Let's consider the function. This function is called: integral as a function of the upper limit. Let us note several properties of this function.
Theorem 2.1. If f(x) is an integrable function, then Ф(x) is continuous on .
Proof. By property 9 definite integral(mean value theorem) we have 
, from where, at , we obtain the required.
Theorem 2.2. If f(x) is a continuous function on , then Ф’(x) = f(x) on .
Proof. By property 10 of the definite integral (second mean value theorem), we have 
Where With– some point of the segment. Due to the continuity of the function f, we obtain
Thus, Ф(x) is one of the antiderivatives of the function f(x), therefore, Ф(x) = F(x) + C, where F(x) is another antiderivative of f(x). Further, since Ф(a) = 0, then 0 = F(a) + C, therefore, C = -F(a) and therefore Ф(x) = F(x) – F(a). Assuming x=b, we obtain the Newton-Leibniz formula
Examples
1. 
Integration by parts in a definite integral
The definite integral preserves the formula for integration by parts. In this case it takes the form
Example.
Changing variables in a definite integral
One of the variants of results on the change of variables in a definite integral is as follows.Theorem 2.3. Let f(x) be continuous on the segment and satisfy the conditions:
1) φ(α) = a
2) φ(β) = b
3) the derivative φ’(t) is defined everywhere on the interval [α, β]
4) for all t from [α, β]
Then
Proof. If F(x) is antiderivative for f(x)dx then F(φ(t)) is antiderivative for Therefore F(b) – F(a) = F(φ(β)) – F(φ(α)). The theorem has been proven.
Comment. If we reject the continuity of the function f(x) under the conditions of Theorem 2.3, we have to require the monotonicity of the function φ(t).
Example. Calculate the integral Let us put Then dx = 2tdt and therefore
Problem 1(about calculating the area of a curved trapezoid).
In the Cartesian rectangular coordinate system xOy, a figure is given (see figure) bounded by the x axis, straight lines x = a, x = b (a by a curvilinear trapezoid. It is required to calculate the area of a curvilinear trapezoid.
Solution. Geometry gives us recipes for calculating the areas of polygons and some parts of a circle (sector, segment). Using geometric considerations, we can only find an approximate value of the required area, reasoning as follows.
Let's split the segment [a; b] (base of a curved trapezoid) into n equal parts; this partition is carried out using 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.
Let us consider the k-th column separately, i.e. a curved trapezoid whose base 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 equal to \(f(x_k) \cdot \Delta x_k \), where \(\Delta x_k \) is the length of the segment; It is natural to consider the resulting product as an approximate value of the area of the kth 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 made up 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; \(\Delta x_0 \) - length of the segment, \(\Delta x_1 \) - length of the segment, etc.; in this case, as we agreed above, \(\Delta x_0 = \dots = \Delta x_(n-1) \)
So, \(S \approx S_n \), and this approximate equality is more accurate, the larger n.
By definition, it is believed that the required area of a curvilinear trapezoid is equal to the limit of the sequence (S n):
$$ S = \lim_(n \to \infty) S_n $$
Problem 2(about moving a point)
Moves in a straight line material point. The dependence of speed on time is expressed by the formula v = v(t). Find the movement of a point over a period of time [a; b].
Solution. If the movement 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 period of time and assume that during this period of time the speed was constant, the same as at time t k. So we assume that v = v(t k).
3) Let’s find the approximate value of the point’s movement over a period of time; we’ll denote this approximate value as s k
\(s_k = v(t_k) \Delta t_k \)
4) Find the approximate value of 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 required displacement is equal to the limit of the sequence (S n):
$$ s = \lim_(n \to \infty) S_n $$
Let's summarize. Solutions to various problems were reduced to the same mathematical model. Many problems from various fields of science and technology lead to the same model in the process of solution. This means that this mathematical model must be specially studied.
The concept of a definite integral
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) 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 proven that this limit exists in the case of a continuous (or piecewise continuous) function. They call him a certain integral of the function y = f(x) over 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 (lower and upper, respectively).
Let's return to the tasks discussed above. The definition of 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 curvilinear trapezoid shown in the figure above. This is geometric meaning definite integral.
The definition of the displacement s of a point moving in a straight line with a speed v = v(t) over the period of time from t = a to t = b, given in Problem 2, can be rewritten as follows:
Newton-Leibniz formula
First, let's answer the question: what is the connection between the definite integral and the 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 period of time from t = a to t = b is calculated by the formula
\(S = \int\limits_a^b v(t) dt \)
On the other hand, the coordinate of a moving point is an antiderivative for speed - let's denote it s(t); This means that 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 of v(t).
The following theorem was proven in the course of mathematical analysis.
Theorem. If the function y = f(x) is continuous on the interval [a; b], then the formula is valid
\(S = \int\limits_a^b f(x) dx = F(b)-F(a) \)
where F(x) is the antiderivative of f(x).
