Integral as a function of the upper limit of integration. Definite integral as a function of the upper limit

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 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 with argument increment  x:

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 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 area rectangle, marked in the figure with 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 meaning, 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 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 (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 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 indefinite integral from 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 at 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 replacing a variable in a definite integral, you do not need to return to the original integration variable.

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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 talking about small increments of an argument or function, we mean the absolute magnitudes of the 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 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 value 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 on the general form of all antiderivatives of the function I(x) = F(x) + C, Where C- not a number. In this case, the right side of 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 a 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 at 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 we can match the number ,

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 shading. At small values of D 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 both positive and negative) this area turns out to be approximately equal to the area of the rectangle marked on double-hatched drawing. 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):

. (1)

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 (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).

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.