Find the area of the feature bounded by lines online. The area of a curvilinear trapezoid is numerically equal to a certain integral. The completion of the solution might look like this
In this article, you will learn how to find the area of a figure bounded by lines using integral calculations. For the first time, we encounter the formulation of such a problem in high school, when the study of certain integrals has just been completed and it is time to start the geometric interpretation of the knowledge gained in practice.
So, what is required to successfully solve the problem of finding the area of \u200b\u200ba figure using integrals:
- Ability to correctly draw drawings;
- Ability to solve a definite integral using the well-known Newton-Leibniz formula;
- The ability to "see" a more profitable solution - i.e. to understand how in this or that case it will be more convenient to carry out the integration? Along the x-axis (OX) or y-axis (OY)?
- Well, where without correct calculations?) This includes understanding how to solve that other type of integrals and correct numerical calculations.
Algorithm for solving the problem of calculating the area of a figure bounded by lines:
1. We build a drawing. It is advisable to do this on a piece of paper in a cage, on a large scale. We sign with a pencil above each graph the name of this function. The signature of the graphs is done solely for the convenience of further calculations. Having received the graph of the desired figure, in most cases it will be immediately clear which integration limits will be used. Thus, we solve the problem graphically. However, it happens that the values of the limits are fractional or irrational. Therefore, you can make additional calculations, go to step two.
2. If the integration limits are not explicitly set, then we find the points of intersection of the graphs with each other, and see if our graphical solution coincides with the analytical one.
3. Next, you need to analyze the drawing. Depending on how the graphs of functions are located, there are different approaches to finding the area of \u200b\u200bthe figure. Consider various examples of finding the area of a figure using integrals.
3.1. The most classic and simplest version of the problem is when you need to find the area of a curvilinear trapezoid. What is a curvilinear trapezoid? This is a flat figure bounded by the x-axis (y=0), straight x = a, x = b and any curve continuous on the interval from a before b. At the same time, this figure is non-negative and is located not lower than the x-axis. In this case, the area of the curvilinear trapezoid is numerically equal to the definite integral calculated using the Newton-Leibniz formula:
Example 1 y = x2 - 3x + 3, x = 1, x = 3, y = 0.
What lines define the figure? We have a parabola y = x2 - 3x + 3, which is located above the axis OH, it is non-negative, because all points of this parabola are positive. Next, given straight lines x = 1 and x = 3 that run parallel to the axis OU, are the bounding lines of the figure on the left and right. Well y = 0, she is the x-axis, which limits the figure from below. The resulting figure is shaded, as seen in the figure on the left. In this case, you can immediately begin to solve the problem. Before us is a simple example of a curvilinear trapezoid, which we then solve using the Newton-Leibniz formula.
3.2. In the previous paragraph 3.1, the case was analyzed when the curvilinear trapezoid is located above the x-axis. Now consider the case when the conditions of the problem are the same, except that the function lies under the x-axis. A minus is added to the standard Newton-Leibniz formula. How to solve such a problem, we will consider further.
Example 2 . Calculate the area of a figure bounded by lines y=x2+6x+2, x=-4, x=-1, y=0.
In this example, we have a parabola y=x2+6x+2, which originates from under the axis OH, straight x=-4, x=-1, y=0. Here y = 0 limits the desired figure from above. Direct x = -4 and x = -1 these are the boundaries within which the definite integral will be calculated. The principle of solving the problem of finding the area of \u200b\u200ba figure almost completely coincides with example number 1. The only difference is that the given function is not positive, and everything is also continuous on the interval [-4; -1] . What does not positive mean? As can be seen from the figure, the figure that lies within the given x has exclusively "negative" coordinates, which is what we need to see and remember when solving the problem. We are looking for the area of \u200b\u200bthe figure using the Newton-Leibniz formula, only with a minus sign at the beginning.
The article is not completed.
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 a flat figure (region of integration). This is the simplest form of the double integral, when the function of two variables is equal to one: .
