Find the area of a flat figure bounded by the graphs. Examples

In this article you will learn how to find the area of a figure bounded by lines using integral calculations. For the first time we encounter the formulation of such a problem in high school, when we have just completed the study of definite integrals and it is time to begin the geometric interpretation of the acquired knowledge in practice.

So, what is required to successfully solve the problem of finding the area of a figure using integrals:

Ability to make competent drawings;
Ability to solve a definite integral using the well-known Newton-Leibniz formula;
The ability to “see” a more profitable solution option - i.e. understand how it will be more convenient to carry out integration in one case or another? Along the x-axis (OX) or the y-axis (OY)?
Well, where would we be without correct calculations?) This includes understanding how to solve that other type of integrals and correct numerical calculations.

Algorithm for solving the problem of calculating the area of a figure bounded by lines:

1. We build a drawing. It is advisable to do this on a checkered piece of paper, on a large scale. We sign the name of this function with a pencil above each graph. Signing the graphs is done solely for the convenience of further calculations. Having received a graph of the desired figure, in most cases it will be immediately clear which limits of integration will be used. Thus, we solve the problem graphically. However, it happens that the values of the limits are fractional or irrational. Therefore, you can make additional calculations, go to step two.

2. If the limits of integration are not explicitly specified, then we find the points of intersection of the graphs with each other and see whether our graphical solution coincides with the analytical one.

3. Next, you need to analyze the drawing. Depending on how the function graphs are arranged, there are different approaches to finding the area of a figure. Let's look at different examples of finding the area of a figure using integrals.

3.1. The most classic and simplest version of the problem is when you need to find the area of a curved trapezoid. What is a curved trapezoid? This is a flat figure limited by the x-axis (y = 0), straight lines x = a, x = b and any curve continuous in the interval from a to b. Moreover, this figure is non-negative and is located not below the x-axis. In this case, the area of the curvilinear trapezoid is numerically equal to a certain integral, calculated using the Newton-Leibniz formula:

Example 1 y = x2 – 3x + 3, x = 1, x = 3, y = 0.

What lines is the figure bounded by? We have a parabola y = x2 - 3x + 3, which is located above the OX axis, it is non-negative, because all points of this parabola have positive values. Next, the straight lines x = 1 and x = 3 are given, which run parallel to the axis of the op-amp and are the boundary lines of the figure on the left and right. Well, y = 0, which is also the x-axis, which limits the figure from below. The resulting figure is shaded, as can be seen from the figure on the left. In this case, you can immediately begin solving the problem. Before us is a simple example of a curved trapezoid, which we then solve using the Newton-Leibniz formula.

3.2. In the previous paragraph 3.1, we examined the case when a curved trapezoid is located above the x-axis. Now consider the case when the conditions of the problem are the same, except that the function lies under the x-axis. A minus is added to the standard Newton-Leibniz formula. We will consider how to solve such a problem below.

Example 2. Calculate the area of the figure bounded by the lines y = x2 + 6x + 2, x = -4, x = -1, y = 0.

In this example we have a parabola y = x2 + 6x + 2, which originates from under the OX axis, straight lines x = -4, x = -1, y = 0. Here y = 0 limits the desired figure from above. The straight lines x = -4 and x = -1 are the boundaries within which the definite integral will be calculated. The principle of solving the problem of finding the area of a figure almost completely coincides with example number 1. The only difference is that the given function is not positive, and is also continuous on the interval [-4; -1] . What do you mean not positive? As can be seen from the figure, the figure that lies within the given x's has exclusively “negative” coordinates, which is what we need to see and remember when solving the problem. We look for the area of the figure using the Newton-Leibniz formula, only with a minus sign at the beginning.

The article is not completed.

We begin to consider the actual process of calculating the double integral and get acquainted with its geometric meaning.

The double integral is numerically equal to the area of the plane figure (the region of integration). This is the simplest form of double integral, when the function of two variables is equal to one: .

