Calculating the area of a figure This is perhaps one of the most difficult problems in area theory. In school geometry, they are taught to find the areas of basic geometric shapes such as, for example, a triangle, a rhombus, a rectangle, a trapezoid, a circle, etc. However, one often has to deal with the calculation of the areas of more complex figures. It is in solving such problems that it is very convenient to use integral calculus.
Definition.
Curvilinear trapezoid some figure G is called, bounded by the lines y = f(x), y = 0, x = a and x = b, and the function f(x) is continuous on the segment [a; b] and does not change its sign on it (Fig. 1). The area of a curvilinear trapezoid can be denoted by S(G).
The definite integral ʃ a b f(x)dx for the function f(x), which is continuous and non-negative on the segment [a; b], and is the area of the corresponding curvilinear trapezoid.
That is, to find the area of the figure G, bounded by the lines y \u003d f (x), y \u003d 0, x \u003d a and x \u003d b, it is necessary to calculate the definite integral ʃ a b f (x) dx.
In this way, S(G) = ʃ a b f(x)dx.
If the function y = f(x) is not positive on [a; b], then the area of the curvilinear trapezoid can be found by the formula S(G) = -ʃ a b f(x)dx.
Example 1
Calculate the area of \u200b\u200bthe figure bounded by the lines y \u003d x 3; y = 1; x = 2.
Solution.
The given lines form the figure ABC, which is shown by hatching on rice. 2.
The desired area is equal to the difference between the areas of the curvilinear trapezoid DACE and the square DABE.
Using the formula S = ʃ a b f(x)dx = S(b) – S(a), we find the limits of integration. To do this, we solve a system of two equations:
(y \u003d x 3,
(y = 1.
Thus, we have x 1 \u003d 1 - the lower limit and x \u003d 2 - the upper limit.
So, S = S DACE - S DABE = ʃ 1 2 x 3 dx - 1 = x 4 /4| 1 2 - 1 \u003d (16 - 1) / 4 - 1 \u003d 11/4 (square units).
Answer: 11/4 sq. units
Example 2
Calculate the area of \u200b\u200bthe figure bounded by lines y \u003d √x; y = 2; x = 9.
Solution.
The given lines form the figure ABC, which is bounded from above by the graph of the function
y \u003d √x, and from below the graph of the function y \u003d 2. The resulting figure is shown by hatching on rice. 3.
The desired area is equal to S = ʃ a b (√x - 2). Let's find the limits of integration: b = 9, to find a, we solve the system of two equations:
(y = √x,
(y = 2.
Thus, we have that x = 4 = a is the lower limit.
So, S = ∫ 4 9 (√x – 2)dx = ∫ 4 9 √x dx –∫ 4 9 2dx = 2/3 x√x| 4 9 - 2x| 4 9 \u003d (18 - 16/3) - (18 - 8) \u003d 2 2/3 (square units).
Answer: S = 2 2/3 sq. units
Example 3
Calculate the area of \u200b\u200bthe figure bounded by the lines y \u003d x 3 - 4x; y = 0; x ≥ 0.
Solution.
Let's plot the function y \u003d x 3 - 4x for x ≥ 0. To do this, we find the derivative y ':
y’ = 3x 2 – 4, y’ = 0 at х = ±2/√3 ≈ 1.1 are critical points.
If we draw the critical points on the real axis and place the signs of the derivative, we get that the function decreases from zero to 2/√3 and increases from 2/√3 to plus infinity. Then x = 2/√3 is the minimum point, the minimum value of the function y is min = -16/(3√3) ≈ -3.
Let's determine the intersection points of the graph with the coordinate axes:
if x \u003d 0, then y \u003d 0, which means that A (0; 0) is the point of intersection with the Oy axis;
if y \u003d 0, then x 3 - 4x \u003d 0 or x (x 2 - 4) \u003d 0, or x (x - 2) (x + 2) \u003d 0, from where x 1 \u003d 0, x 2 \u003d 2, x 3 \u003d -2 (not suitable, because x ≥ 0).
Points A(0; 0) and B(2; 0) are the intersection points of the graph with the Ox axis.
The given lines form the OAB figure, which is shown by hatching on rice. 4.
Since the function y \u003d x 3 - 4x takes on (0; 2) a negative value, then
S = |ʃ 0 2 (x 3 – 4x)dx|.
