The area of the figure bounded by these lines online. Definite integral
a)
Solution.
The first and most important point of the solution is the construction of the drawing.
Let's execute the drawing:
The equation y = 0 sets the x-axis;
- x = -2 and x = 1 - straight lines parallel to the axes OU;
- y = x 2 +2 - a parabola, the branches of which are directed upwards, with apex 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 axis intersection OU and solving the corresponding quadratic equation, find the intersection with the axis Oh .
The vertex of the parabola can be found by the formulas: 
You can draw lines and point by point.
On the segment [-2; 1] the graph of the function y = x 2 +2 situated above the axis Ox , therefore:
Answer: S = 9 square units
After the task is completed, it is always helpful to look at the blueprint and estimate 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 looks like the truth. It is quite clear that if we got, say, the answer: 20 square units, then, obviously, a mistake has been made somewhere - the figure under consideration clearly does not fit 20 cells, at most ten. If the answer is negative, then the task was also solved incorrectly.
What to do if the curved trapezoid is located under the axis Oh?
b) Calculate the area of a shape bounded by lines y = -e x , x = 1 and coordinate axes.
Solution.
Let's complete the drawing.
If the curved trapezoid completely located under the axle Oh , then its area can be found by the formula:
Answer: S = (e-1) sq. units "1.72 sq. units.
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 may 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 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 flat figure bounded by lines y = 2x-x 2, y = -x.
Solution.
First you need to complete the drawing. Generally speaking, when constructing a drawing in problems on an area, we are most interested in the points of intersection of lines. Find the intersection points of the parabola 
and straight 
This can be done in two ways. The first way is analytical.
We solve the equation:
Hence, the lower limit of integration a = 0 , the upper limit of integration b = 3 .
|
We build the given lines: 1. Parabola - apex at the point (1; 1); axis intersection Oh - points (0; 0) and (0; 2). 2. Straight line - bisector of the 2nd and 4th coordinate angles. Now Attention! If on the segment [ a; b] some continuous function f (x) is 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 doesn't 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 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 |
.You can plot the lines point by point, while the limits of integration are clarified as if "by themselves." Nevertheless, the analytical method of finding the limits still has to be applied sometimes if, for example, the graph is large enough, or the precise construction did not reveal the limits of integration (they can be fractional or irrational).
The required figure is bounded by a parabola at the top and a straight line at the bottom.
On the segment 
, according to the corresponding formula:
Answer: S = 4.5 sq. Units
In fact, in order to find the area of a figure, one does not need so much knowledge of the indefinite and definite integral. The task "calculate area using a definite integral" always involves building a drawing, therefore, your knowledge and drawing skills will be a much more pressing issue. In this regard, it is useful to refresh the memory of the graphs of the basic elementary functions, and, at least, to 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 axis:
Then the area of a curvilinear trapezoid is numerically equal to the definite integral... Any definite integral (that exists) has a very good geometric meaning.
From the point of view of geometry, the definite integral is the AREA.
That is, a definite integral (if it exists) geometrically corresponds to the area of some figure. For example, consider a definite integral. The integrand defines a curve on the plane that is located above the axis (those who wish can make a drawing), and the definite integral itself is numerically equal to the area of the corresponding curvilinear trapezoid.
Example 1
This is a typical formulation of the assignment. The first and most important point of the solution is the construction of the drawing... Moreover, the drawing must be built RIGHT.
When building a drawing, I recommend the following order: at first it is better to build all straight lines (if any) and only after- parabolas, hyperbolas, graphs of other functions. It is more profitable to build graphs of functions pointwise.
In this problem, the solution might look like this.
Let's draw a drawing (note that the equation defines the axis):
On the segment, the graph of the function is located above the axis, therefore:
Answer:
After the task is completed, it is always helpful to look at the blueprint and estimate 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 looks like the truth. It is quite clear that if we got, say, the answer: 20 square units, then, obviously, a mistake has been made somewhere - the figure under consideration clearly does not fit 20 cells, at most ten. If the answer is negative, then the task was also solved incorrectly.
