Lecture
This lecture concludes the study of the definite integral. The main thing is that we will understand what we need all this for. We are going to speak about applications, that is, the application of the definite integral.
We are going to start with the area of a curvilinear trapezoid. When we started studying the definite integral, it was the first topic. The problem leading to the concept of the definite integral is the problem of the area of a curvilinear trapezoid.
We calculate this area using the formula (see the video).
f(x) in this case is a continuous, non-negative function set on the interval from a to b.
The curvilinear trapezoid is a configuration limited by the function graph, the Ox axis, and the vertical lines x=a, x=b. We calculate its area using the formula on the screen.
Let's consider other issues as well. What if the plane figure is bounded by graphs of continuous functions f and g on the segment [a,b], and the graph of one function is located above the second one. As for the Ox axis, the location of these graphs is not significant for us. We can also calculate the area in this case using the definite integral (see the video).
Let's consider the following situation. The graph of the function f is located above the graph of the function g. What do we do to calculate the area? The functions f and g are continuous, bounded. Thus, this figure is located between two horizontal lines, in a horizontal bar.
What are we going to do?
Let's raise this flat figure up by r units, so that it is higher than the Ox axis.
The functions f+r, g+r are already nonnegative on the segment [a, b] and are still continuous. It is clear that the area of the figure is equal to the difference in the areas of curvilinear trapezoids.
The first one is for the function f+r, the second one is for g+r. These areas are calculated as definite integrals. We write the difference of integrals (see the video), then use the linearity property to write down the general integral, and we see: this formula is correct.
Let's take a simple example of the area of a flat figure. The problem is as follows: you need to find the area of the ellipse.
Here is its equation (see the video). It is clear that the curve is not the graph of a function. Let's express y, and using the symmetry of the ellipse, it is easy to find the fourth part of this area S with a dash.
This is a curvilinear trapezoid, and it is bounded on top by the graph of a non-negative function (see the video).
The area of the entire ellipse S is 4 multiplied by S with a dash. Thus, the area of a curvilinear trapezoid is calculated using the formula (see the video).
Let's try to calculate this integral using the parametric setting of the ellipse. This is the parametric equation of the ellipse (see the video). The question arises "How does t change?" x changes from 0 to a. If x is 0, the cosine is 0 for t equal to π in half.
The upper value of x is a when the cosine of t is 1. This is when t is 0.
We get a new integral (see the video). We calculate the differential x (see the video). Thus, performing transformations, we get the integral (see the video).
By changing the integration limits, we get a plus sign before the integral.
Applying the formula for reducing the degree, we come to almost tabular integrals. Further calculation is not difficult.
Another generalization of the area. We are going to consider a flat figure called a curvilinear sector. First, we need polar coordinates. What a poetic name, isn't it?
The polar coordinate system is set somewhat differently than the Cartesian one. First, we mark the polar axis – a ray with the origin at point O, with a unit of reference and direction. On this ray, we measure the p-distance.
The point O is the pole.
Two coordinates, ϕ and p, define any point M on the plane other than the pole-point O. ϕ is the polar angle, the angle between the polar axis Oρ and the vector OM, and p is the polar radius, the distance from O to M. of The values of ϕ and p are not completely unambiguous. We can take ϕ, for example, add two π or subtract two π, and we get the same point M. We most often choose the main values.
We count ϕ from minus π to π and consider p to be greater than or equal to 0.
As for the pole, it is set by a single condition: p is zero, the value of ϕ is not defined for it.
What is the relationship between Cartesian and polar coordinates? What do we do?
Let us superpose the polar coordinate system and the Cartesian one, the O point (the pole) and the point (0, 0), and the Ox axis and the polar axis. What do we get? The point M has polar coordinates ϕ and p, while it has Cartesian coordinates x and y.
From a right triangle according to the Pythagorean theorem or from these two formulas, we get that (see the video) x2+y2=P2. This is a great formula that we often use.
What is a curve in polar coordinates? This is a certain equation p, equal to p (ϕ).
Here are some examples (see the video), that is, some function of the variable ϕ, each value of the variable ϕ corresponds to a new value of p, which means a point in the polar coordinate system.
And if the function p(ϕ) is continuous, we get a continuous curve in the polar coordinate system.
The variety of such curves is huge. Here are a few curves that have their own names (see the video), they are well known and often used in solving problems.
What is a curvilinear sector? Let's assume that two rays are drawn in the polar coordinate system: ϕ=α and ϕ=β. A continuous function p(ϕ) is defined for all values of ϕ from the segment [α, β]. In this case, a curve appears in the polar coordinate system. The figure bounded by the rays ϕ=α and ϕ=β and the graph of this function is called a curvilinear sector.
Sometimes it can be a circular sector, that is, it is a broader concept.
The problem of calculating the area of this flat figure arises. What can we begin with?
We assume that the area of the circular sector is known to us. What should we do?
We divide the segment from α to β (these are two real numbers), the angle is given in radians, into partial points (see the video).
We get n partial segments and select a value for each of them (see the video).
Each value ϕi and each point ξi corresponds to a ray. What should we do?
If we define the value of the angle bounded by the rays ϕi and ϕi-1, Δ ϕi (see the video), and construct a circular sector of radius p from ξi with this ray, we'll get a fan-shaped figure like this.
It consists of n circular sectors.