Let us first consider the problem in general terms. Now you will be surprised how simple it really is! Let's calculate the area of a flat figure bounded by lines. For definiteness, we assume that on the interval . The area of this figure is numerically equal to:
Let's depict the area in the drawing:
Let's choose the first way to bypass the area:
In this way:
And immediately an important technical trick: iterated integrals can be considered separately. First the inner integral, then the outer integral. This method is highly recommended for beginners in the topic teapots.
1) Calculate the internal integral, while the integration is carried out over the variable "y":
The indefinite integral here is the simplest, and then the banal Newton-Leibniz formula is used, with the only difference that the limits of integration are not numbers, but functions. First, we substituted the upper limit into the “y” (antiderivative function), then the lower limit
2) The result obtained in the first paragraph must be substituted into the external integral:
A more compact notation for the whole 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! See lesson Calculating area using a definite integral, there she is at every turn!
That is, the problem of calculating the area using a double integral little different from the problem of finding the area using a definite integral! In fact, they are one and the same!
Accordingly, no difficulties should arise! I will not consider very many examples, since you, in fact, have repeatedly encountered this problem.
Example 9
Solution: Let's depict the area in the drawing:
Let's choose the following order of traversal of the region:
Here and below, I won't go into how to traverse an area because the first paragraph was very detailed.
In this way:
As I already noted, it is better for beginners to calculate iterated integrals separately, I will adhere to the same method:
1) First, using the Newton-Leibniz formula, we deal with the internal integral:
2) The result obtained at the first step is substituted into the outer integral:
Point 2 is actually finding the area of a flat figure using a definite integral.
Answer:
Here is such a stupid and naive task.
A curious example for an independent solution:
Example 10
Using the double integral, calculate the area of a plane figure bounded by the lines , ,
An 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 bypassing the area; curious readers, by the way, can change the order of the bypass and calculate the areas in the second way. If you do not make a mistake, then, naturally, the same area values \u200b\u200bare obtained.
But in some cases, the second way to bypass the area is more effective, and in conclusion of the course of a young nerd, we will consider a couple more examples on this topic:
Example 11
Using the double integral, calculate the area of a plane figure bounded by lines.
Solution: we are looking forward to two parabolas with a breeze that lie on their side. No need to smile, similar things in multiple integrals are often encountered.
What is the easiest way to make a drawing?
Let's represent the parabola as two functions:
- upper branch and - lower branch.
Similarly, we represent the parabola as the upper and lower branches.
The area of the figure is calculated using the double integral according to the formula:
What happens if we choose the first way to bypass the area? First, 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-complex level, but ... there is an old mathematical saying: whoever is friendly with the roots does not need a set-off.
Therefore, from the misunderstanding that is given in the condition, we express the inverse functions:
The inverse functions in this example have the advantage that they immediately set the entire parabola without any leaves, acorns, branches and roots.
According to the second method, the area traversal will be as follows:
In this way:
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 embarrassing, if there was a letter "zyu" - 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 revolution, he no longer experiences the slightest embarrassment with integration over "y".
Also pay attention to the first step: the integrand is even, and the integration segment is symmetric about zero. Therefore, the segment can be halved, and the result can be doubled. This technique is commented on in detail in the lesson. Efficient Methods for Computing the Definite Integral.
What to add…. Everything!
Answer:
To test your integration technique, you can try to calculate . The answer should be exactly the same.
Example 12
Using the double integral, calculate the area of a plane figure bounded by lines
This is a do-it-yourself example. It is interesting to note that if you try to use the first way to bypass the area, then the figure will no longer be divided into two, but into three parts! And, accordingly, we get three pairs of iterated 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 the double integral? Solution examples. I'll try not to be so manic in the second article =)
Wish you success!
Solutions and answers:
Example 2:Solution:
Draw an area on the drawing:
Let's choose the following order of traversal of the region:
In this way:
Let's move on to inverse functions:
In this way:
Answer:
Example 4:Solution:
Let's move on to direct functions:
Let's execute the drawing:
Let's change the order of traversal of the area:
Answer:
Area traversal order:
In this way:
1)
2)
Answer:
In the previous section, devoted to the analysis of the geometric meaning of a definite integral, we obtained a number of formulas for calculating the area of a curvilinear trapezoid:
S (G) = ∫ a b f (x) d x for a continuous and non-negative function y = f (x) on the segment [ a ; b] ,
S (G) = - ∫ a b f (x) d x for a continuous and non-positive function y = f (x) on the segment [ a ; b] .
