Definite and indefinite integrals message. Integrals for dummies: how to solve, calculation rules, explanation. Basic properties of the definite integral

Solving integrals is an easy task, but only for a select few. This article is for those who want to learn to understand integrals, but know nothing or almost nothing about them. Integral... Why is it needed? How to calculate it? What are definite and indefinite integrals?

If the only use you know of for an integral is to use a crochet hook shaped like an integral icon to get something useful out of hard-to-reach places, then welcome! Find out how to solve the simplest and other integrals and why you can’t do without it in mathematics.

We study the concept « integral »

Integration was known back in Ancient Egypt. Of course not in modern form, but still. Since then, mathematicians have written many books on this topic. Especially distinguished themselves Newton And Leibniz , but the essence of things has not changed.

How to understand integrals from scratch? No way! To understand this topic you will still need a basic understanding of the basics. mathematical analysis. We already have information about limits and derivatives, necessary for understanding integrals, on our blog.

Indefinite integral

Let us have some function f(x) .

Indefinite integral function f(x) this function is called F(x) , whose derivative is equal to the function f(x) .

In other words, an integral is a derivative in reverse or an antiderivative. By the way, read our article about how to calculate derivatives.

An antiderivative exists for all continuous functions. Also, a constant sign is often added to the antiderivative, since the derivatives of functions that differ by a constant coincide. The process of finding the integral is called integration.

Simple example:

In order not to constantly calculate antiderivatives elementary functions, it is convenient to summarize them in a table and use ready-made values.

Complete table of integrals for students

Definite integral

When dealing with the concept of an integral, we are dealing with infinitesimal quantities. The integral will help to calculate the area of ​​the figure, the mass of the inhomogeneous body, the distance traveled at uneven movement path and much more. It should be remembered that an integral is an infinite sum large quantity infinitesimal terms.

As an example, imagine a graph of some function.

How to find the area of ​​a figure bounded by the graph of a function? Using an integral! Let us divide the curvilinear trapezoid, limited by the coordinate axes and the graph of the function, into infinitesimal segments. This way the figure will be divided into thin columns. The sum of the areas of the columns will be the area of ​​the trapezoid. But remember that such a calculation will give an approximate result. However, the smaller and narrower the segments, the more accurate the calculation will be. If we reduce them to such an extent that the length tends to zero, then the sum of the areas of the segments will tend to the area of ​​the figure. This is a definite integral, which is written like this:

Points a and b are called limits of integration.

« Integral »

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Rules for calculating integrals for dummies

Properties of the indefinite integral

How to solve an indefinite integral? Here we will look at the properties definite integral, which will be useful when solving examples.

• The derivative of the integral is equal to the integrand:

• The constant can be taken out from under the integral sign:

• The integral of the sum is equal to the sum of the integrals. This is also true for the difference:

Properties of a definite integral

• Linearity:

• The sign of the integral changes if the limits of integration are swapped:

• At any points a, b And With:

We have already found out that a definite integral is the limit of a sum. But how to get a specific value when solving an example? For this there is the Newton-Leibniz formula:

Examples of solving integrals

Below we will consider the indefinite integral and examples with solutions. We suggest you figure out the intricacies of the solution yourself, and if something is unclear, ask questions in the comments.

To reinforce the material, watch a video about how integrals are solved in practice. Don't despair if the integral is not given right away. Contact a professional service for students, and any triple or curved integral over a closed surface will be within your power.

In this article we will list the main properties of the definite integral. Most of these properties are proved based on the concepts of the Riemann and Darboux definite integral.

The calculation of the definite integral is very often done using the first five properties, so we will refer to them when necessary. The remaining properties of the definite integral are mainly used to evaluate various expressions.

Before moving on basic properties of the definite integral, let us agree that a does not exceed b.

For the function y = f(x) defined at x = a, the equality is true.

That is, the value of a definite integral with the same limits of integration is equal to zero. This property is a consequence of the definition of the Riemann integral, since in this case each integral sum for any partition of the interval and any choice of points is equal to zero, since, therefore, the limit of integral sums is zero.

For a function integrable on an interval, .

In other words, when the upper and lower limits of integration change places, the value of the definite integral changes to the opposite. This property of a definite integral also follows from the concept of the Riemann integral, only the numbering of the partition of the segment should begin from the point x = b.

for functions integrable on an interval y = f(x) and y = g(x) .

Proof.

Let's write down the integral sum of the function for a given partition of a segment and a given choice of points:

where and are the integral sums of the functions y = f(x) and y = g(x) for a given partition of the segment, respectively.

Going to the limit at we obtain that, by the definition of the Riemann integral, is equivalent to the statement of the property being proved.

