The indefinite integral. Calculation of indefinite integrals

One of the fundamental branches of mathematical analysis is the integral calculus. It covers the widest field of objects, where the first is an indefinite integral. To position it is like a key, that even in high school reveals an increasing number of prospects and opportunities, which is described by higher mathematics.

Appearance

At first glance, the integral seems utterly modern, relevant, but in practice it turns out that it appeared in 1800 BC. Homeland officially considered to be Egypt, since we did not get earlier evidence of its existence. He, because of the lack of information, all this time positioned just as a phenomenon. He once again confirmed the level of development of science among the peoples of those times. Finally, the works of ancient Greek mathematicians dating back to the 4th century BC were found. They described a method where an indefinite integral was applied, the essence of which was to find the volume or area of a curvilinear figure (three-dimensional and two-dimensional planes, respectively). The principle of calculation was based on dividing the original figure into infinitesimal components, provided that the volume (area) of them is already known. Over time, the method grew, Archimedes used it to find the area of the parabola. Analogous calculations at the same time were carried out by scientists in ancient China, moreover they were completely independent of the Greek brothers in science.

Development

The next breakthrough in the 11th century AD was the work of the Arabian "universal" Abu Ali al-Basri, who extended the boundaries of what was already known, by deducing on the basis of the integral the formulas for calculating the sums of series and sums of powers from the first to the fourth, Method of mathematical induction.
Modern minds admire how the ancient Egyptians created amazing architectural monuments, without any special adaptations, except perhaps their own hands, but is not the force of the minds of scientists of that time no less miracle? In comparison with the present times, their life seems almost primitive, but the solution of indefinite integrals was derived everywhere and was used in practice for further development.

The next step occurred in the 16th century, when the Italian mathematician Cavalieri deduced the method of the indivisible, which was picked up by Pierre Fermat. It is these two individuals that laid the foundation for the modern integral calculus that is known at the moment. They linked the concepts of differentiation and integration, which previously were perceived as autonomous units. By and large, mathematics of those times was fragmented, the particles of conclusions existed by themselves, having a limited field of application. The path of unification and the search for common ground was the only correct one at that time, thanks to him modern mathematical analysis was able to grow and develop.

With the passage of time, everything changed, and the designation of the integral as well. By and large, it was denoted by scientists who, what's more, for example, Newton used a square icon in which he put an integrable function or simply put it next to it. This disagreement continued until the 17th century, when the iconic scientist Gottfried Leibniz introduced the symbol so familiar to us for the whole theory of mathematical analysis. The stretched "S" is really based on this letter of the Latin alphabet, since it denotes the sum of the antipodes. The name was given to the integral by Jacob Bernoulli after 15 years.

Formal definition

The indefinite integral depends directly on the definition of the antiderivative, so consider it first.

The primitive is a function inverse to the derivative, in practice it is also called primitive. Otherwise: the antiderivative of the function d is a function D whose derivative is v <=> V '= v. The search for the antiderivative is the calculation of an indefinite integral, and the process itself is called integration.

Example:

The function s (y) = y ³ , and its antiderivative S (y) = (y 4/4).

The set of all antiderivatives of the function under consideration is an indefinite integral; it is denoted as follows: ∫v (x) dx.

Since V (x) is just some primitive of the original function, we have the expression: ∫v (x) dx = V (x) + C, where C is a constant. An arbitrary constant is understood as any constant, since its derivative is zero.

Properties

The properties that an indefinite integral possesses are based on the basic definition and properties of the derivatives.
Let's consider the key points:

• The integral of the antiderivative is itself an antiderivative plus an arbitrary constant C <=> ∫V '(x) dx = V (x) + C;
• The derivative of the integral of the function is the initial function <=> (∫v (x) dx) '= v (x);
• The constant is taken out from the sign of the integral <=> ∫kv (x) dx = k∫v (x) dx, where k is arbitrary;
• The integral that is taken from the sum is identically equal to the sum of the integrals <=> ∫ (v (y) + w (y)) dy = ∫v (y) dy + ∫w (y) dy.

