M: Academy, 2010.
4. Kolmogorov A.N. Algebra and beginnings of analysis: textbook for grades 10-11. educational institutions / A.N. Kolmogorov. - M: Education, 2010.
5. Ostrovsky S.L. How to make a presentation for a lesson?/ S.L. Ostrovsky. – M.: First of September, 2010.

Definite integral. How to calculate the area of ​​a figure

Let's move on to consider applications of integral calculus. In this lesson we will analyze the typical and most common task – how to use a definite integral to calculate the area of ​​a plane figure. Finally, those who are looking for meaning in higher mathematics - may they find it. You never know. We'll have to bring it closer in life country cottage area elementary functions and find its area using a definite integral.

To successfully master the material, you must:

1) Understand indefinite integral at least at an average level. Thus, dummies should first read the lesson Not.

2) Be able to apply the Newton-Leibniz formula and calculate definite integral. Set up warm friendly relations with definite integrals can be found on the page Definite integral. Examples of solutions.

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 much more topical issue will be your knowledge and skills in drawing. 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, parabola and hyperbola. This can be done (for many, it is necessary) using methodological material and articles on geometric transformations of graphs.

Actually, everyone is familiar with the task of finding the area using a definite integral since school, and we will not go much further from school curriculum. This article might not have existed at all, but the fact is that the problem occurs in 99 cases out of 100, when a student suffers from a hated school and enthusiastically masters a course in higher mathematics.

The materials of this workshop are presented simply, in detail and with a minimum of theory.

Let's start with a curved trapezoid.

Curvilinear trapezoid is a flat figure bounded by an axis, straight lines, and the graph of a function continuous on an interval 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. At the lesson Definite integral. Examples of solutions I said that a definite integral is a number. And now it’s time to state one more 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. 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 area corresponding curved trapezoid.

Example 1

This is a typical assignment statement. First and the most important moment solutions - 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, the point-by-point construction technique can be found in 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 draw the drawing (note that the equation defines the axis):

I will not hatch a curved trapezoid, it is obvious here what the area is we're talking about. The solution continues like this:

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

Answer:

Who has difficulties with calculating the definite integral and applying the Newton-Leibniz formula , refer to the 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 , , and axis

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 axle?

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 can be negative.

2) If you are asked to find the area of ​​a figure using a definite integral, then the area is always positive! That is why the minus appears in the formula just discussed.

In practice, most often the figure is located in both the upper and lower half-plane, and therefore, from the simplest school problems Let's 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.” The point-by-point construction technique for various graphs is discussed in detail in the help Graphs and properties of elementary functions. 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:

I repeat that when constructing pointwise, the limits of integration are most often found out “automatically”.

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, limited by schedules given functions and straight 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:

In fact, the school formula for the area of ​​a curvilinear trapezoid in the lower half-plane (see simple example No. 3) is special case formulas . Since the axis is specified by the equation, and the graph of the function is located not higher axes, then

And now a couple of examples for your own solution

Example 5

Example 6

Find the area of ​​the figure bounded by the 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, this is exactly how your humble servant screwed up several times. Here real case from life:

Example 7

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

Solution: First, let's make a drawing:

...Eh, the drawing came out crap, but everything seems to be legible.

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 arises that you need to find the area of ​​​​a figure that is shaded 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:

Answer:

Let's move on to another meaningful task.

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”: .
But what is the lower limit?! It is clear that this is not an integer, but what is it? May be ? But where is the guarantee that the drawing is made with perfect accuracy, it may well turn out that... Or the root. 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 a straight line and a parabola.
To do this, we solve the equation:


,

Really, .

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:

Well, to conclude the lesson, let’s look at two more difficult tasks.

Example 9

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

Solution: Let's depict this figure on the drawing.

Damn, I forgot to sign the schedule, and, sorry, I didn’t want to redo the picture. Not a drawing day, in short, today is the day =)

For point-by-point construction you need to know appearance sinusoids (and generally useful to know graphs of all elementary functions), as well as some sine values, they can be found in trigonometric table. In some cases (as 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 the segment, the graph of the function is located above the axis, therefore: