Examples of calculating the area of a curved trapezoid. Calculation of the areas of figures bounded by given lines
Finished works
DEGREE WORKS
Much has already passed and now you are a graduate, if, of course, you write your thesis on time. But life is such a thing that only now it becomes clear to you that, having ceased to be a student, you will lose all the student joys, many of which you have never tried, putting everything off and putting it off until later. And now, instead of catching up, you're working on your thesis? There is an excellent solution: download the thesis you need from our website - and you will instantly have a lot of free time!
Theses have been successfully defended at leading universities of the Republic of Kazakhstan.
Cost of work from 20,000 tenge
COURSE WORKS
The course project is the first serious practical work. It is with the writing of coursework that preparation for the development of diploma projects begins. If a student learns to correctly present the content of a topic in a course project and format it competently, then in the future he will not have any problems with writing reports, or composing theses, or performing other practical tasks. In order to assist students in writing this type of student work and to clarify questions that arise during its preparation, in fact, this information section was created.
Cost of work from 2,500 tenge
MASTER'S DISSERTATIONS
Currently, in higher educational institutions of Kazakhstan and the CIS countries, the level of higher professional education that follows after bachelor's degree is very common - master's degree. In the master's program, students study with the aim of obtaining a master's degree, which is recognized in most countries of the world more than a bachelor's degree, and is also recognized by foreign employers. The result of master's studies is the defense of a master's thesis.
We will provide you with up-to-date analytical and textual material; the price includes 2 scientific articles and an abstract.
Cost of work from 35,000 tenge
PRACTICE REPORTS
After completing any type of student internship (educational, industrial, pre-graduation), a report is required. This document will be confirmation of the student’s practical work and the basis for forming an assessment for practice. Usually, in order to draw up a report on an internship, you need to collect and analyze information about the enterprise, consider the structure and work routine of the organization in which the internship is taking place, draw up a calendar plan and describe your practical activities.
We will help you write a report on your internship, taking into account the specifics of the activities of a particular enterprise.
In section 4.3 it has already been noted that definite integral () of
non-negative function is numerically equal to the area of the curvilinear trapezoid bounded by the graph of the function = (), straight lines = , = and = 0.
Example 4.24. Calculate the area of the figure enclosed between the axis and the sinusoid = sin (Figure 4.6).
sin = − cos 0 |
= −(cos − cos 0) = 2. |
|||
If the figure is not a curvilinear trapezoid, then they try to represent its area as the sum or difference of the areas of the figures that are curvilinear trapezoids. In particular, the theorem is true.
Theorem 4.13. If a figure is bounded below and above by graphs of continuous functions = 1 (), = 2 () (not necessarily non-negative, ( Figure 4.7 ), then its area can be found using the formula
2 () − 1 () .
Example 4.25. Calculate the area of the figure bounded by the curve = 4 and the lines = and = 4.
y = f2(x) |
|||||||||||
y = f1(x) |
|||||||||||
Figure 4.6 |
Figure 4.7 |
||||||||||
Solution. Let's build |
plane |
(Figure 4.8). Obviously, |
|||||||||
1 () = 4 , 2 () = , |
|||||||||||
= ∫ |
2 − 4 ln |
2 = 8 − 4 ln 4 − (2 − 4 ln 2) = 2(3 − 2 ln 2). |
|||||||||
Part I. Theory
Chapter 4. Theory of integration 4.4. Integral applications. Improper integrals
Figure 4.8 |
4.4.2. Curve arc length
Calculating the lengths of curves also leads to integrals. Let the function = () be continuous on the interval [ ; ] and is differentiable on the interval (;). Its graph represents a certain curve, (; ()), (; ()) (Figure 4.9). We divide the curve with points 0 = , 1 , 2 , . . . , = arbitrary parts. Let's connect two neighboring points −1 and chords = 1, 2, . . . , . We obtain a -link broken line inscribed in the curve. Let
is the length of the chord −1, = 1, 2, . . . , = max16 6 . The length of the broken line will be expressed by the formula
It is natural to define the length of a curve as the limiting value of the lengths of broken lines when → 0, i.e.
