The area of the trapezoid is integral. Definite integral. How to calculate the area of a shape

A figure bounded by the graph of a continuous non-negative on the segment $$ function $ f (x) $ and straight lines $ y = 0, \ x = a $ and $ x = b $, is called a curvilinear trapezoid.

Area corresponding curved trapezoid calculated by the formula:

$ S = \ int \ limits_ (a) ^ (b) (f (x) dx). $ (*)

We will conventionally divide the problems of finding the area of a curvilinear trapezoid into $ 4 $ types. Let's consider each type in more detail.

Type I: a curved trapezoid is explicitly specified. Then we immediately apply the formula (*).

For example, find the area of a curved trapezoid bounded by the graph of the function $ y = 4- (x-2) ^ (2) $, and by the straight lines $ y = 0, \ x = 1 $ and $ x = 3 $.

Let's draw this curved trapezoid.

Applying the formula (*), we find the area of this curved trapezoid.

$ S = \ int \ limits_ (1) ^ (3) (\ left (4- (x-2) ^ (2) \ right) dx) = \ int \ limits_ (1) ^ (3) (4dx) - \ int \ limits_ (1) ^ (3) ((x-2) ^ (2) dx) = 4x | _ (1) ^ (3) - \ left. \ frac ((x-2) ^ (3) ) (3) \ right | _ (1) ^ (3) = $

$ = 4 (3-1) - \ frac (1) (3) \ left ((3-2) ^ (3) - (1-2) ^ (3) \ right) = 4 \ cdot 2 - \ frac (1) (3) \ left ((1) ^ (3) - (- 1) ^ (3) \ right) = 8 - \ frac (1) (3) (1 + 1) = $

$ = 8- \ frac (2) (3) = 7 \ frac (1) (3) $ (unit $ ^ (2) $).

Type II: a curved trapezoid is implicitly specified. In this case, the straight lines $ x = a, \ x = b $ are usually not specified or partially specified. In this case, you need to find the intersection points of the functions $ y = f (x) $ and $ y = 0 $. These points will be the points $ a $ and $ b $.

For example, find the area of a figure bounded by the graphs of the functions $ y = 1-x ^ (2) $ and $ y = 0 $.

Let's find the intersection points. To do this, we equate the right-hand sides of the functions.

So $ a = -1 $ and $ b = 1 $. Let's draw this curved trapezoid.

Let's find the area of this curved trapezoid.

$ S = \ int \ limits _ (- 1) ^ (1) (\ left (1-x ^ (2) \ right) dx) = \ int \ limits _ (- 1) ^ (1) (1dx) - \ int \ limits _ (- 1) ^ (1) (x ^ (2) dx) = x | _ (- 1) ^ (1) - \ left. \ frac (x ^ (3)) (3) \ right | _ (-1) ^ (1) = $

$ = (1 - (- 1)) - \ frac (1) (3) \ left (1 ^ (3) - (- 1) ^ (3) \ right) = 2 - \ frac (1) (3) \ left (1 + 1 \ right) = 2 - \ frac (2) (3) = 1 \ frac (1) (3) $ (unit $ ^ (2) $).

Type III: area of a figure bounded by the intersection of two continuous non-negative functions. This figure will not be a curvilinear trapezoid, which means that you cannot calculate its area using the formula (*). How to be? It turns out that the area of this figure can be found as the difference between the areas of curvilinear trapezoids bounded by the upper function and $ y = 0 $ ($ S_ (uf) $), and the lower function and $ y = 0 $ ($ S_ (lf) $), where the role of $ x = a, \ x = b $ is played by the $ x $ coordinates of the intersection points of these functions, i.e.

$ S = S_ (uf) -S_ (lf) $. (**)

The most important thing when calculating such areas is not to overshoot with the choice of the upper and lower functions.

For example, find the area of a figure bounded by the functions $ y = x ^ (2) $ and $ y = x + 6 $.

