Integrals for dummies: how to solve, calculation rules, explanation. Methods for calculating indefinite integrals

Is it possible to subsume a nonlinear function under the differential sign? Yes, if the integrand is the product of two factors: one factor is a complex function of some nonlinear function, and the other factor is the derivative of this nonlinear function. Let's look at what has been said with examples.

Can't find definite integrals.

Example 1. ∫(2x + 1)(x 2 + x + 2) 5 dx = ∫(x 2 + x + 2) 5 d (x 2 + x + 2) =(x²+x+2) 6 : 6 + C.

What does this integrand represent? Work power function from (x 2 + x + 2) and the multiplier (2x + 1), which is equal to the derivative of the base of the power: (x 2 + x + 2)" = 2x + 1.

This allowed us to put (2x + 1) under the differential sign:

∫u 5 du=u 6 : 6+ C. (Formula 1). )

Examination. (F (x)+ C)" =((x²+x+2) 6 : 6 + C)′=1/6 6 (x 2 + x + 2) 5 (x 2 + x + 2)" =

=(x 2 + x + 2) 5 · (2x + 1) = (2x + 1)(x 2 + x + 2) 5 = f (x).

Example 2.∫(3x 2 – 2x + 3)(x 3 - x 2 + 3x + 1) 5 dx = ∫(x 3 – x 2 + 3x + 1) 5 d (x 3 – x 2 + 3x + 1) =

=(x³- x²+3x+1) 6 : 6+C

And how does this example differ from example 1? Nothing! The same fifth power with the base (x 3 – x 2 + 3x + 1) is multiplied by the trinomial (3x 2 – 2x + 3), which is the derivative of the base of the power: (x 3 – x 2 + 3x + 1)" = 3x 2 – 2x + 3. We brought this base of the degree under the differential sign, from which the value of the integrand did not change, and then applied the same formula 1 (). Integrals)

Example 3.

Here the derivative of (2x 3 – 3x) will give (6x 2 – 3), and with us

there is (12x 2 – 6), that is, the expression in 2 times greater, which means we put (2x 3 – 3x) under the differential sign, and put a factor in front of the integral 2 . Let's apply the formula 2) ( sheet ).

Here's what happens:

Let's check, taking into account that:

Examples. Find indefinite integrals.

1. ∫(6x+5) 3 dx. How will we decide? Looking at the sheet and we reason something like this: the integrand represents a degree, and we have a formula for the integral of the degree (formula 1) ), but in it the basis of the degree u and the integration variable too u.

And we have an integration variable X, and the base of the degree (6x+5). Let's make a change to the integration variable: instead of dx we write d (6x+5). What changed? Since what comes after the differential sign d is, by default, differentiated,

then d (6x+5)=6dx, i.e. when replacing the variable x with the variable (6x+5), the integrand function increased 6 times, so we put the factor 1/6 in front of the integral sign. These arguments can be written like this:

So, we solved this example by introducing a new variable (the variable x was replaced by the variable 6x+5). Where did you write the new variable (6x+5)? Under the differential sign. Therefore, this method of introducing a new variable is often called method ( or way ) summing up(new variable ) under the differential sign.

In the second example, we first obtained a degree with a negative exponent, and then put it under the differential sign (7x-2) and used the formula for the integral of the degree 1) (Integrals ).

Let's look at the solution to the example 3.

The integral is preceded by a coefficient of 1/5. Why? Since d (5x-2) = 5dx, then, by substituting the function u = 5x-2 under the differential sign, we increased the integrand by 5 times, therefore, in order for the value of this expression not to change, we had to divide by 5, i.e. . multiply by 1/5. Next, the formula was used 2) (Integrals) .

All the simplest integral formulas will look like:

∫f (x) dx=F (x)+C, and the equality must be satisfied:

(F (x)+C)"=f (x).

Integration formulas can be obtained by inverting the corresponding differentiation formulas.

Really,

Exponent n may be fractional. Often you have to find the indefinite integral of the function y=√x. Let's calculate the integral of the function f (x)=√x using the formula 1) .

Let's write this example as a formula 2) .

