Let the function be non-negative and continuous on the interval. Then, according to the geometric meaning of a definite integral, the area of ​​a curvilinear trapezoid bounded above by the graph of this function, below by the axis, on the left and right by straight lines and (see Fig. 2) is calculated by the formula

Example 9. Find the area of ​​a figure bounded by a line and axis.

Solution. Function graph is a parabola whose branches are directed downward. Let's build it (Fig. 3). To determine the limits of integration, we find the points of intersection of the line (parabola) with the axis (straight line). To do this, we solve the system of equations

We get: , where , ; hence, , .

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We find the area of ​​the figure using formula (5):

If the function is non-positive and continuous on the segment , then the area of ​​the curvilinear trapezoid bounded below by the graph of this function, above by the axis, on the left and right by straight lines and , is calculated by the formula

. (6)

If the function is continuous on a segment and changes sign at a finite number of points, then the area of ​​the shaded figure (Fig. 4) is equal to the algebraic sum of the corresponding definite integrals:

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Example 10. Calculate the area of ​​the figure bounded by the axis and the graph of the function at .

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Solution. Let's make a drawing (Fig. 5). The required area is the sum of the areas and . Let's find each of these areas. First, we determine the limits of integration by solving the system We get , . Hence:

;

.

Thus, the area of ​​the shaded figure is

(sq. units).

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Finally, let the curvilinear trapezoid be bounded above and below by the graphs of functions continuous on the segment and ,
and on the left and right - straight lines and (Fig. 6). Then its area is calculated by the formula

. (8)

Example 11. Find the area of ​​the figure bounded by the lines and.

Solution. This figure is shown in Fig. 7. Let's calculate its area using formula (8). Solving the system of equations we find, ; hence, , . On the segment we have: . This means that in formula (8) we take as x, and as a quality – . We get:

(sq. units).

More complex problems of calculating areas are solved by dividing the figure into non-overlapping parts and calculating the area of ​​the entire figure as the sum of the areas of these parts.

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Example 12. Find the area of ​​the figure bounded by the lines , , .

Solution. Let's make a drawing (Fig. 8). This figure can be considered as a curvilinear trapezoid, bounded from below by the axis, to the left and right - by straight lines and, from above - by graphs of functions and. Since the figure is limited from above by the graphs of two functions, to calculate its area, we divide this straight line figure into two parts (1 is the abscissa of the point of intersection of the lines and ). The area of ​​each of these parts is found using formula (4):

(sq. units); (sq. units). Hence:

(sq. units).

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X= j ( at)

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In conclusion, we note that if a curvilinear trapezoid is limited by straight lines and , axis and continuous on the curve (Fig. 9), then its area is found by the formula

Volume of a body of rotation

Let a curvilinear trapezoid, bounded by the graph of a function continuous on a segment, by an axis, by straight lines and , rotate around the axis (Fig. 10). Then the volume of the resulting body of rotation is calculated by the formula

. (9)

Example 13. Calculate the volume of a body obtained by rotating around the axis of a curvilinear trapezoid bounded by a hyperbola, straight lines, and axis.

Solution. Let's make a drawing (Fig. 11).

From the conditions of the problem it follows that , . From formula (9) we get

.

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Volume of a body obtained by rotation around an axis OU curvilinear trapezoid bounded by straight lines y = c And y = d, axis OU and a graph of a function continuous on a segment (Fig. 12), determined by the formula

. (10)

X= j ( at)

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Example 14. Calculate the volume of a body obtained by rotating around an axis OU curvilinear trapezoid bounded by lines X 2 = 4at, y = 4, x = 0 (Fig. 13).

Solution. In accordance with the conditions of the problem, we find the limits of integration: , . Using formula (10) we obtain:

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Arc length of a plane curve

Let the curve given by the equation , where , lie in the plane (Fig. 14).

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Definition. The length of an arc is understood as the limit to which the length of a broken line inscribed in this arc tends, when the number of links of the broken line tends to infinity, and the length of the largest link tends to zero.

If a function and its derivative are continuous on the segment, then the arc length of the curve is calculated by the formula

. (11)

Example 15. Calculate the arc length of the curve enclosed between the points for which .

Solution. From the problem conditions we have . Using formula (11) we obtain:

.