The given formula is usually called 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), they use the notation \(\left. F(x)\right|_a^b \) (it is sometimes called double substitution) and, accordingly, rewrite the Newton-Leibniz formula in this form:
\(S = \int\limits_a^b f(x) dx = \left. F(x)\right|_a^b \)
When calculating a definite integral, first find the antiderivative, and then carry out a double substitution.
Based on the Newton-Leibniz formula, we can obtain two properties of the definite integral.
Property 1. Integral of the sum of functions equal to the sum 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 plane figures using a definite integral
Using the integral, you can calculate the areas not only of curvilinear trapezoids, but also of flat figures more complex type, for example the one shown in the figure. The figure P is limited 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 a figure bounded by straight lines x = a, x = b and graphs of functions y = f(x), y = g(x), continuous on the segment and such that for any x from the segment [a; b] the inequality \(g(x) \leq f(x) \) is satisfied, calculated by the formula
\(S = \int\limits_a^b (f(x)-g(x))dx \)
Table of indefinite integrals (antiderivatives) of some functions
$$ \int 0 \cdot dx = C $$ $$ \int 1 \cdot dx = x+C $$ $$ \int x^n dx = \frac(x^(n+1))(n+1 ) +C \;\; (n \neq -1) $$ $$ \int \frac(1)(x) dx = \ln |x| +C $$ $$ \int e^x dx = e^x +C $$ $$ \int a^x dx = \frac(a^x)(\ln a) +C \;\; (a>0, \;\; a \neq 1) $$ $$ \int \cos x dx = \sin x +C $$ $$ \int \sin x dx = -\cos x +C $$ $ $ \int \frac(dx)(\cos^2 x) = \text(tg) x +C $$ $$ \int \frac(dx)(\sin^2 x) = -\text(ctg) x +C $$ $$ \int \frac(dx)(\sqrt(1-x^2)) = \text(arcsin) x +C $$ $$ \int \frac(dx)(1+x^2 ) = \text(arctg) x +C $$ $$ \int \text(ch) x dx = \text(sh) x +C $$ $$ \int \text(sh) x dx = \text(ch ) x +C $$Newton - Leibniz formula
Main theorem of analysis or Newton - Leibniz formula gives a relationship between two operations: taking a definite integral and calculating the antiderivative
Formulation
Consider the integral of the function y = f(x) ranging from constant number a up to the number x, which we will consider variable. Let's write the integral in the following form:
This type integral is called an integral with a variable upper limit. Using the mean value theorem in a definite integral, it is easy to show that this function continuous and differentiable. And also the derivative of a given function at point x is equal to the integrable function itself. From this it follows that any continuous function has an antiderivative in the form of a quadrature: . And since the class of antiderivative functions of the function f differs by a constant, it is easy to show that: the definite integral of the function f is equal to the difference in the values of the antiderivatives at points b and a
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Let some continuous function f be given on a certain segment of the Ox axis. Let us assume that this function does not change its sign throughout the entire segment.
If f is a continuous and non-negative function on a certain segment, and F is some antiderivative of it on this segment, then the area of the curvilinear trapezoid S is equal to the increment of the antiderivative on this segment.
This theorem can be written as follows:
S = F(b) - F(a)
The integral of the function f(x) from a to b will be equal to S. Here and further, to denote the definite integral of some function f(x), with the limits of integration from a to b, we will use the following notation (a;b)∫f( x). Below is an example of how it will look.
Newton-Leibniz formula
This means that we can equate these two results. We obtain: (a;b)∫f(x)dx = F(b) - F(a), provided that F is an antiderivative for the function f on . This formula is called Newton - Leibniz formulas. It will be true for any continuous function f on an interval.
The Newton-Leibniz formula is used to calculate integrals. Let's look at a few examples:
Example 1: calculate the integral. Find the antiderivative for the integrand function x 2 . One of the antiderivatives will be the function (x 3)/3.
Now we use the Newton-Leibniz formula:
(-1;2)∫x 2 dx = (2 3)/3 - ((-1) 3)/3 = 3
Answer: (-1;2)∫x 2 dx = 3.
Example 2: calculate the integral (0;pi)∫sin(x)dx.
Find the antiderivative for the integrand function sin(x). One of the antiderivatives will be the -cos(x) function. Let's use the Newton-Leibniz formula:
(0;pi)∫cos(x)dx = -cos(pi) + cos(0) = 2.
Answer: (0;pi)∫sin(x)dx=2
Sometimes, for simplicity and convenience of recording, the increment of the function F on the segment (F(b)-F(a)) is written as follows:
Using this notation for the increment, the Newton-Leibniz formula can be rewritten as follows:
As noted above, this is only an abbreviation for ease of recording; this recording does not affect anything else. This notation and the formula (a;b)∫f(x)dx = F(b) - F(a) will be equivalent.
