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Let the function f(t) is defined and continuous on some interval containing the point a. Then each number x from this interval you can match the number 
,
thereby defining on the interval the function I(x), which is usually called a definite integral with a variable upper limit. Note that at the point x = a this function is equal to zero. Let's calculate the derivative of this function at the point x. To do this, first consider the increment of the function at the point x when incrementing argument D x:
D I(x) = I(x+ D x) – I(x) =
.
As shown in Fig. 4, the value of the last integral in the formula for the increment D I(x) is equal to the area of the curvilinear trapezoid, marked by hatching. At small values of D x(here, as elsewhere in this course, when speaking about small increments of an argument or function, we mean absolute values increments, since the increments themselves can be both positive and negative), this area turns out to be approximately equal to the area of the rectangle marked in the figure with double hatching. The area of a rectangle is given by the formula f(x)D x. From here we get the relation
.
In the last approximate equality, the accuracy of the approximation is higher, the smaller the value of D x.
From the above it follows the formula for the derivative of the function I(x):
.
The derivative of the definite integral with respect to the upper limit at point x is equal to the value of the integrand at point x. It follows that the function 
is an antiderivative of the function f(x), and such an antiderivative that takes at the point x = a meaning, equal to zero. This fact makes it possible to represent the definite integral in the form
. (1)
Let F(x) is also an antiderivative of the function f(x), then by the theorem about general view all antiderivatives of functions I(x) = F(x) + C, Where C- not a number. Wherein right part formula (1) takes the form
I(x) – I(a) = F(x) + C– (F(a) +C) = F(x) – F(a). (2)
From formulas (1) and (2) after replacement x on b follows the formula for calculating the definite integral of the function f(t) along the interval [ a;b]:
,
which is usually called the formula Newton-Leibniz. Here F(x)- any antiderivative of function f(x).
In order to calculate the definite integral of the function f(x) along the interval [ a;b], you need to find some antiderivative F(x) functions f(x) and calculate the difference between the values of the antiderivative at the points b And a. The difference between these antiderivative values is usually denoted by the symbol , ᴛ.ᴇ. 
.
Let us give examples of calculating definite integrals using the Newton-Leibniz formula.
Example 1. 
.
When calculating definite integrals, you can use variable replacement formula:
.
Here a And b are determined, respectively, from the equations j(a) = a; j(b) = b, and the functions f,j, j¢ must be continuous over appropriate intervals.
Example 2..
Let's make a replacement: ln x = t or x = e t, then if x = 1, then t = 0, and if x = e, That t = 1. As a result we get:
.
However, when calculating a definite integral using a change of variables, it is not extremely important to return to the previous integration variable. It is enough just to introduce new limits of integration.
Let the function f(t) is defined and continuous on some interval containing the point a. Then each number x from this interval you can match the number
I(x) = I(x+x) – I(x) =
As shown in Figure 23, the value of the last integral in the formula for the increment I(x) is equal to the area curved trapezoid, marked with shading. At small values x(here, as elsewhere in this course, when talking about small increments of an argument or function, we mean the absolute magnitudes of the increments, since the increments themselves can be positive and negative) this area turns out to be approximately equal to the area of the rectangle marked in the figure double hatching. The area of a rectangle is given by the formula f(x)x. From here we get the relation
.
In the last approximate equality, the accuracy of the approximation is higher, the smaller the value x.
From the above it follows the formula for the derivative of the function I(x):
.
Derivative of the definite integral with respect to the upper limit at the pointx
equal to the value of the integrand at the pointx. It follows that the function 
is an antiderivative of the function f(x), and such an antiderivative that takes at the point x = a value equal to zero. This fact makes it possible to represent a definite integral in the form
. (9)
Let F(x) is also an antiderivative of the function f(x), then by the theorem on the general form of all antiderivatives of the function I(x) = F(x) + C, Where C- some number. In this case, the right side of formula (9) takes the form
I(x) – I(a) = F(x) + C– (F(a) +C) = F(x) – F(a). (10)
From formulas (9) and (10) after replacement x on b follows the formula for calculating the definite integral of the function f(t) along the interval [ a;b]:
,
which is called the formula Newton-Leibniz. Here F(x)- any antiderivative of a function f(x).
To calculate the definite integral of a function f(x) along the interval [ a;b], you need to find some antiderivative F(x) functions f(x) and calculate the difference between the values of the antiderivative at the points b And a. The difference between these antiderivative values is usually denoted by the symbol 
.
Let us give examples of calculating definite integrals using the Newton-Leibniz formula.
Examples. 1. 
.
2.
.
First, let's calculate the indefinite integral of the function f(x) = xe x. Using the method of integration by parts, we obtain: 
. As an antiderivative function f(x) choose a function e x (x– 1) and apply the Newton-Leibniz formula:
I = e x (x – 1)= 1.