First, let's look at the problem in general form. Now you will be quite surprised how simple everything really is! Let's calculate the area of a flat figure bounded by lines. For definiteness, we assume that on the segment . The area of this figure is numerically equal to:

Let's depict the area in the drawing:

Let's choose the first way to traverse the area:

Thus:

And immediately an important technical trick: repeated integrals can be calculated separately. First the inner integral, then the outer integral. I highly recommend this method to beginners in the subject.

1) Let's calculate the internal integral, and the integration is carried out over the variable “y”:

The indefinite integral here is the simplest, and then the banal Newton-Leibniz formula is used, with the only difference that the limits of integration are not numbers, but functions. First, we substituted the upper limit into the “y” (antiderivative function), then the lower limit

2) The result obtained in the first paragraph must be substituted into the external integral:

A more compact representation of the entire solution looks like this:

The resulting formula is exactly the working formula for calculating the area of a plane figure using the “ordinary” definite integral! See the lesson Calculating area using a definite integral, there it is at every step!

That is, the problem of calculating area using double integral not much different from the problem of finding the area using a definite integral! In fact, it's the same thing!

Accordingly, no difficulties should arise! I won’t look at very many examples, since you, in fact, have repeatedly encountered this task.

Example 9

Solution: Let's depict the area in the drawing:

Let us choose the following order of traversal of the area:

Here and further I will not dwell on how to traverse the area, since very detailed explanations were given in the first paragraph.

Thus:

As I already noted, it is better for beginners to calculate iterated integrals separately, and I will stick to the same method:

1) First, using the Newton-Leibniz formula, we deal with the internal integral:

2) The result obtained in the first step is substituted into the external integral:

Point 2 is actually finding the area of a plane figure using a definite integral.

Answer:

This is such a stupid and naive task.

An interesting example for an independent solution:

Example 10

Using a double integral, calculate the area of a plane figure bounded by the lines , ,

An approximate example of a final solution at the end of the lesson.

In Examples 9-10, it is much more profitable to use the first method of traversing the area; curious readers, by the way, can change the order of traversal and calculate the areas using the second method. If you do not make a mistake, then, naturally, you will get the same area values.

But in some cases, the second method of traversing the area is more effective, and at the end of the young nerd’s course, let’s look at a couple more examples on this topic:

Example 11

Using a double integral, calculate the area of a plane figure bounded by lines,

Solution: we are looking forward to two parabolas with a quirk that lie on their sides. There is no need to smile; similar things occur quite often in multiple integrals.

What is the easiest way to make a drawing?

Let's imagine a parabola in the form of two functions:
– the upper branch and – the lower branch.

Similarly, imagine a parabola in the form of upper and lower branches.

Next, point-wise plotting of graphs rules, resulting in such a bizarre figure:

We calculate the area of the figure using the double integral according to the formula:

What happens if we choose the first method of traversing the area? Firstly, this area will have to be divided into two parts. And secondly, we will observe this sad picture: . Integrals, of course, are not of a super-complicated level, but... there is an old mathematical saying: those who are close to their roots do not need a test.

Therefore, from the misunderstanding given in the condition, we express the inverse functions:

Inverse functions in this example have the advantage that they specify the entire parabola at once without any leaves, acorns, branches and roots.

According to the second method, the area traversal will be as follows:

Thus:

As they say, feel the difference.

1) We deal with the internal integral:

We substitute the result into the outer integral:

Integration over the variable “y” should not be confusing; if there were a letter “zy”, it would be great to integrate over it. Although anyone who has read the second paragraph of the lesson How to calculate the volume of a body of rotation no longer experiences the slightest awkwardness with integration using the “Y” method.

Also pay attention to the first step: the integrand is even, and the interval of integration is symmetrical about zero. Therefore, the segment can be halved, and the result can be doubled. This technique is commented on in detail in the lesson Effective methods for calculating a definite integral.

What to add…. All!

