The mean value formula for a definite integral. Definite integral and methods for its calculation

\int\limits_(a)^(b)f(x)\,dx= -\int\limits_(b)^(a)f(x)\,dx\,.

This definition is natural, since when the direction of the integration interval changes, all differences \Delta x_k=x_(k+1)-x_k change their sign, and then they change the signs and Darboux sums and, thus, the number separating them, i.e. integral.

Since for a=b all \Delta x_k vanish, we put

\int\limits_(b)^(a)f(x)\,dx=0.

We have obtained two definitions of the concept of a definite integral: as the difference between the values ​​of the antiderivative and as a separating number for Darboux sums. These definitions lead to the same result in the most important cases:

Theorem 2. If the function y=f(x) is bounded on a segment and has an antiderivative y=F(x) on it, and there is a single number separating the lower and upper Darboux sums, then this number is equal to F(b)-F(a) .

Proof. We proved above that the number F(a)-F(b) separates the sets \(s_P\) and \(S_P\) . Since the separating number is uniquely determined by the condition, it coincides with F(b)-F(a) .

From now on, we will use the notation \textstyle(\int\limits_(a)^(b)f(x)\,dx) only for a single number separating the sets \(s_P\) and \(S_P\) . It follows from the proved theorem that in this case there is no contradiction with the understanding of this notation that we used above.

Properties of lower and upper Darboux sums

For the definition of the integral given earlier to make sense, we must prove that the set of upper Darboux sums is indeed located to the right of the set of lower Darboux sums.

Lemma 1. For every partition P, the corresponding lower Darboux sum is at most the upper Darboux sum, s_P\leqslant S_P .

Proof. Consider some partition P of the segment :

a=x_0 "

Obviously, for any k and for any chosen partition P, the inequality s_P\leqslant S_P holds. Hence, m_k\cdot\Delta x_k\leqslant M_k\cdot\Delta x_k, and that's why

s_P= \sum_(k=0)^(n-1)(m_k\cdot\Delta x_k)\leqslant \sum_(k=0)^(n-1)(M_k\cdot\Delta x_k)=S_P.

Inequality (4) is valid only for a fixed partition P . Therefore, it is not yet possible to assert that the lower Darboux sum of one partition cannot exceed the upper Darboux sum of another partition. To prove this assertion, we need the following lemma:

Lemma 2. By adding a new division point, the lower Darboux sum cannot decrease, and the upper sum cannot increase.

Proof. Let's choose some partition P of the segment and add a new division point to it (x^(\ast)) . Denote the new partition P^(\ast) . The partition P^(\ast) is a refinement of the partition P , i.e. each split point of P is, at the same time, a split point of P^(\ast) .

Let the point (x^(\ast)) fall on the segment \colon\, x_k . Consider the two formed segments and and denote the corresponding exact lower bounds of the function values ​​by m_(k)^(\ast) and m_(k)^(\ast\ast) , and the exact upper bounds by M_(k)^(\ast) and M_(k )^(\ast\ast) .

term m_k(x_(k+1)-m_(k)) The original lower Darboux sum in the new lower Darboux sum corresponds to two terms:

m_(k)^(\ast)(x^(\ast)-x_k)+ m_(k)^(\ast\ast)(x_(k+1)-x^(\ast)).

Wherein m_k\leqslant m_(k)^(\ast) and m_k\leqslant m_(k)^(\ast\ast), since m_k is the exact lower bound of the values ​​of the function f(x) on the entire interval , and m_(k)^(\ast) and m_(k)^(\ast\ast) only on its parts and respectively.

Let us estimate the sum of the obtained terms from below:

\begin(aligned) m_(k)^(\ast)\bigl(x^(\ast)-x_(k)\bigr)+ m_(k)^(\ast\ast)\bigl(x_(k+ 1)-x^(\ast)\bigr) \geqslant & \,\,m_k \bigl(x^(\ast)-x_k)+m_k(x_(k+1)-x^(\ast)\bigr )=\\ &=m_k\bigl(x^(\ast)-x_k+x_(k+1)-x^(\ast)\bigr)=\\ &=m_k\bigl(x_(k+1) -x_k\bigr).\end(aligned)

Since the rest of the terms in both the old and the new lower Darboux sums remained unchanged, the lower Darboux sum did not decrease after adding a new division point, s_P\leqslant S_P .

The proved assertion remains valid even when adding any finite number of points to the partition P .

The assertion about the upper Darboux sum is proved similarly: S_(P^(\ast))\leqslant S_(P).

Let us proceed to comparing the Darboux sums for any two partitions.

Lemma 3. No lower Darboux sum exceeds any upper Darboux sum (at least corresponding to another partition of the segment ).