Since (x+C)"=1, then ∫dx=x+C.

3) ∫dx=x+C.

Replacing 1/x² with x -2, we calculate the integral of 1/x².

And it was possible to get this answer by inverting the well-known differentiation formula:

Let us write our reasoning in the form of a formula 4).

Multiplying both sides of the resulting equality by 2, we obtain the formula 5).

Let's find the integrals of the main ones trigonometric functions, knowing their derivatives: (sinx)"=cosx; (cosx)"=-sinx; (tgx)"=1/cos²x; (ctgx)"=-1/sin²x. We obtain the integration formulas 6) — 9).

6) ∫cosxdx=sinx+C;

7) ∫sinxdx=-cosx+C;

After studying the exponential and logarithmic functions, let's add a few more formulas.

Basic properties of the indefinite integral.

I. The derivative of the indefinite integral is equal to the integrand .

(∫f (x) dx)"=f (x).

II. The differential of an indefinite integral is equal to the integrand.

d∫f (x) dx=f (x) dx.

III. Indefinite integral of the differential (derivative) of some function equal to the sum this function and an arbitrary constant C.

∫dF (x)=F (x)+C or ∫F"(x) dx=F (x)+C.

Please note: in properties I, II and III, the signs of the differential and integral (integral and differential) “eat” each other!

IV. The constant factor of the integrand can be taken out of the integral sign.

∫kf (x) dx=k ∫f (x) dx, Where k- a constant value that is not equal to zero.

V. The integral of the algebraic sum of functions is equal to algebraic sum integrals of these functions.

∫(f (x)±g (x)) dx=∫f (x) dx±∫g (x) dx.

VI. If F (x) is an antiderivative of f (x), and k And b are constant values, and k≠0, then (1/k)·F (kx+b) is an antiderivative for f (kx+b). Indeed, according to the rule for calculating the derivative of a complex function, we have:

You can write:

For every mathematical action there is an inverse action. For the action of differentiation (finding derivatives of functions), there is also an inverse action - integration. Through integration, a function is found (reconstructed) from its given derivative or differential. The found function is called antiderivative.

Definition. Differentiable function F(x) is called the antiderivative of the function f(x) on a given interval, if for all X from this interval the following equality holds: F′(x)=f (x).

Examples. Find antiderivatives for the functions: 1) f (x)=2x; 2) f (x)=3cos3x.

1) Since (x²)′=2x, then, by definition, the function F (x)=x² will be an antiderivative of the function f (x)=2x.

2) (sin3x)′=3cos3x. If we denote f (x)=3cos3x and F (x)=sin3x, then, by definition of an antiderivative, we have: F′(x)=f (x), and, therefore, F (x)=sin3x is an antiderivative for f ( x)=3cos3x.

Note that (sin3x +5 )′= 3cos3x, and (sin3x -8,2 )′= 3cos3x, ... V general view can be written: (sin3x +C)′= 3cos3x, Where WITH- some constant value. These examples indicate the ambiguity of the action of integration, in contrast to the action of differentiation, when any differentiable function has a single derivative.

Definition. If the function F(x) is an antiderivative of the function f(x) on a certain interval, then the set of all antiderivatives of this function has the form:

F(x)+C, where C is any real number.

The set of all antiderivatives F (x)+C of the function f (x) on the interval under consideration is called the indefinite integral and is denoted by the symbol ∫ (integral sign). Write down: ∫f (x) dx=F (x)+C.

Expression ∫f(x)dx read: “integral ef from x to de x.”

f(x)dx- integrand expression,

f(x)- integrand function,

X is the integration variable.

F(x)- antiderivative of a function f(x),

WITH- some constant value.

Now the considered examples can be written as follows:

1) ∫ 2xdx=x²+C. 2) ∫ 3cos3xdx=sin3x+C.

What does the sign d mean?

d— differential sign - has a dual purpose: firstly, this sign separates the integrand from the integration variable; secondly, everything that comes after this sign is differentiated by default and multiplied by the integrand.

Examples. Find the integrals: 3) ∫ 2pxdx; 4) ∫ 2pxdp.