Let's find the intersection points of these graphs:

By Vieta's theorem,

$ x_ (1) = - 2, \ x_ (2) = 3. $

That is, $ a = -2, \ b = 3 $. Let's draw a figure:

So the upper function is $ y = x + 6 $, and the lower one is $ y = x ^ (2) $. Next, find $ S_ (uf) $ and $ S_ (lf) $ by the formula (*).

$ S_ (uf) = \ int \ limits _ (- 2) ^ (3) ((x + 6) dx) = \ int \ limits _ (- 2) ^ (3) (xdx) + \ int \ limits _ (- 2 ) ^ (3) (6dx) = \ left. \ Frac (x ^ (2)) (2) \ right | _ (- 2) ^ (3) + 6x | _ (- 2) ^ (3) = 32 , 5 $ (units $ ^ (2) $).

$ S_ (lf) = \ int \ limits _ (- 2) ^ (3) (x ^ (2) dx) = \ left. \ Frac (x ^ (3)) (3) \ right | _ (- 2) ^ (3) = \ frac (35) (3) $ (unit $ ^ (2) $).

Substitute the found in (**) and get:

$ S = 32,5- \ frac (35) (3) = \ frac (125) (6) $ (unit $ ^ (2) $).

Type IV: area of a figure bounded by a function (s) that does not satisfy the condition of non-negativity. In order to find the area of such a figure, you need to symmetrically about the axis $ Ox $ ( in other words, put “minuses” in front of the functions) display the area and, using the methods described in types I - III, find the area of the displayed area. This area will be the required area. Previously, you may have to find the intersection points of the function graphs.

For example, find the area of a figure bounded by the graphs of the functions $ y = x ^ (2) -1 $ and $ y = 0 $.

Let's find the intersection points of the graphs of the functions:

those. $ a = -1 $ and $ b = 1 $. Let's draw the area.

Display the area symmetrically:

$ y = 0 \ \ Rightarrow \ y = -0 = 0 $

$ y = x ^ (2) -1 \ \ Rightarrow \ y = - (x ^ (2) -1) = 1-x ^ (2) $.

You get a curvilinear trapezoid bounded by the graph of the function $ y = 1-x ^ (2) $ and $ y = 0 $. This is the problem of finding a curvilinear trapezoid of the second type. We have already solved it. The answer was: $ S = 1 \ frac (1) (3) $ (unit $ ^ (2) $). Hence, the area of the required curvilinear trapezoid is equal to:

$ S = 1 \ frac (1) (3) $ (unit $ ^ (2) $).

Definite integral. How to calculate the area of a shape

We now turn to consideration of applications of integral calculus. In this lesson we will analyze a typical and most common task. - how to calculate the area of a flat figure using a definite integral... Finally meaning seekers in higher mathematics - may they find it. You never know. I'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 familiarize themselves with the lesson Not.

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

In fact, in order to find the area of a figure, one does not need so much knowledge of the indefinite and definite integral. The task "calculate area using a definite integral" always involves building a drawing so much more topical issue will be your knowledge and drawing skills. In this regard, it is useful to refresh the memory of the graphs of the basic elementary functions, and, at least, to be able to build a straight line, a parabola and a hyperbola. This can be done (many need) using methodological material and articles on geometric transformations of graphs.

Actually, everyone is familiar with the problem of finding the area using a definite integral since school, and we will go a little further from school curriculum... This article might not exist at all, but the fact is that the problem occurs in 99 cases out of 100, when a student suffers from the hated tower with enthusiasm to master the course of 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.

Curved trapezoid is called a flat figure bounded by an axis, straight lines, and a graph of a continuous function on a segment, which does not change sign on this interval. Let this figure be located not less abscissa axis:

Then the area of a curvilinear trapezoid is numerically equal to the 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 the AREA.

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

Example 1

This is a typical formulation of the assignment. First and the most important moment solutions - drawing building... Moreover, the drawing must be built RIGHT.

When building a drawing, I recommend next order: first it is better to build all straight lines (if any) and only Then- parabolas, hyperbolas, graphs of other functions. It is more profitable to build graphs of functions pointwise, the technique of point-by-point construction can be found in reference material Graphs and properties of elementary functions... There you can also find very useful material in relation to our lesson - how to quickly build a parabola.

