How to find the area of a curved shape. Integral. Calculating areas using an integral. Find an "external" reason that does not allow you to calculate the area of the figure
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TwitterIt follows from the definition that for a nonnegative function f (x) the definite integral is equal to the area of a curvilinear trapezoid bounded by the curve y = f (x), straight lines x = a, x = b and the abscissa axis y = 0 (Figure 4.1).
If the function - f (x) is non-positive, then the definite integral is equal to the area of the corresponding curved trapezoid, taken with a minus sign (Figure 4.7).
Figure 4.7 - Geometric meaning definite integral for non-positive function
For an arbitrary continuous function f (x), the definite integral 
is equal to the sum of the areas of curvilinear trapeziums lying under the graph of the function f (x) and above the abscissa axis, minus the sum of the areas of curvilinear trapeziums lying above the graph of the function f (x) and below the abscissa (Figure 4.8).
Figure 4.8 - The geometric meaning of a definite integral for an arbitrary continuous function f (x) (the plus sign marks the area that is being added, and the minus sign marks the area that is subtracted).
When calculating in practice the areas of curved shapes, the following formula is often used: 
, where S is the area of the figure enclosed between the curves y = f 1 (x) and y = f 2 (x) on the segment [a, b], and f 1 (x) and f 2 (x) are continuous functions defined on this segment such that f 1 (x) ≥ f 2 (x) (see Figures 4.9, 4.10).
When studying the economic meaning of the derivative, it was found that the derivative acts as the rate of change of some economic object or process in time or relative to another investigated factor. To establish the economic meaning of a definite integral, it is necessary to consider this speed itself as a function of time or other factor. Then, since a certain integral is a change in the antiderivative, we get that in economics it evaluates the change in this object (process) over a certain period of time (or with a certain change in another factor).
For example, if the function q = q (t) describes labor productivity as a function of time, then a definite integral of this function 
represents the volume of products Q for the period from t 0 to t 1.
Methods for calculating definite integrals are based on the previously considered integration methods (we will not carry out proofs).
To find the indefinite integral, we used the variable change method based on the formula: f (x) dx = f ( (t)) ` (t) dt, where x = (t) is a function differentiable on the considered interval. For a definite integral, the variable change formula takes the form 
, where 
and for everyone.
Example 1... To find
Let t = 2 –x 2. Then dt = -2xdx and xdx = - ½dt.
For x = 0 t = 2 - 0 2 = 2. For x = 1t = 2 - 1 2 = 1. Then
Example 2... To find
Example 3... To find
The formula for integration by parts for a definite integral will take the form: 
, where 
.
Example 1... To find
Let u = ln (1 + x), dv = dx. Then
Example 2... To find
Calculating the areas of planar figures using a definite integral
Example 1. Find the area of a figure, limited by lines y = x 2 - 2 and y = x.
The graph of the function y = x 2 - 2 is a parabola with a minimum point at x = 0, y = -2; the abscissa axis intersects at points 
... The graph of the function y = x is a straight line, the bisector of a non-negative coordinate quarter.
Find the coordinates of the intersection points of the parabola y = x 2 - 2 and the straight line y = x, solving the system of these equations:
x 2 - x - 2 = 0
x = 2; y = 2 or x = -1; y = -1
Thus, the figure whose area needs to be found can be represented in Figure 4.9.
Figure 4.9 - The figure bounded by the lines y = x 2 - 2 and y = x
On the segment [-1, 2] x ≥ x 2 - 2.
Let's use the formula 
, setting f 1 (x) = x; f 2 (x) = x 2 - 2; a = -1; b = 2.
Example 2. Find the area of the figure bounded by the lines y = 4 - x 2 and y = x 2 - 2x.
The graph of the function y = 4 - x 2 is a parabola with a maximum point at x = 0, y = 4; the abscissa axis intersects at points 2 and -2. The graph of the function y = x 2 - 2x is a parabola with a minimum point at 2x- 2 = 0, x = 1; y = -1; the abscissa axis intersects at points 0 and 2.
Find the coordinates of the intersection points of the curves:
4 - x 2 = x 2 - 2x
2x 2 - 2x - 4 = 0
x 2 - x - 2 = 0
x = 2; y = 0 or x = -1; y = 3
Thus, the figure whose area is to be found can be represented in Figure 4.10.
