Consider a curvilinear trapezoid bounded by the Ox axis, the curve y \u003d f (x) and two straight lines: x \u003d a and x \u003d b (Fig. 85). Let's take an arbitrary value of x (but not a and not b). We give it an increment h \u003d 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. This strip will be called an elementary strip. The area of \u200b\u200ban elementary strip differs from the area of \u200b\u200bthe rectangle ACQB by a curvilinear triangle BQD, and the area of \u200b\u200bthe latter is less than the area of \u200b\u200bthe rectangle BQDM with sides BQ \u003d h \u003d dx) QD \u003d Ay and an area equal to hAy \u003d Ay dx. As the side h decreases, the side Du also decreases and simultaneously with h tends to zero. Therefore, the area of \u200b\u200bthe BQDM is second-order infinitesimal. The area of \u200b\u200ban elementary strip is the increment of the area, and the area of \u200b\u200bthe rectangle ACQB, equal to AB-AC \u003d\u003d / (x) dx\u003e is the area differential. Consequently, we find the area itself by integrating its differential. Within the considered figure, the independent variable l: varies from a to b, so the required area 5 will be equal to 5 \u003d \\ f (x) dx. (I) Example 1. Let us calculate the area bounded by the parabola y - 1 -x *, straight lines X \u003d - Fj-, x \u003d 1 and the axis O * (Fig. 86). Fig. 87. Fig. 86.1 Here f (x) \u003d 1 - n?, The limits of integration are a \u003d - and t \u003d 1, therefore J [* -m] \\ - -fl - T -1- ± A_ 1V1 -ll-Ii- 3) | _ 2 3V 2 / J 3 24 24 * Example 2. Calculate the area bounded by the sinusoid y \u003d sinXy by the Ox axis and the straight line (Fig. 87). Applying formula (I), we obtain Л 2 S \u003d J sinxdx \u003d [-cos x] Q \u003d 0 - (- 1) \u003d lf Example 3. Calculate the area bounded by the arc of the sinusoid ^ y \u003d 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 it is clear from geometric considerations that this area will be twice the area of \u200b\u200bthe previous example. However, let's do the calculations: I 5 \u003d | s \\ nxdx \u003d [- cosx) * - - cos i - (- cos 0) \u003d 1 + 1 \u003d 2. o Indeed, our assumption turned out to be true. Example 4. Calculate the area bounded by a sinusoid and ^ by the Ox axis in one pe-x period (Fig. 88). Preliminary considerations allow us to assume that the area will be four times larger than in pr. 2. However, after making calculations, we get "i G, * i S - \\ sin x dx \u003d [- cos x] 0 \u003d \u003d - cos 2l - (- cos 0) \u003d - 1 + 1 \u003d 0. This result requires clarification. To clarify the essence of the matter, we also calculate the area bounded by the same sinusoid y \u003d sin l: and the Ox axis in the range from l to 2i. Applying formula (I), we get 2l $ 2l sin xdx \u003d [- cosx] l \u003d -cos 2ya ~) -c05ya \u003d - 1-1 \u003d -2. Thus, we see that this area turned out to be negative. Comparing it with the area calculated in pr. 3, we find that their absolute values \u200b\u200bare the same, but the signs are different. If we apply property V (see Ch. XI, § 4), then we get 2l i 2l J sin xdx \u003d J sin * dx [sin x dx \u003d 2 + (- 2) \u003d 0 What happened in this example is not by chance. Always the area below the Ox axis, provided that the independent variable changes from left to right, is obtained by calculating negative integrals. In this course, we will always consider areas without signs. Therefore, the answer in the just analyzed example will be: the required area is equal to 2 + | -2 | \u003d 4. Example 5. Let us calculate the area of \u200b\u200bthe BA shown in fig. 89. This area is bounded by the Ox axis, the parabola y \u003d - xr and the straight line y - \u003d -x + \\. Curvilinear trapezoid area The search area OAV consists of two parts: OAM and MAV. Since point A is the intersection point of the parabola and the straight line, we will find its coordinates by solving the system of equations 3 2 Y \u003d mx. (we only need to find the abscissa of point A). Solving the system, we find l; \u003d ~. 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 and does not change its sign on it (fig. 1).The area of \u200b\u200ba curved trapezoid can be designated S (G).