We have: ʃ 0 2 (x 3 - 4x)dx = (x 4 /4 - 4x 2 /2)| 0 2 \u003d -4, from where S \u003d 4 square meters. units
Answer: S = 4 sq. units
Example 4
Find the area of the figure bounded by the parabola y \u003d 2x 2 - 2x + 1, the straight lines x \u003d 0, y \u003d 0 and the tangent to this parabola at the point with the abscissa x 0 \u003d 2.
Solution.
First, we compose the equation of the tangent to the parabola y \u003d 2x 2 - 2x + 1 at the point with the abscissa x₀ \u003d 2.
Since the derivative y' = 4x - 2, then for x 0 = 2 we get k = y'(2) = 6.
Find the ordinate of the touch point: y 0 = 2 2 2 – 2 2 + 1 = 5.
Therefore, the tangent equation has the form: y - 5 \u003d 6 (x - 2) or y \u003d 6x - 7.
Let's build a figure bounded by lines:
y \u003d 2x 2 - 2x + 1, y \u003d 0, x \u003d 0, y \u003d 6x - 7.
Г y \u003d 2x 2 - 2x + 1 - parabola. Points of intersection with the coordinate axes: A(0; 1) - with the Oy axis; with the Ox axis - there are no intersection points, because the equation 2x 2 - 2x + 1 = 0 has no solutions (D< 0). Найдем вершину параболы:
x b \u003d 2/4 \u003d 1/2;
y b \u003d 1/2, that is, the vertex of the parabola point B has coordinates B (1/2; 1/2).
So, the figure whose area is to be determined is shown by hatching on rice. five.
We have: S O A B D \u003d S OABC - S ADBC.
Find the coordinates of point D from the condition:
6x - 7 = 0, i.e. x \u003d 7/6, then DC \u003d 2 - 7/6 \u003d 5/6.
We find the area of triangle DBC using the formula S ADBC = 1/2 · DC · BC. In this way,
S ADBC = 1/2 5/6 5 = 25/12 sq. units
S OABC = ʃ 0 2 (2x 2 - 2x + 1)dx = (2x 3 /3 - 2x 2 /2 + x)| 0 2 \u003d 10/3 (square units).
Finally we get: S O A B D \u003d S OABC - S ADBC \u003d 10/3 - 25/12 \u003d 5/4 \u003d 1 1/4 (sq. units).
Answer: S = 1 1/4 sq. units
We have reviewed examples finding the areas of figures bounded by given lines. To successfully solve such problems, you need to be able to build lines and graphs of functions on a plane, find the points of intersection of lines, apply a formula to find the area, which implies the ability and skills to calculate certain integrals.
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In this lesson we will learn how to calculate areas of flat figures, which are called curvilinear trapezoids .
Examples of such figures are in the figure below.
On the one hand, finding the area of a flat figure using a definite integral is extremely simple. We are talking about the area of \u200b\u200bthe figure, which is limited from above by a certain curve, from below - by the abscissa axis ( Ox), and on the left and right are some straight lines. The simplicity is that the definite integral of the function to which the curve is given, and there is the area of such a figure(curvilinear trapezoid).
To calculate the area of a figure, we need:
- Definite integral of the function defining the curve , which limits the curvilinear trapezoid from above. And here comes the first significant nuance: a curvilinear trapezoid can be limited by a curve not only from above, but also from below . How to act in this case? Simple but important to remember: the integral in this case is taken with a minus sign .
- Limits of integration a And b, which we find from the equations of lines that bound the figure on the left and right: x = a , x = b, where a And b- numbers.
Separately, some more nuances.
The curve that limits the curvilinear trapezoid from above (or below) must be graph of a continuous and non-negative function y = f(x) .
X values must belong to the segment [a, b] . That is, such, for example, lines as a section of a mushroom are not taken into account, in which the leg fits perfectly into this segment, and the hat is much wider.
Side segments can degenerate into points . If you saw such a figure in the drawing, this should not confuse you, since this point always has its own value on the x-axis. So everything is in order with the limits of integration.
Now you can move on to formulas and calculations. So the area s curvilinear trapezoid can be calculated by the formula
If f(x) ≤ 0 (the graph of the function is located below the axis Ox), then area of a curved trapezoid can be calculated by the formula
There are also cases when both the upper and lower boundaries of the figure are functions, respectively y = f(x) And y = φ (x) , then the area of such a figure is calculated by the formula
. (3)
We solve problems together
Let's start with cases where the area of a figure can be calculated using formula (1).
