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 looking for meaning in 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 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 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.

Materials of this workshop presented simply, in detail and with a minimum of theory.

Let's start with curved trapezoid.

Curvilinear trapezoid called flat figure, limited by the axis, straight lines, and the graph of a function continuous on the segment, which does not change sign on this interval. Let this figure be located not lower 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. In class 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, with technology point-by-point construction 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 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 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 no 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 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, analytical method finding limits still sometimes has to be used if, for example, the graph is quite large, 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 corresponding formula:

Answer:

In fact 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 specified by the equation, and the graph of the function is located no 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 occurs that you need to find the area of a figure that is shaded green!

This example is also useful because 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 extra 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:

Knowledge of how to measure the Earth appeared in ancient times and gradually took shape in the science of geometry. WITH Greek language This word is translated as “land surveying”.

The measure of the extent of a flat section of the Earth in length and width is area. In mathematics it is usually denoted Latin letter S (from the English “square” - “area”, “square”) or Greek letterσ (sigma). S denotes the area of a figure on a plane or the surface area of a body, and σ is the area cross section wires in physics. These are the main symbols, although there may be others, for example, in the field of strength of materials, A is the cross-sectional area of the profile.

Calculation formulas

Knowing the area simple figures, you can find parameters of more complex. Ancient mathematicians developed formulas that can be used to easily calculate them. Such figures are triangle, quadrangle, polygon, circle.

To find the area of a complex plane figure, it is broken down into many simple figures such as triangles, trapezoids or rectangles. Then mathematical methods derive a formula for the area of this figure. A similar method is used not only in geometry, but also in mathematical analysis to calculate the areas of figures bounded by curves.

Triangle

Let's start with the simplest figure - a triangle. They are rectangular, isosceles and equilateral. Let's take any triangle ABC with sides AB=a, BC=b and AC=c (∆ ABC). To find its area, recall the well-known school course mathematics theorems of sines and cosines. Letting go of all calculations, we come to the following formulas:

S=√ - Heron’s formula, known to everyone, where p=(a+b+c)/2 is the semi-perimeter of the triangle;
S=a h/2, where h is the height lowered to side a;
S=a b (sin γ)/2, where γ is the angle between sides a and b;
S=a b/2, if ∆ ABC is rectangular (here a and b are legs);
S=b² (sin (2 β))/2, if ∆ ABC is isosceles (here b is one of the “hips”, β is the angle between the “hips” of the triangle);
S=a² √¾, if ∆ ABC is equilateral (here a is a side of the triangle).

Quadrangle

Let there be a quadrilateral ABCD with AB=a, BC=b, CD=c, AD=d. To find the area S of an arbitrary 4-gon, you need to divide it diagonally into two triangles, the areas of which are S1 and S2 in general case not equal.

Then use the formulas to calculate them and add them, i.e. S=S1+S2. However, if a 4-gon belongs to a certain class, then its area can be found using previously known formulas:

S=(a+c) h/2=e h, if the tetragon is a trapezoid (here a and c are the bases, e are midline trapezoid, h - height lowered to one of the bases of the trapezoid;
S=a h=a b sin φ=d1 d2 (sin φ)/2, if ABCD is a parallelogram (here φ is the angle between sides a and b, h is the height dropped to side a, d1 and d2 are diagonals);
S=a b=d²/2, if ABCD is a rectangle (d is a diagonal);
S=a² sin φ=P² (sin φ)/16=d1 d2/2, if ABCD is a rhombus (a is the side of the rhombus, φ is one of its angles, P is the perimeter);
S=a²=P²/16=d²/2, if ABCD is a square.

Polygon

To find the area of an n-gon, mathematicians break it down into the simplest equal figures-triangles, find the area of each of them and then add them. But if the polygon belongs to the class of regular, then use the formula:

S=a n h/2=a² n/=P²/, where n is the number of vertices (or sides) of the polygon, a is the side of the n-gon, P is its perimeter, h is the apothem, i.e. a segment drawn from the center of the polygon to one of its sides at an angle of 90°.

Circle

A circle is a perfect polygon with infinite number parties. We need to calculate the limit of the expression on the right in the formula for the area of a polygon with the number of sides n tending to infinity. In this case, the perimeter of the polygon will turn into the length of a circle of radius R, which will be the boundary of our circle, and will become equal to P=2 π R. Substitute this expression into the above formula. We will receive:

S=(π² R² cos (180°/n))/(n sin (180°/n)).