4. Improper integrals
with infinite limits of integration

When introducing the concept of a definite integral, it was assumed that the following two conditions were satisfied:

a) limits of integration A and are finite;

b) the integrand is bounded on the interval.

If at least one of these conditions is not satisfied, then the integral is called not your own.

Let us first consider improper integrals with infinite limits of integration.

Definition. Let the function be defined and continuous on the interval, then and unlimited on the right (Fig. 15).

If the improper integral converges, then this area is finite; if the improper integral diverges, then this area is infinite.

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An improper integral with an infinite lower limit of integration is defined similarly:

. (13)

This integral converges if the limit on the right side of equality (13) exists and is finite; otherwise the integral is said to be divergent.

An improper integral with two infinite limits of integration is defined as follows:

, (14)

where с is any point of the interval. The integral converges only if both integrals on the right side of equality (14) converge.

;

G) = [select a complete square in the denominator: ] = [replacement:

] =

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

Any definite integral (that exists) has a very good geometric meaning. In class 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 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 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 assignment statement. The first and most important point in the decision is the construction of a 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 the reference material.

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 shade the curved trapezoid; it is obvious here 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 difficulties with calculating the definite integral and applying the Newton-Leibniz formula , refer to the lecture Definite integral. Examples of solutions.

After the task is completed, it is always useful to look at the drawing and figure out whether the answer is real. In this case, we count the number of cells in the drawing “by eye” - well, there will be about 9, it seems to be true. It is absolutely clear that if we got, say, the answer: 20 square units, then it is obvious that a mistake was made somewhere - 20 cells obviously do not fit into the figure in question, at most a dozen. If the answer is negative, then the task was also solved incorrectly.

Example 2

Calculate the area of ​​a figure bounded by lines , , and axis

This is an example for you to solve on your own. Full solution and answer at the end of the lesson.

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

Example 3

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

Solution: Let's make a drawing:

If a curved trapezoid completely located under the 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 not to use this method, if possible.

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 on a segment there is some continuous function greater than or equal to some continuous function, then the area of ​​the corresponding figure 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 a special case of the formula . Since the axis is specified by the equation and the graph of the function is located below the axis, 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 is a real life case:

Example 7

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

First let's make a drawing:

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, it often arises 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:

Answer:

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:

Hence, .

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.

To construct a point-by-point drawing, you need to know the appearance of a sinusoid (and in general it’s 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:

(1) You can see how sines and cosines are integrated in odd powers in the lesson Integrals of trigonometric functions. This is a typical technique, we pinch off one sinus.

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

(3) Let’s change the variable , then:

New areas of integration:

Anyone who is really bad with substitutions, please take a lesson. Substitution method in indefinite integral. For those who do not quite understand the replacement algorithm in a definite integral, visit the page Definite integral. Examples of solutions.

Example1 . Calculate the area of ​​the figure bounded by the lines: x + 2y – 4 = 0, y = 0, x = -3, and x = 2

Let's construct a figure (see figure) We construct a straight line x + 2y – 4 = 0 using two points A(4;0) and B(0;2). Expressing y through x, we get y = -0.5x + 2. Using formula (1), where f(x) = -0.5x + 2, a = -3, b = 2, we find

S = = [-0.25=11.25 sq. units

Example 2. Calculate the area of ​​the figure bounded by the lines: x – 2y + 4 = 0, x + y – 5 = 0 and y = 0.

Solution. Let's construct the figure.

Let's construct a straight line x – 2y + 4 = 0: y = 0, x = - 4, A(-4; 0); x = 0, y = 2, B(0; 2).

Let's construct a straight line x + y – 5 = 0: y = 0, x = 5, C(5; 0), x = 0, y = 5, D(0; 5).

Let's find the point of intersection of the lines by solving the system of equations:

x = 2, y = 3; M(2; 3).

To calculate the required area, we divide the triangle AMC into two triangles AMN and NMC, since when x changes from A to N, the area is limited by a straight line, and when x changes from N to C - by a straight line

For triangle AMN we have: ; y = 0.5x + 2, i.e. f(x) = 0.5x + 2, a = - 4, b = 2.

For triangle NMC we have: y = - x + 5, i.e. f(x) = - x + 5, a = 2, b = 5.

By calculating the area of ​​each triangle and adding the results, we find:

sq. units

sq. units

9 + 4, 5 = 13.5 sq. units Check: = 0.5AC = 0.5 sq. units

Example 3. Calculate the area of ​​a figure bounded by lines: y = x 2 , y = 0, x = 2, x = 3.