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

I will not hatch a curved trapezoid, here it is obvious about what area in question... The solution continues like this:

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

Answer:

Who has difficulty calculating a 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 helpful to look at the blueprint and estimate if the answer is real. In this case, "by eye" we count the number of cells in the drawing - well, about 9 will be typed, it looks like the truth. It is quite clear that if we got, say, the answer: 20 square units, then, obviously, a mistake has been made somewhere - the figure under consideration clearly does not fit 20 cells, at most ten. If the answer is negative, then the task was also solved incorrectly.

Example 2

Calculate the area of a shape bounded by lines, and an axis

This is an example for independent decision. Complete solution and the answer at the end of the lesson.

What to do if the curved trapezoid is located under the axis?

Example 3

Calculate the area of a shape bounded by lines and coordinate axes.

Solution: Let's execute the drawing:

If the curved trapezoid is located under the axis(or at least not higher given axis), then its area can be found by the formula:
In this case:

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

1) If you are asked to solve just 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 minus appears in the formula just considered.

In practice, most often the figure is located in both the upper and lower half-planes, and therefore, from the simplest school problems, we move on to more meaningful examples.

Example 4

Find the area of a flat figure bounded by lines,.

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

Hence, the lower limit of integration, the upper limit of integration.
It is better not to use this method, if possible..

It is much more profitable and faster to construct the lines point by point, while the limits of integration become clear, as it were, "by themselves." The technique of point-by-point plotting for various charts is discussed in detail in the help. Graphs and properties of elementary functions... Nevertheless, the analytical method of finding the limits still has to be applied sometimes if, for example, the graph is large enough, or the precise construction did not reveal the limits of integration (they can be fractional or irrational). And we will also consider such an example.

Returning to our problem: it is more rational to first construct a straight line and only then a parabola. Let's execute the drawing:

I repeat that in the case of a pointwise construction, the limits of integration are most often found out by an “automaton”.

And now the working formula: If on a segment some continuous function greater than or equal of some continuous function, then the area of the figure, bounded by the graphs of these functions and straight lines, can be found by 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 is important which schedule is ABOVE(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 completion of the solution might look like this:

The required figure is bounded by a parabola at the top and a straight line at the bottom.
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) - special case formulas ... Since the axis is given by the equation, and the graph of the function is located not higher axis, then

And now a couple of examples for self-solution

Example 5

Example 6

Find the area of the figure bounded by lines,.

In the course of solving problems for calculating the area using a definite integral, a funny incident sometimes happens. The drawing is done correctly, the calculations are correct, but through inattention ... the area of the wrong figure is found, this is how your humble servant screwed up several times. Here's a real life case:

Example 7

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

Solution: First, let's execute the drawing:

... Eh, a lousy drawing came out, but everything seems to be legible.

The figure whose area we need to find is shaded in blue(carefully look at the condition - what the figure is limited to!). But in practice, due to inattention, a "glitch" often arises, that you 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) A line graph is located on the segment above the axis;

2) The hyperbola graph is located on the segment above the axis.

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

Answer:

Let's move on to one more meaningful task.

Example 8

Calculate the area of a shape bounded by lines,
Let's represent the equations in a "school" form, and we will execute a point-by-point drawing:

It can be seen from the drawing that our upper limit is "good":.
But what is the lower limit ?! It is clear that this is not an integer, but which one? May be ? But where is the guarantee that the drawing is made with perfect accuracy, it may well be that. Or root. What if we plotted the graph incorrectly at all?

In such cases, you have to spend additional time and refine the limits of integration analytically.

Find the intersection points of the line and the 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 are not the easiest here.

On the segment , according to the corresponding formula:

Answer:

Well, in conclusion of the lesson, we will consider two more difficult tasks.