The constant factor can be taken out of the sign of the definite integral. That is, for a function y = f(x) integrable on an interval and an arbitrary number k, the following equality holds: .

The proof of this property of the definite integral is absolutely similar to the previous one:


Let the function y = f(x) be integrable on the interval X, and and then .

This property is true for both , and or .

The proof can be carried out based on the previous properties of the definite integral.

If a function is integrable on an interval, then it is integrable on any internal interval.

The proof is based on the property of Darboux sums: if new points are added to an existing partition of a segment, then the lower Darboux sum will not decrease, and the upper one will not increase.

If the function y = f(x) is integrable on the interval and for any value of the argument, then .

This property is proven through the definition of the Riemann integral: any integral sum for any choice of points of partition of the segment and points at will be non-negative (not positive).

Consequence.

For functions y = f(x) and y = g(x) integrable on an interval, the following inequalities hold:


This statement means that integration of inequalities is permissible. We will use this corollary to prove the following properties.

Let the function y = f(x) be integrable on the interval , then the inequality holds .

Proof.

It's obvious that . In the previous property, we found out that the inequality can be integrated term by term, therefore, it is true . This double inequality can be written as .

Let the functions y = f(x) and y = g(x) be integrable on the interval and for any value of the argument , then , Where And .

The proof is carried out similarly. Since m and M are the smallest and highest value function y = f(x) on the segment , then . Multiplying the double inequality by a non-negative function y = g(x) leads us to the following double inequality. Integrating it on the interval , we arrive at the statement being proved.

Consequence.

If we take g(x) = 1, then the inequality takes the form .

First average formula.

Let the function y = f(x) be integrable on the interval, And , then there is a number such that .

Consequence.

If the function y = f(x) is continuous on the interval, then there is a number such that .

The first average value formula in generalized form.

Let the functions y = f(x) and y = g(x) be integrable on the interval, And , and g(x) > 0 for any value of the argument . Then there is a number such that .

Second average formula.

If on an interval the function y = f(x) is integrable, and y = g(x) is monotonic, then there exists a number such that the equality .

These properties are used to transform the integral in order to reduce it to one of the elementary integrals and further calculation.

1. The derivative of the indefinite integral is equal to the integrand:

2. The differential of the indefinite integral is equal to the integrand:

3. The indefinite integral of the differential of a certain function is equal to the sum of this function and an arbitrary constant:

4. The constant factor can be taken out of the integral sign:

Moreover, a ≠ 0

5. The integral of the sum (difference) is equal to the sum (difference) of the integrals:

6. Property is a combination of properties 4 and 5:

Moreover, a ≠ 0 ˄ b ≠ 0

7. Invariance property of the indefinite integral:

If , then

8. Property:

If , then

Actually this property represents a special case of integration using the variable change method, which is discussed in more detail in the next section.

Let's look at an example:

First we applied property 5, then property 4, then we used the table of antiderivatives and got the result.

The algorithm of our online integral calculator supports all the properties listed above and can easily find detailed solution for your integral.

These properties are used to transform the integral in order to reduce it to one of the elementary integrals and further calculation.

1. The derivative of the indefinite integral is equal to the integrand:

2. The differential of the indefinite integral is equal to the integrand:

3. The indefinite integral of the differential of a certain function is equal to the sum of this function and an arbitrary constant:

4. The constant factor can be taken out of the integral sign:

Moreover, a ≠ 0

5. The integral of the sum (difference) is equal to the sum (difference) of the integrals:

6. Property is a combination of properties 4 and 5:

Moreover, a ≠ 0 ˄ b ≠ 0

7. Invariance property of the indefinite integral:

If , then

8. Property:

If , then

In fact, this property is a special case of integration using the variable change method, which is discussed in more detail in the next section.

Let's look at an example:

First we applied property 5, then property 4, then we used the table of antiderivatives and got the result.

The algorithm of our online integral calculator supports all the properties listed above and will easily find a detailed solution for your integral.

In differential calculus the problem is solved: under this function ƒ(x) find its derivative(or differential). Integral calculus solves the inverse problem: find the function F(x), knowing its derivative F "(x)=ƒ(x) (or differential). The sought function F(x) is called the antiderivative of the function ƒ(x).

The function F(x) is called antiderivative function ƒ(x) on the interval (a; b), if for any x є (a; b) the equality

F " (x)=ƒ(x) (or dF(x)=ƒ(x)dx).

For example, the antiderivative of the function y = x 2, x є R, is the function, since

Obviously, any functions will also be antiderivatives

where C is a constant, since

Theorem 29. 1. If the function F(x) is an antiderivative of the function ƒ(x) on (a;b), then the set of all antiderivatives for ƒ(x) is given by the formula F(x)+C, where C is a constant number.