Example 3
Calculate the area of a shape bounded by lines and coordinate axes.
Solution: Let's execute the drawing:
If the curved trapezoid is located under the axis(or at least not higher given axis), 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 may 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 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 problems on an area, we are most interested in the points of intersection of lines. Find the intersection points 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 construct the lines point by point, while the limits of integration become clear, as it were, "by themselves." Nevertheless, the analytical method of finding the limits still has to be applied sometimes if, for example, the graph is large enough, or the precise construction did not reveal the limits of integration (they can be fractional or irrational). And we will also consider such an example.
Returning to our problem: it is more rational to first construct a straight line and only then a parabola. Let's execute the drawing:
And now the working formula: If on a segment some continuous function greater than or equal of 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 you no longer need to think about where the figure is located - above the axis or below the axis, and, roughly speaking, it is important which schedule 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 required figure is bounded by a parabola at the top and a straight line at the bottom.
On the segment, according to the corresponding formula:
Answer:
Example 4
Calculate the area of the figure bounded by the lines,,,.
Solution: First, let's execute the drawing:
The figure whose area we need to find is shaded in blue(carefully look at the condition - what the figure is limited to!). But in practice, due to inattention, a "glitch" often arises, that you need to find the area of the figure, which is shaded in green!
This example is also useful in that it calculates the area of a figure using two definite integrals.
Really:
1) A line graph is located on the segment above the axis;
2) The hyperbola graph is located on the segment above the axis.
It is quite obvious that the areas can (and should) be added, therefore:
We now turn to consideration of applications of integral calculus. In this lesson we will analyze a typical and most common task. calculating the area of a flat figure using a definite integral... Finally, all those seeking meaning in higher mathematics - may they find it. You never know. We'll have to bring the suburban area closer in life with elementary functions and find its area using a definite integral.
To successfully master the material, you must:
1) Understand the indefinite integral at least at the middle level. Thus, dummies should first familiarize themselves with the lesson Not.
2) Be able to apply the Newton-Leibniz formula and calculate a definite integral. You can build warm friendships with definite integrals on the page Definite integral. Examples of solutions. The task "calculate area using a definite integral" always involves building a drawing, therefore, your knowledge and skills in building drawings will also be an urgent issue. At a minimum, you need to be able to build a straight line, a parabola and a hyperbola.
Let's start with a curved trapezoid. A curved trapezoid is a flat figure bounded by the graph of some function y = f(x), the axis OX and lines x = a; x = b.
The area of a curved trapezoid is numerically equal to the definite integral
Any definite integral (that exists) has a very good geometric meaning. At the lesson Definite integral. Examples of solutions we 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, a definite integral (if it exists) geometrically corresponds to the area of some figure... Consider the definite integral
Integrand
defines a curve on the plane (it can 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 formulation of the assignment. The most important point of the solution is the construction of the drawing... Moreover, the drawing must be built RIGHT.
When building a drawing, I recommend the following order: at first it is better to build all straight lines (if any) and only after- parabolas, hyperbolas, graphs of other functions. The technique of point-by-point construction can be found in the reference material. Graphs and properties of elementary functions... There you can also find very useful material in relation to our lesson - how to quickly build a parabola.
In this problem, the solution might look like this.
Let's complete the drawing (note that the equation y= 0 specifies the axis OX):
We will not hatch the curvilinear trapezoid, here it is obvious what area we are talking about. The solution continues like this:
On the segment [-2; 1] function graph y = x 2 + 2 located above the axisOX, therefore:
Answer: .