Let's find the limit of this expression as n→∞. To do this, we take into account that lim (cos (180°/n)) for n→∞ is equal to cos 0°=1 (lim is the sign of the limit), and lim = lim for n→∞ is equal to 1/π (we translated degree measure into a radian, using the relation π rad=180°, and applied the first remarkable limit lim(sin x)/x=1 at x→∞). Substituting the obtained values into the last expression for S, we arrive at well-known formula:

S=π² R² 1 (1/π)=π R².

Units of measurement

Systemic and non-systemic units of measurement are used. System units belong to the SI (System International). This is a square meter (sq. meter, m²) and units derived from it: mm², cm², km².

In square millimeters (mm²), for example, the cross-sectional area of wires in electrical engineering is measured, in square centimeters (cm²) - the cross-section of a beam in structural mechanics, in square meters (m²) - apartments or houses, in square kilometers (km²) - territories in geography.

However, sometimes non-systemic units of measurement are used, such as: weave, ar (a), hectare (ha) and acre (as). Let us present the following relations:

1 hundred square meters=1 a=100 m²=0.01 hectares;
1 ha=100 a=100 acres=10000 m²=0.01 km²=2.471 ac;
1 ac = 4046.856 m² = 40.47 a = 40.47 acres = 0.405 hectares.

Lesson on the topic: "Formulas for determining the area of a triangle, rectangle, square"

Additional materials
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Educational aids and simulators in the Integral online store for grade 5
Simulator for the textbook by I.I. Zubareva and A.G. Mordkovich
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Definition and concept of area of a figure

To better understand what the area of a figure is, consider the figure.
This arbitrary figure is divided into 12 small squares. The side of each square is 1 cm. And the area of each square is 1 square centimeter, which is written as follows: 1 cm 2.

Then the area of the figure is 12 square centimeters. In mathematics, area is denoted by the Latin letter S.
This means that the area of our figure is: S shape = 12 cm 2.

The area of the figure is equal to the area of all the small squares that make it up!

Guys, remember!
Area is measured in square units of length. Area units:
1. Square kilometer- km 2 (when the areas are very large, for example, a country or sea).
2. Square meter- m2 (quite suitable for measuring the area of a plot or apartment).
3. Square centimeter- cm 2 (usually used in mathematics lessons when drawing figures in a notebook).
4. Square millimeter - mm 2.

Area of a triangle

Let's consider two types of triangles: right-angled and arbitrary.

To find the area of a right triangle you need to know the length of the base and the height. In a right triangle, the height is replaced by one of the sides. Therefore, in the formula for the area of a triangle, instead of the height, we substitute one of the sides.
In our example, the sides are 7 cm and 4 cm. The formula for calculating the area of a triangle is written as follows:
S rectangular triangle ABC= BC * SA: 2

S of right triangle ABC = 7 cm * 4 cm: 2 = 14 cm 2

Now consider an arbitrary triangle.

For such a triangle, you need to draw the height to the base.
In our example, the height is 6 cm and the base is 8 cm. As in the previous example, we calculate the area using the formula:
S arbitrary triangle ABC = BC * h: 2.

Let's substitute our data into the formula and get:
S of an arbitrary triangle ABC = 8 cm * 6 cm: 2 = 24 cm 2.

Area of a rectangle and square

Take a rectangle ABCD with sides 5 cm and 8 cm.

The formula for calculating the area of a rectangle is written as follows:
S rectangle ABCD = AB * BC.

S rectangle ABCD = 8 cm * 5 cm = 40 cm 2.

Now let's calculate the area of the square. Unlike a rectangle and a triangle, to find the area of a square you only need to know one side. In our example, the side of the square ABCD is 9 cm. S square ABCD = AB * BC = AB 2.

Let's substitute our data into the formula and get:
S square ABCD = 9 cm * 9 cm = 81 cm 2.

Square geometric figure - numerical characteristic a geometric figure showing the size of this figure (part of the surface limited by the closed contour of this figure). The size of the area is expressed by the number of square units contained in it.