When calculating definite integrals, you can use formula for changing a variable in a definite integral:
.
Here And are determined, respectively, from the equations ( ) = a; ( ) = b, and the functions f, , must be continuous over appropriate intervals.
Example: 
.
Let's make a replacement: ln x = t or x = e t, then if x = 1, then t = 0, and if x = e, That t = 1. As a result we get:
.
When changing a variable in a definite integral, you do not need to return to the original integration variable.
Integral with variable upper limit. The value of a definite integral does not depend on what letter the integration variable is denoted by: (to verify this, it is enough to write out the integral sums; they coincide). In this section integration variable we will denote by the letter t
, and the letter x
let us denote the upper limit of integration. We will assume that the upper limit of the integral can change, i.e. What x
- variable, as a result the integral will be a function Ф( x
) his upper limit: 
. It is easy to prove that if f
(t
) is integrable, then Ф( x
) is continuous, but the following fundamental theorem is more important for us:
Integral theorem with variable upper limit. If the function f
(t
) is continuous in a neighborhood of the point t
= x
, then at this point the function Ф( x
) is differentiable, and 
.
In other words, the derivative of a definite integral of a continuous function with respect to the upper limit is equal to the value of the integrand in this limit.
Document. Let's give the upper limit x
increment . Then 
, Where c
- a point lying between x
and (the existence of such a point is stated by the mean value theorem; the numbers above the equal sign are the number of the applied property of the definite integral). . Let's rush. Wherein ( c
- a point located between x
And ). Because f
(t
) is continuous at the point t
= x
, That 
. Therefore there is 
, And 
. The theorem has been proven.
Let's note the first important consequence this theorem. Essentially, we have proven that any continuous function f
(x
) has an antiderivative, and this antiderivative is determined by the formula 
36. Newton-Leibniz formula.
If f
(x
) is continuous on the interval [ a
, b
], And F
(x
) is some antiderivative of the function, then 
.
Doc. We have established that the function - antiderivative of continuous f
(x
). Because F
(x
) is also antiderivative, then Ф( x
) = F
(x
) + C
. Let us put in this equality x
= a
. Because 
, That . In equality 
let's redesignate the variables: for the integration variable t
let's return to the notation x
, upper limit x
let's denote b
. Finally, 
.
The difference on the right side of the Newton-Leibniz formula is denoted by a special symbol: 
(here reads as "substitution from a
before b
"), so the Newton-Leibniz formula is usually written like this: 
.
37. Integration by parts and change of variable in a definite integral.
If u(x) And v(x) - two functions defined on the interval [ a, b] and having continuous derivatives there, then
Formula (24) is formula for integration by parts for definite integrals.
The proof is very simple. Exactly,
Since according to the integration by parts formula it will be
then this is where (24) follows.
Let f(zp, q], A φ (x) is a continuous function defined on the interval [ a, b], which has a continuous derivative there φ "(x) and satisfying the inequality p ≤ φ (x) ≤ q.
In this case
Formula (22) expresses the rule for changing a variable in a definite integral. It resembles the rule for replacing a variable in an indefinite integral, but differs from it in that there is no need to return to the old variable, since formula (22) represents the equality of two constant numbers. Let us also note that for the case of definite integrals, this formula replaces both types of substitution rules in indefinite integrals; only, when applying it in practice, sometimes you have to read it from left to right, and sometimes from right to left.
Moving on to the proof of the theorem, we denote the integrals included in the left and right sides of formula (22), respectively, by I lion and I right
Let F(z) is an antiderivative function for f(z). Then, according to the Newton-Leibniz formula/p>
I rights = F[φ (b)] - F[φ (a)]. (23)
As for I lion then
But according to the theorem it will be
I lion = F[φ (b)] - F[φ (a)].
From here and from (23) it follows that I lion = I right
38. Integrals of even, odd and periodic functions.
Theory 1. Let f(x) be integrable on the interval [-a,a] even function:
To prove this, let us present the original integral as a sum of two integrals:
The statement has been proven.
Theory 2. Let f(x) be an odd function integrable on the interval [-a,a]:
The theorem is proven in a similar way:
does not depend on λ. In particular,
Let us calculate the derivative with respect to λ from the expression on the right side of this equality:
Improper integrals
Improper integral with infinite limit(s) of integration
Sometimes such an improper integral is also called improper integral of the first kind. In general, an improper integral with an infinite limit most often looks like this: . How is it different from a definite integral? At the upper limit. It is endless: .
Less common are integrals with an infinite lower limit or two infinite limits: .
We will consider the most popular case. The technique for working with other varieties is similar, and at the end of the paragraph there will be a link to such examples.
