Definite integral as a function of the upper limit. Variable Replacement Method

Let the function f(t) is defined and continuous on some interval containing the point a. Then each number x from this interval we can match the number ,

thereby defining on the interval the function I(x), which is 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 curved trapezoid, marked with shading. 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 a 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 the function I(x) = F(x) + C, Where C- a certain number. At the same time right side 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 called Newton-Leibniz formula. 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, i.e. .

Change of variable in a definite integral. When calculating definite integrals using the Newton-Leibniz formula, it is preferable not to strictly differentiate the stages of solving the problem (finding the antiderivative of the integrand, finding the increment of the antiderivative). This approach, which uses, in particular, formulas for change of variable and integration by parts for a definite integral, usually makes it possible to simplify the writing of the solution.

THEOREM. Let the function φ(t) have a continuous derivative on the interval [α,β], a=φ(α), β=φ(β) and the function f(x) be continuous at each point x of the form x=φ(t), where t [α,β].

Then the following equality is true:

This formula is called the formula for changing a variable in a definite integral.

Just as was the case with the indefinite integral, using a change of variable allows us to simplify the integral, bringing it closer to the tabular one(s). Moreover, in contrast to the indefinite integral in in this case there is no need to return to the original integration variable. It is enough just to find the limits of integration of α and β over a new variable t as a solution to the variable t of the equations φ(t)=a and φ(t)=b. In practice, when performing a variable replacement, they often start by indicating the expression t=ψ(x) of the new variable in terms of the old one. In this case, finding the limits of integration over the variable t is simplified: α=ψ(a), β=ψ(b).

Example 19. Calculate

Let's put t=2-x 2. Then dt=d(2-x 2)=(2-x 2)"dx=-2xdx and xdx=- dt. If x=0, then t=2-0 2 =2, and if x=1, then t=2-1 2 =1. Therefore:

Integration by parts. The method of integration by parts allows us to reduce the original indefinite integral to more simple view or to a table integral. This method is most often used if the integrand contains logarithmic, exponential, inverse trigonometric, trigonometric functions, as well as their combinations.

The formula for integration by parts is as follows.

That is, integrand f(x)dx represent it as a product of the function u(x) on d(v(x))- differential function v(x). Next we find the function v(x)(most often by method direct integration) And d(u(x))- differential function u(x). We substitute the found expressions into the integration by parts formula and the original indefinite integral is reduced to the difference . The last indefinite integral can be taken using any integration method, including the method of integration by parts.

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 integration variable through t. Then integral (4) will be written in the form The value of this integral is a function upper limit x and is denoted by Ф(х):

. (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 connection between indefinite and definite integrals:

where C is an arbitrary constant.

Question 4. Calculation of a definite integral. Newton-Leibniz formula

The calculation of definite integrals by a method based on the definition of the integral as the limit of integral sums is usually associated with great difficulties. There is a more convenient method for calculating definite integrals, which is based on the established connection between the indefinite and definite integrals.

T.4.1. If the function f(x) is continuous on an interval and F(x) is any antiderivative for the function f(x) on , then the formula is valid

. (6)

Formula (6) is called Newton–Leibniz formula.

If you enter the designation then the Newton-Leibniz formula (6) can be rewritten in the form

The Newton–Leibniz formula gives convenient way calculations of definite integrals. To calculate the definite integral, it is necessary to find any antiderivative function F(x) for f(x) and take the difference F(b) ‒ F(a) at the ends of the segment.

Example

Question 5. Change of variable and integration by parts in a definite integral

Variable Replacement Method

When calculating definite integrals, the substitution method or the variable change method is widely used.

T.5.1. (change of variable in a definite integral)

Let the function y = f(x) be continuous on the interval. Then if:

1) the function x = j(t) and its derivative x′ = j′(t) are continuous on the interval;

2) the set of values of the function x = j(t) is the segment ;

3) j(a) = a, j(b) = b,

then it's fair formula for changing a variable in a definite integral:

Comment

1. When calculating a definite integral using the substitution method, there is no need to return to the old variable.

2. Often, instead of the substitution x = j(t), the substitution t = g(x) is used.

3. When using the formula, it is necessary to check the fulfillment of the conditions listed in the theorem. If these conditions are violated, then an incorrect result may be obtained.

Example. Calculate

Integration by parts

T.5.2. (integration by parts in a definite integral)

If the functions u = u(x) and v = v(x) have continuous derivatives on the interval , then formula for integration by parts in a definite integral:

Example. Calculate integral

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.

Definite integral of algebraic sum two functions is equal to the algebraic sum of the 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 highest value function f(x) on the interval , then .

T.2.1. (mean value theorem))

If the function f(x) is continuous on the segment , then there is a point c on this segment 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(х):

. (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 segment , then the function Ф(x) is differentiable at any interior point x of this segment, and the equality

Consequence

If the function f(x) is continuous on the interval, then for this function there is an antiderivative on this interval, 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

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 of the curvilinear trapezoid, marked by 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- a certain 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).

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 antiderivative functionf(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 at appropriate intervals.