Proof. Consider two arbitrary partitions P_1 and P_2 of the segment and form the third partition P_3 , consisting of all points of the partitions P_1 and P_2 . Thus, partition P_3 is a refinement of both partition P_1 and partition P_2 (Fig. 7).

Let us denote the lower and upper Darboux sums for these partitions, respectively s_1,~S_1.~s_2,~S_2 and prove that s_1\leqslant S_2 .

Since P_3 is a refinement of the partition of P_1 , then s_1\leqslant s_3 . Next, s_3\leqslant S_3 , since the sums of s_3 and S_3 correspond to the same partition. Finally, S_3\leqslant S_2 , since P_3 is a refinement of the partition of P_2 .

Thus, s_1\leqslant s_3\leqslant S_3\leqslant S_2, i.e. s_1\leqslant S_2 , which was to be proved.

Lemma 3 implies that the numerical set X=\(s_P\) of the lower Darboux sums lies to the left of the numerical set Y=\(S_P\) of the upper Darboux sums.

By virtue of the theorem on the existence of a separating number for two numerical sets1, there is at least one number / separating the sets X and Y , i.e. such that for any partition of the segment, the double inequality holds:

s_P= \sum_(k=0)^(n-1)\bigl(m_k\cdot\Delta x_k\bigr) \leqslant I\leqslant \sum_(k=0)^(n-1)\bigl(M_k\ cdot\Delta x_k\bigr)=S_P.

If this number is unique, then \textstyle(I= \int\limits_(a)^(b) f(x)\,dx).

Let us give an example showing that such a number I , generally speaking, is not uniquely determined. Recall that the Dirichlet function is the function y=D(x) on the interval defined by the equalities:

D(x)= \begin(cases)0,& \text(if)~~ x~~\text(is irrational number);\\1,& \text(if)~~ x~~ \text(is rational number).\end(cases)

Whatever segment we take, there are both rational and irrational points on it, i.e. and points where D(x)=0 , and points where D(x)=1 . Therefore, for any partition of the segment, all values ​​of m_k are equal to zero, and all values ​​of M_k are equal to one. But then all the lower Darboux sums \textstyle(\sum\limits_(k=0)^(n-1)\bigl(m_k\cdot\Delta x_k\bigr)) are equal to zero, and all upper Darboux sums \textstyle(\sum\limits_(k=0)^(n-1)\bigl(M_k\cdot\Delta x_k\bigr)) are equal to one,

Theorem. If the function f(x) integrable on the interval [ a, b], where a< b , and for all x ∈ the inequality

Using the inequalities from the theorem, one can estimate the definite integral, i.e. indicate the boundaries between which its meaning is enclosed. These inequalities express an estimate for a definite integral.

Theorem [Mean value theorem]. If the function f(x) integrable on the interval [ a, b] and for all x ∈ the inequalities m ≤ f(x) ≤ M, then

where m ≤ μ ≤ M.

Comment. In the case where the function f(x) continuous on the segment [ a, b], the equality from the theorem takes the form

where c ∈. Number μ=f(c) defined by this formula is called average functions f(x) on the segment [ a, b]. This equality has the following geometric sense: area of ​​a curvilinear trapezoid bounded by a continuous line y=f(x) (f(x) ≤ 0) is equal to the area of ​​a rectangle with the same base and a height equal to the ordinate of some point on this line.

Existence of an antiderivative for a continuous function

First, we introduce the concept of an integral with a variable upper limit.

Let the function f(x) integrable on the interval [ a, b]. Then whatever the number x from [ a, b], function f(x) integrable on the interval [ a, b]. Therefore, on the segment [ a, b] function defined

which is called an integral with a variable upper limit.

Theorem. If the integrand is continuous on the interval [ a, b], then the derivative of a definite integral with a variable upper limit exists and is equal to the value of the integrand for this limit, i.e.

Consequence. The definite integral with a variable upper limit is one of the antiderivatives for a continuous integrand. In other words, for any function continuous on an interval, there exists an antiderivative.

Remark 1. Note that if the function f(x) integrable on the interval [ a, b], then the integral with a variable upper limit is a continuous function of the upper limit on this interval. Indeed, from St. 2 and the mean value theorem we have

Remark 2. The integral with a variable upper limit of integration is used in the definition of many new functions, for example, . These functions are not elementary; as already noted, the antiderivatives of the indicated integrands cannot be expressed in terms of elementary functions.

Basic integration rules

Newton-Leibniz formula

Since any two antiderivative functions f(x) differ by a constant, then, according to the previous theorem, it can be argued that any antiderivative Φ(x) continuous on the segment [ a, b] functions f(x) has the form

where C is some constant.

