Figure, limited by schedule continuous non-negative function $f(x)$ on the segment $$ and the straight lines $y=0, \ x=a$ and $x=b$ is called a curvilinear trapezoid.
Area corresponding curved trapezoid calculated by the formula:
$S=\int\limits_(a)^(b)(f(x)dx).$ (*)
We will conditionally divide problems to find the area of a curvilinear trapezoid into $4$ types. Let's look at each type in more detail.
Type I: a curved trapezoid is specified explicitly. Then immediately apply the formula (*).
For example, find the area of a curvilinear trapezoid bounded by the graph of the function $y=4-(x-2)^(2)$ and the lines $y=0, \ x=1$ and $x=3$.
Let's draw this curved trapezoid.
Using formula (*), we find the area of this curvilinear trapezoid.
$S=\int\limits_(1)^(3)(\left(4-(x-2)^(2)\right)dx)=\int\limits_(1)^(3)(4dx)- \int\limits_(1)^(3)((x-2)^(2)dx)=4x|_(1)^(3) – \left.\frac((x-2)^(3) )(3)\right|_(1)^(3)=$
$=4(3-1)-\frac(1)(3)\left((3-2)^(3)-(1-2)^(3)\right)=4 \cdot 2 – \frac (1)(3)\left((1)^(3)-(-1)^(3)\right) = 8 – \frac(1)(3)(1+1) =$
$=8-\frac(2)(3)=7\frac(1)(3)$ (units$^(2)$).
Type II: the curved trapezoid is specified implicitly. In this case, the straight lines $x=a, \ x=b$ are usually not specified or partially specified. In this case, you need to find the intersection points of the functions $y=f(x)$ and $y=0$. These points will be points $a$ and $b$.
For example, find the area of a figure bounded by the graphs of the functions $y=1-x^(2)$ and $y=0$.
Let's find the intersection points. To do this, we equate the right-hand sides of the functions.
Thus, $a=-1$ and $b=1$. Let's draw this curved trapezoid.
Let's find the area of this curved trapezoid.
$S=\int\limits_(-1)^(1)(\left(1-x^(2)\right)dx)=\int\limits_(-1)^(1)(1dx)-\int \limits_(-1)^(1)(x^(2)dx)=x|_(-1)^(1) – \left.\frac(x^(3))(3)\right|_ (-1)^(1)=$
$=(1-(-1))-\frac(1)(3)\left(1^(3)-(-1)^(3)\right)=2 – \frac(1)(3) \left(1+1\right) = 2 – \frac(2)(3) = 1\frac(1)(3)$ (units$^(2)$).
Type III: the area of a figure limited by the intersection of two continuous non-negative functions. This figure will not be a curved trapezoid, which means you cannot calculate its area using formula (*). How to be? It turns out that the area of this figure can be found as the difference between the areas of curvilinear trapezoids bounded by the upper function and $y=0$ ($S_(uf)$), and the lower function and $y=0$ ($S_(lf)$), where the role of $x=a, \ x=b$ is played by the $x$ coordinates of the points of intersection of these functions, i.e.
$S=S_(uf)-S_(lf)$. (**)
The most important thing when calculating such areas is not to “miss” with the choice of the upper and lower functions.
For example, find the area of a figure bounded by the functions $y=x^(2)$ and $y=x+6$.
Let's find the intersection points of these graphs:
According to Vieta's theorem,
$x_(1)=-2,\x_(2)=3.$
That is, $a=-2,\b=3$. Let's draw a figure:
Thus, the top function is $y=x+6$, and the bottom function is $y=x^(2)$. Next, we find $S_(uf)$ and $S_(lf)$ using formula (*).
$S_(uf)=\int\limits_(-2)^(3)((x+6)dx)=\int\limits_(-2)^(3)(xdx)+\int\limits_(-2 )^(3)(6dx)=\left.\frac(x^(2))(2)\right|_(-2)^(3) + 6x|_(-2)^(3)= 32 .5$ (units$^(2)$).
$S_(lf)=\int\limits_(-2)^(3)(x^(2)dx)=\left.\frac(x^(3))(3)\right|_(-2) ^(3) = \frac(35)(3)$ (units$^(2)$).
Let's substitute what we found into (**) and get:
$S=32.5-\frac(35)(3)= \frac(125)(6)$ (units$^(2)$).