Example 1Ox) and direct x = 1 , x = 3 .
Solution. Because y = 1/x> 0 on the segment , then the area of the curvilinear trapezoid is found by the formula (1):
.
Example 2 Find the area of the figure bounded by the graph of the function , straight line x= 1 and the x-axis ( Ox ).
Solution. The result of applying formula (1):
If then s= 1/2; if then s= 1/3 , etc.
Example 3 Find the area of \u200b\u200bthe figure bounded by the graph of the function, the x-axis ( Ox) and direct x = 4 .
Solution. The figure corresponding to the condition of the problem is a curvilinear trapezoid, in which the left segment has degenerated into a point. The integration limits are 0 and 4. Since, according to formula (1), we find the area of the curvilinear trapezoid:
.
Example 4 Find the area of the figure bounded by the lines , , and located in the 1st quarter.
Solution. To use formula (1), we represent the area of the figure given by the conditions of the example as the sum of the areas of a triangle OAB and curvilinear trapezoid ABC. When calculating the area of a triangle OAB the limits of integration are the abscissas of the points O And A, and for the figure ABC- abscissas of points A And C (A is the point of intersection of the line OA and parabolas, and C- point of intersection of the parabola with the axis Ox). Solving jointly (as a system) the equations of a straight line and a parabola, we obtain (the abscissa of the point A) and (the abscissa of another point of intersection of the line and the parabola, which is not needed for the solution). Similarly, we obtain , (abscissas of points C And D). Now we have everything to find the area of the figure. We find:
Example 5 Find the area of a curvilinear trapezoid ACDB, if the equation of the curve CD and abscissa A And B respectively 1 and 2.
Solution. We express this equation of the curve through Y: The area of the curvilinear trapezoid is found by the formula (1):
.
Let's move on to cases where the area of a figure can be calculated using formula (2).
Example 6 Find the area of the figure bounded by the parabola and the x-axis ( Ox ).
Solution. This figure is located below the x-axis. Therefore, to calculate its area, we use formula (2). The limits of integration are the abscissas and points of intersection of the parabola with the axis Ox. Consequently,
Example 7 Find the area between the x-axis ( Ox) and two neighboring sine waves.
Solution. The area of this figure can be found by the formula (2):
.
Let's find each term separately:
.
.
Finally we find the area:
.
Example 8 Find the area of the figure enclosed between the parabola and the curve.
Solution. Let's express the equations of the lines in terms of Y:
The area according to the formula (2) will be obtained as
,
where a And b- abscissas of points A And B. We find them by solving the equations together:
Finally we find the area:
And, finally, there are cases when the area of a figure can be calculated using formula (3).
Example 9 Find the area of the figure enclosed between the parabolas 
And .
Calculate the area of a figure bounded by lines.
Solution.
We find the points of intersection of the given lines. To do this, we solve the system of equations:
To find the abscissas of the points of intersection of the 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 \u200b\u200bwhich is calculated using the above formula:
According to the Newton-Leibniz formula, we find:
Find the area of an area bounded by an ellipse.
Solution.
From the ellipse equation for the I quadrant we have . From here, according to the formula, we obtain
Let's apply the 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.
We find one fourth of the required area
From here S = pab.
Find the area of a figure bounded by linesy = - x 2 + x + 4 andy = - x + 1.
Solution.
Find the intersection points 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. Find the roots x 1 = -1, x 2 = 3 and their corresponding ordinates y 1 = 2, y 2 = -2.
Using the figure area formula, we get
Find the area enclosed by the parabolay = x 2 + 1 and directx + y = 3.
Solution.
Solving the system of equations
find the abscissas of the intersection points x 1 = -2 and x 2 = 1.
Assuming 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 the figure bounded by the arc of the curve r = f(φ ) and two polar radii φ 1 = ʅ And φ 2 = ʆ , is expressed by the integral
Due to the symmetry of the curve, we first determine one-fourth of the desired area
Therefore, the total area is S = a 2 .
Calculate the arc length of an astroidx 2/3 + y 2/3 = a 2/3 .
Solution.
We 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 .
In view of the symmetry of the curve (*), it suffices 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 in the range from 0 to π /2, we get
From here L = 6a.