Triangle area formulas

Formula for the area of a triangle by side and height
Area of a triangle equal to half the product of the length of a side of a triangle and the length of the altitude drawn to this side
Formula for the area of a triangle based on three sides and the radius of the circumcircle
Formula for the area of a triangle based on three sides and the radius of the inscribed circle
Area of a triangle is equal to the product of the semi-perimeter of the triangle and the radius of the inscribed circle.

Square area formulas

Formula for the area of a square by side length
Square area equal to the square of the length of its side.
Formula for the area of a square along the diagonal length
Square area equal to half the square of the length of its diagonal.
S= 1 2
2

Rectangle area formula

Area of a rectangle

Parallelogram area formulas

Formula for the area of a parallelogram based on side length and height
Area of a parallelogram
Formula for the area of a parallelogram based on two sides and the angle between them
Area of a parallelogram is equal to the product of the lengths of its sides multiplied by the sine of the angle between them.
a b sin α

Formulas for the area of a rhombus

Formula for the area of a rhombus based on side length and height
Area of a rhombus equal to the product of the length of its side and the length of the height lowered to this side.
Formula for the area of a rhombus based on side length and angle
Area of a rhombus is equal to the product of the square of the length of its side and the sine of the angle between the sides of the rhombus.
Formula for the area of a rhombus based on the lengths of its diagonals
Area of a rhombus equal to half the product of the lengths of its diagonals.

Trapezoid area formulas

Heron's formula for trapezoid
Where S is the area of the trapezoid,
- lengths of the bases of the trapezoid,
- lengths of the sides of the trapezoid,

We have to deal with such a concept as area in our daily lives. So, for example, when building a house you need to know it in order to calculate the amount required material. Size garden plot will also be characterized by area. Even renovations in an apartment cannot be done without this definition. Therefore, the question of how to find the area of a rectangle comes up very often and is important not only for schoolchildren.

For those who don't know, a rectangle is a flat figure that has opposite sides are equal and the angles are 90°. To denote area in mathematics we use English letter S. It is measured in square units: meters, centimeters and so on.

Now we will try to give a detailed answer to the question of how to find the area of a rectangle. There are several ways to determine this value. Most often we come across a method of determining area using width and length.

Let's take a rectangle with width b and length k. To calculate the area of a given rectangle, you need to multiply the width by the length. All this can be represented in the form of a formula that will look like this: S = b * k.

Now let's look at this method specific example. It is necessary to determine the area of a garden plot with a width of 2 meters and a length of 7 meters.

S = 2 * 7 = 14 m2

In mathematics, especially in mathematics, we have to determine the area in other ways, since in many cases we do not know either the length or width of the rectangle. At the same time, other known quantities exist. How to find the area of a rectangle in this case?

If we know the length of the diagonal and one of the angles that makes up the diagonal with any side of the rectangle, then in this case we will need to remember the area. After all, if you look at it, the rectangle consists of two equal right triangles. So, let's return to the determined value. First you need to determine the cosine of the angle. Multiply the resulting value by the length of the diagonal. As a result, we get the length of one of the sides of the rectangle. Similarly, but using the definition of sine, you can determine the length of the second side. How to find the area of a rectangle now? Yes, it’s very simple, multiply the resulting values.

In formula form it will look like this:

S = cos(a) * sin(a) * d2, where d is the length of the diagonal

Another way to determine the area of a rectangle is through the circle inscribed in it. It is used if the rectangle is a square. To use this method need to know How to calculate the area of a rectangle in this way? Of course, according to the formula. We will not prove it. And it looks like this: S = 4 * r2, where r is the radius.

It happens that instead of the radius, we know the diameter of the inscribed circle. Then the formula will look like this:

S=d2, where d is the diameter.

If one of the sides and the perimeter are known, then how to find out the area of the rectangle in this case? To do this, it is necessary to make a series simple calculations. As we know, the opposite sides of a rectangle are equal, so the known length multiplied by two must be subtracted from the perimeter value. Divide the result by two and get the length of the second side. Well, then the standard technique is to multiply both sides and get the area of the rectangle. In formula form it will look like this:

S=b* (P - 2*b), where b is the length of the side, P is the perimeter.

As you can see, the area of a rectangle can be determined in various ways. It all depends on what quantities we know before considering this issue. Of course, the latest calculus methods are practically never encountered in life, but they can be useful for solving many problems in school. Perhaps this article will be useful for solving your problems.