From the last two properties it can be concluded that the indefinite integral is linear. Because of this, we have: ∫ (kv (y) dy + ∫ lw (y)) dy = k∫v (y) dy + l∫w (y) dy.

For fixing, we consider examples of solutions of indefinite integrals.

It is necessary to find the integral ∫ (3sinx + 4cosx) dx:

• ∫ (3sinx + 4cosx) dx = ∫3sinxdx + ∫4cosxdx = 3∫sinxdx + 4∫cosxdx = 3 (-cosx) + 4sinx + C = 4sinx - 3cosx + C.

From the example we can conclude: do not know how to solve the indefinite integrals? Just find all the antitypical! And here are the principles of search below.

Methods and examples

In order to solve the integral, we can resort to the following methods:

• Use the finished table;
• Integrate by parts;
• Integrate by changing a variable;
• Summation under the sign of the differential.

Tables

The easiest and most enjoyable way. At the moment, mathematical analysis can boast of fairly extensive tables, in which the basic formulas of indeterminate integrals are prescribed. In other words, there are templates, derived before you and for you, it remains only to use them. Here is a list of the main table positions to which almost every example having a solution can be derived:

• ∫0dy = C, where C is a constant;
• ∫dy = y + C, where C is a constant;
• ∫y ⁿ dy = (y ^{n + 1} ) / (n + 1) + C, where C is a constant, and n is a nonzero number;
• ∫ (1 / y) dy = ln | y | + C, where C is a constant;
• ∫e ^y dy = e ^y + C, where C is a constant;
• ∫k ^y dy = (k ^y / ln k) + C, where C is a constant;
• ∫cosydy = siny + C, where C is a constant;
• ∫sinydy = -cosy + C, where C is a constant;
• ∫dy / cos ² y = tgy + C, where C is a constant;
• ∫dy / sin ² y = -ctgy + C, where C is a constant;
• ∫dy / (1 + y ² ) = arctgy + C, where C is a constant;
• ∫chydy = shy + C, where C is a constant;
• ∫shydy = chy + C, where C is a constant.

If necessary, take a couple of steps, bring the integrand to the table view and enjoy the victory. Example: ∫cos (5x -2) dx = 1 / 5∫cos (5x-2) d (5x-2) = 1/5 x sin (5x-2) + C.

By decision it is clear that for the table example the integrand does not have a multiplier of 5. We add it, multiplying by 1/5 in parallel, so that the general expression does not change.

Integration by parts

Consider two functions - z (y) and x (y). They must be continuously differentiable on the whole domain of definition. By one of the differentiation properties we have: d (xz) = xdz + zdx. Integrating both sides of the equality, we obtain: ∫d (xz) = ∫ (xdz + zdx) => zx = ∫zdx + ∫xdz.

Rewriting the resulting equation, we obtain a formula that describes the method of integration by parts: ∫zdx = zx - ∫xdz.

Why is it needed? The fact is that some examples have the opportunity to simplify, relatively speaking, reduce ∫zdx to ∫xdz, if the latter is close to the tabular form. Also, this formula can be applied more than once, achieving the optimal result.

How to solve indefinite integrals in this way:

• It is necessary to calculate ∫ (s + 1) e ^2s ds

∫ (x + 1) e ^2s ds = {z = s + 1, dz = ds, y = 1 / 2e ^2s , dy = e ^2x ds} = ((s + 1) e ^2s ) / 2-1 / 2 ∫e ^2s dx = ((s + 1) e ^2s ) / 2-e ^2s / 4 + C;

• You need to calculate ∫lnsds

∫lnsds = {z = lns, dz = ds / s, y = s, dy = ds} = slns - ∫s x ds / s = slns - ∫ds = slns -s + C = s (lns-1) + C.

Variable replacement

This principle of solving indefinite integrals is no less in demand than the previous two, although it is more complicated. The method consists in the following: let V (x) be the integral of some function v (x). In the event that the integral itself in the example is complex, there is a great chance of getting confused and going the wrong way. To avoid this, the transition from the variable x to z is practiced, in which the general expression is visually simplified when the dependence of z on x is maintained.