3) After the differential icon d costs XX, A R

∫ 2хрdx=рх²+С. Compare with example 1).

Let's do a check. F′(x)=(px²+C)′=p·(x²)′+C′=p·2x=2px=f (x).

4) After the differential icon d costs R. This means that the integration variable R, and the multiplier X should be considered some constant value.

∫ 2хрдр=р²х+С. Compare with examples 1) And 3).

Let's do a check. F′(p)=(p²x+C)′=x·(p²)′+C′=x·2p=2px=f (p).

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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 integrals and why you can't do without it.

We study the concept of "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 basic knowledge basics mathematical analysis. It is this fundamental information that you will find 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.

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 of elementary functions, it is convenient to put them in a table and use ready-made values:

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, limited by schedule functions?

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.

Bari Alibasov and the group "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 of the 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:

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 several examples of finding indefinite integrals. We invite you to 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. Ask and they will tell you everything they know about calculating integrals. With our help, any triple or curved integral over a closed surface will be within your power.

Previously we given function, guided by various formulas and rules, found its derivative. The derivative has numerous uses: it is the speed of movement (or, more generally, the speed of any process); slope tangent to the graph of a function; using the derivative, you can examine a function for monotonicity and extrema; it helps solve optimization problems.

But along with the problem of finding the speed according to a known law of motion, there is also an inverse problem - the problem of restoring the law of motion according to a known speed. Let's consider one of these problems.

Example 1. A material point moves in a straight line, its speed at time t is given by the formula v=gt. Find the law of motion.
Solution. Let s = s(t) be the desired law of motion. It is known that s"(t) = v(t). This means that to solve the problem you need to select a function s = s(t), the derivative of which is equal to gt. It is not difficult to guess that $s(t) = \frac(gt^ 2)(2)$.
$s"(t) = \left(\frac(gt^2)(2) \right)" = \frac(g)(2)(t^2)" = \frac(g)(2) \ cdot 2t = gt$
Answer: $s(t) = \frac(gt^2)(2) $

Let us immediately note that the example is solved correctly, but incompletely. We got $s(t) = \frac(gt^2)(2) $. In fact, the problem has infinitely many solutions: any function of the form $s(t) = \frac(gt^2)(2) + C$, where C is an arbitrary constant, can serve as a law of motion, since $\left (\frac(gt^2)(2) +C \right)" = gt $

To make the problem more specific, we had to fix the initial situation: indicate the coordinate of a moving point at some point in time, for example at t = 0. If, say, s(0) = s 0, then from the equality s(t) = (gt 2)/2 + C we get: s(0) = 0 + C, i.e. C = s 0. Now the law of motion is uniquely defined: s(t) = (gt 2)/2 + s 0.

In mathematics, mutually inverse operations are assigned different names, come up with special notations, for example: squaring (x 2) and extracting square root($\sqrt(x) $), sine (sin x) and arcsine (arcsin x), etc. The process of finding the derivative of a given function is called differentiation, and the inverse operation, i.e. the process of finding a function from a given derivative, is integration.

The term “derivative” itself can be justified “in everyday life”: the function y = f(x) “produces” new feature y" = f"(x). The function y = f(x) acts as a “parent”, but mathematicians, naturally, do not call it a “parent” or “producer”; they say that it is, in relation to the function y" = f"(x) , primary image, or primitive.

Definition. The function y = F(x) is called antiderivative for the function y = f(x) on the interval X if the equality F"(x) = f(x) holds for $x \in X$

In practice, the interval X is usually not specified, but is implied (as the natural domain of definition of the function).

Let's give examples.
1) The function y = x 2 is antiderivative for the function y = 2x, since for any x the equality (x 2)" = 2x is true
2) The function y = x 3 is antiderivative for the function y = 3x 2, since for any x the equality (x 3)" = 3x 2 is true
3) The function y = sin(x) is antiderivative for the function y = cos(x), since for any x the equality (sin(x))" = cos(x) is true

When finding antiderivatives, as well as derivatives, not only formulas are used, but also some rules. They are directly related to the corresponding rules for calculating derivatives.

We know that the derivative of a sum is equal to the sum of its derivatives. This rule generates the corresponding rule for finding antiderivatives.

