Let's move on to consider applications of integral calculus. In this lesson we will analyze the typical and most common task calculating the area of a plane figure using a definite integral. Finally everything searching 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 the indefinite integral at least at an intermediate 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. The task “calculate the area using a definite integral” always involves constructing a drawing, so your knowledge and drawing skills will also be a relevant issue. At a minimum, you need to be able to construct a straight line, parabola and hyperbola.
Let's start with curved trapezoid. A curved trapezoid is a flat figure bounded by the graph of some function y = f(x), axis OX and lines x = a; x = b.
The area of a curvilinear trapezoid is numerically equal to a definite integral
Any definite integral (that exists) has a very good geometric meaning. At the lesson Definite integral. Examples of solutions we 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. Consider the definite integral
Integrand
defines a curve on the plane (it can be drawn if desired), and the definite integral itself is numerically equal to area corresponding curved trapezoid.
Example 1
, , , .
This is a typical assignment statement. The most important point in the decision is the construction of the 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. The point-by-point construction technique can be found in reference material Graphs and properties of elementary functions. There you can also find very useful material for our lesson - how to quickly build a parabola.
In this problem, the solution might look like this.
Let's do the drawing (note that the equation y= 0 specifies the axis OX):
We will not shade the curved trapezoid; here it is obvious what area we are talking about. The solution continues like this:
On the segment [-2; 1] function graph y = x 2 + 2 located above the axisOX, That's why:
Answer: .
Who has difficulties with calculating the definite integral and applying the Newton-Leibniz formula
,
refer to 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 completely clear that if we got, say, the answer: 20 square units, then it is obvious that a mistake was made somewhere - 20 cells clearly do not fit into the figure in question, at most a dozen. If the answer is negative, then the task was also solved incorrectly.
Example 2
Calculate the area of the figure, limited by lines xy = 4, x = 2, x= 4 and axis OX.
This is an example for independent decision. Full solution and answer at the end of the lesson.
What to do if the curved trapezoid is located under the axleOX?
Example 3
Calculate the area of a figure bounded by lines y = e-x, x= 1 and coordinate axes.
Solution: Let's make a drawing:
If a curved trapezoid completely located under the axis OX , 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 we move on to more meaningful examples.
Example 4
Find the area of a plane figure bounded by lines y = 2x – x 2 , y = -x.
Solution: First you need to make a drawing. 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 y = 2x – x 2 and straight y = -x. This can be done in two ways. The first method is analytical. We solve the equation:
This means that the lower limit of integration a= 0, upper limit of integration b= 3. It is often more profitable and faster to construct lines point by point, and the limits of integration become clear “by themselves.” 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). 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:
Let us repeat that when constructing pointwise, the limits of integration are most often determined “automatically”.
And now working formula:
If on the segment [ a; b] some continuous function f(x) greater than or equal to some continuous function g(x), 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, but 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 from 2 x – x 2 must be subtracted – x.
The completed solution might look like this:
The desired figure is limited by a parabola y = 2x – x 2 on top and straight y = -x below.
On segment 2 x – x 2 ≥ -x. According to the corresponding formula:
Answer: .
In fact, the school formula for the area of a curvilinear trapezoid in the lower half-plane (see example No. 3) is special case formulas
.
Because the axis OX given by the equation y= 0, and the graph of the function g(x) located below the axis OX, That
.
And now a couple of examples for your own solution
Example 5
Example 6
Find the area of a figure bounded by 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.
Example 7
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, they often decide that they need to find the area of the figure that is shaded green!
This example is also useful in that it calculates the area of a figure using two definite integrals. Really:
1) On the segment [-1; 1] above the axis OX the graph is located straight y = x+1;
2) On a segment above the axis OX the graph of a hyperbola is located y = (2/x).
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”: b = 1.
But what is the lower limit?! It is clear that this is not an integer, but what is it?