In mathematical language, it looks like this: ∫v (x) dx = ∫v (y (z)) y '(z) dz = V (z) = V (y ^-1 (x)), where x = y ( Z) is a permutation. And, of course, the inverse function z = y ^-1 (x) fully describes the dependence and interrelation of variables. An important observation is that the differential dx is necessarily replaced by the new differential dz, since the replacement of a variable in an indefinite integral implies the replacement of it everywhere, and not only in the integrand.

Example:

• It is necessary to find ∫ (s + 1) / (s ² + 2s - 5) ds

We apply the substitution z = (s + 1) / (s ² + 2s-5). Then dz = 2sds = 2 + 2 (s + 1) ds <=> (s + 1) ds = dz / 2. As a result, we get the following expression, which is very easy to calculate:

∫ (s + 1) / (s ² + 2s-5) ds = ∫ (dz / 2) / z = 1 / 2ln | z | + C = 1 / 2ln | s ² + 2s-5 | + C;

• It is necessary to find the integral ∫2 ^s e ^s dx

For the solution, we rewrite the expression in the following form:

∫2 ^s e ^s ds = ∫ (2e) ^s ds.

We denote by a = 2e (by replacing the argument this step is not, it is still s), we give our, at first glance, complex integral, to the elementary tabular form:

∫ (2e) ^s ds = ∫a ^s ds = a ^s / lna + C = (2e) ^s / ln (2e) + C = 2 ^s e ^s / ln (2 + lne) + C = 2 ^s e ^s / (Ln2 + 1) + C.

Drawing under the sign of the differential

By and large, this method of indefinite integrals is a twin brother of the variable replacement principle, but there are differences in the design process. Let's consider more in detail.

If ∫v (x) dx = V (x) + C and y = z (x), then ∫v (y) dy = V (y) + C.

At the same time, one should not forget trivial integral transformations, among which:

• Dx = d (x + a), where a is any constant;
• Dx = (1 / a) d (ax + b), where a is again a constant, but not equal to zero;
• Xdx = 1 / 2d (x2 + b);
• Sinxdx = -d (cosx);
• Cosxdx = d (sinx).

If we consider the general case when we compute an indefinite integral, the examples can be reduced to the general formula w '(x) dx = dw (x).

Examples:

• It is necessary to find ∫ (2s + 3) ² ds, ds = 1 / 2d (2s + 3)

∫ (2s + 3) ² ds = 1 / 2∫ (2s + 3) ² d (2s + 3) = (1/2) x ((2s + 3) ² ) / 3 + C = (1/6) X (2s + 3) ² + C;

∫tgsds = ∫sins / cossds = ∫d (coss) / coss = -ln | coss | + C.

Online help

In some cases, the guilt that can be either laziness or an urgent need, you can use online tips, or rather, use the calculator of uncertain integrals. Despite all the apparent complexity and controversy of integrals, their solution is subject to a certain algorithm, which is built on the principle "if not ..., then ...".

Of course, especially intricate examples such a calculator can not be mastered, since there are cases when the solution has to be found artificially, "forcibly" introducing certain elements in the process, because obvious ways of the result can not be achieved. Despite all the controversy of this statement, it is true, since mathematics, in principle, is abstract science, and considers its primary task to expand the boundaries of possibilities. Indeed, it is extremely difficult to move up and develop on smoothly run-in theories, so do not assume that examples of solving the indefinite integrals that we gave are the top of the possibilities. However, let us return to the technical side of the matter. At least to check the calculations you can use the services in which everything was written before us. If there is a need for automatic calculation of a complex expression, they can not be dispensed with, you will have to resort to more serious software. It is worth paying attention first of all to the MatLab environment.

Application

The solution of indefinite integrals at first glance seems completely divorced from reality, since it is difficult to see the obvious application planes. Indeed, they can not be directly used anywhere, but they are considered to be an indispensable intermediate element in the process of drawing decisions that are used in practice. Thus, integration is inversely differentiated, due to which it actively participates in the process of solving equations.
In turn, these equations have a direct impact on the solution of mechanical problems, the calculation of trajectories and thermal conductivity - in short, everything that makes up the present and shapes the future. The indefinite integral, the examples of which we considered above, is trivial only at first glance, as it is the basis for making more and more new discoveries.