The definite integral ʃ а b f (x) dx for the function f (x), which is continuous and non-negative on the interval [а; b], and is the area of \u200b\u200bthe corresponding curved trapezoid.

That is, to find the area of \u200b\u200bthe figure G, bounded by the lines y \u003d f (x), y \u003d 0, x \u003d a and x \u003d b, it is necessary to calculate the definite integral ʃ a b f (x) dx.

In this way, S (G) \u003d ʃ a b f (x) dx.

If the function y \u003d f (x) is not positive on [a; b], then the area of \u200b\u200ba curved trapezoid can be found by the formula S (G) \u003d -ʃ a b f (x) dx.

Example 1.

Calculate the area of \u200b\u200bthe figure bounded by the lines y \u003d x 3; y \u003d 1; x \u003d 2.

Decision.

The specified lines form the ABC figure, which is shown by hatching on fig. 2.

The required area is equal to the difference between the areas of the DACE curved trapezoid and the DABE square.

Using the formula S \u003d ʃ and b f (x) dx \u003d S (b) - S (a), we find the limits of integration. To do this, we solve a system of two equations:

(y \u003d x 3,
(y \u003d 1.

Thus, we have x 1 \u003d 1 - the lower limit and x \u003d 2 - the upper limit.

So, S \u003d S DACE - S DABE \u003d ʃ 1 2 x 3 dx - 1 \u003d x 4/4 | 1 2 - 1 \u003d (16 - 1) / 4 - 1 \u003d 11/4 (sq. Units).

Answer: 11/4 sq. units

Example 2.

Calculate the area of \u200b\u200bthe figure bounded by the lines y \u003d √x; y \u003d 2; x \u003d 9.

Decision.

The given lines form an ABC figure, which is bounded from above by the graph of the function

y \u003d √x, and below the graph of the function y \u003d 2. The resulting figure is shown by shading on fig. 3.

The required area is S \u003d ʃ a b (√x - 2). Let us find the limits of integration: b \u003d 9, to find a, we solve the system of two equations:

(y \u003d √x,
(y \u003d 2.

Thus, we have that x \u003d 4 \u003d a is the lower limit.

So, S \u003d ∫ 4 9 (√x - 2) dx \u003d ∫ 4 9 √x dx –∫ 4 9 2dx \u003d 2/3 x√x | 4 9 - 2x | 4 9 \u003d (18 - 16/3) - (18 - 8) \u003d 2 2/3 (sq. Units).

Answer: S \u003d 2 2/3 sq. units

Example 3.

Calculate the area of \u200b\u200bthe figure bounded by the lines y \u003d x 3 - 4x; y \u003d 0; x ≥ 0.

Decision.

Let's build a graph of the function y \u003d x 3 - 4x for x ≥ 0. To do this, find the derivative y ':

y '\u003d 3x 2 - 4, y' \u003d 0 at х \u003d ± 2 / √3 ≈ 1.1 are critical points.

If we depict the critical points on the numerical axis and arrange the signs of the derivative, then we find that the function decreases from zero to 2 / √3 and increases from 2 / √3 to plus infinity. Then x \u003d 2 / √3 is the minimum point, the minimum value of the function is min \u003d -16 / (3√3) ≈ -3.

Let's define the points of intersection of the graph with the coordinate axes:

if x \u003d 0, then y \u003d 0, which means that A (0; 0) is the point of intersection with the Oy axis;

if y \u003d 0, then x 3 - 4x \u003d 0 or x (x 2 - 4) \u003d 0, or x (x - 2) (x + 2) \u003d 0, whence x 1 \u003d 0, x 2 \u003d 2, x 3 \u003d -2 (not suitable, since x ≥ 0).