Find the area bounded by the spiral of Archimedesr = aφ and two radius vectors that correspond to polar anglesφ 1 Andφ 2 (φ 1 < φ 2 ).
Solution.
Area bounded by a curve r = f(φ ) is calculated by the formula , where α And β - limits of change of the polar angle.
Thus, we get
(*)
From (*) it follows that the area bounded by the polar axis and the first turn of the Archimedes spiral ( φ 1 = 0; φ 2 = 2π ):
Similarly, we find the area bounded 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 figure bounded by parabolasy = x 2 Andx = y 2 .
Solution.
Let's solve the system of equations
and 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 desired volume of the body of revolution is equal to the difference between the two volumes formed by rotation around the axis Ox curvilinear trapezoids OCBA And ODBA:
Calculate the area bounded by the axisOx and sinusoidy = sinx on segments: a); b) .
Solution.
a) On the segment, the function sin x preserves the sign, and therefore by the formula , assuming y= sin x, we find
b) On the segment , function sin x changes sign. For the correct solution of the problem, it is necessary to divide the segment into two and [ π , 2π ], in each of which the function retains its sign.
According to the rule of signs, on the segment [ π , 2π ] area is taken with a minus sign.
As a result, the desired area is equal to
Determine the volume of the body bounded by the surface obtained from the rotation of the ellipsearound the major axisa .
Solution.
Given that the ellipse is symmetrical about 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 the body of revolution through 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 )
Find the area bounded by parabolasy 2 = 2 px Andx 2 = 2 py .
Solution.
First, we find the coordinates of the intersection points of the parabolas in order to determine the integration interval. 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.
We find the roots of the equations:
Considering the fact that the point A the intersection of the parabolas is in the first quarter, then the limits of integration x= 0 and x = 2p.
The desired area is found by the formula
but)
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.
from) 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
In fact, in order to find the area of \u200b\u200ba figure, you do not need so much knowledge of the indefinite and definite integral. The task "calculate the area using a definite integral" always involves the construction of a drawing, so your knowledge and drawing skills will be a much more relevant issue. In this regard, it is useful to refresh the memory of the graphs of the main elementary functions, and, at a minimum, be able to build a straight line, and a hyperbola.
A curvilinear trapezoid is a flat figure bounded by an axis, straight lines, and a graph of a continuous function on a segment that does not change sign on this interval. Let this figure be located not less abscissa:
Then the area of a curvilinear trapezoid is numerically equal to a certain integral. Any definite integral (that exists) has a very good geometric meaning.
In terms of geometry, the definite integral is the AREA.
I.e, the definite integral (if it exists) corresponds geometrically to the area of some figure. For example, consider the definite integral . The integrand defines a curve on the plane that is located above the axis (those who wish can complete the drawing), 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: at first it is better to construct all lines (if any) and only Then- parabolas, hyperbolas, graphs of other functions. Function graphs are more profitable to build pointwise.
In this problem, the solution might look like this.
Let's make a drawing (note that the equation defines the axis):
On the segment, the graph of the function is located over axis, that's why:
Answer:
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.
Example 3
Calculate the area of the figure bounded by lines and coordinate axes.
Solution: Let's make a drawing:
If the curvilinear trapezoid is located under axle(or at least not higher given axis), then its area can be found by the formula:
In this case:
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, and therefore, from the simplest school problems, we move on to more meaningful examples.
Example 4
Find the area of a flat figure bounded by lines , .
Solution: First you need to complete the drawing. Generally speaking, when constructing a drawing in 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 best 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”. 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:
And now the working formula: If there is some continuous function on the interval greater than or equal some continuous function, then the area of the figure bounded by the graphs of these functions and straight lines, can be found by the formula:
Here it is no longer necessary to think 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.
On the segment , according to the corresponding formula:
Answer:
Example 4
Calculate the area of the figure bounded by the lines , , , .
Solution: Let's make a drawing 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, a “glitch” 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:
How to calculate the volume of a body of revolutionusing a definite integral?
Imagine some flat figure on the coordinate plane. We have already found its area. But, in addition, this figure can also be rotated, and rotated in two ways:
Around the x-axis;
Around the y-axis .
In this article, both cases will be discussed. The second method of rotation is especially interesting, it causes the greatest difficulties, but in fact the solution is almost the same as in the more common rotation around the x-axis.
Let's start with the most popular type of rotation.
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