Figure 4.10 - The figure bounded by the lines y = 4 - x 2 and y = x 2 - 2x
On the segment [-1, 2] 4 - x 2 ≥ x 2 - 2x.
Let's use the formula 
, setting f 1 (x) = 4 - - x 2; f 2 (x) = x 2 - 2x; a = -1; b = 2.
Example 3. Find the area of the figure bounded by the lines y = 1 / x; y = x 2 and y = 4 in a non-negative coordinate quarter.
The graph of the function y = 1 / x is a hyperbola, for positive x it is convex downward; the coordinate axes are asymptotes. The graph of the function y = x 2 in a non-negative coordinate quarter is a branch of a parabola with a minimum point at the origin. These graphs intersect at 1 / x = x 2; x 3 = 1; x = 1; y = 1.
The straight line y = 4, the graph of the function y = 1 / x intersects at x = 1/4, and the graph of the function y = x 2 at x = 2 (or -2).
Thus, the figure whose area needs to be found can be represented in Figure 4.11.
Figure 4.11 - The figure bounded by the lines y = 1 / x; y = x 2 and y = 4 in non-negative coordinate quarter
The sought area of the figure ABC is equal to the difference between the area of the rectangle ABHE, which is 4 * (2 - ¼) = 7, and the sum of the areas of two curvilinear trapezoids ACFE and CBHF. Let's calculate the area ACFE:
We calculate the area of CBHF:
.
So, the required area is 7 - (ln4 + 7/3) = 14/3 –ln43.28 (unit 2).
Calculating the area of a shape- this is perhaps one of the most difficult problems in the theory of areas. In school geometry, they teach how to find the areas of the main geometric shapes such as, for example, triangle, rhombus, rectangle, trapezoid, circle, etc. However, you often have to deal with calculating the areas of more complex shapes. It is when solving such problems that it is very convenient to use integral calculus.
Definition.
Curved trapezoid is called some figure G, bounded by the lines y = f (x), y = 0, x = a and x = b, and the function f (x) is continuous on the segment [a; b] and does not change its sign on it (fig. 1). The area of a 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 the corresponding curved trapezoid.
That is, to find the area of the figure G, bounded by the lines y = f (x), y = 0, x = a and x = b, it is necessary to calculate the definite integral ʃ a b f (x) dx.
Thus, S (G) = ʃ a b f (x) dx.
If the function y = f (x) is not positive on [a; b], then the area of a curved trapezoid can be found by the formula S (G) = -ʃ a b f (x) dx.
Example 1.
Calculate the area of the figure bounded by the lines y = x 3; y = 1; x = 2.
Solution.
The specified lines form the ABC figure, which is shown by hatching on rice. 2.
The desired area is equal to the difference between the areas of the DACE curved trapezoid and the DABE square.
Using the formula S = ʃ and b f (x) dx = S (b) - S (a), we find the limits of integration. To do this, we will solve a system of two equations:
(y = x 3,
(y = 1.
Thus, we have x 1 = 1 - the lower limit and x = 2 - the upper limit.
So, S = S DACE - S DABE = ʃ 1 2 x 3 dx - 1 = x 4/4 | 1 2 - 1 = (16 - 1) / 4 - 1 = 11/4 (sq. Units).
Answer: 11/4 sq. units
Example 2.
Calculate the area of the figure bounded by the lines y = √x; y = 2; x = 9.
Solution.
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 rice. 3.
The required area is S = ʃ a b (√x - 2). Let us find the limits of integration: b = 9, to find a, we solve the system of two equations:
(y = √x,
(y = 2.
Thus, we have that x = 4 = a - this is the lower limit.
So, S = ∫ 4 9 (√x - 2) dx = ∫ 4 9 √x dx –∫ 4 9 2dx = 2/3 x√x | 4 9 - 2x | 4 9 = (18 - 16/3) - (18 - 8) = 2 2/3 (sq. Units).
Answer: S = 2 2/3 sq. units
Example 3.
Calculate the area of the figure bounded by the lines y = x 3 - 4x; y = 0; x ≥ 0.
Solution.
Let's construct a graph of the function y = x 3 - 4x for x ≥ 0. To do this, we find the derivative y ':
y ’= 3x 2 - 4, y’ = 0 at x = ± 2 / √3 ≈ 1.1 are critical points.