Points A (0; 0) and B (2; 0) are the points of intersection of the graph with the Ox axis.

The specified lines form an OAB figure, which is shown by hatching on fig. 4.

Since the function y \u003d x 3 - 4x takes a negative value on (0; 2), then

S \u003d | ʃ 0 2 (x 3 - 4x) dx |.

We have: ʃ 0 2 (x 3 - 4x) dx \u003d (x 4/4 - 4x 2/2) | 0 2 \u003d -4, whence S \u003d 4 sq. units

Answer: S \u003d 4 sq. units

Example 4.

Find the area of \u200b\u200bthe figure bounded by the parabola y \u003d 2x 2 - 2x + 1, the straight lines x \u003d 0, y \u003d 0 and the tangent to this parabola at the point with the abscissa x 0 \u003d 2.

Decision.

First, we compose the equation of the tangent to the parabola y \u003d 2x 2 - 2x + 1 at the point with the abscissa x₀ \u003d 2.

Since the derivative y ’\u003d 4x - 2, then at x 0 \u003d 2 we get k \u003d y’ (2) \u003d 6.

Find the ordinate of the touching point: y 0 \u003d 2 2 2 - 2 2 + 1 \u003d 5.

Therefore, the tangent equation has the form: y - 5 \u003d 6 (x - 2) or y \u003d 6x - 7.

Let's draw a shape bounded by lines:

y \u003d 2x 2 - 2x + 1, y \u003d 0, x \u003d 0, y \u003d 6x - 7.

G y \u003d 2x 2 - 2x + 1 - parabola. Points of intersection with the coordinate axes: A (0; 1) - with the Oy axis; with the Ox axis - there are no intersection points, because the equation 2x 2 - 2x + 1 \u003d 0 has no solutions (D< 0). Найдем вершину параболы:

x b \u003d 2/4 \u003d 1/2;

y b \u003d 1/2, that is, the vertex of the parabola point B has coordinates B (1/2; 1/2).

So, the figure whose area you want to determine is shown by hatching on fig. five.

We have: S О A В D \u003d S OABC - S ADBC.

Find the coordinates of point D from the condition:

6x - 7 \u003d 0, i.e. x \u003d 7/6, so DC \u003d 2 - 7/6 \u003d 5/6.

The area of \u200b\u200bthe triangle DBC is found by the formula S ADBC \u200b\u200b\u003d 1/2 DC BC. In this way,

S ADBC \u200b\u200b\u003d 1/2 5/6 5 \u003d 25/12 sq. units

S OABC \u003d ʃ 0 2 (2x 2 - 2x + 1) dx \u003d (2x 3/3 - 2x 2/2 + x) | 0 2 \u003d 10/3 (sq. Units).

Finally, we get: S О A В D \u003d S OABC - S ADBC \u200b\u200b\u003d 10/3 - 25/12 \u003d 5/4 \u003d 1 1/4 (square units).

Answer: S \u003d 1 1/4 sq. units

We have analyzed examples finding the areas of figures bounded by specified lines... To successfully solve such problems, you need to be able to build lines and graphs of functions on the plane, find the intersection points of lines, apply a formula to find the area, which implies the presence of skills and abilities to calculate certain integrals.

site, with full or partial copying of the material, a link to the source is required.

Any definite integral (that exists) has a very good geometric meaning. In the lesson, I said that a definite integral is a number. And now it's time to state another useful fact. From the point of view of geometry, the definite integral is the AREA.

I.e, a definite integral (if it exists) geometrically corresponds to the area of \u200b\u200bsome figure... For example, consider a definite integral. The integrand defines a certain curve on the plane (it can always be drawn if desired), and the definite integral itself is numerically equal to the area of \u200b\u200bthe corresponding curvilinear trapezoid.

Example 1

This is a typical task statement. The first and most important point of the solution is the construction of the drawing... Moreover, the drawing must be built CORRECTLY.

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

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

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

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

On the segment, the function graph is located over axis, so:

Answer:

Who has difficulty calculating a definite integral and applying the Newton-Leibniz formula , refer to the lecture Definite integral. Solution examples.

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

Example 2

Calculate the area of \u200b\u200ba shape bounded by lines, and an axis

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

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

Example 3

Calculate the area of \u200b\u200ba shape bounded by lines and coordinate axes.

Solution: Let's execute the drawing:

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

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

1) If you are asked to solve just a definite integral without any geometric meaning, then it can be negative.

2) If you are asked to find the area of \u200b\u200ba figure using a definite integral, then the area is always positive! That is why minus appears in the formula just considered.

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

Example 4

Find the area of \u200b\u200ba flat figure bounded by lines,.

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

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

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

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

I repeat that in the case of a point-by-point construction, the limits of integration are most often found out “automatically”.

And now the working formula: If on a segment some continuous function greater than or equal some continuous function, then the area of \u200b\u200bthe corresponding figure can be found by the formula:

Here you no longer need to think about where the figure is located - above the axis or below the axis, and, roughly speaking, it is important which schedule is ABOVE(relative to another graph), and which one is BELOW.

In the example under consideration, it is obvious that on the segment the parabola is located above the straight line, and therefore it is necessary to subtract from

Completing the solution might look like this:

The required figure is bounded by a parabola at the top and a straight line at the bottom.
On the segment, according to the corresponding formula:

Answer:

In fact, the school formula for the area of \u200b\u200ba curved trapezoid in the lower half-plane (see simple example No. 3) is a special case of the formula ... Since the axis is given by the equation, and the graph of the function is located below the axis, then

And now a couple of examples for an independent solution

Example 5

Example 6

Find the area of \u200b\u200bthe figure bounded by lines,.

In the course of solving problems of calculating the area using a definite integral, a funny incident sometimes happens. The drawing is done correctly, the calculations are correct, but through inattention ... the area of \u200b\u200bthe wrong figure is found, this is how your humble servant screwed up several times. Here's a real life case:

Example 7

Calculate the area of \u200b\u200bthe figure bounded by the lines,,,.

First, let's execute the drawing:

The figure whose area we need to find is shaded in blue(carefully look at the condition - what the figure is limited by!). But in practice, due to inattention, it often arises that you need to find the area of \u200b\u200bthe figure, which is shaded in green!

This example is also useful in that it calculates the area of \u200b\u200ba figure using two definite integrals. Really:

1) On the segment above the axis there is a line graph;

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

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

Answer:

Example 8

Calculate the area of \u200b\u200ba shape bounded by lines
Let's represent the equations in the "school" form, and execute a point-by-point drawing:

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

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

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

Hence, .

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

On the segment , according to the corresponding formula:

Answer:

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

Example 9

Calculate the area of \u200b\u200ba figure bounded by lines,

Solution: Draw this figure in the drawing.

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

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

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

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

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

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

New integration redistributions:

Who is really bad with substitutions, please go to the lesson Replacement method in the indefinite integral... To whom the replacement algorithm in a definite integral is not very clear, visit the page Definite integral. Solution examples.

Introduction

Finding the derivative f "(x) or the differential df \u003d f" (x) dx of the function f (x) is the main problem of differential calculus. In integral calculus, the inverse problem is solved: for a given function f (x), it is required to find a function F (x) such that F "(x) \u003d f (x) or F (x) \u003d F" (x) dx \u003d f (x ) dx. Thus, the main task of the integral calculus is to restore the function F (x) from the known derivative (differential) of this function. Integral calculus has numerous applications in geometry, mechanics, physics, and engineering. It provides a general method for finding areas, volumes, centers of gravity, etc.

The course of mathematical analysis contains a variety of material, however, one of its central sections is the definite integral. The integration of many kinds of functions is sometimes one of the most difficult problems in mathematical analysis.

The calculation of a definite integral is not only of theoretical interest. Sometimes tasks associated with the practical activity of a person are reduced to its calculation.

Also, the concept of a definite integral is widely used in physics.