These formulas are applicable for solving relatively simple problems. In fact, we often have to work with more complex shapes. In this regard, we will devote this section to the analysis of algorithms for calculating the area of figures, which are limited by functions in an explicit form, i.e. like y = f(x) or x = g(y) .
TheoremLet the functions y = f 1 (x) and y = f 2 (x) be defined and continuous on the segment [ a ; b ] , and f 1 (x) ≤ f 2 (x) for any value x from [ a ; b] . Then the formula for calculating the area of \u200b\u200ba figure Gbounded by lines x \u003d a, x \u003d b, y \u003d f 1 (x) and y \u003d f 2 (x) will look like S (G) \u003d ∫ a b f 2 (x) - f 1 (x) d x .
A similar formula will be applicable for the area of \u200b\u200bthe figure bounded by the lines y \u003d c, y \u003d d, x \u003d g 1 (y) and x \u003d g 2 (y): S (G) \u003d ∫ c d (g 2 (y) - g 1 (y) d y .
Proof
We will analyze three cases for which the formula will be valid.
In the first case, taking into account the additivity property of the area, the sum of the areas of the original figure G and the curvilinear trapezoid G 1 is equal to the area of the figure G 2 . It means that
Therefore, S (G) = S (G 2) - S (G 1) = ∫ a b f 2 (x) d x - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) d x .
We can perform the last transition using the third property of the definite integral.
In the second case, the equality is true: S (G) = S (G 2) + S (G 1) = ∫ a b f 2 (x) d x + - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) d x
The graphic illustration will look like:
If both functions are non-positive, we get: S (G) = S (G 2) - S (G 1) = - ∫ a b f 2 (x) d x - - ∫ a b f 1 (x) d x = ∫ a b (f 2 (x) - f 1 (x)) d x . The graphic illustration will look like:
Let's move on to the consideration of the general case when y = f 1 (x) and y = f 2 (x) intersect the axis O x .
We will denote the intersection points as x i , i = 1 , 2 , . . . , n - 1 . These points break the segment [ a ; b ] into n parts x i - 1 ; x i , i = 1 , 2 , . . . , n , where α = x 0< x 1 < x 2 < . . . < x n - 1 < x n = b . Фигуру G можно представить объединением фигур G i , i = 1 , 2 , . . . , n . Очевидно, что на своем интервале G i попадает под один из трех рассмотренных ранее случаев, поэтому их площади находятся как S (G i) = ∫ x i - 1 x i (f 2 (x) - f 1 (x)) d x , i = 1 , 2 , . . . , n
Consequently,
S (G) = ∑ i = 1 n S (G i) = ∑ i = 1 n ∫ x i x i f 2 (x) - f 1 (x)) d x = = ∫ x 0 x n (f 2 (x) - f ( x)) d x = ∫ a b f 2 (x) - f 1 (x) d x
We can make the last transition using the fifth property of the definite integral.
Let us illustrate the general case on the graph.
The formula S (G) = ∫ a b f 2 (x) - f 1 (x) d x can be considered proven.
And now let's move on to the analysis of examples of calculating the area of \u200b\u200bfigures that are limited by the lines y \u003d f (x) and x \u003d g (y) .
Considering any of the examples, we will begin with the construction of a graph. The image will allow us to represent complex shapes as combinations of simpler shapes. If you are having trouble plotting graphs and figures on them, you can study the section on basic elementary functions, geometric transformation of graphs of functions, as well as plotting while examining a function.
Example 1
It is necessary to determine the area of \u200b\u200bthe figure, which is limited by the parabola y \u003d - x 2 + 6 x - 5 and straight lines y \u003d - 1 3 x - 1 2, x \u003d 1, x \u003d 4.