By a definite integral from a continuous function f(x) on the final segment [ a, b] (where ) is the increment of some of its antiderivatives on this segment. (In general, understanding will be noticeably easier if you repeat the topic of the indefinite integral) In this case, the notation is used
As can be seen in the graphs below (the increment of the antiderivative function is indicated by ), a definite integral can be either a positive or a negative number(It is calculated as the difference between the value of the antiderivative in the upper limit and its value in the lower limit, i.e. as F(b) - F(a)).
Numbers a And b are called the lower and upper limits of integration, respectively, and the segment [ a, b] – segment of integration.
Thus, if F(x) – some antiderivative function for f(x), then, according to the definition,
(38)
Equality (38) is called Newton-Leibniz formula . Difference F(b) – F(a) is briefly written as follows:
Therefore, we will write the Newton-Leibniz formula like this:
(39)
Let us prove that the definite integral does not depend on which antiderivative of the integrand is taken when calculating it. Let F(x) and F( X) are arbitrary antiderivatives of the integrand. Since these are antiderivatives of the same function, they differ by a constant term: Ф( X) = F(x) + C. That's why
This establishes that on the segment [ a, b] increments of all antiderivatives of the function f(x) match.
Thus, to calculate a definite integral, it is necessary to find any antiderivative of the integrand, i.e. First you need to find the indefinite integral. Constant WITH excluded from subsequent calculations. Then the Newton-Leibniz formula is applied: the value of the upper limit is substituted into the antiderivative function b , further - the value of the lower limit a and the difference is calculated F(b) - F(a) . The resulting number will be a definite integral..
At a = b by definition accepted
Example 1.
Solution. First, let's find the indefinite integral:
Applying the Newton-Leibniz formula to the antiderivative
(at WITH= 0), we get
However, when calculating a definite integral, it is better not to find the antiderivative separately, but to immediately write the integral in the form (39).
Example 2. Calculate definite integral
Solution. Using formula
Properties of the definite integral
Theorem 2.The value of the definite integral does not depend on the designation of the integration variable, i.e.
(40)
Let F(x) – antiderivative for f(x). For f(t) the antiderivative is the same function F(t), in which the independent variable is only designated differently. Hence,
Based on formula (39), the last equality means the equality of the integrals
Theorem 3.The constant factor can be taken out of the sign of the definite integral, i.e.
(41)
Theorem 4.The definite integral of an algebraic sum of a finite number of functions is equal to the algebraic sum of definite integrals of these functions, i.e.
(42)
Theorem 5.If a segment of integration is divided into parts, then the definite integral over the entire segment is equal to the sum of definite integrals over its parts, i.e. If
(43)
Theorem 6.When rearranging the limits of integration, the absolute value of the definite integral does not change, but only its sign changes, i.e.
(44)
Theorem 7(mean value theorem). A definite integral is equal to the product of the length of the integration segment and the value of the integrand at some point inside it, i.e.
(45)
Theorem 8.If the upper limit of integration is greater than the lower one and the integrand is non-negative (positive), then the definite integral is also non-negative (positive), i.e. If
Theorem 9.If the upper limit of integration is greater than the lower one and the functions and are continuous, then the inequality
can be integrated term by term, i.e.
(46)
The properties of the definite integral make it possible to simplify the direct calculation of integrals.
Example 5. Calculate definite integral
Using Theorems 4 and 3, and when finding antiderivatives - table integrals (7) and (6), we obtain
Definite integral with variable upper limit
Let f(x) – continuous on the segment [ a, b] function, and F(x) is its antiderivative. Consider the definite integral
(47)
and through t the integration variable is designated so as not to confuse it with the upper bound. When changing X the definite integral (47) also changes, i.e. it is a function of the upper limit of integration X, which we denote by F(X), i.e.
(48)
Let us prove that the function F(X) is an antiderivative for f(x) = f(t). Indeed, differentiating F(X), we get
because F(x) – antiderivative for f(x), A F(a) is a constant value.
Function F(X) – one of the infinite number of antiderivatives for f(x), namely the one that x = a goes to zero. This statement is obtained if in equality (48) we put x = a and use Theorem 1 of the previous paragraph.
Calculation of definite integrals by the method of integration by parts and the method of change of variable
where, by definition, F(x) – antiderivative for f(x). If we change the variable in the integrand
then, in accordance with formula (16), we can write
In this expression
antiderivative function for
In fact, its derivative, according to rule of differentiation of complex functions, is equal
Let α and β be the values of the variable t, for which the function
takes values accordingly a And b, i.e.
But, according to the Newton-Leibniz formula, the difference F(b) – F(a) There is