We denote the area of the fan-shaped figure, by S with an asterisk, this is the sum of the areas Si with an asterisk (see the video), n circular sectors. The area of the i-th sector is (see the video). Then the area of the entire fan-shaped figure, S with an asterisk, is calculated using the formula (see the video). What is λ? This is the largest of Δ ϕi. If λ tends to 0, this is the partition step, then S with an asterisk becomes closer and closer to the area S of the curvilinear sector each time.
But what is an S with an asterisk?
S with an asterisk is an integral sum for a function (see the video) that is continuous, which means that it is integrable on the segment [α, β]. The limit of these integral sums is a certain integral.
Thus, the area S of a curvilinear sector is (see video).
Here's another task. We've already seen the cardioid. Let's write down the cardioid equation (see the video). The task is to find the area of a flat figure bounded by a cardioid. Thus, we have the formula, the values of ϕ are from 0 to two π, and under the sign of the integral, we write the square of the function p (see the video). The formula for S is as follows (see the video). The next step is to perform some conversions (see the video).
We use the formula of decreasing the power while expanding the last summand in the square of the sum. You are to analyze thoroughly the details of this calculation on your own.
You can see the answer here.
One more application of the definite integral is the volume of the body of rotation.
There are two bodies here. Let us have a curvilinear trapezoid set on the segment [a, b] by a continuous non-negative function.
If we start rotating this trapezoid around the Ox axis, we get a body of rotation, and we use V with the index OX for marking its volume.
You can see the formula for calculating it on the screen (see the video).
Let's try to find out how we can get this formula.
Let us have a curvilinear trapezoid on the segment [a, b]. We get the body of rotation at rotating. We do not have the possibility to determine the volume of this body right away. What can we begin with?
We know the volume of the cylinder.
Let us begin with this formula.
What will we do?
Just as for calculating the area of a curvilinear trapezoid, we divide the segment [a, b] into partial segments.
We take the point ξi on each of the partial segments and draw a rectangle.
We draw a rectangle on each partial segment as on a base with a height f from ξ.
We get a stepped configuration.
Make it spin.
What do we get? A new body of rotation.
It is made up as if from disks, these are circular cylinders. We can easily calculate their volumes. Let's try to do this.
The volume of the resulting body is equal to the sum of partial volumes (see the video). This is a cylinder, its height Δxi, the length of the i-th partial segment, and the radius f of ξi. We can compose a formula (see the video).
If the partition step tends to 0, that is, the largest of the lengths of the partial segments, then the volume of this body tends to the value of the VOX volume.
Note that the sum is integral for a function continuous on the segment [a, b] (see the video).
Therefore, the limit of integral sums is a certain integral of this function. Therefore, the volume of the body of rotation is calculated by the formula (see the video), the verity of which we have established.
The next point: a curvilinear trapezoid can rotate not only relative to the Ox axis, but also relative to the Oy axis. The only requirement is for it to be in the first coordinate quarter.
The segment [a, b] must be to the right of the point 0, and a is greater than or equal to 0.
Our reasoning can lead us to the proof of this formula. However, we are not going to do this (see the video). Thus, when a curvilinear trapezoid rotates relative to the axis Oy, the volume of the body of rotation is calculated using this formula.
Let's take a simple problem.
A flat figure is bounded by a hyperbola xy=6, a horizontal line y=0, the Ox axis, and two vertical lines - x=1 and x=4.
The first task is to find the volume of VOX if we make this flat figure rotate relative to the Ox axis. The calculation formula (see the video) and the calculation turns out to be quite simple (see the video).
We substitute the value y=6/x instead of f(x) and the simple integral leads us to the answer. We calculate VOY using a new formula (see the video).
We use y=6/x instead of f(x) and get a simple integral, its calculation leads us to the answer.
By the way, the volume of this body can be calculated using the formula and VOX, only by re-assigning the axes. You can try to solve this problem. The new approach makes the solution a little bit longer, but the answer is the same.
If flat figures that have a finite area are called quadrirable, then the curves that have a finite length expressed by a number are called rectifiable.
And the new application that we consider is calculating the length of a curve. We are not going to derive these formulas (see the video). By the way, the curve length differential is used in other applications as well.
We are going to use these formulas later on.
Pay attention to these formulas.
These are in Cartesian coordinates, in polar coordinates, and if the curve is set parametrically. These are different cases.
If the curve is given as a graph of a continuous and continuously differentiable function, then l is the integral of dl, the formula is as follows (see the video).
If the curve is set by parametric continuous differentiable functions, that is, the function itself is continuous, and its derivative is continuous, then the length is calculated using the following formula (see the video).
If the curve is set in polar coordinates by the function p, which is continuous and continuously differentiable, its length is calculated using the following formula (see the video).
Another approach is to calculate the surface area of the body of rotation. If a curvilinear trapezoid rotates around the Ox axis, then there is also a rotation surface.
It appears when the curve of the function graph rotates.
We calculate the area using the following formula (see the video).
If the curve is a graph of the function f(x), then the formula is transformed as follows (see the video).
Besides, the definite integral is applied in physics.
Here are just some physical problems that apply the concept of the definite integral.
The applications of the definite integral in physics are the calculation of the distance traversed by a body in the time interval from t1 to t2, if the instantaneous velocity at each point is known; the work of a variable force F(x); and the mass of an inhomogeneous rod, if at each point p(x) it is the density value.
To have some practice, you can solve these geometric problems and check your answers.