▲ The function F(x)+C is an antiderivative of ƒ(x).

Indeed, (F(x)+C) " =F " (x)=ƒ(x).

Let Ф(х) be some other antiderivative of the function ƒ(x), different from F(x), i.e. Ф "(x)=ƒ(х). Then for any x є (а; b) we have

And this means (see Corollary 25.1) that

where C is a constant number. Therefore, Ф(x)=F(x)+С.▼

The set of all antiderivative functions F(x)+С for ƒ(x) is called indefinite integral of the function ƒ(x) and is denoted by the symbol ∫ ƒ(x) dx.

Thus, by definition

∫ ƒ(x)dx= F(x)+C.

Here ƒ(x) is called integrand function, ƒ(x)dx — integrand, X - integration variable, ∫ -sign indefinite integral .

The operation of finding the indefinite integral of a function is called integrating this function.

Geometrically, the indefinite integral is a family of “parallel” curves y=F(x)+C (each numerical value of C corresponds to a specific curve of the family) (see Fig. 166). The graph of each antiderivative (curve) is called integral curve.

Does every function have an indefinite integral?

There is a theorem stating that “every function continuous on (a;b) has an antiderivative on this interval,” and, consequently, an indefinite integral.

Let us note a number of properties of the indefinite integral that follow from its definition.

1. The differential of the indefinite integral is equal to the integrand, and the derivative of the indefinite integral is equal to the integrand:

d(∫ ƒ(x)dx)=ƒ(x)dх, (∫ ƒ(x)dx) " =ƒ(x).

Indeed, d(∫ ƒ(x) dx)=d(F(x)+C)=dF(x)+d(C)=F " (x) dx =ƒ(x) dx

(∫ ƒ (x) dx) " =(F(x)+C)"=F"(x)+0 =ƒ (x).

Thanks to this property, the correctness of integration is checked by differentiation. For example, equality

∫(3x 2 + 4) dx=х з +4х+С

true, since (x 3 +4x+C)"=3x 2 +4.

2. The indefinite integral of the differential of a certain function is equal to the sum of this function and an arbitrary constant:

∫dF(x)= F(x)+C.

Really,

3. The constant factor can be taken out of the integral sign:

α ≠ 0 is a constant.

Really,

(put C 1 / a = C.)

4. The indefinite integral of the algebraic sum of a finite number of continuous functions is equal to the algebraic sum of the integrals of the summands of the functions:

Let F"(x)=ƒ(x) and G"(x)=g(x). Then

where C 1 ±C 2 =C.

5. (Invariance of the integration formula).

If , where u=φ(x) is an arbitrary function with a continuous derivative.

▲ Let x be an independent variable, ƒ(x) - continuous function and F(x) is its antigen. Then

Let us now set u=φ(x), where φ(x) is a continuously differentiable function. Consider the complex function F(u)=F(φ(x)). Due to the invariance of the form of the first differential of the function (see p. 160), we have

From here▼

Thus, the formula for the indefinite integral remains valid regardless of whether the variable of integration is the independent variable or any function of it that has a continuous derivative.

So, from the formula by replacing x with u (u=φ(x)) we get

In particular,

Example 29.1. Find the integral

where C=C1+C 2 +C 3 +C 4.

Example 29.2. Find the integral Solution:

• 29.3. Table of basic indefinite integrals

Taking advantage of the fact that integration is the inverse action of differentiation, one can obtain a table of basic integrals by inverting the corresponding formulas of differential calculus (table of differentials) and using the properties of the indefinite integral.

For example, because

d(sin u)=cos u . du

The derivation of a number of formulas in the table will be given when considering the basic methods of integration.

The integrals in the table below are called tabular. They should be known by heart. In integral calculus there are no simple and universal rules for finding antiderivatives of elementary functions, as in differential calculus. Methods for finding antiderivatives (i.e., integrating a function) are reduced to indicating techniques that bring a given (sought) integral to a tabular one. Therefore, it is necessary to know table integrals and be able to recognize them.

Note that in the table of basic integrals, the integration variable can denote both an independent variable and a function of the independent variable (according to the invariance property of the integration formula).

The validity of the formulas below can be verified by taking the differential on the right side, which will be equal to the integrand on the left side of the formula.

Let us prove, for example, the validity of formula 2. The function 1/u is defined and continuous for all values ​​of and other than zero.

If u > 0, then ln|u|=lnu, then That's why

If u<0, то ln|u|=ln(-u). НоMeans

So, formula 2 is correct. Similarly, let's check formula 15:

Table of main integrals

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