Rule 1. The antiderivative of a sum is equal to the sum of the antiderivatives.

We know that the constant factor can be taken out of the sign of the derivative. This rule generates the corresponding rule for finding antiderivatives.

Rule 2. If F(x) is an antiderivative for f(x), then kF(x) is an antiderivative for kf(x).

Theorem 1. If y = F(x) is an antiderivative for the function y = f(x), then the antiderivative for the function y = f(kx + m) is the function $y=\frac(1)(k)F(kx+m) $

Theorem 2. If y = F(x) is an antiderivative for the function y = f(x) on the interval X, then the function y = f(x) has infinitely many antiderivatives, and they all have the form y = F(x) + C.

Integration methods

Variable replacement method (substitution method)

The method of integration by substitution involves introducing a new integration variable (that is, substitution). In this case, the given integral is reduced to a new integral, which is tabular or reducible to it. There are no general methods for selecting substitutions. The ability to correctly determine substitution is acquired through practice.
Let it be necessary to calculate the integral $\textstyle \int F(x)dx $. Let's make the substitution $x= \varphi(t) $ where $\varphi(t) $ is a function that has a continuous derivative.
Then $dx = \varphi " (t) \cdot dt $ and based on the invariance property of the integration formula for the indefinite integral, we obtain the integration formula by substitution:
$\int F(x) dx = \int F(\varphi(t)) \cdot \varphi " (t) dt $

Integration of expressions of the form $\textstyle \int \sin^n x \cos^m x dx $

If m is odd, m > 0, then it is more convenient to make the substitution sin x = t.
If n is odd, n > 0, then it is more convenient to make the substitution cos x = t.
If n and m are even, then it is more convenient to make the substitution tg x = t.

Integration by parts

Integration by parts - applying the following formula for integration:
$\textstyle \int u \cdot dv = u \cdot v - \int v \cdot du $
or:
$\textstyle \int u \cdot v" \cdot dx = u \cdot v - \int v \cdot u" \cdot dx $

Table of indefinite integrals (antiderivatives) of some functions

$$ \int 0 \cdot dx = C $$ $$ \int 1 \cdot dx = x+C $$ $$ \int x^n dx = \frac(x^(n+1))(n+1 ) +C \;\; (n \neq -1) $$ $$ \int \frac(1)(x) dx = \ln |x| +C $$ $$ \int e^x dx = e^x +C $$ $$ \int a^x dx = \frac(a^x)(\ln a) +C \;\; (a>0, \;\; a \neq 1) $$ $$ \int \cos x dx = \sin x +C $$ $$ \int \sin x dx = -\cos x +C $$ $ $ \int \frac(dx)(\cos^2 x) = \text(tg) x +C $$ $$ \int \frac(dx)(\sin^2 x) = -\text(ctg) x +C $$ $$ \int \frac(dx)(\sqrt(1-x^2)) = \text(arcsin) x +C $$ $$ \int \frac(dx)(1+x^2 ) = \text(arctg) x +C $$ $$ \int \text(ch) x dx = \text(sh) x +C $$ $$ \int \text(sh) x dx = \text(ch ) x +C $$

The process of solving integrals in the science called mathematics is called integration. Using integration we can find some physical quantities: area, volume, mass of bodies and much more.

Integrals can be indefinite or definite. Let's consider the form of the definite integral and try to understand its physical meaning. It is represented in this form: $$ \int ^a _b f(x) dx $$. Distinctive feature writing a definite integral of an indefinite integral is that there are limits of integration of a and b. Now we’ll find out why they are needed, and what a definite integral actually means. In a geometric sense, such an integral equal to area a figure bounded by the curve f(x), lines a and b, and the Ox axis.

From Fig. 1 it is clear that the definite integral is the same area that is shaded gray. Let's check this with a simple example. Let's find the area of the figure in the image below using integration, and then calculate it in the usual way of multiplying the length by the width.