If we depict the critical points on the numerical axis and arrange the signs of the derivative, we get that the function decreases from zero to 2 / √3 and increases from 2 / √3 to plus infinity. Then x = 2 / √3 is a minimum point, the minimum value of the function is min = -16 / (3√3) ≈ -3.
Let's define the points of intersection of the graph with the coordinate axes:
if x = 0, then y = 0, which means that A (0; 0) is the point of intersection with the Oy axis;
if y = 0, then x 3 - 4x = 0 or x (x 2 - 4) = 0, or x (x - 2) (x + 2) = 0, whence x 1 = 0, x 2 = 2, x 3 = -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 shape, which is shown by hatching on rice. 4.
Since the function y = x 3 - 4x takes on (0; 2) negative meaning, then
S = | ʃ 0 2 (x 3 - 4x) dx |.
We have: ʃ 0 2 (x 3 - 4x) dx = (x 4/4 - 4x 2/2) | 0 2 = -4, whence S = 4 sq. units
Answer: S = 4 sq. units
Example 4.
Find the area of the figure bounded by the parabola y = 2x 2 - 2x + 1, the straight lines x = 0, y = 0 and the tangent to this parabola at the point with the abscissa x 0 = 2.
Solution.
First, we compose the equation of the tangent to the parabola y = 2x 2 - 2x + 1 at the point with the abscissa x₀ = 2.
Since the derivative y ’= 4x - 2, then at x 0 = 2 we get k = y’ (2) = 6.
Find the ordinate of the touching point: y 0 = 2 2 2 - 2 2 + 1 = 5.
Therefore, the equation of the tangent has the form: y - 5 = 6 (x - 2) or y = 6x - 7.
Let's draw a shape bounded by lines:
y = 2x 2 - 2x + 1, y = 0, x = 0, y = 6x - 7.
G y = 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 = 0 has no solutions (D< 0). Найдем вершину параболы:
x b = 2/4 = 1/2;
y b = 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 rice. five.
We have: S О A В D = S OABC - S ADBC.
Find the coordinates of point D from the condition:
6x - 7 = 0, i.e. x = 7/6, so DC = 2 - 7/6 = 5/6.
The area of the triangle DBC is found by the formula S ADBC = 1/2 DC BC. Thus,
S ADBC = 1/2 5/6 5 = 25/12 sq. units
S OABC = ʃ 0 2 (2x 2 - 2x + 1) dx = (2x 3/3 - 2x 2/2 + x) | 0 2 = 10/3 (sq. Units).
Finally, we get: S О A В D = S OABC - S ADBC = 10/3 - 25/12 = 5/4 = 1 1/4 (square units).
Answer: S = 1 1/4 sq. units
We have analyzed examples finding the areas of figures limited given 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 for finding the area, which implies the presence of skills and abilities to calculate certain integrals.
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Definite integral. How to calculate the area of a shape
We now turn to the consideration of applications of integral calculus. In this lesson we will analyze a typical and most common task. - how to calculate the area of a flat figure using a definite integral... Finally, those who are looking for meaning in higher mathematics - may they find it. You never know. We'll have to bring the suburban area closer in life with 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 familiarize themselves with the lesson Not.
2) Be able to apply the Newton-Leibniz formula and calculate a definite integral. Establish warm friendly relations with definite integrals you can on the page Definite integral. Examples of solutions.
In fact, in order to find the area of a figure, one does not need so much knowledge of the indefinite and definite integral. The task "calculate area using a definite integral" always involves building a drawing, therefore, your knowledge and drawing skills will be a much more pressing issue. In this regard, it is useful to refresh the memory of the graphs of the basic elementary functions, and, at least, to be able to construct a straight line, a parabola and a hyperbola. This can be done (many people need) using methodological material and articles on geometric transformations of graphs.
Actually, everyone is familiar with the problem of finding the area using a definite integral since school, and we will go a little further from school curriculum... This article might not exist at all, but the fact is that the problem occurs in 99 cases out of 100, when a student suffers from the hated tower with enthusiasm mastering the course of 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.