Who has difficulty calculating a definite integral and applying the Newton-Leibniz formula
,
refer to the lecture Definite integral. Examples of solutions... After the task is completed, it is always helpful to look at the blueprint and estimate 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 looks like the truth. It is quite clear that if we got, say, the answer: 20 square units, then, obviously, a mistake has been made somewhere - the figure under consideration clearly does not fit 20 cells, at most ten. If the answer is negative, then the task was also solved incorrectly.
Example 2
Calculate the area of a shape bounded by lines xy = 4, x = 2, x= 4 and axis OX.
This is an example for a do-it-yourself solution. Complete solution and answer at the end of the tutorial.
What to do if the curved trapezoid is located under the axisOX?
Example 3
Calculate the area of a shape bounded by lines y = e - x, x= 1 and coordinate axes.
Solution: Let's execute the drawing:
If the curved trapezoid completely located under the axle OX , 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 may 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 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 y = 2x – x 2 , y = -x.
Solution: First you need to complete the drawing. When constructing a drawing in problems on the area, we are most interested in the points of intersection of lines. Find the intersection points of the parabola y = 2x – x 2 and straight y = -x... This can be done in two ways. The first way is analytical. We solve the equation:
Hence, the lower limit of integration a= 0, the upper limit of integration b= 3. It is often more profitable and faster to construct the lines point by point, while the limits of integration become clear as if “by themselves”. Nevertheless, the analytical method of finding the limits still has to be applied sometimes if, for example, the graph is large enough, or the precise construction did not reveal the limits of integration (they can be fractional or irrational). Returning to our problem: it is more rational to first construct a straight line and only then a parabola. Let's execute the drawing:
Let us repeat that in the case of a pointwise construction, the limits of integration are most often found out “automatically”.
And now the working formula:
If on the segment [ a; b] some continuous function f(x) greater than or equal some continuous function g(x), then the area of the corresponding figure can be found by the formula:
Here you no longer need to think where the figure is located - above the axis or below the axis, but it is important which schedule 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 from 2 x – x 2 must be subtracted - x.
The completion of the solution might look like this:
The sought-after figure is bounded by a parabola y = 2x – x 2 top and straight y = -x from below.
On segment 2 x – x 2 ≥ -x... According to the corresponding formula:
Answer: .
In fact, the school formula for the area of a curvilinear trapezoid in the lower half-plane (see example No. 3) is a special case of the formula
.
Since the axis OX given by the equation y= 0, and the graph of the function g(x) is located below the axis OX, then
.
And now a couple of examples for self-solution
Example 5
Example 6
Find the area of a shape bounded by lines
In the course of solving problems for calculating the area using a definite integral, a funny incident sometimes happens. The drawing is done correctly, the calculations are correct, but, inadvertently, ... the area of the wrong figure was found.
Example 7
First, let's execute the drawing:
The figure whose area we need to find is shaded in blue(carefully look at the condition - what the figure is limited to!). But in practice, through inattention, they often decide that it is necessary to find the area of the figure, which is shaded in green!
This example is also useful in that it calculates the area of a figure using two definite integrals. Really:
1) On the segment [-1; 1] above the axis OX the graph is straight y = x+1;
2) On a segment above the axis OX the graph of hyperbole is located y = (2/x).
It is quite obvious that the areas can (and should) be added, therefore:
Answer:
Example 8
Calculate the area of a shape bounded by lines
Let's represent the equations in the "school" form
and execute a point-by-point drawing:
It can be seen from the drawing that our upper limit is "good": b = 1.
But what is the lower limit ?! It is clear that this is not an integer, but which one?
May be, a= (- 1/3)? But where is the guarantee that the drawing is made with perfect accuracy, it may well turn out that a= (- 1/4). What if we plotted the graph incorrectly at all?
In such cases, you have to spend additional time and refine the limits of integration analytically.
Find the intersection points of the graphs
To do this, we solve the equation:
.
Hence, a=(-1/3).
The further solution is trivial. The main thing is not to get confused in substitutions and signs. The calculations here are not the easiest ones. On the segment
, 
,
according to the corresponding formula:
Answer: 
At the end of the lesson, we will consider two more difficult tasks.