Putting in this formula x=a and x=b, using St.1 definite integrals, we find

From these equalities follows the relation

which is called Newton-Leibniz formula.

Thus we have proved the following theorem:

Theorem. The definite integral of a continuous function is equal to the difference between the values ​​of any of its antiderivatives for the upper and lower integration limits.

The Newton-Leibniz formula can be rewritten as

Change of variable in a definite integral

Theorem. If a

• function f(x) continuous on the segment [ a, b];
• line segment [ a, b] is the set of function values φ(t) defined on the interval α ≤ t ≤ β and having a continuous derivative on it;
• φ(α)=a, φ(β)=b

then the formula is valid

Integration by parts formula

Theorem. If functions u=u(x), v=v(x) have continuous derivatives on the interval [ a, b], then the formula

Applied value mean value theorems consists in the possibility of obtaining a qualitative estimate of the value of a certain integral without calculating it. We formulate : if the function is continuous on the interval , then inside this interval there is such a point that .

This formula is quite suitable for a rough estimate of the integral of a complex or cumbersome function. The only moment that makes the formula approximate , is a necessity self-selection points . If we take the simplest path - the middle of the integration interval (as suggested in a number of textbooks), then the error can be quite significant. For more accurate results recommend carry out the calculation in the following sequence:

Construct a function graph on the interval ;

Draw the upper border of the rectangle in such a way that the cut off parts of the graph of the function are approximately equal in area (this is exactly how it is shown in the above figure - two curvilinear triangles are almost the same);

Determine from figure ;

Use the mean value theorem.

As an example, let's calculate a simple integral:

Exact value ;

For the middle of the interval we will also obtain an approximate value , i.e. clearly inaccurate result;

Having built a graph with drawing the upper side of the rectangle in accordance with the recommendations, we get , from where and the approximate value of . Quite satisfactory result, the error is 0.75%.

Trapezoidal formula

The accuracy of calculations using the mean value theorem essentially depends, as was shown, on visual purpose point chart. Indeed, by choosing, in the same example, points or , you can get other values ​​of the integral, and the error may increase. Subjective factors, the scale of the graph and the quality of the drawing greatly affect the result. This is unacceptably in critical calculations, so the mean value theorem applies only to fast quality integral estimates.

In this section, we will consider one of the most popular methods of approximate integration - trapezoid formula . The basic idea of ​​constructing this formula comes from the fact that the curve can be approximately replaced by a broken line, as shown in the figure.

Let us assume, for definiteness (and in accordance with the figure), that the integration interval is divided into equal (this is optional, but very convenient) parts. The length of each of these parts is calculated by the formula and is called step . The abscissas of the split points, if specified, are determined by the formula , where . It is easy to calculate ordinates from known abscissas. Thus,

This is the trapezoid formula for the case. Note that the first term in brackets is the half-sum of the initial and final ordinates, to which all intermediate ordinates are added. For an arbitrary number of partitions of the integration interval general formula of trapezoids looks like: quadrature formulas: rectangles, simpson, gauss, etc. They are based on the same idea of ​​representing a curvilinear trapezoid by elementary areas of various shapes, therefore, after mastering the trapezoid formula, it will not be difficult to understand similar formulas. Many formulas are not as simple as the trapezoid formula, but allow you to get a high accuracy result with a small number of partitions.

With the help of the trapezoid formula (or similar ones), it is possible to calculate, with the accuracy required in practice, both "non-taking" integrals and integrals of complex or cumbersome functions.

Trapezoidal method

Main article:Trapezoidal method

If the function on each of the partial segments is approximated by a straight line passing through the final values, then we obtain the trapezoid method.

The area of ​​the trapezoid on each segment:

Approximation error on each segment:

where

The full formula for trapezoids in the case of dividing the entire integration interval into segments of the same length:

where

Trapezoidal formula error:

where

Simpson method.

Integrand f(x) is replaced by an interpolation polynomial of the second degree P(x)– a parabola passing through three nodes, for example, as shown in the figure ((1) is a function, (2) is a polynomial).

Consider two steps of integration ( h= const = x i+1 – x i), that is, three nodes x0, x1, x2, through which we draw a parabola, using Newton's equation:

Let be z = x - x0,
then

Now, using the obtained relation, we calculate the integral over this interval:

.
For uniform grid and even number of steps n Simpson's formula becomes:

Here , a under the assumption that the fourth derivative of the integrand is continuous.

[edit] Increasing Accuracy

Approximation of a function by one polynomial over the entire interval of integration, as a rule, leads to a large error in estimating the value of the integral.

To reduce the error, the integration segment is divided into parts and a numerical method is used to evaluate the integral on each of them.

As the number of partitions tends to infinity, the estimate of the integral tends to its true value for analytic functions for any numerical method.

