Properties of formula integrals. Basic properties of the indefinite integral. Rules for calculating integrals for dummies

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 the segment and given choice 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 largest values ​​of the 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 carry out transformations of 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 special case 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.

This article talks in detail about the main properties of the definite integral. They are proved using the concept of the Riemann and Darboux integral. The calculation of a definite integral takes place thanks to 5 properties. The remaining ones are used to evaluate various expressions.

Before moving on to the main properties of the definite integral, it is necessary to make sure that a does not exceed b.

Basic properties of the definite integral

Definition 1

The function y = f (x) defined at x = a is similar to the fair equality ∫ a a f (x) d x = 0.

Evidence 1

From this we see that the value of the integral with coinciding limits is equal to zero. This is a consequence of the Riemann integral, because every integral sum σ for any partition on the interval [ a ; a ] and any choice of points ζ i equals zero, because x i - x i - 1 = 0 , i = 1 , 2 , . . . , n , which means we find that the limit of integral functions is zero.

Definition 2

For a function that is integrable on the interval [a; b ] , the condition ∫ a b f (x) d x = - ∫ b a f (x) d x is satisfied.

Evidence 2

In other words, if you swap the upper and lower limits of integration, the value of the integral will change to the opposite value. This property is taken from the Riemann integral. However, the numbering of the partition of the segment starts from the point x = b.

Definition 3

∫ a b f x ± g (x) d x = ∫ a b f (x) d x ± ∫ a b g (x) d x applies to integrable functions of type y = f (x) and y = g (x) defined on the interval [ a ; b ] .

Evidence 3

Write down the integral sum of the function y = f (x) ± g (x) for partitioning into segments with a given choice of points ζ i: σ = ∑ i = 1 n f ζ i ± g ζ i · x i - x i - 1 = = ∑ i = 1 n f (ζ i) · x i - x i - 1 ± ∑ i = 1 n g ζ i · x i - x i - 1 = σ f ± σ g

where σ f and σ g are the integral sums of the functions y = f (x) and y = g (x) for partitioning the segment. After passing to the limit at λ = m a x i = 1, 2, . . . , n (x i - x i - 1) → 0 we obtain that lim λ → 0 σ = lim λ → 0 σ f ± σ g = lim λ → 0 σ g ± lim λ → 0 σ g .

From Riemann's definition, this expression is equivalent.

Definition 4

Extending the constant factor beyond the sign of the definite integral. Integrated function from the interval [a; b ] with an arbitrary value k has a fair inequality of the form ∫ a b k · f (x) d x = k · ∫ a b f (x) d x .

Proof 4

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

σ = ∑ i = 1 n k · f ζ i · (x i - x i - 1) = = k · ∑ i = 1 n f ζ i · (x i - x i - 1) = k · σ f ⇒ lim λ → 0 σ = lim λ → 0 (k · σ f) = k · lim λ → 0 σ f ⇒ ∫ a b k · f (x) d x = k · ∫ a b f (x) d x

Definition 5

If a function of the form y = f (x) is integrable on an interval x with a ∈ x, b ∈ x, we obtain that ∫ a b f (x) d x = ∫ a c f (x) d x + ∫ c b f (x) d x.

Evidence 5

The property is considered valid for c ∈ a; b, for c ≤ a and c ≥ b. The proof is similar to the previous properties.

Definition 6

When a function can be integrable from the segment [a; b ], then this is feasible for any internal segment c; d ∈ a ; b.

Proof 6

The proof is based on the Darboux property: if 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.

Definition 7

When a function is integrable on [a; b ] from f (x) ≥ 0 f (x) ≤ 0 for any value x ∈ a ; b , then we get that ∫ a b f (x) d x ≥ 0 ∫ a b f (x) ≤ 0 .

The property can be proven using the definition of the Riemann integral: any integral sum for any choice of points of partition of the segment and points ζ i with the condition that f (x) ≥ 0 f (x) ≤ 0 is non-negative.

Evidence 7

If the functions y = f (x) and y = g (x) are integrable on the interval [ a ; b ], then the following inequalities are considered valid:

∫ a b f (x) d x ≤ ∫ a b g (x) d x , f (x) ≤ g (x) ∀ x ∈ a ; b ∫ a b f (x) d x ≥ ∫ a b g (x) d x , f (x) ≥ g (x) ∀ x ∈ a ; b

Thanks to the statement, we know that integration is permissible. This corollary will be used in the proof of other properties.

Definition 8

For an integrable function y = f (x) from the interval [ a ; b ] we have a fair inequality of the form ∫ a b f (x) d x ≤ ∫ a b f (x) d x .

Proof 8

We have that - f (x) ≤ f (x) ≤ f (x) . From the previous property we found that the inequality can be integrated term by term and it corresponds to an inequality of the form - ∫ a b f (x) d x ≤ ∫ a b f (x) d x ≤ ∫ a b f (x) d x . This double inequality can be written in another form: ∫ a b f (x) d x ≤ ∫ a b f (x) d x .