Solution
Let's plot the lines on the graph in the Cartesian coordinate system.
On the interval [ 1 ; 4] the graph of the parabola y = - x 2 + 6 x - 5 is located above the straight line y = - 1 3 x - 1 2 . In this regard, to obtain an answer, we use the formula obtained earlier, as well as the method for calculating a definite integral using the Newton-Leibniz formula:
S (G) = ∫ 1 4 - x 2 + 6 x - 5 - - 1 3 x - 1 2 d x = = ∫ 1 4 - x 2 + 19 3 x - 9 2 d x = - 1 3 x 3 + 19 6 x 2 - 9 2 x 1 4 = = - 1 3 4 3 + 19 6 4 2 - 9 2 4 - - 1 3 1 3 + 19 6 1 2 - 9 2 1 = = - 64 3 + 152 3 - 18 + 1 3 - 19 6 + 9 2 = 13
Answer: S (G) = 13
Let's look at a more complex example.
Example 2
It is necessary to calculate the area of the figure, which is limited by the lines y = x + 2 , y = x , x = 7 .
Solution
In this case, we have only one straight line parallel to the x-axis. This is x = 7 . This requires us to find the second integration limit ourselves.
Let's build a graph and put on it the lines given in the condition of the problem.
Having a graph in front of our eyes, we can easily determine that the lower limit of integration will be the abscissa of the intersection point of the graph with a straight line y \u003d x and a semi-parabola y \u003d x + 2. To find the abscissa, we use the equalities:
y = x + 2 O DZ: x ≥ - 2 x 2 = x + 2 2 x 2 - x - 2 = 0 D = (- 1) 2 - 4 1 (- 2) = 9 x 1 = 1 + 9 2 = 2 ∈ O D G x 2 = 1 - 9 2 = - 1 ∉ O D G
It turns out that the abscissa of the intersection point is x = 2.
We draw your attention to the fact that in the general example in the drawing, the lines y = x + 2 , y = x intersect at the point (2 ; 2) , so such detailed calculations may seem redundant. We have provided such a detailed solution here only because in more complex cases the solution may not be so obvious. This means that it is better to always calculate the coordinates of the intersection of lines analytically.
On the interval [ 2 ; 7 ] the graph of the function y = x is located above the graph of the function y = x + 2 . Apply the formula to calculate the area:
S (G) = ∫ 2 7 (x - x + 2) d x = x 2 2 - 2 3 (x + 2) 3 2 2 7 = = 7 2 2 - 2 3 (7 + 2) 3 2 - 2 2 2 - 2 3 2 + 2 3 2 = = 49 2 - 18 - 2 + 16 3 = 59 6
Answer: S (G) = 59 6
Example 3
It is necessary to calculate the area of \u200b\u200bthe figure, which is limited by the graphs of the functions y \u003d 1 x and y \u003d - x 2 + 4 x - 2.
Solution
Let's draw lines on the graph.
Let's define the limits of integration. To do this, we determine the coordinates of the points of intersection of the lines by equating the expressions 1 x and - x 2 + 4 x - 2 . Provided that x is not equal to zero, the equality 1 x \u003d - x 2 + 4 x - 2 becomes equivalent to the equation of the third degree - x 3 + 4 x 2 - 2 x - 1 \u003d 0 with integer coefficients. You can refresh the memory of the algorithm for solving such equations by referring to the section “Solution of cubic equations”.
The root of this equation is x = 1: - 1 3 + 4 1 2 - 2 1 - 1 = 0.