Type IV: figure area, limited function(s) that do not satisfy the non-negativity condition. In order to find the area of such a figure, you need to be symmetrical about the $Ox$ axis ( in other words, put “minuses” in front of the functions) display the area and, using the methods outlined in types I – III, find the area of the displayed area. This area will be the required area. First, you may have to find the intersection points of the function graphs.
For example, find the area of a figure bounded by the graphs of the functions $y=x^(2)-1$ and $y=0$.
Let's find the intersection points of the function graphs:
those. $a=-1$, and $b=1$. Let's draw the area.
Let's display the area symmetrically:
$y=0 \ \Rightarrow \ y=-0=0$
$y=x^(2)-1 \ \Rightarrow \ y= -(x^(2)-1) = 1-x^(2)$.
The result is a curvilinear trapezoid bounded by the graph of the function $y=1-x^(2)$ and $y=0$. This is a problem to find a curved trapezoid of the second type. We have already solved it. The answer was: $S= 1\frac(1)(3)$ (units $^(2)$). This means that the area of the required curvilinear trapezoid is equal to:
$S=1\frac(1)(3)$ (units$^(2)$).
It is required to calculate the area of a curved trapezoid bounded by straight lines, 
,
and curve 
.
Let's split the segment 
dotmina 
elementary segments, length 
th segment 
. Let us restore perpendiculars from the points of division of the segment to the intersection with the curve 
, let 
. As a result we get 
elementary trapezoids, the sum of their areas is obviously equal to the sum of a given curvilinear trapezoid.
Let us determine the largest and smallest values of the function on each elementary interval; on the first interval this is 
, on the second 
and so on. Let's calculate the amounts
The first sum represents the area of all described, the second is the area of all rectangles inscribed in a curved trapezoid.
It is clear that the first sum gives an approximate value of the area of the trapezoid “with an excess”, the second - “with a deficiency”. The first sum is called the upper Darboux sum, the second – accordingly, the lower Darboux sum. Thus, the area of a curved trapezoid is 
satisfies the inequality 
. Let us find out how the Darboux sums behave as the number of points of partition of the segment increases 
. Let the number of partition points increase by one, and it is in the middle of the interval 
. Now the number is like
inscribed and circumscribed rectangles increased by one. Let us consider how the lower Darboux sum changed. Instead of a square 
th inscribed rectangle, equal to 
we get the sum of the areas of two rectangles 
, since the length 
can't be less 
the smallest value of the function at 
. On the other side, 
, because the 
there can't be more 
the largest value of the function on the interval 
. So, adding new points to split a segment increases the value of the lower Darboux sum and decreases the upper Darboux sum. In this case, the lower Darboux sum, with any increase in the number of partition points, cannot exceed the value of any upper sum, since the sum of the areas of the described rectangles is always more than the amount areas of rectangles inscribed in a curved trapezoid.
Thus, the sequence of lower Darboux sums increases with the number of points of partition of the segment and is bounded from above; according to the well-known theorem, it has a limit. This limit is the area of a given curved trapezoid.
Similarly, the sequence of upper Darboux sums decreases with increasing number of points of partition of the interval and is limited from below by any lower Darboux sum, which means that it also has a limit, and it is also equal to the area of the curvilinear trapezoid.
Therefore, to calculate the area of a curved trapezoid, it is enough to 
partitions of the interval, determine either the lower or upper Darboux sum, and then calculate 
, or 
.
However, such a solution to the problem presupposes for any, arbitrarily large number partitions 
, finding the largest or smallest value of a function on each elementary interval, which is a very labor-intensive task.
A simpler solution is obtained using the Riemann integral sum, which is
Where 
some point of each elementary interval, that is 
. Consequently, the Riemann integral sum is the sum of the areas of all possible rectangles, and 
. As shown above, the limits of the upper and lower Darboux sums are the same and equal to the area of the curved trapezoid. Using one of the properties of the limit of a function (the two-police rule), we obtain that for any partition of the segment 
and selecting points 
The area of a curved trapezoid can be calculated using the formula 
.
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Keywords: integral, curvilinear trapezoid, area of figures bounded by lilies
Equipment: marker board, computer, multimedia projector
Lesson type: lesson-lecture
Lesson Objectives:
- educational: shape the culture mental labor, create a situation of success for each student, 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 different situations, 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 and computational skills, developing students’ thinking in the course of completing proposed tasks, developing an algorithmic culture.