May be, a=(-1/3)? But where is the guarantee that the drawing is made with perfect accuracy, it may well turn out that a=(-1/4). 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 the graphs
To do this, we solve the equation:
.
Hence, a=(-1/3).
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 appropriate formula:
Answer: 
To conclude the lesson, let's look at two more difficult tasks.
Example 9
Calculate the area of a figure bounded by lines
Solution: Let's depict this figure in the drawing.
To draw a point-by-point drawing you need to know appearance sinusoids. In general, it is useful to know the graphs of all elementary functions, as well as some sine values. They can be found in the table of values trigonometric functions . In some cases (for example, 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 a segment, the graph of a function y= sin 3 x located above the axis OX, That's why:
(1) You can see how sines and cosines are integrated in odd powers in the lesson Integrals of trigonometric functions. We pinch off one sinus.
(2) We use the main trigonometric identity in the form
(3) Let's change the variable t=cos x, then: is located above the axis, therefore:
.
.
Note: note how the integral of the tangent in cube is taken; a corollary of the main one is used here trigonometric identity
.
Calculate the area of a figure bounded by lines.
Solution.
We find the intersection points of the given lines. To do this, we solve the system of equations:
To find the abscissa of the intersection points of given lines, we solve the equation:
We find: x 1 = -2, x 2 = 4.
So, these lines, which are a parabola and a straight line, intersect at points A(-2; 0), B(4; 6).
These lines form a closed figure, the area of which is calculated using the above formula:
Using the Newton-Leibniz formula we find:
Find the area of the region bounded by the ellipse.
Solution.
From the equation of the ellipse for the first quadrant we have. From here, using the formula, we get
Let's apply substitution x = a sin t, dx = a cos t dt. New limits of integration t = α And t = β are determined from the equations 0 = a sin t, a = a sin t. Can be put α = 0 and β = π /2.
Find one fourth of the required area
From here S = πab.
Find the area of a figure bounded by linesy = - x 2 + x + 4 andy = - x + 1.
Solution.
Let's find the points of intersection of the lines y = -x 2 + x + 4, y = -x+ 1, equating the ordinates of the lines: - x 2 + x + 4 = -x+ 1 or x 2 - 2x- 3 = 0. Finding the roots x 1 = -1, x 2 = 3 and their corresponding ordinates y 1 = 2, y 2 = -2.
Using the formula for the area of a figure, we get
Determine the area enclosed by a parabolay = x 2 + 1 and straightx + y = 3.
Solution.
Solving a system of equations
find the abscissa of the intersection points x 1 = -2 and x 2 = 1.
Believing y 2 = 3 - x And y 1 = x 2 + 1, based on the formula we get
Calculate the area contained within the Bernoulli lemniscater 2 = a 2 cos 2 φ .
Solution.
In the polar coordinate system, the area of a figure bounded by an arc of a curve r = f(φ ) and two polar radii φ 1 = ʅ And φ 2 = ʆ , will be expressed by the integral
Due to the symmetry of the curve, we first determine one-fourth of the required area
Therefore, the entire area is equal to S = a 2 .
Calculate the arc length of the astroidx 2/3 + y 2/3 = a 2/3 .
Solution.
Let us write the equation of the astroid in the form
(x 1/3) 2 + (y 1/3) 2 = (a 1/3) 2 .
Let's put x 1/3 = a 1/3 cos t, y 1/3 = a 1/3 sin t.
From here we obtain the parametric equations of the astroid
x = a cos 3 t, y = a sin 3 t, (*)
where 0 ≤ t ≤ 2π .
Due to the symmetry of the curve (*), it is enough to find one fourth of the arc length L, corresponding to the parameter change t from 0 to π /2.
We get
dx = -3a cos 2 t sin t dt, dy = 3a sin 2 t cos t dt.
From here we find
Integrating the resulting expression from 0 to π /2, we get
From here L = 6a.