Dividing the expression - x 3 + 4 x 2 - 2 x - 1 by the binomial x - 1, we get: - x 3 + 4 x 2 - 2 x - 1 ⇔ - (x - 1) (x 2 - 3 x - 1) = 0
We can find the remaining roots from the equation x 2 - 3 x - 1 = 0:
x 2 - 3 x - 1 = 0 D = (- 3) 2 - 4 1 (- 1) = 13 x 1 = 3 + 13 2 ≈ 3 . 3; x 2 \u003d 3 - 13 2 ≈ - 0. 3
We have found an interval x ∈ 1; 3 + 13 2 , where G is enclosed above the blue line and below the red line. This helps us determine the area of the figure:
S (G) = ∫ 1 3 + 13 2 - x 2 + 4 x - 2 - 1 x d x = - x 3 3 + 2 x 2 - 2 x - ln x 1 3 + 13 2 = = - 3 + 13 2 3 3 + 2 3 + 13 2 2 - 2 3 + 13 2 - ln 3 + 13 2 - - - 1 3 3 + 2 1 2 - 2 1 - ln 1 = 7 + 13 3 - ln 3 + 13 2
Answer: S (G) \u003d 7 + 13 3 - ln 3 + 13 2
Example 4
It is necessary to calculate the area of the figure, which is limited by the curves y \u003d x 3, y \u003d - log 2 x + 1 and the x-axis.
Solution
Let's put all the lines on the graph. We can get the graph of the function y = - log 2 x + 1 from the graph y = log 2 x if we place it symmetrically about the x-axis and move it up one unit. The equation of the x-axis y \u003d 0.
Let's denote the points of intersection of the lines.
As can be seen from the figure, the graphs of the functions y \u003d x 3 and y \u003d 0 intersect at the point (0; 0) . This is because x \u003d 0 is the only real root of the equation x 3 \u003d 0.
x = 2 is the only root of the equation - log 2 x + 1 = 0 , so the graphs of the functions y = - log 2 x + 1 and y = 0 intersect at the point (2 ; 0) .
x = 1 is the only root of the equation x 3 = - log 2 x + 1 . In this regard, the graphs of the functions y \u003d x 3 and y \u003d - log 2 x + 1 intersect at the point (1; 1) . The last statement may not be obvious, but the equation x 3 \u003d - log 2 x + 1 cannot have more than one root, since the function y \u003d x 3 is strictly increasing, and the function y \u003d - log 2 x + 1 is strictly decreasing.
The next step involves several options.
Option number 1
We can represent the figure G as the sum of two curvilinear trapezoids located above the abscissa axis, the first of which is located below the midline on the segment x ∈ 0; 1 , and the second one is below the red line on the segment x ∈ 1 ; 2. This means that the area will be equal to S (G) = ∫ 0 1 x 3 d x + ∫ 1 2 (- log 2 x + 1) d x .
Option number 2
The figure G can be represented as the difference of two figures, the first of which is located above the x-axis and below the blue line on the segment x ∈ 0; 2 , and the second one is between the red and blue lines on the segment x ∈ 1 ; 2. This allows us to find the area like this:
S (G) = ∫ 0 2 x 3 d x - ∫ 1 2 x 3 - (- log 2 x + 1) d x
In this case, to find the area, you will have to use a formula of the form S (G) \u003d ∫ c d (g 2 (y) - g 1 (y)) d y. In fact, the lines that bound the shape can be represented as functions of the y argument.
Let's solve the equations y = x 3 and - log 2 x + 1 with respect to x:
y = x 3 ⇒ x = y 3 y = - log 2 x + 1 ⇒ log 2 x = 1 - y ⇒ x = 2 1 - y
We get the required area:
S (G) = ∫ 0 1 (2 1 - y - y 3) d y = - 2 1 - y ln 2 - y 4 4 0 1 = = - 2 1 - 1 ln 2 - 1 4 4 - - 2 1 - 0 ln 2 - 0 4 4 = - 1 ln 2 - 1 4 + 2 ln 2 = 1 ln 2 - 1 4
Answer: S (G) = 1 ln 2 - 1 4
Example 5
It is necessary to calculate the area of the figure, which is limited by the lines y \u003d x, y \u003d 2 3 x - 3, y \u003d - 1 2 x + 4.
Solution
Draw a line on the chart with a red line, given by the function y = x . Draw the line y = - 1 2 x + 4 in blue, and mark the line y = 2 3 x - 3 in black.
Note the intersection points.