In this case, you need to calculate the area of ​​a curved trapezoid bounded by the parabola y = x 2 , straight lines x = 2 and x = 3 and the Ox axis (see figure) Using formula (1) we find the area of ​​the curvilinear trapezoid

= = 6 sq. units

Example 4. Calculate the area of ​​the figure bounded by the lines: y = - x 2 + 4 and y = 0

Let's construct the figure. The required area is enclosed between the parabola y = - x 2 + 4 and the Ox axis.

Let's find the intersection points of the parabola with the Ox axis. Assuming y = 0, we find x = Since this figure is symmetrical about the Oy axis, we calculate the area of ​​the figure located to the right of the Oy axis, and double the result obtained: = +4x]sq. units 2 = 2 sq. units

Example 5. Calculate the area of ​​a figure bounded by lines: y 2 = x, yx = 1, x = 4

Here you need to calculate the area of ​​a curvilinear trapezoid bounded by the upper branch of the parabola 2 = x, Ox axis and straight lines x = 1 and x = 4 (see figure)

According to formula (1), where f(x) = a = 1 and b = 4, we have = (= sq. units.

Example 6 . Calculate the area of ​​the figure bounded by the lines: y = sinx, y = 0, x = 0, x= .

The required area is limited by the half-wave of the sinusoid and the Ox axis (see figure).

We have - cosx = - cos = 1 + 1 = 2 sq. units

Example 7. Calculate the area of ​​the figure bounded by the lines: y = - 6x, y = 0 and x = 4.

The figure is located under the Ox axis (see figure).

Therefore, we find its area using formula (3)

= =

Example 8. Calculate the area of ​​the figure bounded by the lines: y = and x = 2. Construct the y = curve from the points (see figure). Thus, we find the area of ​​the figure using formula (4)

Example 9 .

X 2 + y 2 = r 2 .

Here you need to calculate the area enclosed by the circle x 2 + y 2 = r 2 , i.e. the area of ​​a circle of radius r with the center at the origin. Let's find the fourth part of this area by taking the limits of integration from 0

before; we have: 1 = = [

Hence, 1 =

Example 10. Calculate the area of ​​a figure bounded by lines: y= x 2 and y = 2x

This figure is limited by the parabola y = x 2 and the straight line y = 2x (see figure) To determine the intersection points of the given lines, we solve the system of equations: x 2 – 2x = 0 x = 0 and x = 2

Using formula (5) to find the area, we obtain

= \- -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 the area of ​​​​the 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 function. A figure bounded by a curve? (?) and rays? = ?, ? = ?, is called a curvilinear sector. The area of ​​the curvilinear sector is equal to

Finding the Arc Length of a Curve

Rectangular coordinates

Let a plane curve AB be given in rectangular coordinates, the equation of which is y = f(x), where a? x? b. (Figure 2)

The length of the arc AB is understood as the limit to which the length of a broken line inscribed in this arc tends when the number of links of the broken line increases indefinitely, and the length of its largest link tends to zero.

Let's apply scheme I (sum method).

Using points X = a, X, …, X = b (X ? X? … ? X) we divide the segment into n parts. Let these points correspond to the points M = A, M, …, M = B on the curve AB. Let us draw chords MM, MM, …, MM, the lengths of which will be denoted by ?L, ?L, …, ?L, respectively.

We obtain a broken line MMM ... MM, the length of which is equal to L = ?L+ ?L+ ... + ?L = ?L.

The length of a chord (or a broken line link) ?L can be found using the Pythagorean theorem from a triangle with legs ?X and ?Y:

L = , where?X = X - X, ?Y = f(X) - f(X).

By Lagrange's theorem on the finite increment of a function

Y = (C) ?X, where C (X, X).

and the length of the entire broken line MMM...MM is equal to

The length of the curve AB, by definition, is equal to

Note that when ?L 0 also ?X 0 (?L = and therefore | ?X |< ?L). Функция непрерывна на отрезке , так как, по условию, непрерывна функция f (X). Следовательно, существует предел интегральной суммы L=?L= , кода max ?X 0:

Thus L = dx.

Example: Find the circumference of a circle of radius R. (Figure 3)

Shall we find it? part of its length from point (0; R) to point (R; 0). Because