Find the area enclosed by the Archimedes spiralr = aφ and two radius vectors that correspond to polar anglesφ 1 Andφ 2 (φ 1 < φ 2 ).
Solution.
Area enclosed by a curve r = f(φ ) is calculated by the formula, where α And β - limits of polar angle change.
Thus, we get
(*)
From (*) it follows that the area limited by the polar axis and the first turn of the Archimedes spiral ( φ 1 = 0; φ 2 = 2π ):
Similarly, we find the area limited by the polar axis and the second turn of the Archimedes spiral ( φ 1 = 2π ; φ 2 = 4π ):
The required area is equal to the difference of these areas
Calculate the volume of a body obtained by rotation around an axisOx figures bounded by parabolasy = x 2 Andx = y 2 .
Solution.
Let's solve the system of equations
and we get x 1 = 0, x 2 = 1, y 1 = 0, y 2 = 1, whence the intersection points of the curves O(0; 0), B(eleven). As can be seen in the figure, the required volume of a body of revolution is equal to the difference between two volumes formed by rotation around an axis Ox curvilinear trapezoids O.C.B.A. And ODBA:
Calculate the area enclosed by an axisOx and sinusoidy = sinx on segments: a) ; b) .
Solution.
a) On a segment sin function x preserves the sign, and therefore according to the formula, assuming y= sin x, we find
b) On the segment, function sin x changes sign. To solve the problem correctly, it is necessary to divide the segment into two and [ π , 2π ], in each of which the function preserves its sign.
According to the rule of signs, on the segment [ π , 2π ] the area is taken with a minus sign.
As a result, the required area is equal to
Determine the volume of a body bounded by a surface obtained from the rotation of an ellipsearound the major axisa .
Solution.
Considering that the ellipse is symmetrical relative to the coordinate axes, it is enough to find the volume formed by rotation around the axis Ox area OAB, equal to one quarter of the area of the ellipse, and double the result.
Let us denote the volume of a body of revolution by V x; then based on the formula we have , where 0 and a- abscissas of points B And A. From the equation of the ellipse we find . From here
Thus, the required volume is equal to . (When the ellipse rotates around the minor axis b, the volume of the body is equal to )
Find the area bounded by parabolasy 2 = 2 px Andx 2 = 2 py .
Solution.
First, we find the coordinates of the points of intersection of the parabolas to determine the segment of integration. Transforming the original equations, we obtain and . Equating these values, we get or x 4 - 8p 3 x = 0.
x 4 - 8p 3 x = x(x 3 - 8p 3) = x(x - 2p)(x 2 + 2px + 4p 2) = 0.
Finding the roots of the equations:
Considering the fact that the point A intersection of parabolas is in the first quarter, then the limits of integration x= 0 and x = 2p.
We find the required area using the formula
A)
Solution.
The first and most important point of the decision is the construction of the drawing.
Let's make the drawing:
The equation y=0 sets the “x” axis;
- x=-2 And x=1 - straight, parallel to the axis OU;
- y=x 2 +2 - a parabola, the branches of which are directed upward, with the vertex at the point (0;2).
Comment. To construct a parabola, it is enough to find the points of its intersection with the coordinate axes, i.e. putting x=0 find the intersection with the axis OU and deciding accordingly quadratic equation, find the intersection with the axis Oh .
The vertex of a parabola can be found using the formulas:
You can also build lines point by point.
On the interval [-2;1] the graph of the function y=x 2 +2 located above the axis Ox , That's why:
Answer: S =9 sq. units
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, “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.
What to do if the curved trapezoid is located under the axle Oh?
b) Calculate the area of a figure bounded by lines y=-e x , x=1 and coordinate axes.
Solution.
Let's make a drawing.
If a curved trapezoid completely located under the axis Oh , then its area can be found using the formula:
Answer: S=(e-1) sq. units" 1.72 sq. units
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.