From Fig. 2 it is clear that $ y=f(x)=3 $, $ a=1, b=2 $. Now we substitute them into the definition of the integral, we get that $$ S=\int _a ^b f(x) dx = \int _1 ^2 3 dx = $$ $$ =(3x) \Big|_1 ^2=(3 \ cdot 2)-(3 \cdot 1)=$$ $$=6-3=3 \text(units)^2 $$ Let's do the check in the usual way. In our case, length = 3, width of the figure = 1. $$ S = \text(length) \cdot \text(width) = 3 \cdot 1 = 3 \text(units)^2 $$ As you can see, everything fits perfectly .

The question arises: how to solve indefinite integrals and what is their meaning? Solving such integrals is finding antiderivative functions. This process is the opposite of finding the derivative. In order to find the antiderivative, you can use our help in solving problems in mathematics, or you need to independently memorize the properties of integrals and the table of integration of the simplest elementary functions. The finding looks like this: $$ \int f(x) dx = F(x) + C \text(where) F(x) $ is the antiderivative of $ f(x), C = const $.

To solve the integral, you need to integrate the function $ f(x) $ over a variable. If the function is tabular, then the answer is written in the appropriate form. If not, then the process comes down to obtaining a tabular function from the function $ f(x) $ through tricky mathematical transformations. For this there is various methods and properties that we will consider further.

So, now let’s create an algorithm for solving integrals for dummies?

Algorithm for calculating integrals

Let's find out the definite integral or not.
If undefined, then you need to find the antiderivative function $ F(x) $ of the integrand $ f(x) $ using mathematical transformations leading to a tabular form of the function $ f(x) $.
If defined, then you need to perform step 2, and then substitute the limits $ a $ and $ b $ into the antiderivative function $ F(x) $. You will find out what formula to use to do this in the article “Newton-Leibniz Formula”.

Examples of solutions

So, you have learned how to solve integrals for dummies, examples of solving integrals have been sorted out. We learned their physical and geometric meaning. The solution methods will be described in other articles.

Application

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We definitely guarantee this to you, but we will be able to calculate your integral only after you enter the correct expression into the domain of the integrand. Don’t be afraid to make mistakes, everything can be corrected in this matter! Usually, solving integrals comes down to using tabular functions from well-known textbooks or encyclopedias. Like any other indefinite integral, it will be calculated using the standard formula without any major criticisms. First-year students easily and naturally grasp the material they have studied on the spot, and for them finding an integral sometimes takes no more than two minutes. And if a student has learned the table of integrals, then he can generally determine the answers in his head. Expanding functions by variables relative to surfaces initially means the correct vector direction at some abscissa point. The unpredictable behavior of surface lines takes definite integrals as a basis in the response source of mathematical functions. The left edge of the ball does not touch the cylinder in which the circle is inscribed, if you look at the cut in a plane. The sum of small areas divided into hundreds of piecewise continuous functions is the online integral of a given function. The mechanical meaning of the integral lies in many applied problems, such as determining the volume of bodies and calculating the mass of a body. Triple and double integrals are involved in these calculations. We insist that the solution of integrals online is carried out only under the supervision of experienced teachers and through numerous checks. We are often asked about the performance of students who do not attend lectures, skip them for no reason, and how they manage to find the integral themselves. We answer that students are free people and are quite capable of studying externally, preparing for a test or exam in the comfort of their own home. In a matter of seconds, our service will help anyone calculate the integral of any given function over a variable. The result obtained should be checked by taking the derivative of the antiderivative function. In this case, the constant from the solution of the integral becomes zero. This rule obviously applies to everyone. As multidirectional operations are justified, the indefinite integral is often reduced to dividing the domain into small parts. However, some students and schoolchildren neglect this requirement. As always, online integrals can be solved in detail by our service website and there are no restrictions on the number of requests, everything is free and available to everyone. There are not many sites that provide step-by-step answers in a matter of seconds, and most importantly, with high accuracy and in a convenient form. In the last example on page five homework I came across one that indicates the need to calculate the integral step by step. But we must not forget about how it is possible to find the integral using a ready-made service, time-tested and tested on thousands of solved examples online. How such an integral determines the motion of the system is clearly and clearly demonstrated to us by the nature of the motion of the viscous fluid, which is described by this system of equations.