The above methods allow for a simple procedure of halving the step, while at each step it is required to calculate the function values ​​only at newly added nodes. The Runge rule is used to estimate the calculation error.

Application of Runge's rule

edit] Estimating the accuracy of computing a definite integral

The integral is calculated using the chosen formula (rectangles, trapezoids, Simpson's parabolas) with the number of steps equal to n, and then with the number of steps equal to 2n. The error in calculating the value of the integral with the number of steps equal to 2n is determined by the Runge formula:
, for the formulas of rectangles and trapezoids, and for the Simpson formula.
Thus, the integral is calculated for successive values ​​of the number of steps, where n 0 is the initial number of steps. The calculation process ends when the next value N will satisfy the condition , where ε is the specified accuracy.

Features of the behavior of the error.

It would seem that why analyze different methods of integration if we can achieve high accuracy by simply reducing the value of the integration step. However, consider the graph of the behavior of the a posteriori error R results of numerical calculation depending on and from the number n interval partitions (that is, at step . In section (1), the error decreases due to a decrease in step h. But in section (2), the computational error begins to dominate, accumulating as a result of numerous arithmetic operations. Thus, for each method there is its own Rmin, which depends on many factors, but primarily on the a priori value of the error of the method R.

Refinement formula of Romberg.

The Romberg method consists in successive refinement of the value of the integral with a multiple increase in the number of partitions. The formula of trapezoids with a uniform step can be taken as the base h.
Denote the integral with the number of partitions n= 1 as .
Decreasing the step by half, we get .
If we successively reduce the step by 2 n times, we get a recursive relation for calculating .

Mean theorem. If f(x) is continuous on the segment , then there exists a point such that . Doc. A function that is continuous on a segment takes its smallest m and largest M values ​​on this segment. Then . Number is between the minimum and maximum values ​​of the function on the interval. One of the properties of a function continuous on an interval is that this function takes any value between m and M. Thus, there is a point such that . This property has a simple geometric interpretation: if is continuous on the segment , then there is a point such that the area of ​​the curvilinear trapezoid ABCD is equal to the area of ​​the rectangle with base and height f(c) (highlighted in the figure).

7. Integral with a variable upper limit. Its continuity and differentiability.

Consider a function f (x) that is Riemann-integrable on the interval . Since it is integrable on , then it is also integrable on ∀x ∈ . Then for each x ∈ the expression makes sense, and for each x it is equal to some number.

Thus, each x ∈ is associated with some number ,

those. function is given:

(3.1)

Definition:

The function F (x) given in (3.1), as well as the expression itself, is called

integral with a variable upper limit. It is defined on the entire segment

integrability of the function f (x).

Condition: f (t) is continuous on , and the function F (x) is given by formula (3.1).

Statement: The function F(x) is differentiable on , and F (x) = f (x).

(At a it is right differentiable, and at b it is left differentiable.)

Proof:

Since for a function of one variable F (x) differentiability is equivalent to the existence of a derivative at all points (at point a on the right, and at point b on the left), then we will find the derivative F (x). Consider the difference

Thus,

moreover, the point ξ lies on the segment (or if ∆x< 0).

Now recall that the derivative of the function F(x) at a given point x ∈ is equal to the limit of the difference relation: . From equality we have:

,

Letting ∆x → 0 now, on the left side of this equality we obtain F’(x), and on the right

Recall the definition of the continuity of the function f (t) at the point x:

Let x1 be equal to ξ in this definition. Since ξ ∈ (ξ ∈ ) and

∆x → 0, then |x − ξ| → 0, and by the definition of continuity, f (ξ) → f (x). Hence we have:

F'(x) = f(x).

Consequence:

Condition: f (x) is continuous on .

Statement: Any antiderivative of the function f (x) has the form

where C ∈ R is some constant.

Proof. By Theorem 3.1, the function is a prototype for f(x). Suppose that G(x) is another antiderivative f (x). Then G'(x) = f(x) and for the function F(x) − G(x) we have: (F (x) + G(x))' = F'(x)−G'(x) = f (x)−f(x) ≡ 0. Hence, the derivative of the function F (x)−G(x)

is equal to zero, therefore, this function is a constant: F(x) − G(x) = const.

8. Newton-Leibniz formula for a definite integral.

Theorem:

Condition: f(t) is continuous on , and F(x) is its any antiderivative.

Statement:

Proof: Consider some antiderivative F (x) of the function f (x). According to the Corollary of the Theorem “On the Differentiability of an Integral with a Variable Upper Limit” (see the previous question), it has the form . From here

=> c= F(a) , and

Let's move F(a) in the last equality to the left side, redesignate the integration variable again as x and get the Newton-Leibniz formula: