Topic: Calculating area flat figure using a definite integral

Objectives: learn the definition and formulas for finding area curved trapezoid;

consider various cases finding the area of ​​a curvilinear trapezoid;

Be able to calculate the area of ​​a curved trapezoid.

Plan:

Curvilinear trapezoid.

Formulas for calculating the area of ​​a curved trapezoid.

Curvilinear trapezoid is a figure that is limited by the graph of a continuous, non-negative function f(x) on the interval, line segments x=a and x=b, as well as a segment of the x-axis between points a and b.

Images of curved trapezoids:

Now let's move on to possible options the location of figures whose area must be calculated on the coordinate plane.

First there will be the simplest option (the first picture), the usual curved trapezoid, as in the definition. There’s no need to invent anything here, just take the integral of a before b from function f(x). If we find the integral, we will also know the area of ​​this trapezoid.


In second option, our figure will be limited not by the x-axis, but by another function g(x). Therefore, to find the area CEFD, we need to first find the area AEFB(using the integral of f(x)), then find the area ACDB(using the integral of g(x)). And the required area of ​​the figure CEFD, there will be a difference between the first and second areas of the curved trapezoid. Since the boundaries of integration are the same here, all this can be written under one integral (see the formulas below the figure), it all depends on the complexity of the functions, in which case it will be easier to find the integral.



Third very similar to the first one, but only our trapezoid is placed, not above x-axis, and under it. Therefore, here we need to take the same integral, only with a minus sign, because the value of the integral will be negative, and the value of the area should be positive. If instead of a function f(x) take function –f(x), then its graph will be the same, simply symmetrically displayed relative to the x-axis.


AND fourth option when part of our figure is above the x-axis, and part below it. Therefore, we first need to find the area of ​​the figure AEFB, as in the first option, and then the area of ​​the figure ABCD, as in the third option and then fold them. As a result, we get the area of ​​the figure DEFC. Since the boundaries of integration are the same here, all this can be written under one integral (see the formulas below the figure), it all depends on the complexity of the functions, in which case it will be easier to find the integral.

Self-test questions:

What figure is called a curved trapezoid?

How to find the area of ​​a curved trapezoid?

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

Let's move on to applications 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 searching for meaning V higher mathematics- may they find him. You never know. In real life, you will have to approximate a dacha plot using elementary functions and find its area using a definite integral.

To successfully master the material, you must:

1) Understand 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 the definite integral. You can establish warm friendly relations with certain integrals on the page Definite integral. Examples of solutions.

In fact, in order to find the area of ​​a figure, you don’t need so 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 the main 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 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 completely clear that if we got, say, the answer: 20 square units, then it is obvious that a mistake was made somewhere - 20 cells clearly 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 ​​the figure, limited by lines, , and 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 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 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 upper limit integration
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 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:

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 in 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:

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

The area of ​​the corresponding curvilinear trapezoid is calculated by the formula:

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

We will conditionally divide problems to find the area of ​​a curvilinear trapezoid into $4$ types. Let's look at each type in more detail.

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

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

Let's draw this curved trapezoid.

Using formula (*), we find the area of ​​this curvilinear trapezoid.

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

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

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

Type II: the curved trapezoid is specified implicitly. 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 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.

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

Let's find the area of ​​this curved trapezoid.

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

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

Type III: the area of ​​a figure limited by the intersection of two continuous non-negative functions. This figure will not be a curved trapezoid, which means you cannot calculate its area using 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 points of intersection of these functions, i.e.

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

The most important thing when calculating such areas is not to “miss” 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:

According to Vieta's theorem,

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

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

Thus, the top function is $y=x+6$, and the bottom function is $y=x^(2)$. Next, we find $S_(uf)$ and $S_(lf)$ using 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$^(2)$).

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

Let's substitute what we found into (**) and get:

$S=32.5-\frac(35)(3)= \frac(125)(6)$ (units$^(2)$).

Type IV: figure area, limited function(s) that do not satisfy the non-negativity condition. In order to find the area of ​​such a figure, you need to be symmetrical about the $Ox$ axis ( in other words, put “minuses” in front of the functions) display the area and, using the methods outlined in types I – III, find the area of ​​the displayed area. This area will be the required area. First, 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 function graphs:

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

Let's display the area symmetrically:

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

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

The result is a curvilinear trapezoid bounded by the graph of the function $y=1-x^(2)$ and $y=0$. This is a problem to find a curved trapezoid of the second type. We have already solved it. The answer was: $S= 1\frac(1)(3)$ (units $^(2)$). This means that the area of ​​the required curvilinear trapezoid is equal to:

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

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 ​​rectangle ACQB by curvilinear triangle BQD, and the area of ​​the latter less area 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 us 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+\. Area of ​​a curvilinear trapezoid The required area 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:

] =

Means, improper integral converges and its value is equal to .