Curved trapezoid is called a flat figure bounded by an axis, straight lines, and a graph of a continuous function on a segment, which does not change sign on this interval. Let this figure be located not less abscissa axis:
Then the area of a curved trapezoid is numerically equal to the definite integral... Any definite integral (that exists) has a very good geometric meaning. In the classroom Definite integral. Examples of solutions 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 some figure... For example, consider a definite integral. The integrand sets a curve on the plane that is located above the axis (those who wish can make a drawing), and the definite integral itself is numerically equal to the area of the corresponding curvilinear trapezoid.
Example 1
This is a typical formulation of the assignment. First and the most important moment solutions - drawing building... Moreover, the drawing must be built RIGHT.
When building a drawing, I recommend the following order: at first it is better to build all straight 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 reference material Graphs and properties of elementary functions... 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 graph of the function is located above the axis, so:
Answer:
Who has difficulty calculating a 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 helpful 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 obviously 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 a shape bounded by lines, and an axis
This is an example for a do-it-yourself solution. Complete solution and the answer at the end of the lesson.
What to do if the curved trapezoid is located under the axis?
Example 3
Calculate the area of the shape bounded by lines and coordinate axes.
Solution: Let's execute the drawing: 
If the curved trapezoid is located under the axis(or at least not higher given axis), 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 a 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 a 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 it were, "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 for finding the limits still has to be used 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.
We return 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 pointwise construction, the limits of integration are most often found out by an “automaton”.
And now the working formula: If on a segment some continuous function greater than or equal of some continuous function, then the area of the figure, bounded by the graphs of these functions and straight lines, 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
The completion of 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 a curvilinear 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 not higher axis, then 
And now a couple of examples for an independent solution
Example 5
Example 6
Find the area of the figure bounded by lines,.
In the course of solving problems for calculating the area using a definite integral, a funny incident sometimes happens. The drawing is done correctly, the calculations are correct, but inadvertently ... the area of the 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 the figure bounded by the lines,,,.
Solution: First, let's execute the drawing: 
... Eh, a lousy drawing came out, but everything seems to be legible.
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, a "glitch" often arises, that you need to find the area of the figure, which is shaded in green!
This example is also useful in that it calculates the area of a figure using two definite integrals. Really:
1) A line graph is located on the segment above the axis;
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:
Let's move on to one more meaningful task.
Example 8
Calculate the area of a shape bounded by lines,
Let's represent the equations in the "school" form, and execute a point-by-point drawing: 
It can be seen from the drawing that our upper limit 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 at all?
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: 
,
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 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 a figure bounded by lines,
Solution: Let's depict this figure in the drawing. 
Damn, I forgot to sign the schedule, but to redo the picture, sorry, not hotz. Not drawing, in short, today is the day =)
For point-by-point construction, you need to know appearance sinusoids (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 the 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:
Class: 11
Lesson presentation
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Lesson objectives: derive a formula for calculating the areas of flat figures using a definite integral; develop the skill of calculating the areas of flat figures using a definite integral; repeat known and report new information from the history of integral calculus; exam preparation; continue work on the development of attention, speech, logical thinking, accuracy in recording; improve the graphic culture; continue development work creativity students; to increase interest in the study of mathematics;
Equipment: multimedia projector, screen, presentation on the topic, developed in the Power Point environment.
During the classes
I. Organizational moment, communication of the topic and purpose of the lesson.
II. Homework check.
Checking additional homework(the teacher shows the solution in a previously prepared drawing, the solution with back side boards):
Calculate the area of the figure bounded by the graphs of the functions y = 1+ 3cos (x / 2), x = -π / 2, x = 3π / 2, y = 0
III. Updating basic knowledge.
1. Oral work(Slides 3-4)
- Express using the integral of the areas of the figures shown in the figures:
- Calculate the integrals:
2. A bit of history. ( Slides 5-9)
Fragment of a computer project of students on the topic "From the history of integral calculus."
1 student
Integral- one of the most important concepts of mathematics, which arose in connection with the need, on the one hand, to find functions by their derivatives, and on the other, to measure areas, volumes, lengths of arcs, the work of forces over a certain period of time, etc.
The very word integral came up with J. Bernoulli(1690). It comes from Latin integero, translated as to restore, restore.
Other terms you know about integral calculus appeared much later. Now used name antiderivative function replaced the earlier "Primitive function" introduced by Joseph Louis Lagrange(1797). Latin word primitivus translates as "initial".