Definition 9

When the functions y = f (x) and y = g (x) are integrated from the interval [ a ; b ] for g (x) ≥ 0 for any x ∈ a ; b , we obtain an inequality of the form m · ∫ a b g (x) d x ≤ ∫ a b f (x) · g (x) d x ≤ M · ∫ a b g (x) d x , where m = m i n x ∈ a ; b f (x) and M = m a x x ∈ a ; b f (x) .

Evidence 9

The proof is carried out in a similar way. M and m are considered to be the largest and lowest value function y = f (x) defined from the segment [ a ; b ] , then m ≤ f (x) ≤ M . It is necessary to multiply the double inequality by the function y = g (x), which will give the value of the double inequality of the form m g (x) ≤ f (x) g (x) ≤ M g (x). It is necessary to integrate it on the interval [a; b ] , then we get the statement to be proved.

Consequence: For g (x) = 1, the inequality takes the form m · b - a ≤ ∫ a b f (x) d x ≤ M · (b - a) .

First average formula

Definition 10

For y = f (x) integrable on the interval [ a ; b ] with m = m i n x ∈ a ; b f (x) and M = m a x x ∈ a ; b f (x) there is a number μ ∈ m; M , which fits ∫ a b f (x) d x = μ · b - a .

Consequence: When the function y = f (x) is continuous from the interval [ a ; b ], then there is a number c ∈ a; b, which satisfies the equality ∫ a b f (x) d x = f (c) b - a.

The first average formula in generalized form

Definition 11

When the functions y = f (x) and y = g (x) are integrable from the interval [ a ; b ] with m = m i n x ∈ a ; b f (x) and M = m a x x ∈ a ; b f (x) , and g (x) > 0 for any value x ∈ a ; b. From here we have that there is a number μ ∈ m; M , which satisfies the equality ∫ a b f (x) · g (x) d x = μ · ∫ a b g (x) d x .

Second average formula

Definition 12

When the function y = f (x) is integrable from the interval [ a ; b ], and y = g (x) is monotonic, then there is a number that c ∈ a; b , where we obtain a fair equality of the form ∫ a b f (x) · g (x) d x = g (a) · ∫ a c f (x) d x + g (b) · ∫ c b f (x) d x

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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 , 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 about how in our article.

The antiderivative exists for everyone 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 calculate the area of ​​the figure, mass inhomogeneous body, passed at uneven movement path and much more. It should be remembered that an integral is the sum of an infinitely large number of 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 indefinite 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:

• Integral of the sum equal to the sum 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.

The main task of differential calculus is to find the derivative f'(x) or differential df=f'(x)dx functions f(x). In integral calculus the inverse problem is solved. By given function f(x) you need to find such a function F(x), What F'(x)=f(x) or dF(x)=F'(x)dx=f(x)dx.

Thus, the main task of integral calculus is the restoration of function F(x) by the known derivative (differential) of this function. Integral calculus has numerous applications in geometry, mechanics, physics and technology. It gives general method finding areas, volumes, centers of gravity, etc.

Definition. FunctionF(x), , is called an antiderivative for the functionf(x) on the set X if it is differentiable for any andF'(x)=f(x) ordF(x)=f(x)dx.

Theorem. Any continuous line on the interval [a;b] functionf(x) has an antiderivative on this segmentF(x).

Theorem. IfF 1 (x) andF 2 (x) – two different antiderivatives of the same functionf(x) on the set x, then they differ from each other by a constant term, i.e.F 2 (x)=F 1x)+C, where C is a constant.

Indefinite integral, its properties.

Definition. TotalityF(x)+From all antiderivative functionsf(x) on the set X is called an indefinite integral and is denoted:

- (1)

In formula (1) f(x)dx called integrand expression,f(x) – integrand function, x – integration variable, A C – integration constant.

Let us consider the properties of the indefinite integral that follow from its definition.

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

And .

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

3. The constant factor a (a≠0) can be taken out as the sign of the indefinite integral:

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

5. IfF(x) – antiderivative of the functionf(x), then:

6 (invariance of integration formulas). Any integration formula retains its form if the integration variable is replaced by any differentiable function of this variable:

Whereu is a differentiable function.

Table of indefinite integrals.

Let's give basic rules for integrating functions.

Let's give table of basic indefinite integrals.(Note that here, as in differential calculus, the letter u can be designated as an independent variable (u=x), and a function of the independent variable (u=u(x)).)

(n≠-1). (a >0, a≠1). (a≠0). (a≠0). (|u| > |a|).(|u|< |a|).

Integrals 1 – 17 are called tabular.

Some of the above formulas in the table of integrals, which do not have an analogue in the table of derivatives, are verified by differentiating their right-hand sides.

Change of variable and integration by parts in the indefinite integral.

Integration by substitution (variable replacement). Let it be necessary to calculate the integral

, which is not tabular. The essence of the substitution method is that in the integral the variable X replace with a variable t according to the formula x=φ(t), where dx=φ’(t)dt.

Theorem. Let the functionx=φ(t) is defined and differentiable on a certain set T and let X be the set of values ​​of this function on which the function is definedf(x). Then if on the set X the functionf(