Find the intersection points of the graphs of functions y = x and y = - 1 2 x + 4:
x = - 1 2 x + 4 O DZ: x ≥ 0 x = - 1 2 x + 4 2 ⇒ x = 1 4 x 2 - 4 x + 16 ⇔ x 2 - 20 x + 64 = 0 D = (- 20) 2 - 4 1 64 \u003d 144 x 1 \u003d 20 + 144 2 \u003d 16; x 2 = 20 - 144 2 = 4 i is the solution to the equation x 2 = 4 = 2 , - 1 2 x 2 + 4 = - 1 2 4 + 4 = 2 ⇒ x 2 = 4 is the solution to the equation ⇒ (4 ; 2) point of intersection i y = x and y = - 1 2 x + 4
Find the intersection point of the graphs of functions y = x and y = 2 3 x - 3:
x = 2 3 x - 3 O DZ: x ≥ 0 x = 2 3 x - 3 2 ⇔ x = 4 9 x 2 - 4 x + 9 ⇔ 4 x 2 - 45 x + 81 = 0 D = (- 45 ) 2 - 4 4 81 = 729 x 1 = 45 + 729 8 = 9, x 2 45 - 729 8 = 9 4 Check: x 1 = 9 = 3, 2 3 x 1 - 3 \u003d 2 3 9 - 3 \u003d 3 ⇒ x 1 \u003d 9 is the solution to the equation ⇒ (9; 3) point and intersection y = x and y = 2 3 x - 3 x 2 = 9 4 = 3 2 , 2 3 x 1 - 3 = 2 3 9 4 - 3 = - 3 2 ⇒ x 2 = 9 4 is not a solution to the equation
Find the point of intersection of the lines y = - 1 2 x + 4 and y = 2 3 x - 3:
1 2 x + 4 = 2 3 x - 3 ⇔ - 3 x + 24 = 4 x - 18 ⇔ 7 x = 42 ⇔ x = 6 - 1 2 6 + 4 = 2 3 6 - 3 = 1 ⇒ (6 1) point of intersection y = - 1 2 x + 4 and y = 2 3 x - 3
Method number 1
We represent the area of the desired figure as the sum of the areas of individual figures.
Then the area of the figure is:
S (G) = ∫ 4 6 x - - 1 2 x + 4 d x + ∫ 6 9 x - 2 3 x - 3 d x = = 2 3 x 3 2 + x 2 4 - 4 x 4 6 + 2 3 x 3 2 - x 2 3 + 3 x 6 9 = = 2 3 6 3 2 + 6 2 4 - 4 6 - 2 3 4 3 2 + 4 2 4 - 4 4 + + 2 3 9 3 2 - 9 2 3 + 3 9 - 2 3 6 3 2 - 6 2 3 + 3 6 = = - 25 3 + 4 6 + - 4 6 + 12 = 11 3
Method number 2
The area of the original figure can be represented as the sum of the other two figures.
Then we solve the line equation for x, and only after that we apply the formula for calculating the area of \u200b\u200bthe figure.
y = x ⇒ x = y 2 red line y = 2 3 x - 3 ⇒ x = 3 2 y + 9 2 black line y = - 1 2 x + 4 ⇒ x = - 2 y + 8 s i n i i l i n i i
So the area is:
S (G) = ∫ 1 2 3 2 y + 9 2 - - 2 y + 8 d y + ∫ 2 3 3 2 y + 9 2 - y 2 d y = = ∫ 1 2 7 2 y - 7 2 d y + ∫ 2 3 3 2 y + 9 2 - y 2 d y = = 7 4 y 2 - 7 4 y 1 2 + - y 3 3 + 3 y 2 4 + 9 2 y 2 3 = 7 4 2 2 - 7 4 2 - 7 4 1 2 - 7 4 1 + + - 3 3 3 + 3 3 2 4 + 9 2 3 - - 2 3 3 + 3 2 2 4 + 9 2 2 = = 7 4 + 23 12 = 11 3
As you can see, the values match.
Answer: S (G) = 11 3
Results
To find the area of a figure that is limited by given lines, we need to draw lines on a plane, find their intersection points, and apply the formula for finding the area. In this section, we have reviewed the most common options for tasks.