Example 9
Calculate the area of a shape bounded by lines
Solution: Draw this figure in the drawing.
For point-by-point construction of the drawing, you need to know the appearance of the sinusoid. In general, it is useful to know the graphs of all elementary functions, as well as some values of the sine. They can be found in the table of values trigonometric functions... In a number of cases (for example, in this one), it is allowed to construct a schematic drawing, on which the graphs and limits of integration should be displayed correctly in principle.
There are no problems with the limits of integration, they follow directly from the condition:
- "x" changes from zero to "pi". We make a further decision:
On a segment, the graph of the function y= sin 3 x located above the axis OX, therefore:
(1) How sines and cosines are integrated in odd powers, you can see in the lesson Integrals of trigonometric functions... We pinch off one sinus.
(2) We use the basic trigonometric identity in the form
(3) Change the variable t= cos x, then: is located above the axis, therefore:
.
.
Note: note how the integral of the tangent in the cube is taken, here a consequence of the main trigonometric identity is used
.
Definite integral. How to calculate the area of a shape
We now turn to consideration of applications of integral calculus. In this lesson we will analyze a typical and most common task. - how to calculate the area of a flat figure using a definite integral... Finally, those who are looking for meaning in higher mathematics - may they find it. You never know. We'll have to bring the suburban area closer in life with elementary functions and find its area using a definite integral.
To successfully master the material, you must:
1) Understand the indefinite integral at least at the middle level. Thus, dummies should first familiarize themselves with the lesson Not.
2) Be able to apply the Newton-Leibniz formula and calculate a definite integral. You can build warm friendships with definite integrals on the page Definite integral. Examples of solutions.
In fact, in order to find the area of a figure, one does not need so much knowledge of the indefinite and definite integral. The task "calculate area using a definite integral" always involves building a drawing, therefore, your knowledge and drawing skills will be a much more pressing issue. In this regard, it is useful to refresh the memory of the graphs of the basic elementary functions, and, at least, to be able to build a straight line, a parabola and a hyperbola. This can be done (many people need it) with the help of methodological material and an article on geometric transformations of graphs.
Actually, everyone is familiar with the problem of finding the area using a definite integral since school, and we will not go far ahead of the school curriculum. This article might not exist at all, but the fact is that the problem occurs in 99 cases out of 100, when a student suffers from the hated tower with enthusiasm to master the course of higher mathematics.
The materials of this workshop are presented simply, in detail and with a minimum of theory.
Let's start with a curved trapezoid.
Curved trapezoid is called a flat figure bounded by an axis, straight lines, and a graph of a continuous function on a segment, which does not change sign on this interval. Let this figure be located not less abscissa axis:
Then the area of a curvilinear trapezoid is numerically equal to the definite integral... Any definite integral (that exists) has a very good geometric meaning. At the lesson Definite integral. Examples of solutions 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, a definite integral (if it exists) geometrically corresponds to the area of some figure... For example, consider a definite integral. The integrand sets a curve on the plane that is located above the axis (those who wish can make a drawing), and the definite integral itself is numerically equal to the area of the corresponding curvilinear trapezoid.
Example 1
This is a typical formulation of the assignment. The first and most important point of the solution is the construction of the drawing... Moreover, the drawing must be built RIGHT.
When building a drawing, I recommend the following order: at first it is better to build all straight lines (if any) and only after- parabolas, hyperbolas, graphs of other functions. It is more profitable to build graphs of functions pointwise, the technique of point-by-point construction can be found in the reference material Graphs and properties of elementary functions... There you can also find very useful material in relation to our lesson - how to quickly build a parabola.
In this problem, the solution might look like this.
Let's draw a drawing (note that the equation defines the axis):
I will not hatch a curved trapezoid, here it is obvious what area we are talking about. The solution continues like this:
On the segment, the graph of the function is located above the axis, therefore:
Answer:
Who has difficulty calculating a definite integral and applying the Newton-Leibniz formula 
, refer to the lecture Definite integral. Examples of solutions.