Finding the area of \u200b\u200ba curved trapezoid

A curved trapezoid is a figure located in a rectangular coordinate system and bounded by the abscissa axis, straight x \u003d a and x \u003d b and a curve, and is non-negative on a segment. The approximate area of \u200b\u200ba curved trapezoid can be found as follows:

1.divide the abscissa segment by n equal segments;

2. draw through the dividing points the segments perpendicular to the abscissa axis until the intersection with the curve;

3.replace the resulting columns with rectangles with base and height equal to the value of the function f at the left end of each segment;

4. find the sum of the areas of these rectangles.

But you can find the curvilinear area differently: by the Newton-Leibniz formula. To prove the formula bearing their names, we will prove that the area of \u200b\u200ba curvilinear trapezoid is equal to, where is any of the antiderivatives of the function whose graph bounds the curvilinear trapezoid.

The calculation of the area of \u200b\u200ba curved trapezoid is written as follows:

1. find any of the antiderivatives of the function.

2. is recorded. is the Newton-Leibniz formula.

Finding the area of \u200b\u200ba curved sector

Consider a curve? \u003d? (?) in a polar coordinate system, where? (?) - continuous and non-negative on [?; ?] function. A shape bounded by a curve? (?) and rays? \u003d?,? \u003d?, is called a curved sector. The area of \u200b\u200bthe curved sector is

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 \u003d f (x), where a? x? b. (fig 2)

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

Let us apply Scheme I (the method of sums).

The points X \u003d a, X,…, X \u003d b (X? X?…? X) divide the segment into n parts. Let these points correspond to points M \u003d A, M,…, M \u003d 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 get a broken line MMM ... MM, the length of which is L \u003d? L +? L + ... +? L \u003d? L.

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

L \u003d, where? X \u003d X - X,? Y \u003d f (X) - f (X).

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

Y \u003d (C)? X, where C (X, X).

and the length of the entire broken MMM ... MM is

The length of the curve AB, by definition, is

Note that for? L 0 also? X 0 (? L \u003d and hence |? 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:

So L \u003d dx.

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

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

No. ____ Date ________

Theme:Curved trapezoid and its area b

Lesson objectives: Give definitions of a curved trapezoid and its area, learn how to calculate the area of \u200b\u200ba curved trapezoid.

DURING THE CLASSES

1. Organizational moment.

Greeting students, checking the readiness of the class for the lesson, organizing the attention of students, disclosing the general goals of the lesson and its plan.

2. Stage of homework checking.

Tasks: To establish the correctness, completeness and awareness of the fulfillment of d / s by all students, to identify gaps in the knowledge and methods of student activity. Determine the causes of difficulties, eliminate the detected gaps.

3. Stage of actualization.

Tasks: providing motivation for the learning of schoolchildren, inclusion in joint activities to determine the goals of the lesson. To actualize the subjective experience of students.

Let's recall the basic concepts and formulas.

Definition. Function y \u003df(x), x (a, b), is called the antiderivative for the function y \u003d f (x), x (a, b), if for everyone x (a, b) equality holds

F (x) \u003d f (x).

Comment. If a f(x) there is an antiderivative for the function f (x), then for any constant FROM, F (x) + C is also an antiderivative for f (x).

The problem of finding all antiderivatives of a function f (x) is called integration, and the set of all antiderivatives is called the indefinite integral for the function f (x) by dx and denoted

The properties take place:

1. ;

2. If a C \u003dConst, then
;

3.
.

Comment. The school mathematics course does not use the term "indefinite integral", instead they say "the set of all antiderivatives."

Here is a table of indefinite integrals.

Example 1. Find the Antiderivative for a Function
passing through the point M(2;4).

Decision. Set of all antiderivatives
there is an indefinite integral
... We calculate it using the properties of the integral 1 and 2. We have:

We got that the set of all antiderivatives is given by the family of functions y \u003d F (x) + C, i.e y \u003d x 3 – 2x + Cwhere FROM Is an arbitrary constant.