With) Find the area of a plane figure bounded by lines y=2x-x 2, y=-x.
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 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 a=0 , upper limit of integration b=3 .
|
We build the given lines: 1. Parabola - vertex at point (1;1); axis intersection Oh - points (0;0) and (0;2). 2. Straight line - bisector of the 2nd and 4th coordinate angles. And now Attention! If on the segment [ a;b] some continuous function f(x) greater than or equal to some continuous function g(x), then the area of the corresponding figure can be found using the formula: . And it doesn’t matter where the figure is located - above the axis or below the axis, but what matters is which graph is HIGHER (relative to another graph), and which 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 |
You can construct lines point by point, and the limits of integration become clear “by themselves.” 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).
The desired figure is limited by a parabola above and a straight line below.
On the segment, according to the corresponding formula:
Answer: S =4.5 sq. units
Back forward
Attention! Slide previews are for informational purposes only and may not represent all the features of the presentation. If you are interested in this work, please download the full version.
Keywords: integral, curvilinear trapezoid, area of figures bounded by lilies
Equipment: marker board, computer, multimedia projector
Lesson type: lesson-lecture
Lesson Objectives:
- educational: to create a culture of mental work, create a situation of success for each student, and create positive motivation for learning; develop the ability to speak and listen to others.
- developing: formation of independent thinking of the student in applying knowledge in various situations, the ability to analyze and draw conclusions, development of logic, development of the ability to correctly pose questions and find answers to them. Improving the formation of computational skills, developing students’ thinking in the course of completing proposed tasks, developing an algorithmic culture.
- educational: to form concepts about a curvilinear trapezoid, about an integral, to master the skills of calculating the areas of plane figures
Teaching method: explanatory and illustrative.
During the classes
In previous classes we learned to calculate the areas of figures whose boundaries are broken lines. In mathematics, there are methods that allow you to calculate the areas of figures bounded by curves. Such figures are called curvilinear trapezoids, and their area is calculated using antiderivatives.
Curvilinear trapezoid ( slide 1)
A curved trapezoid is a figure bounded by the graph of a function, ( sh.m.), straight x = a And x = b and x-axis
Various types of curved trapezoids ( slide 2)
We consider various types of curvilinear trapezoids and notice: one of the straight lines is degenerate to a point, the role of the limiting function is played by the straight line
Area of a curved trapezoid (slide 3)
Fix the left end of the interval A, and the right one X we will change, i.e., we move the right wall of the curvilinear trapezoid and get a changing figure. The area of a variable curvilinear trapezoid bounded by the graph of the function is an antiderivative F for function f
And on the segment [ a; b] area of a curved trapezoid, formed by the function f, is equal to the increment of the antiderivative of this function:
Exercise 1:
Find the area of a curvilinear trapezoid bounded by the graph of the function: f(x) = x 2 and straight y = 0, x = 1, x = 2.
Solution: ( according to the algorithm slide 3)
Let's draw a graph of the function and lines
Let's find one of the antiderivatives of the function f(x) = x 2 :
Slide self-test
Integral
Consider a curvilinear trapezoid defined by the function f on the segment [ a; b]. Let's break this segment into several parts. The area of the entire trapezoid will be divided into the sum of the areas of smaller curved trapezoids. ( slide 5). Each such trapezoid can be approximately considered a rectangle. The sum of the areas of these rectangles gives an approximate idea of the entire area of the curved trapezoid. The smaller we divide the segment [ a; b], the more accurately we calculate the area.
Let us write these arguments in the form of formulas.
Divide the segment [ a; b] into n parts by dots x 0 =a, x1,...,xn = b. Length k- th denote by xk = xk – xk-1. Let's make a sum
Geometrically, this sum represents the area of the figure shaded in the figure ( sh.m.)
Sums of the form are called integral sums for the function f. (sh.m.)