If you notice a mistake in the text, please highlight it and press Ctrl+Enter
a)
Solution.
The first and most important moment of the decision is the construction of a drawing.
Let's make a drawing:
The equation y=0 sets the x-axis;
- x=-2 and x=1 - straight, parallel to the axis OU;
- y \u003d x 2 +2 - a parabola whose branches are directed upwards, with a 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 solving the corresponding quadratic equation, find the intersection with the axis Oh .
The vertex of a parabola can be found using the formulas:
You can draw lines and point by point.
On the interval [-2;1] the graph of the function y=x 2 +2 located over axis Ox , that's why:
Answer: S \u003d 9 square units
After the task is completed, it is always useful to look at the drawing and figure out 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 seems to be true. It is quite clear that if we had, say, the answer: 20 square units, then, obviously, a mistake was made somewhere - 20 cells clearly do not fit into the figure in question, at most a dozen. If the answer turned out to be negative, then the task was also solved incorrectly.
What to do if the curvilinear trapezoid is located under 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 curvilinear trapezoid completely under the axle Oh , then its area can be found by the formula:
Answer: S=(e-1) sq. unit" 1.72 sq. unit
Attention! Don't confuse the two types of tasks:
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 the minus appears in the formula just considered.
In practice, most often the figure is located in both the upper and lower half-planes.
With) Find the area of a plane figure bounded by lines y \u003d 2x-x 2, y \u003d -x.
Solution.
First you need to make a drawing. Generally speaking, when constructing a drawing in area problems, we are most interested in the intersection points of lines. Let's find the points of intersection of the parabola and the line. This can be done in two ways. The first way is analytical.
We solve the equation:
So the lower limit of integration a=0 , the upper limit of integration b=3 .
|
We build the given lines: 1. Parabola - vertex at the point (1;1); axis intersection Oh - points(0;0) and (0;2). 2. Straight line - the 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 by the formula: . And it does not matter where the figure is located - above the axis or below the axis, but it is important which chart is HIGHER (relative to another chart), 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 |
It is possible to construct lines point by point, while the limits of integration are found out as if "by themselves". Nevertheless, the analytical method of finding the limits still sometimes has to be used if, for example, the graph is large enough, or the threaded construction did not reveal the limits of integration (they can be fractional or irrational).
The desired figure is limited by a parabola from above and a straight line from below.
On the segment , according to the corresponding formula:
Answer: S \u003d 4.5 sq. units
Any definite integral (that exists) has a very good geometric meaning. In class, 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.
That is, the definite integral (if it exists) geometrically corresponds to the area of some figure. For example, consider the definite integral . The integrand defines a certain curve on the plane (it can always be drawn if desired), and the definite integral itself is numerically equal to the area of the corresponding curvilinear trapezoid.
Example 1
This is a typical task statement. The first and most important moment of the decision is the construction of a drawing. Moreover, the drawing must be built RIGHT.
When building a blueprint, I recommend the following order: first it is better to construct all lines (if any) and only after- parabolas, hyperbolas, graphs of other functions. Function graphs are more profitable to build point by point, the technique of pointwise construction can be found in the reference material.
There you can also find material that is very useful in relation to our lesson - how to quickly build a parabola.
In this problem, the solution might look like this.
Let's make a drawing (note that the equation defines the axis):
I will not hatch a curvilinear trapezoid, it is obvious what area we are talking about here. The solution continues like this:
On the segment, the graph of the function is located over axis, that's why:
Answer:
For those who have difficulty calculating the definite integral and applying the Newton-Leibniz formula, please refer to the lecture Definite integral. Solution examples.
After the task is completed, it is always useful to look at the drawing and figure out 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 seems to be true. It is quite clear that if we had, say, the answer: 20 square units, then, obviously, a mistake was made somewhere - 20 cells obviously do not fit into the figure in question, at most a dozen. If the answer turned out to be negative, then the task was also solved incorrectly.