After the task is completed, it is always helpful to look at the blueprint and estimate 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 looks like the truth. It is quite clear that if we got, say, the answer: 20 square units, then, obviously, a mistake has been made somewhere - the figure under consideration clearly does not fit 20 cells, at most ten. If the answer is negative, then the task was also solved incorrectly.
Example 2
Calculate the area of a shape bounded by lines, and an axis
This is an example for a do-it-yourself solution. Complete solution and answer at the end of the tutorial.
What to do if the curved trapezoid is located under the axis?
Example 3
Calculate the area of a shape bounded by lines and coordinate axes.
Solution: Let's execute the drawing: 
If the curved trapezoid is located under the axis(or at least not higher given axis), 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 may 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 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 problems on an area, we are most interested in the points of intersection of lines. Find the intersection points 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 construct the lines point by point, while the limits of integration become clear, as it were, "by themselves." The technique of point-by-point plotting 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 has to be applied sometimes if, for example, the graph is large enough, or the precise construction did not reveal the limits of integration (they can be fractional or irrational). And we will also consider such an example.
Returning to our problem: it is more rational to first construct a straight line and only then a parabola. Let's execute the drawing: 
I repeat that in the case of a pointwise construction, the limits of integration are most often found out by an “automaton”.
And now the working formula: If on a segment some continuous function greater than or equal of 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 you no longer need to think about where the figure is located - above the axis or below the axis, and, roughly speaking, it is important which schedule 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 required figure is bounded by a parabola at the top and a straight line at the bottom.
On the segment, according to the corresponding formula:
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 not higher axis, then 
And now a couple of examples for self-solution
Example 5
Example 6
Find the area of the figure bounded by lines,.
In the course of solving problems for calculating the area using a definite integral, a funny incident sometimes happens. The drawing is done correctly, the calculations are correct, but through inattention ... the area of the wrong figure is found, this is how your humble servant screwed up several times. Here's a real life case:
Example 7
Calculate the area of the figure bounded by the lines,,,.
Solution: First, let's execute the drawing: 
... Eh, a lousy drawing came out, but everything seems to be legible.
The figure whose area we need to find is shaded in blue(carefully look at the condition - what the figure is limited to!). But in practice, due to inattention, a "glitch" often arises, that you need to find the area of the figure, which is shaded in green!
This example is also useful in that it calculates the area of a figure using two definite integrals. Really:
1) A line graph is located on the segment above the axis;
2) The hyperbola graph is located on the segment above the axis.
It is quite obvious that the areas can (and should) be added, therefore:
Answer:
Let's move on to one more meaningful task.
Example 8
Calculate the area of a shape bounded by lines,
Let's represent the equations in a "school" form, and we will execute 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 which one? May be ? But where is the guarantee that the drawing is made with perfect accuracy, it may well be that. Or root. What if we plotted the graph incorrectly at all?
In such cases, you have to spend additional time and refine the limits of integration analytically.
Find the intersection points of the line and the parabola.
To do this, we solve the equation: 
,
Really, .
The further solution is trivial, the main thing is not to get confused in substitutions and signs, the calculations are not the easiest here.
On the segment 
, according to the corresponding formula: 
Answer: 
Well, in conclusion of the lesson, we will consider two more difficult tasks.
Example 9
Calculate the area of a figure bounded by lines,
Solution: Let's depict this figure in the drawing. 
Damn, I forgot to sign the schedule, but to redo the picture, sorry, not hotts. Not drawing, in short, today is the day =)
For point-by-point construction, you need 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 a number of cases (as in this one), it is allowed to construct a schematic drawing, on which the graphs and limits of integration should be displayed correctly in principle.
There are no problems with the limits of integration, 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:
Problem 1(on calculating the area of a curved trapezoid).