Example 9

Calculate the area of a figure bounded by lines,

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

Damn, I forgot to sign the schedule, but to redo the picture, sorry, not hotts. Not drawing, in short, today is the day =)

For a point-by-point construction, you need to know appearance sinusoids (and in general it is useful to know graphs of all elementary functions), as well as some sine values, they can be found in trigonometric table... In a number of cases (as in this one), it is allowed to construct a schematic drawing, on which the graphs and limits of integration should be displayed correctly in principle.

There are no problems with the limits of integration, they follow directly from the condition: - "x" changes from zero to "pi". We make a further decision:

On the segment, the graph of the function is located above the axis, therefore:

The area of a curved trapezoid is numerically equal to the definite integral

Any definite integral (that exists) has a very good geometric meaning. In the lesson, I 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 the AREA.

That is, a definite integral (if it exists) geometrically corresponds to the area of some figure... For example, consider a definite integral. The integrand defines a certain curve on the plane (it can always 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 formulation of the assignment. The first and most important point of the solution is the construction of the drawing... Moreover, the drawing must be built RIGHT.

When building a drawing, I recommend the following order: first it is better to build all straight lines (if any) and only Then- parabolas, hyperbolas, graphs of other functions. It is more profitable to build graphs of functions pointwise, the technique of point-by-point construction can be found in the reference material.

There you can also find very useful material in relation to our lesson - how to quickly build a parabola.

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

I will not hatch a curved trapezoid, here it is obvious what area we are 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 difficulty calculating a definite integral and applying the Newton-Leibniz formula , refer to the lecture Definite integral. Examples of solutions.

Example 2

Calculate the area of a shape bounded by lines, and an axis

This is an example for a do-it-yourself solution. Complete solution and answer at the end of the tutorial.

What to do if the curved trapezoid is located under the axis?

Example 3

Calculate the area of a shape bounded by lines and coordinate axes.

Solution: Let's execute the drawing:

If the curved trapezoid completely located under the axle, then its area can be found by the formula:
In this case:

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

1) If you are asked to solve just 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 minus appears in the formula just considered.

In practice, most often the figure is located in both the upper and lower half-planes, and therefore, from the simplest school problems, we move on to more meaningful examples.

Example 4

Find the area of a flat figure bounded by lines,.

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

Hence, the lower limit of integration, the upper limit of integration.
It is better not to use this method, if possible.

And now the working formula: If on a segment some continuous function greater than or equal of some continuous function, then the area of the corresponding figure can be found by the formula:

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 completion of the solution might look like this:

The required figure is bounded by a parabola at the top and a straight line at the bottom.
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 a special case of the formula ... Since the axis is given by the equation, and the graph of the function is located below the axis, then

And now a couple of examples for self-solution

Example 5

Example 6

Find the area of the figure bounded by lines,.

Example 7

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

First, let's execute the drawing:

The figure whose area we need to find is shaded in blue(carefully look at the condition - what the figure is limited by!). But in practice, due to inattention, it often arises that you need to find the area of the figure, which is shaded in green!

This example is also useful in that it calculates the area of a figure using two definite integrals. Really:

1) A line graph is located on the segment above the axis;

2) The hyperbola graph is located on the segment above the axis.

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

Answer:

Example 8

In such cases, you have to spend additional time and refine the limits of integration analytically.

Find the intersection points of the line and the parabola.
To do this, we solve the equation:

Hence, .

The further solution is trivial, the main thing is not to get confused in substitutions and signs, the calculations are not the easiest here.

On the segment , according to the corresponding formula:

Answer:

Well, in conclusion of the lesson, we will consider two more difficult tasks.

Example 9

Calculate the area of a figure bounded by lines,

Solution: Draw this figure in the drawing.

For point-by-point construction of a drawing, you need to know the appearance of a sinusoid (and in general it is useful to know graphs of all elementary functions), as well as some sine values, they can be found in trigonometric table ... In a number of cases (as in this one), it is allowed to construct a schematic drawing, on which the graphs and limits of integration should be displayed correctly in principle.

There are no problems with the limits of integration, they follow directly from the condition: - "x" changes from zero to "pi". We make a further decision:

On the segment, the graph of the function is located above the axis, therefore:

(1) How to integrate sines and cosines in odd powers can be seen in the lesson Integrals from trigonometric functions ... This is a typical technique, we pinch off one sinus.