Example 2
Calculate the area of the figure bounded by the lines , , and the axis
This is a do-it-yourself example. Full solution and answer at the end of the lesson.
What to do if the curvilinear trapezoid is located under axle?
Example 3
Calculate the area of the figure bounded by lines and coordinate axes.
Solution: Let's make a drawing:
If a curvilinear trapezoid completely under the axle, then its area can be found by the formula:
In this case:
Attention! The two types of tasks should not be confused:
1) If you are asked to solve just a definite integral without any geometric meaning, then it can be negative.
2) If you are asked to find the area of a figure using a definite integral, then the area is always positive! That is why the 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 make a drawing. Generally speaking, when constructing a drawing in area problems, we are most interested in the intersection points of lines. Let's find the points of intersection 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 build the lines point by point, while the limits of integration are found out as if “by themselves”. The point-by-point construction technique for various charts is discussed in detail in the help Graphs and properties of elementary functions. Nevertheless, the analytical method of finding the limits still sometimes has to be used if, for example, the graph is large enough, or the threaded 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 task: it is more rational to first construct a straight line and only then a parabola. Let's make a drawing:
I repeat that with pointwise construction, the limits of integration are most often found out “automatically”.
And now the working formula: If on a segment some continuous function greater than or equal some continuous function, then the area of the corresponding figure can be found by the formula:
Here it is no longer necessary to think about where the figure is located - above the axis or below the axis, and, roughly speaking, it matters which chart 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 desired figure is limited by a parabola from above and a straight line from below.
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 below the axis, then
And now a couple of examples for an independent decision
Example 5
Example 6
Find the area of the figure enclosed by the lines , .
In the course of solving problems for calculating the area using a certain integral, a funny incident sometimes happens. The drawing was made correctly, the calculations were correct, but due to inattention ... found the area of the wrong figure, that's how your obedient servant screwed up several times. Here is a real life case:
Example 7
Calculate the area of the figure bounded by the lines , , , .
Let's draw first:
The figure whose area we need to find is shaded in blue.(carefully look at the condition - how the figure is limited!). But in practice, due to inattention, it often occurs that you need to find the area of \u200b\u200bthe figure that is shaded in green!
This example is also useful in that in it the area of \u200b\u200bthe figure is calculated using two definite integrals. Really:
1) On the segment above the axis there is a straight line graph;
2) On the segment above the axis is a hyperbola graph.
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 a "school" form, and perform 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 what? May be ? But where is the guarantee that the drawing is made with perfect accuracy, it may well turn out that. Or root. What if we didn't get the graph right at all?
In such cases, one has to spend additional time and refine the limits of integration analytically.
Let's find the points of intersection of the line and the parabola.
To do this, we solve the equation:
Consequently, .
The further solution is trivial, the main thing is not to get confused in substitutions and signs, the calculations here are not the easiest.
On the segment , according to the corresponding formula:
Well, in conclusion of the lesson, we will consider two tasks more difficult.
Example 9
Calculate the area of the figure bounded by lines , ,
Solution: Draw this figure in the drawing.
For point-by-point construction of a drawing, it is necessary to know the appearance of the sinusoid (and in general it is useful to know graphs of all elementary functions), as well as some sine values, they can be found in trigonometric table. In some cases (as in this case), it is allowed to construct a schematic drawing, on which graphs and integration limits must be displayed in principle correctly.
There are no problems with the integration limits here, they follow directly from the condition: - "x" changes from zero to "pi". We make a further decision:
On the segment, the graph of the function is located above the axis, therefore:
(1) How sines and cosines are integrated in odd powers can be seen in the lesson Integrals of trigonometric functions. This is a typical technique, we pinch off one sine.
(2) We use the basic trigonometric identity in the form
(3) Let's change the variable , then:
New redistributions of integration:
Who is really bad business with substitutions, please go to the lesson Replacement method in indefinite integral. For those who are not very clear about the replacement algorithm in a definite integral, visit the page Definite integral. Solution examples. Example 5: Solution: so:
Answer:
Note: note how the integral of the tangent in the cube is taken, the corollary of the basic trigonometric identity is used here.