Knowing that the antiderivative goes through the point M(2; 4), substitute its coordinates into the previous expression and find FROM.

4=2 3 –2 2+FROM  FROM=4–8+4; FROM=0.

Answer: F (x) \u003d x 3 - 2x Is the desired antiderivative.

4. Formation of new concepts and methods of action.

Tasks: Provide the perception, comprehension and memorization of the studied material by students. To ensure the assimilation by students of the methods of reproducing the studied material, to promote philosophical comprehension of the concepts, laws, rules, formulas being mastered. To establish the correctness and awareness of the studied material by the students, to identify the gaps in the primary comprehension, to carry out the correction. Ensure that students relate their subjective experience to the signs of scientific knowledge.

Finding the Areas of Plane Figures

The problem of finding the area of \u200b\u200ba flat figure is closely related to the problem of finding antiderivatives (integration). Namely: the area of \u200b\u200bthe curvilinear trapezoid limited by the graph of the functiony \u003d f (x) (f (x)\u003e 0) straightx \u003d a; x \u003d b; y \u003d 0 is equal to the difference between the values \u200b\u200bof the antiderivative for the functiony \u003d f (x) in pointsb anda :

S \u003d F (b) –F (a)

Let us give the definition of a definite integral.

ABOUT
definition. Let the function y \u003d f (x) is defined and integrable on the interval [ a, b] let it go F (x) - some of its antiderivative. Then the number F (b) –F (a) is called the integral of and before b functions f (x) and denoted

.

Equality
is called the Newton – Leibniz formula.

This formula connects the problem of finding the area of \u200b\u200ba flat figure with an integral.

In general, if the figure is limited by the graphs of functions y \u003d f (x); y \u003d g (x) (f (x)\u003e g (x)) and straight lines x \u003d a; x \u003d b, then its area is equal to:

.

Example 2. At what point in the graph of the function y \u003d x 2 + 1, you need to draw a tangent so that it cuts off from the figure formed by the graph of this function and straight lines y \u003d0, x \u003d0, x \u003d1 trapezoid of the largest area?

Decision. Let be M 0 (x 0 , y 0 ) - function graph point y \u003d x 2 + 1, in which the required tangent is drawn.

Find the tangent equation y \u003d y 0 + f (x 0 ) (x – x 0 ) .

We have:

therefore

.

Find the area of \u200b\u200bthe trapezoid OABS.

.

B - point of intersection of a tangent with a straight line x \u003d1 

The task was reduced to finding the largest function value

S(x)\u003d –X 2 + x +1 on the segment. Find S (x)=– 2x +1. Find the critical point from the condition S (x)= 0  x \u003d.

We see that the function reaches its maximum value at x \u003d... Find
.

Answer: the tangent must be drawn at the point
.

Note that the problem of finding the integral based on its geometric meaning is often encountered. Let us show by an example how such a problem is solved.

Example 4. Using the geometric meaning of the integral, calculate

and )
; b)
.

Decision.

and)
- equal to the area of \u200b\u200bthe curved trapezoid bounded by lines.

P transform

- the upper half of the circle with the center R(1; 0) and radius R \u003d1.

therefore
.

Answer:
.

b) Arguing similarly, we construct the area bounded by the graphs. 2 – 2x +2 tangent to it at points A
, B(4;2)

y \u003d–9x–59, parabola y \u003d3x 2 + ax +1 if it is known that the tangent to the parabola at the point x \u003d -2 make up with axle Ox angle of arctg6.

To find and, if it is known that the area of \u200b\u200ba curved trapezoid bounded by lines y \u003d3x 3 + 2x, x \u003d a, y \u003d0 is equal to one.

Find the smallest value of the area of \u200b\u200ba figure bounded by a parabola y \u003d x 2 + 2x–3 and straight y \u003d kx +1.

6. The homework information stage.

Objectives: To provide students with an understanding of the purpose, content and ways of doing homework. # 18, 19, 20, 21 odd

7. Summing up the results of the lesson.

Objective: To give a qualitative assessment of the work of the class and individual students.