(2) We use the basic trigonometric identity in the form

(3) Let's change the variable, then:

New redistributions of integration:

Whoever is doing very badly with substitutions, please go to the lesson Replacement method in the indefinite integral... For those who do not really understand the replacement algorithm in a certain integral, visit the page Definite integral. Examples of solutions.

Back forward

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

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

Equipment: whiteboard, computer, multimedia projector

Lesson type: lesson-lecture

Lesson objectives:

educational: to form a culture of mental work, to create a situation of success for each student, to form positive motivation for learning; develop the ability to speak and listen to others.
developing: the formation of student independence of thinking on the application of 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 computing, calculating skills, the development of students' thinking in the course of completing the proposed tasks, the development of algorithmic culture.
educational: to form the concept of a curvilinear trapezoid, an integral, master the skills of calculating the areas of flat figures

Teaching method: explanatory and illustrative.

During the classes

In the previous classes, we learned how to calculate the areas of shapes, the boundaries of which are polygonal lines. There are methods in mathematics that allow you to calculate the areas of shapes that are bounded by curves. Such shapes are called curvilinear trapezoids, and their area is calculated using antiderivatives.

Curved trapezoid ( slide 1)

A curvilinear trapezoid is a figure bounded by the graph of a function, ( schm.), straight x = a and x = b and the abscissa

Various types of curved trapezoids ( slide 2)

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

Curved trapezoid area (slide 3)

Fix the left end of the gap a, and right X we will change, that is, we move the right wall of the curved trapezoid and get a changing shape. The area of a variable curvilinear trapezoid, limited by the graph of the function, is the antiderivative F for function f

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

Exercise 1:

Find the area of a curved 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 antiderivatives f (x) = x 2 :

Self-test by slide

Integral

Consider a curved trapezoid given 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 roughly 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 split the segment [ a; b], the more accurately we calculate the area.

Let us write this reasoning in the form of formulas.

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

Geometrically, this sum is the area of the figure shaded in the figure ( m.)

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

Integral sums give an approximate value of the area. The exact value is obtained by going to the limit. Imagine that we refine 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 curvilinear trapezoid is equal to the limit of integral sums, Sk.t. (schm.) or an integral, i.e.,

Definition:

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

= (schm.)

Newton-Leibniz formula.

Remember that the limit of integral sums is equal to the area of a curvilinear trapezoid, which means you can write:

Sk.t. = (schm.)

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

S K. t. (schm.)

Comparing these formulas, we get:

= (schm.)

This equality is called the Newton-Leibniz formula.

For the convenience of calculations, the formula is written in the form:

= = (schm.)

Assignments: (schm.)

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

2. Make up the integrals according to the drawing ( check 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 flat figures ( slide 8)

How do you find the area of shapes that are not curved trapezoids?

Let there be given two functions, the graphs of which you see on the slide ... (schm.) It is necessary to find the area of the filled figure ... (schm.)... The figure in question is a curved trapezoid? And how can you find its area using the property of area additivity? Consider two curved trapezoids and subtract the area of the other from the area of one of them ( schm.)

Let's compose an algorithm for finding the area by animation on a slide:

Plot function graphs
Project the intersection points of the graphs on the abscissa axis
Shade the figure obtained at the intersection of the graphs
Find curved trapezoids whose intersection or union is a given figure.
Calculate the area of each of them
Find the difference or sum of areas

Oral assignment: How to get the area of a shaded figure (tell with the help of animation, slide 8 and 9)

Homework: Work out the synopsis, No. 353 (a), No. 364 (a).

Bibliography

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

The area of ​​the trapezoid is integral. Definite integral. How to calculate the area of ​​a shape

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

The area of ​​a curved trapezoid is numerically equal to the definite integral

The area of the trapezoid is integral. Definite integral. How to calculate the area of a shape

Definite integral. How to calculate the area of a shape

The area of a curved trapezoid is numerically equal to the definite integral