The emergence of problems of integral calculus is associated with finding areas and volumes. A number of problems of this kind have been solved by mathematicians ancient greece... The first known method for calculating integrals is the Eudoxus exhaustion method ( about 370 BC BC), who tried to find areas and volumes, breaking them into an infinite number of parts for which the area or volume is already known. This method was picked up and developed by Archimedes, and was used to calculate the areas of parabolas and approximate the area of a circle.
However, Archimedes did not identify the general content of integration methods and concepts of the integral, and even more so did not create an algorithm for integral calculus.
The works of Archimedes, first written in 1544, were one of the most important starting points for the development of integral calculus.
2 student
The concept of an integral is directly related to integral calculus - a branch of mathematics dealing with the study of integrals, their properties and methods of calculation.
The integral came closer and more precisely to the concept Isaac Newton... He was the first to construct differential and integral calculus and called it "The method of fluxions ..." (1670-1671, publ. 1736). Variables Newton called fluents(current values, from lat... fluo - flow). The rate of change of the Newton fluent is fluxia, and the infinitesimal changes in the fluent necessary for calculating the fluxions are " moments"(Leibniz called them differentials.) Thus, Newton based the concepts of fluxions (derivative) and fluents (antiderivatives, or indefinite integrals).
This immediately made it possible to solve a wide variety of mathematical and physical problems.
Simultaneously with Newton, another outstanding scientist came to similar ideas - Gottfried Wilhelm Leibniz.
Reflecting on philosophical and mathematical questions, Leibniz became convinced that mathematics can become the most reliable means of seeking and finding truth in science. The integral sign (∫) was first used by Leibniz at the end of the 17th century. This symbol was formed from the letter S - an abbreviation of the word lat. summa(sum).
Newton and Leibniz developed two interpretations of the notion of an ordinary definite integral.
Newton interpreted a definite integral as the difference between the corresponding values antiderivative function:
,
where F` (x) = f (x).
For Leibniz, the definite integral was the sum of all infinitesimal differentials.
The formula that Newton and Leibniz discovered independently of each other was called Newton - Leibniz formula.
Thus, the concept of an integral was associated with the names of famous scientists: Newton, Leibniz, Bernoulli, who laid the foundation for modern mathematical analysis.
IV. Explanation of the new material.
Using the integral, you can calculate the areas of not only curvilinear trapezoids, but also plane figures of a more complex type.
Let the figure P limited to direct NS = a, x = b and function graphs y = f(x) and y = g(x), and on the segment [ a;b] the inequality g(x)f(x).
To calculate the area of a figure, we will argue as follows. Perform a parallel transfer of the figure P on the m units up so that the figure P ended up located in coordinate plane above the abscissa axis.
Now it is bounded above and below by graphs of functions y = f(x)+m and
y = g(x)+m, and both functions are continuous and non-negative on the interval [ a;b].
The resulting figure will be denoted ABCD... Its area can be found as the difference between the areas of the figures:
S ABCD = S aDCb - S aABb =
=
=
Thus, the area of the figure S bounded by straight lines NS = a, x = b and function graphs y = f(x) and y = g(x) continuous on the segment [ a;b] and such that for everyone NS from the segment [ a;b] the inequality g(x)f(x), is calculated by the formula
Example.(Slide 11) Calculate the area of the figure bounded by lines y = x, y = 5 – x, x = 1, x = 2.
Select from these formulas for calculating the area of a shape that fits one of the six drawings. (Slide 14)
Task 3.(Slide 15) Calculate the area of the figure bounded by the graph of the function y = 0,5x 2+ 2, tangent to this graph at a point with an abscissa NS= -2 and straight NS = 0.
1. Let's compose the equation of the tangent to the graph of the function y = 0,5x 2+ 2 at the point with the abscissa NS = -2:
y = f(x 0) + f"(x 0)(x - x 0)
f(-2) = 0,5∙(-2) 2 + 2 = 4
f"(x) = (0,5x 2 + 2)"= x
f"(-2) = -2
y = 4 – 2(x + 2)
y = -2x
2. Let's build graphs of functions.
3. Find the area of the figure ABC.
Vi. Summarizing.
- formula for calculating the areas of flat figures;
- writing the formulas for the areas of flat figures using a definite integral;
- repeating the equation of the tangent to the graph of the function and solving the equation with the modulus;
- grading students.
Vii. Homework.