Integral sums give an approximate value of the area. The exact value is obtained by passing to the limit. Let's imagine that we are refining the partition of the segment [ a; b] so that the lengths of all small segments tend to zero. Then the area of the composed figure will approach the area of the curved trapezoid. We can say that the area of a curved trapezoid is equal to the limit of integral sums, Sc.t. (sh.m.) or integral, i.e.,
Definition:
Integral of a function f(x) from a before b called the limit of integral sums
= (sh.m.)
Newton-Leibniz formula.
We remember that the limit of integral sums is equal to the area of a curvilinear trapezoid, which means we can write:
Sc.t. = (sh.m.)
On the other hand, the area of a curved trapezoid is calculated by the formula
S k.t. (sh.m.)
Comparing these formulas, we get:
= (sh.m.)This equality is called the Newton-Leibniz formula.
For ease of calculation, the formula is written as:
= = (sh.m.)Tasks: (sh.m.)
1. Calculate the integral using the Newton-Leibniz formula: ( check on slide 5)
2. Compose integrals according to the drawing ( check on slide 6)
3. Find the area of the figure bounded by the lines: y = x 3, y = 0, x = 1, x = 2. ( Slide 7)
Finding the areas of plane figures ( slide 8)
How to find the area of figures that are not curved trapezoids?
Let two functions be given, the graphs of which you see on the slide . (sh.m.) Find the area of the shaded figure . (sh.m.). Is the figure in question a curved trapezoid? How can you find its area using the property of additivity of area? Consider two curved trapezoids and subtract the area of the other from the area of one of them ( sh.m.)
Let's create an algorithm for finding the area using animation on a slide:
- Graph functions
- Project the intersection points of the graphs onto the x-axis
- Shade the figure obtained when the graphs intersect
- Find curvilinear trapezoids whose intersection or union is the given figure.
- Calculate the area of each of them
- Find the difference or sum of areas
Oral task: How to obtain the area of a shaded figure (tell using animation, slide 8 and 9)
Homework: Work through the notes, No. 353 (a), No. 364 (a).
Bibliography
- Algebra and the beginnings of analysis: a textbook for grades 9-11 of evening (shift) school / ed. G.D. Glaser. - M: Enlightenment, 1983.
- Bashmakov M.I. Algebra and the beginnings of analysis: a textbook for 10-11 grades of secondary school / Bashmakov M.I. - M: Enlightenment, 1991.
- Bashmakov M.I. Mathematics: textbook for institutions beginning. and Wednesday prof. education / M.I. Bashmakov. - M: Academy, 2010.
- Kolmogorov A.N. Algebra and beginnings of analysis: textbook for grades 10-11. educational institutions / A.N. Kolmogorov. - M: Education, 2010.
- Ostrovsky S.L. How to make a presentation for a lesson?/ S.L. Ostrovsky. – M.: September 1st, 2010.
We begin to consider the actual process of calculating the double integral and get acquainted with its geometric meaning.
The double integral is numerically equal to the area of the plane figure (the region of integration). This simplest form double integral, when the function of two variables is equal to one: .
Let's first consider the problem in general view. Now you will be quite surprised how simple everything really is! Let's calculate the area of a flat figure bounded by lines. For definiteness, we assume that on the segment . The area of this figure is numerically equal to:
Let's depict the area in the drawing:
Let's choose the first way to traverse the area:
Thus:
And immediately an important technical technique: iterated integrals can be calculated separately. First the inner integral, then the outer integral. This method I highly recommend it to beginners in the subject.
1) Let's calculate the internal integral, and the integration is carried out over the variable “y”:
Indefinite integral here is the simplest one, and then the banal Newton-Leibniz formula is used, with the only difference that the limits of integration are not numbers, but functions. First they put it in “Y” ( antiderivative function) upper limit, then lower limit
2) The result obtained in the first paragraph must be substituted into the external integral:
A more compact representation of the entire solution looks like this:
The resulting formula is exactly the working formula for calculating the area of a flat figure using the “ordinary” definite integral! Watch the lesson Calculating Area Using a Definite Integral, there she is at every step!