Does an improper integral always exist? No, not always. The integrand must be continuous on the interval
Help: strictly speaking, the statement is false: if there are discontinuities in the function, then in some cases it is possible to split the half-interval into several parts and calculate several improper integrals. For simplicity, hereinafter I will say that an improper integral does not exist.
Let us depict in the drawing the graph of the integrand function. Typical graph and curved trapezoid for this case looks like that:
Everything is fine here, the integrand is continuous on the half-interval, and, therefore, the improper integral exists. Please note that our curved trapezoid is endless(not limited to the right) figure.
Improper integral numerically equal to area shaded figure, two cases are possible:
1) First, the thought that comes to mind: “since the figure is infinite, then 
", in other words, the area is also infinite. It may be so. In this case they say that the improper integral diverges.
2) But. As paradoxical as it may sound, the area of an infinite figure can be equal to... a finite number! For example: 
. Could this be true? Easily. In the second case, the improper integral converges.
In what cases does an improper integral diverge and in what cases does it converge? It depends on the integrand, and specific examples we'll look into it very soon.
What happens if an infinite curved trapezoid is located below the axis? In this case, the improper integral 
(diverges) or is equal to a finite negative number.
An improper integral can be negative.
Important! When ANY improper integral is offered to you for a solution, then, generally speaking, there is no talk of any area and there is no need to construct a drawing. Your task is to find the NUMBER or prove that the improper integral diverges. I explained the geometric meaning of the improper integral only to make it easier to understand the material.
Since the improper integral is very similar to the definite integral, then recall the formula Newton-Leibniz: 
. In fact, the formula also applies to improper integrals, it just needs to be modified a little. What is the difference? At the infinite upper limit of integration: . Probably, many guessed that this already smacks of the application of the theory of limits, and the formula will be written like this: 
.
In today's lecture we will continue to study the definite integral and obtain a formula for calculating it. As we will see later, the definite integral is equal to the increment of the antiderivative, and represents constant number, equal to area curvilinear trapezoid. Therefore, all methods for calculating the indefinite integral are also valid for the definite integral.
Question 1. Basic properties of the definite integral
Integral
was introduced for case a< b. Обобщим понятие определенного интеграла на случай, когда пределы интегрирования совпадают или нижний предел больше верхнего.
Property 1. .
This formula is obtained from (1) provided that all Δx i = 0.
Property 2. 
.
This formula is obtained from (1) provided that the segment is run in the opposite direction (from b to a), i.e. all Δx i< 0.
Property 3. (additivity property)
If the function f(x) is integrable on an interval and a< c < b, то
. (2)
Equality (2) is valid for any location of points a, b and c (we assume that the function f(x) is integrable on the larger of the resulting segments).
Property 4.
Constant multiplier can be taken out as a sign of a definite integral, i.e.
,
where k = const.
Property 5.
The definite integral of the algebraic sum of two functions is equal to algebraic sum integrals of these functions, i.e.
.
Comment
- Property 5 applies to the amount of any finite number terms.
- Properties 4 and 5 together represent linearity property definite integral.
Question 2. Estimates of the integral. Mean value theorem
1. If the function f(x) ≥ 0 everywhere on the interval, then .
2.
If f(x) ≥ g(x) everywhere on the interval, then 
.
3.
For a function f(x) defined on the interval , the inequality holds 
.
In particular, if everywhere on the interval then 
And .
4.
If m and M are, respectively, the smallest and largest values of the function f(x) on the segment , then 
.
T.2.1. (mean value theorem))
If the function f(x) is continuous on the segment , then on this segment there exists a point c such that
. (3)
Equality (3) is called average value formula, and the value f(c) is called mean value of the function f(x) on the segment .
Question 3: Definite integral as a function of the upper limit
If the function y = f(x) is integrable on the interval , then it is integrable on any smaller interval, i.e. for "xО there is an integral
In order not to confuse the designations of the limit and the integration variable, we denote the integration variable by t. Then integral (4) will be written in the form The value of this integral is a function of the upper limit x and is denoted by Ф(x):
. (5)
The function Ф(х) is called integral with a variable upper limit.
Let's consider some properties of the function Ф(х).
T.3.1.(continuity of function Ф(х))
If the function f(x) is continuous on the interval, then the function Ф(x) will also be continuous on the interval.
T.3.2.(differentiation of function Ф(х))
If the function f(x) is continuous on the interval, then the function Ф(x) is differentiable at any internal point x of this segment, and the equality is true
.
Consequence
If the function f(x) is continuous on the interval, then for this function there is an antiderivative on this segment, and the function Ф(x) - an integral with a variable upper limit - is an antiderivative for the function f(x).
Since every other antiderivative for the function f(x) differs from Ф(x) only by a constant term, we can establish