- p. 4 pp. 228-230;
- No. 1025 (c, d), No. 1037 (c, d), No. 1038 (c, d)
textbook: A. G. Mordkovich "Algebra and the beginning of analysis 10-11"
26. Calculation of the areas of plane figures using a definite integral
Definition 1.Curved trapezoid generated by the graph of the non-negative function f on a segment is a figure bounded by a segment 
abscissa axis, line segments 
,
and the function graph 
on the 
.
1. We split the segment 
points to partial line segments.
2. In each segment 
(where k=1,2,...,n) choose an arbitrary point 
.
3. Let's calculate the areas of rectangles, which bases have line segments 
the abscissa axes, and the heights have lengths 
... Then the area of the stepped figure formed by these rectangles is 
.
Note that the smaller the lengths of the partial segments, the more the stepped figure is close in position to a given curvilinear trapezoid. Therefore, it is natural to give the following definition.
Definition 2.The area of a curved trapezoid, generated by the graph of the non-negative function f on the segment 
, is called the limit (as the lengths of all partial segments tend to 0) of the areas of stepped figures if:
1) this limit exists and is finite;
2) does not depend on the method of dividing the segment 
into partial segments;
3) does not depend on the choice of points 
.
Theorem 1.If the function
continuous and non-negative on the segment 
, then curved trapezoidF,graph-generated functionfon the 
, has an area, which is calculated by the formula 
.
Using a definite integral, you can calculate the areas of flat figures and more complex types.
If f and g- continuous and non-negative on the segment
functions, and for all x from segment
the inequality holds 
, then the area of the figure F limited by straight lines 
,
and function graphs 
,
, calculated by the formula
.
Comment. If we drop the condition of non-negativity of functions f and g, the last formula remains true.
Topic 9. Differential equations
27. The concept of a differential equation. General and specific solution. Cauchy problem. The problem of building a mathematical model of the demographic process
Theory differential equations arose at the end of the 17th century under the influence of the needs of mechanics and other natural science disciplines, essentially simultaneously with integral and differential calculus.
Definition 1.n-th order is an equation of the form in which 
- unknown function.
Definition 2. Function 
is called the solutions of the differential equation on the interval I if the substitution of this function and its derivatives, the differential equation turns into an identity.
Solve differential equation is to find all his solutions.
Definition 3. The graph of the solution to a differential equation is called integral curve differential equation.
Definition 4.Ordinary differential equation 1-th order is called an equation of the form 
.
Definition 5. Equation of the form 
called differential equation 1-th order,permitted with respect to the derivative.
As a rule, any differential equation has infinitely many solutions. In order to single out one of the totality of all decisions, additional conditions must be imposed.
Definition 6. Condition of the form 
imposed on the solution of a first-order differential equation is called initial condition, or the Cauchy condition.
Geometrically, this means that the corresponding integral curve passes through the point 
.
Definition 7.By general decision 1st order differential equation 
on a flat area D is called a one-parameter family of functions 
satisfying the conditions:
1) for any 
function 
is a solution to the equation;
2) for each point 
there is such a parameter value 
that the corresponding function 
is a solution to the equation satisfying the initial condition 
.
Definition 8. The solution obtained from the general solution for a certain value of the parameter is called by private decision differential equation.
Definition 9.A special solution differential equation is any solution that cannot be obtained from the general solution for any value of the parameter.
Solving differential equations is a very difficult task, and, generally speaking, the higher the order of the equation, the more difficult it is to indicate ways of solving the equation. Even for differential equations of the first order, it is possible only in a small number of special cases to indicate methods for finding a general solution. Moreover, in these cases, the desired solution is not always an elementary function.
One of the main problems of the theory of differential equations, first studied by O. Cauchy, is to find a solution to a differential equation that satisfies given initial conditions.
For example, is there always a solution to the differential equation 
satisfying the initial condition 
, and will it be the only one? Generally speaking, the answer is no. Indeed, the equation 
, the right-hand side of which is continuous on the entire plane, has solutions y= 0 and y=(x+C) 3 ,CR
... Therefore, through any point of the O axis NS passes through two integral curves.
Thus, the function must satisfy some requirements. The following theorem contains one of the options for sufficient conditions for the existence and uniqueness of a solution to the differential equation 
satisfying the initial condition 
.
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