That is, problem of calculating area using double integral not much different from the problem of finding the area using a definite integral! In fact, it's the same thing!
Accordingly, no difficulties should arise! I won’t look at very many examples, since you, in fact, have repeatedly encountered this task.
Example 9
Solution: Let's depict the area in the drawing:
Let us choose the following order of traversal of the area:
Here and further I will not dwell on how to traverse the area, since very detailed explanations were given in the first paragraph.
Thus:
As I already noted, it is better for beginners to calculate iterated integrals separately, and I will stick to the same method:
1) First, using the Newton-Leibniz formula, we deal with the internal integral:
2) The result obtained in the first step is substituted into the external integral:
Point 2 is actually finding the area of a plane figure using a definite integral.
Answer:
This is such a stupid and naive task.
An interesting example for an independent solution:
Example 10
Using a double integral, calculate the area of a plane figure bounded by the lines , ,
An approximate example of a final solution at the end of the lesson.
In Examples 9-10, it is much more profitable to use the first method of traversing the area; curious readers, by the way, can change the order of traversal and calculate the areas using the second method. If you do not make a mistake, then, naturally, you will get the same area values.
But in some cases, the second method of traversing the area is more effective, and at the end of the young nerd’s course, let’s look at a couple more examples on this topic:
Example 11
Using a double integral, calculate the area of a plane figure bounded by lines,
Solution: We are looking forward to two parabolas with a quirk that lie on their sides. There is no need to smile; similar things occur quite often in multiple integrals.
What is the easiest way to make a drawing?
Let's imagine a parabola in the form of two functions:
– the upper branch and – the lower branch.
Similarly, imagine a parabola in the form of upper and lower branches.
We calculate the area of the figure using the double integral according to the formula:
What happens if we choose the first method of traversing the area? Firstly, this area will have to be divided into two parts. And secondly, we will observe this sad picture: . Integrals, of course, are not of a super-complicated level, but... there is an old mathematical saying: those who are close to their roots do not need a test.
Therefore, from the misunderstanding given in the condition, we express the inverse functions:
Inverse functions V in this example have the advantage that they specify the entire parabola at once without any leaves, acorns, branches and roots.
According to the second method, the area traversal will be as follows:
Thus:
As they say, feel the difference.
1) We deal with the internal integral:
We substitute the result into the outer integral:
Integration over the variable “y” should not be confusing; if there were a letter “zy”, it would be great to integrate over it. Although who read the second paragraph of the lesson How to calculate the volume of a body of rotation, he no longer experiences the slightest awkwardness with integration according to the “Y” method.
Also pay attention to the first step: the integrand is even, and the interval of integration is symmetrical about zero. Therefore, the segment can be halved, and the result can be doubled. This technique is commented in detail in the lesson. Effective methods calculation of a definite integral.
What to add…. All!
Answer:
To test your integration technique, you can try calculating . The answer should be exactly the same.
Example 12
Using a double integral, calculate the area of a plane figure bounded by lines
This is an example for you to solve on your own. It is interesting to note that if you try to use the first method of traversing the area, the figure will no longer have to be divided into two, but into three parts! And, accordingly, we get three pairs of repeated integrals. Sometimes it happens.
The master class has come to an end, and it’s time to move on to the grandmaster level - How to calculate double integral? Examples of solutions. I’ll try not to be so maniacal in the second article =)
I wish you success!
Solutions and answers:
Example 2:Solution:
Let's depict the area on the drawing:
Let us choose the following order of traversal of the area:
Thus:
Let's move on to inverse functions:
Thus:
Answer:
Example 4:Solution:
Let's move on to direct functions:
Let's make the drawing:
Let's change the order of traversing the area:
Answer:
The order of walking around the area:
Thus:
1)
2)
Answer:
