The simplest properties of integrals. Basic properties of an indefinite integral We study the concept of "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 of an integral you know of is to crochet something useful from hard-to-reach places with a crochet in the shape of an integral icon, then you are welcome! Learn how to solve elementary and other integrals and why you can't do without it in mathematics.

Exploring the concept « integral »

Integration has been known since ancient Egypt. Of course, not in its 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 still need a basic knowledge of the basics of calculus. We already have information about, necessary for understanding integrals, in our blog.

Indefinite integral

Suppose we have some kind of function f (x) .

Indefinite integral of a function f (x) such a function is called F (x) whose derivative is equal to the function f (x) .

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

The antiderivative exists for all continuous functions. Also, the sign of a constant 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 the antiderivatives of elementary functions, it is convenient to bring them down to 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 a figure, the mass of an inhomogeneous body, the path traveled with uneven movement, and much more. It should be remembered that the integral is the sum of an infinitely large number of infinitely small terms.

As an example, let's imagine a graph of some function.

How to find the area of a shape bounded by the graph of a function? Using the integral! We divide the curvilinear trapezoid, bounded by the coordinate axes and the graph of the function, into infinitely small segments. Thus, 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 the limits of integration.

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Integral computation rules for dummies

Indefinite integral properties

How to solve indefinite integral? Here we will look at the properties of the indefinite integral, which will come in handy 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. It is also true for the difference:

Properties of the definite integral

Linearity:

The integral sign changes if the integration limits are reversed:

At any points a, b and with:

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

Integral solutions examples

Below we will consider an indefinite integral and examples with a solution. We offer you to independently figure out the intricacies of the solution, and if something is not clear, ask questions in the comments.

To consolidate the material, watch the video on how integrals are solved in practice. Don't be discouraged if the integral isn't given right away. Contact the professional student service and you can handle any triple or curvilinear integral over a closed surface.

This article details the basic properties of a definite integral. They are proved using the concept of the Riemann and Darboux integral. The calculation of the definite integral takes place due to 5 properties. The rest of them are used to evaluate various expressions.

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

Basic properties of a definite integral

Definition 1

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

Proof 1

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

Definition 2

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

Proof 2

In other words, if the upper and lower limits of integration are changed in places, then the value of the integral will change its value to the opposite. This property is taken from the Riemann integral. However, the numbering of the division of the segment comes 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 is used for integrable functions of the type y = f (x) and y = g (x) defined on the interval [a; b].

Proof 3

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

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

Proof 4

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

σ = ∑ i = 1 nk f ζ i (xi - xi - 1) = = k ∑ i = 1 nf ζ i (xi - xi - 1) = k σ f ⇒ lim λ → 0 σ = lim λ → 0 (k σ f) = k lim λ → 0 σ f ⇒ ∫ abk f (x) dx = k ∫ abf (x) dx

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.

Proof 5

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

Definition 6

When the function has the ability to be integrable from the segment [a; b], then it is doable for any inner segment c; d ∈ a; b.

Proof 6

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

Definition 7

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

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

Proof 7

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

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

Thanks to the statement, we know that integration is admissible. This corollary will be used to prove other properties.

Definition 8

With an integrable function y = f (x) from the segment [a; b] we have a valid 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 obtained 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 segment [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).

Proof 9

The proof is carried out in a similar way. M and m are considered the largest and the smallest value of the function y = f (x), determined 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 segment [a; b], then we obtain the assertion to be proved.

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

First mean value formula

Definition 10

For y = f (x), integrable on the segment [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.

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

First mean value formula in generalized form

Definition 11

When the functions y = f (x) and y = g (x) are integrable from the segment [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 of x ∈ a; b. Hence 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 mean value formula

Definition 12

When the function y = f (x) is integrable from the segment [a; b], and y = g (x) is monotone, then there is a number that c ∈ a; b, where we obtain a valid 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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These properties are used to carry out transformations of the integral with the aim of reducing 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 some 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. The property is a combination of properties 4 and 5:

Moreover, a ≠ 0 ˄ b ≠ 0

7. Property of invariance 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 consider an example:

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

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

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 definite integral of Riemann and Darboux.

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

Before moving on to the 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, equality is true.

That is, the value of a definite integral with coinciding limits of integration is 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 the integral sums is zero.

For a function integrable on a segment, .

In other words, when changing the upper and lower limits of integration in 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 a segment should be started from the point x = b.

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

Proof.

We write the integral sum of the function for a given division 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 the given partition of the segment, respectively.

Passing to the limit at we obtain that, by the definition of the Riemann integral, it is equivalent to the assertion of the property being proved.

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

The proof of this property of a 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 for or.

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

If a function is integrable on a segment, then it is integrable on any inner segment as well.

The proof is based on the property of Darboux sums: if you add new points to the existing partition of the 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 an interval and for any value of the argument, then .

This property is proved through the definition of the Riemann integral: any integral sum for any choice of partition points of a 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 the integration of inequalities is admissible. We will use this corollary to prove the following properties.

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

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 an interval and for any value of the argument, then , where and .

The proof is similar. 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 the nonnegative function y = g (x) leads us to the following double inequality. Integrating it on a segment, we arrive at the assertion being proved.

Consequence.

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

First average value formula.

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

Consequence.

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

The first formula for the mean in generalized form.

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

Second formula for the average.

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