Let there be abscissas of points, = 1, 2, . . . , |
||||||||
< < . . . < = . |
||||||||
Then the coordinates of the points are (; ()), and, using formula for the distance between two points, we'll find
Cn−1 |
|||
C k 1C k |
|||
Consequently, there is an integral sum for the function √ 1 + (′ ())2 on the interval [ ; ]. Then, based on equalities (4.31), we have:
= ∫ |
|||||||
1 + (′ ())2 |
|||||||
Example 4.26. Find the length of the graph = 2 |
between = 0 and = 3. |
||||||
Solution. Let's build a graph of the specified function (Figure 4.10).
y=2 |
√x 3 |
|
Figure 4.10
Using formula (4.33) we find: |
|||||||||||||||||||
= ∫ 3 |
= ∫ 3 √ |
= ∫ 3 √ |
|||||||||||||||||
1 + (2 1 )2 |
|||||||||||||||||||
1 + (′ ())2 |
|||||||||||||||||||
(+ 1)2 |
3 (+ 1)2 0 = 3 (8 − 1) = 3 . |
||||||||||||||||||
Let's consider a curved trapezoid bounded by the Ox axis, the curve y=f(x) and two straight lines: x=a and x=b (Fig. 85). Let's take an arbitrary value of x (just not a and not b). Let's give it an increment h = 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. We will call this strip an elementary strip. The area of an elementary strip differs from the area of the rectangle ACQB by the curvilinear triangle BQD, and the area of the latter is less than the area of the rectangle BQDM with sides BQ = =h=dx) QD=Ay and area equal to hAy = Ay dx. As side h decreases, side Du also decreases and simultaneously with h tends to zero. Therefore, the area of the BQDM is second-order infinitesimal. The area of an elementary strip is the increment of the area, and the area of the rectangle ACQB, equal to AB-AC ==/(x) dx> is the differential of the area. Consequently, we find the area itself by integrating its differential. Within the figure under consideration, the independent variable l: changes from a to b, so the required area 5 will be equal to 5= \f(x) dx. (I) Example 1. Let's calculate the area bounded by the parabola y - 1 -x*, straight lines X =--Fj-, x = 1 and the O* axis (Fig. 86). at Fig. 87. Fig. 86. 1 Here f(x) = 1 - l?, the limits of integration are a = - and £ = 1, therefore J [*-t]\- -fl -- Г -1-±Л_ 1V1 -l-l-Ii-^ 3) |_ 2 3V 2 / J 3 24 24* Example 2. Let's calculate the area limited by the sinusoid y = sinXy, the Ox axis and the straight line (Fig. 87). Applying formula (I), we obtain A 2 S= J sinxdx= [-cos x]Q =0 -(-1) = lf Example 3. Calculate the area limited by the arc of the sinusoid ^у = sin jc, enclosed between two adjacent intersection points with the Ox axis (for example, between the origin and the point with the abscissa i). Note that from geometric considerations it is clear that this area will be twice the area of the previous example. However, let's do the calculations: I 5= | s\nxdx= [ - cosх)* - - cos i-(-cos 0)= 1 + 1 = 2. o Indeed, our assumption turned out to be correct. Example 4. Calculate the area bounded by the sinusoid and the Ox axis at one period (Fig. 88). Preliminary calculations suggest that the area will be four times larger than in Example 2. However, after making calculations, we obtain “i Г,*i S - \ sin x dx = [ - cos x]0 = = - cos 2l -(-cos 0) = - 1 + 1 = 0. This result requires clarification. To clarify the essence of the matter, we also calculate the area limited by the same sinusoid y = sin l: and the Ox axis in the range from l to 2i. Applying formula (I), we obtain 2l $2l sin xdx=[ - cosх]l = -cos 2i~)-c05i=- 1-1 =-2. Thus, we see that this area turned out to be negative. Comparing it with the area calculated in exercise 3, we find that their absolute values are the same, but the signs are different. If we apply property V (see Chapter XI, § 4), we get 2l I 2l J sin xdx= J sin * dx [ sin x dx = 2 + (- 2) = 0What happened in this example is not an accident. Always the area located below the Ox axis, provided that the independent variable changes from left to right, is obtained when calculated using integrals. In this course we will always consider areas without signs. Therefore, the answer in the example just discussed will be: the required area is 2 + |-2| = 4. Example 5. Let's calculate the area of the BAB shown in Fig. 89. This area is limited by the Ox axis, the parabola y = - xr and the straight line y - = -x+\. Area of a curvilinear trapezoid The required area OAB consists of two parts: OAM and MAV. Since point A is the intersection point of a parabola and a straight line, we will find its coordinates by solving the system of equations 3 2 Y = mx. (we only need to find the abscissa of point A). Solving the system, we find l; = ~. Therefore, the area has to be calculated in parts, first square. OAM and then pl. MAV: .... G 3 2, 3 G xP 3 1/2 U 2. QAM-^x area of a curvilinear 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 :
Self-test on slide
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 using 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.