- educational: formulate concepts about a curvilinear trapezoid, an integral, master the skills of calculating areas flat 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 are considering different kinds curvilinear trapezoids and notice: one of the lines degenerates into a point, the role of the limiting function is played by the 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 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 antiderivative functions 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. 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, limited by 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.
Example1 . Calculate the area of the figure bounded by the lines: x + 2y – 4 = 0, y = 0, x = -3, and x = 2
Let's construct a figure (see figure) We construct a straight line x + 2y – 4 = 0 using two points A(4;0) and B(0;2). Expressing y through x, we get y = -0.5x + 2. Using formula (1), where f(x) = -0.5x + 2, a = -3, b = 2, we find
S = = [-0.25=11.25 sq. units
Example 2. Calculate the area of the figure bounded by the lines: x – 2y + 4 = 0, x + y – 5 = 0 and y = 0.
Solution. Let's construct the figure.
Let's construct a straight line x – 2y + 4 = 0: y = 0, x = - 4, A(-4; 0); x = 0, y = 2, B(0; 2).
Let's construct a straight line x + y – 5 = 0: y = 0, x = 5, C(5; 0), x = 0, y = 5, D(0; 5).
Let's find the point of intersection of the lines by solving the system of equations:
x = 2, y = 3; M(2; 3).
To calculate the required area, we divide the triangle AMC into two triangles AMN and NMC, since when x changes from A to N, the area is limited by a straight line, and when x changes from N to C - by a straight line
For triangle AMN we have: ; y = 0.5x + 2, i.e. f(x) = 0.5x + 2, a = - 4, b = 2.
For triangle NMC we have: y = - x + 5, i.e. f(x) = - x + 5, a = 2, b = 5.
By calculating the area of each triangle and adding the results, we find:
sq. units
sq. units
9 + 4, 5 = 13.5 sq. units Check: = 0.5AC = 0.5 sq. units
Example 3. Calculate the area of a figure bounded by lines: y = x 2 , y = 0, x = 2, x = 3.
IN in this case you need to calculate the area of a curved trapezoid, limited by a parabola y = x 2 , straight lines x = 2 and x = 3 and the Ox axis (see figure) Using formula (1) we find the area of the curvilinear trapezoid
= = 6 sq. units
Example 4. Calculate the area of the figure bounded by the lines: y = - x 2 + 4 and y = 0
Let's construct the figure. The required area is enclosed between the parabola y = - x 2 + 4 and the Ox axis.
Let's find the intersection points of the parabola with the Ox axis. Assuming y = 0, we find x = Since this figure is symmetrical about the Oy axis, we calculate the area of the figure located to the right of the Oy axis, and double the result obtained: = +4x]sq. units 2 = 2 sq. units
Example 5. Calculate the area of a figure bounded by lines: y 2 = x, yx = 1, x = 4
Here you need to calculate the area of a curvilinear trapezoid bounded by the upper branch of the parabola 2 = x, Ox axis and straight lines x = 1 and x = 4 (see figure)
According to formula (1), where f(x) = a = 1 and b = 4, we have = (= sq. units.
Example 6 . Calculate the area of the figure bounded by the lines: y = sinx, y = 0, x = 0, x= .
The required area is limited by the half-wave of the sinusoid and the Ox axis (see figure).
We have - cosx = - cos = 1 + 1 = 2 sq. units
Example 7. Calculate the area of the figure bounded by the lines: y = - 6x, y = 0 and x = 4.
The figure is located under the Ox axis (see figure).
Therefore, we find its area using formula (3)
= =
Example 8. Calculate the area of the figure bounded by the lines: y = and x = 2. Construct the y = curve from the points (see figure). Thus, we find the area of the figure using formula (4)
Example 9 .
X 2 + y 2 = r 2 .
Here you need to calculate the area, bounded by a circle X 2 + y 2 = r 2 , i.e. the area of a circle of radius r with the center at the origin. Let's find the fourth part of this area by taking the limits of integration from 0
before; we have: 1 = = [
Hence, 1 =
Example 10. Calculate the area of a figure bounded by lines: y= x 2 and y = 2x
This figure limited by the parabola y=x 2 and the straight line y = 2x (see figure) To determine the intersection points of the given lines, we solve the system of equations: x 2 – 2x = 0 x = 0 and x = 2
Using formula (5) to find the area, we obtain
= }