In the Cartesian rectangular coordinate system xOy, a figure is given (see the figure), bounded by the x-axis, by the straight lines x = a, x = b (a by a curved trapezoid. It is required to calculate the area of a curved trapezoid.
Solution. Geometry gives us recipes for calculating the areas of polygons and some parts of a circle (sector, segment). Using geometric considerations, we will be able to find only an approximate value of the required area, arguing as follows.
We split the segment [a; b] (base of a curved trapezoid) into n equal parts; this partition is realizable using the points x 1, x 2, ... x k, ... x n-1. Let us draw straight lines through these points parallel to the y-axis. Then the given curvilinear trapezoid will be divided into n parts, into n narrow columns. The area of the entire trapezoid is equal to the sum of the areas of the columns.
Consider the k-th column separately, i.e. a curvilinear trapezoid, the base of which is a segment. Let's replace it with a rectangle with the same base and height equal to f (x k) (see figure). The area of the rectangle is \ (f (x_k) \ cdot \ Delta x_k \), where \ (\ Delta x_k \) is the length of the segment; it is natural to consider the compiled product as an approximate value of the area of the k-th column. 
If we now do the same with all the other columns, we will arrive at the following result: the area S of a given curvilinear trapezoid is approximately equal to the area S n of a stepped figure composed of n rectangles (see figure):
\ (S_n = f (x_0) \ Delta x_0 + \ dots + f (x_k) \ Delta x_k + \ dots + f (x_ (n-1)) \ Delta x_ (n-1) \)
Here, for the sake of uniformity of notation, we assume that a = x 0, b = x n; \ (\ Delta x_0 \) - segment length, \ (\ Delta x_1 \) - segment length, etc. at the same time, as we agreed above, \ (\ Delta x_0 = \ dots = \ Delta x_ (n-1) \)
So, \ (S \ approx S_n \), and this approximate equality is the more accurate, the larger n.
By definition, it is assumed that the required area of a curvilinear trapezoid is equal to the limit of the sequence (S n):
$$ S = \ lim_ (n \ to \ infty) S_n $$
Task 2(about moving point)
A material point moves in a straight line. The dependence of speed on time is expressed by the formula v = v (t). Find the displacement of a point over a period of time [a; b].
Solution. If the motion were uniform, then the problem would be solved very simply: s = vt, i.e. s = v (b-a). For uneven movement, you have to use the same ideas on which the solution to the previous problem was based.
1) Divide the time interval [a; b] into n equal parts.
2) Consider a time interval and assume that during this time interval the speed was constant, such as at the time t k. So, we consider that v = v (t k).
3) Find the approximate value of the displacement of the point over a period of time, this approximate value will be denoted by s k
\ (s_k = v (t_k) \ Delta t_k \)
4) Find the approximate value of the displacement s:
\ (s \ approx S_n \) where
\ (S_n = s_0 + \ dots + s_ (n-1) = v (t_0) \ Delta t_0 + \ dots + v (t_ (n-1)) \ Delta t_ (n-1) \)
5) The desired displacement is equal to the sequence limit (S n):
$$ s = \ lim_ (n \ to \ infty) S_n $$
Let's summarize. The solutions to various problems have been reduced to the same mathematical model. Many problems from various fields of science and technology lead in the process of solving to the same model. This means that this mathematical model must be specially studied.
Definitive integral concept
Let us give a mathematical description of the model that was built in the three considered problems for the function y = f (x), continuous (but not necessarily non-negative, as was assumed in the considered problems) on the interval [a; b]:
1) we split the segment [a; b] into n equal parts;
2) make up the sum $$ S_n = f (x_0) \ Delta x_0 + f (x_1) \ Delta x_1 + \ dots + f (x_ (n-1)) \ Delta x_ (n-1) $$
3) calculate $$ \ lim_ (n \ to \ infty) S_n $$
In the course of mathematical analysis, it was proved that this limit exists in the case of a continuous (or piecewise continuous) function. He's called a definite integral of the function y = f (x) along the segment [a; b] and denoted as follows:
\ (\ int \ limits_a ^ b f (x) dx \)
The numbers a and b are called the limits of integration (respectively, lower and upper).