Area of a curved trapezoid
Curvilinear trapezoid is a figure bounded by a graph given on the segment [ a, b] continuous and non-negative function f(x), ordinates drawn at points a And b, and axis segment Ox between points a And b(see Fig. 2).
Let us prove the following statement.
A curved trapezoid is a squared figure, area P
Proof. Since continuous on the segment [ a, b] the function is integrable, then for any positive number ε you can specify such a partition T segment [ a, b], what is the difference S - s < ε , Where S And s- the upper and lower sums of the partition, respectively T. But S And s are equal respectively S d And S i, Where S d And S i- areas of stepped figures (polygons), the first of which contains a curvilinear trapezoid, and the second is contained in a curvilinear trapezoid (Fig. 2 also shows these stepped figures). Because S d - S i < ε , then, by virtue of Theorem 1, the curvilinear trapezoid is squarable. Since the limit for Δ → 0 of the upper and lower sums is equal to s ≤ P ≤ S, then the area P curved trapezoid can be found using formula (1).
Comment. If the function f(x) is continuous and non-positive on the segment [ a, b], then the value of the integral is equal to the area of the curvilinear trapezoid taken with a negative sign, limited by the graph of the function f(x), ordinates at points a And b and axis segment Ox between points a And b. Therefore, if f(x) changes sign, then it is equal to the sum of the areas of curvilinear trapezoids located above and below the axis taken with a certain sign Ox, and the areas of the former are taken with the + sign, and of the latter with the - sign.
Area of a curved sector
Let the curve L is given in the polar coordinate system by the equation r = r(θ ), α ≤ θ ≤ β (see Fig. 3), and the function r(θ ) is continuous and non-negative on the segment [ α , β ]. A flat figure bounded by a curve L and two rays making angles with the polar axis α And β , we will call curvilinear sector.
Let us prove the following statement. A curvilinear sector is a squared figure, area P which can be calculated using the formula
Proof. Consider the partition T segment [ α , β ] dots α = θ 0 < θ 1 < ... < θ n = β and for each partial segment [ θ i -1 , θ i] construct circular sectors whose radii are equal to the minimum r i and maximum R i values r(θ ) on the segment [ θ i -1 , θ i]. As a result, we obtain two fan-shaped figures, the first of which is contained in the curvilinear sector, and the second contains the curvilinear sector (these fan-shaped figures are shown in Fig. 3). The areas of and of the indicated fan-shaped figures are equal to and respectively. Note that the first of these sums is the lower sum s for a function for a specified partition T segment [ α , β ], and the second sum is the top sum S for the same function and the same partition. Since the function is integrable on the segment [ α , β ], then the difference can be as small as desired. For example, for any fixed ε > 0 this difference can be made smaller ε /2. Let us now inscribe a polygon into the inner fan-shaped figure Q i with area S i, for which , and we describe a polygon around the external fan-shaped figure Q d area S d, for which * . Obviously, the first of these polygons is inscribed in a curvilinear sector, and the second is circumscribed around it. Since the inequalities are true