Answer:

To test your integration technique, you can try to calculate . The answer should be exactly the same.

Example 12

Using a double integral, calculate the area of a plane figure bounded by lines

This is an example for you to solve on your own. It is interesting to note that if you try to use the first method of traversing the area, the figure will no longer have to be divided into two, but into three parts! And, accordingly, we get three pairs of repeated integrals. Sometimes it happens.

The master class has come to an end, and it’s time to move on to the grandmaster level - How to calculate a double integral? Examples of solutions. I’ll try not to be so maniacal in the second article =)

I wish you success!

Solutions and answers:

Example 2:Solution: Let's depict the area on the drawing:

Let us choose the following order of traversal of the area:

Thus:
Let's move on to inverse functions:

Thus:
Answer:

Example 4:Solution: Let's move on to direct functions:

Let's make the drawing:

Let's change the order of traversing the area:

Answer:

Solution.

The first and most important point in the decision is drawing.

Let's make the drawing:

The equation y=0 sets the “x” axis;

- x=-2 And x=1- straight, parallel to the axis OU;

- y=x 2 +2 - a parabola, the branches of which are directed upward, with the vertex at the point (0;2).

Comment. To construct a parabola, it is enough to find the points of its intersection with the coordinate axes, i.e. putting x=0 find the intersection with the axis OU and solving the corresponding quadratic equation, find the intersection with the axis Oh .

The vertex of a parabola can be found using the formulas:

You can also build lines point by point.

On the interval [-2;1] the graph of the function y=x 2 +2 located above the axis Ox, That's why:

Answer: S=9 sq. units

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.

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

b) Calculate the area of the figure bounded by the lines y=-e x , x=1 and coordinate axes.

Solution.

Let's make a drawing.

If a curved trapezoid is completely located under the axis Oh , then its area can be found using the formula:

Answer: S=(e-1) sq. units" 1.72 sq. units

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.

c) Find the area of a flat figure bounded by lines y=2x-x 2, y=-x.

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 straight 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 .

We build the given lines: 1. Parabola - vertex at point (1;1); axis intersection Oh - points (0;0) and (0;2). 2. Straight line - bisector of the 2nd and 4th coordinate angles. And now Attention! 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: .

And it doesn’t matter where the figure is located - above the axis or below the axis, but what matters is which graph is HIGHER (relative to another graph), and which 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

You can 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).

The desired figure is limited by a parabola above and a straight line below.

On the segment , according to the corresponding formula:

Answer: S=4.5 sq. units

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Any fractal is constructed according to a certain rule, which is consistently applied an unlimited number of times. Each such time is called an iteration.

The iterative algorithm for constructing a Menger sponge is quite simple: the original cube with side 1 is divided by planes parallel to its faces into 27 equal cubes. One central cube and 6 cubes adjacent to it along the faces are removed from it. The result is a set consisting of the remaining 20 smaller cubes. Doing the same with each of these cubes, we get a set consisting of 400 smaller cubes. Continuing this process endlessly, we get a Menger sponge.

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 skills in constructing drawings will be a much more pressing question. 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 the curvilinear trapezoid is numerically equal to the definite integral. Any definite integral (that exists) has a very good geometric meaning.

From a geometry point of view, 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 in the decision is drawing. Moreover, the drawing must be constructed CORRECTLY.

When constructing a drawing, I recommend the following order: first, it is better to construct all the straight lines (if any) and only then - parabolas, hyperbolas, and 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, therefore:

Answer:

Example 3

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

Solution: Let's make a drawing:

If the curved trapezoid is located under the axis (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 make a 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 .

It is better, if possible, 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 on a segment some continuous function is greater than or equal to some continuous function, then the area of the figure limited by the graphs of these 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 is important which graph is HIGHER (relative to another graph) and which 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 in 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:

Find the area of ​​a flat figure bounded by the graphs. Examples

Find the area of a flat figure bounded by the graphs. Examples