Let's return to the tasks discussed above. The definition of the area given in Problem 1 can now be rewritten as follows:
\ (S = \ int \ limits_a ^ b f (x) dx \)
here S is the area of the curved trapezoid shown in the figure above. This is geometric meaning of a definite integral.
The definition of the displacement s of a point moving in a straight line with a speed v = v (t) over the time interval from t = a to t = b, given in Problem 2, can be rewritten as follows:
Formula of Newton - Leibniz
To begin with, let's answer the question: what is the connection between a definite integral and an antiderivative?
The answer can be found in Problem 2.On the one hand, the displacement s of a point moving in a straight line with a speed v = v (t) over the time interval from t = a to t = b and is calculated by the formula
\ (S = \ int \ limits_a ^ b v (t) dt \)
On the other hand, the coordinate of the moving point is the antiderivative for the velocity - let us denote it by s (t); hence, the displacement s is expressed by the formula s = s (b) - s (a). As a result, we get:
\ (S = \ int \ limits_a ^ b v (t) dt = s (b) -s (a) \)
where s (t) is the antiderivative for v (t).
In the course of mathematical analysis, the following theorem was proved.
Theorem. If the function y = f (x) is continuous on the segment [a; b], then the following formula is valid
\ (S = \ int \ limits_a ^ b f (x) dx = F (b) -F (a) \)
where F (x) is the antiderivative for f (x).
The above formula is usually called by the Newton - Leibniz formula in honor of the English physicist Isaac Newton (1643-1727) and the German philosopher Gottfried Leibniz (1646-1716), who received it independently of each other and almost simultaneously.
In practice, instead of writing F (b) - F (a), use the notation \ (\ left. F (x) \ right | _a ^ b \) (sometimes called double substitution) and, accordingly, rewrite the Newton - Leibniz formula in the following form:
\ (S = \ int \ limits_a ^ b f (x) dx = \ left. F (x) \ right | _a ^ b \)
Calculating a definite integral, first find the antiderivative, and then perform double substitution.
Based on the Newton - Leibniz formula, two properties of a definite integral can be obtained.
Property 1. The integral of the sum of functions is equal to the sum of integrals:
\ (\ int \ limits_a ^ b (f (x) + g (x)) dx = \ int \ limits_a ^ b f (x) dx + \ int \ limits_a ^ b g (x) dx \)
Property 2. The constant factor can be taken out of the integral sign:
\ (\ int \ limits_a ^ b kf (x) dx = k \ int \ limits_a ^ b f (x) dx \)
Calculating the areas of planar figures using a definite integral
Using the integral, it is possible to calculate the areas of not only curvilinear trapezoids, but also plane figures of a more complex type, for example, the one shown in the figure. The figure P is bounded by straight lines x = a, x = b and graphs of continuous functions y = f (x), y = g (x), and on the segment [a; b] the inequality \ (g (x) \ leq f (x) \) holds. To calculate the area S of such a figure, we will proceed as follows:
\ (S = S_ (ABCD) = S_ (aDCb) - S_ (aABb) = \ int \ limits_a ^ b f (x) dx - \ int \ limits_a ^ b g (x) dx = \)
\ (= \ int \ limits_a ^ b (f (x) -g (x)) dx \)
So, the area S of the figure bounded by the straight lines x = a, x = b and the graphs of the functions y = f (x), y = g (x), continuous on the segment and such that for any x from the segment [a; b] the inequality \ (g (x) \ leq f (x) \) holds, calculated by the formula
\ (S = \ int \ limits_a ^ b (f (x) -g (x)) dx \)
