One of the fundamental sections of mathematical analysis is integral calculus. It covers the widest field of objects, where the first is an indefinite integral. It is worth positioning it as a key, which even in high school reveals an increasing number of prospects and opportunities that are described by higher mathematics.
Appearance
At first glance, the integral seems very modern, relevant, but in practice it turns out that it appeared back in 1800 BC. Egypt is officially considered the homeland, since earlier evidence of its existence has not reached us. He, due to lack of information, all this time was positioned simply 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 were found dating back to the 4th century BC. They described a method where an indefinite integral was used, the essence of which was to search for the volume or area of a curved figure (three-dimensional and two-dimensional planes, respectively). The principle of calculation was based on dividing the original figure into infinitesimal components, provided that their volume (area) is already known. Over time, the method grew, Archimedes used it to search for the area of the parabola. Similar calculations were carried out at the same time by scientists in ancient China, moreover, they were completely independent of their Greek counterparts in science.
Development
The next breakthrough in the 11th century AD was the work of the Arab universal scientist Abu Ali al-Basri, who pushed the boundaries of the already known one, deriving, on the basis of the integral, formulas for calculating the sums of series and sums of degrees from the first to the fourth, using the known to us method of mathematical induction.
The minds of modernity admire the way the ancient Egyptians created amazing architectural monuments without any special adaptations, with the possible exception of their own hands, but is not the power of mind of scientists of that time a no lesser miracle? Compared to modern times, their life seems almost primitive, but the solution of indefinite integrals was deduced everywhere and used in practice for further development.
The next step occurred in the XVI century, when the Italian mathematician Cavalieri developed the indivisible method, which was adopted by Pierre Fermat. It is these two personalities that laid the foundation for modern integral calculus, which is currently known. They linked the concepts of differentiation and integration, which were previously perceived as autonomous units. By and large, the mathematics of those times was fragmented, the particles of conclusions existed on their own, having a limited scope. The path to unification and the search for common ground was the only right one at that time, thanks to it, modern mathematical analysis got the opportunity to grow and develop.
Over time, everything changed, and the designation of the integral as well. By and large, scientists designated him who was much more than that, for example, Newton used a square icon in which he placed an integrable function or simply put it nearby.
This disagreement continued until the 17th century, when the scientist Gottfried Leibniz, a landmark for the whole theory of mathematical analysis, introduced a symbol so familiar to us. The elongated "S" is really based on this letter of the Latin alphabet, since it denotes the sum of the antiderivatives. The name is integral thanks to
Jacob Bernoulli after 15 years.
Formal definition
The indefinite integral directly depends on the definition of the antiderivative, so we will consider it first of all.
The antiderivative is a function inverse to the derivative; in practice, it is also called primitive. Otherwise: the antiderivative of the function d is such a function D whose derivative is equal to v <=> V '= v. The search for the antiderivative is the calculation of the indefinite integral, and this process itself is called integration.
Example:
The function s (y) = y 3 , and its antiderivative S (y) = (y 4/4).
The set of all the antiderivatives of the function under consideration is this indefinite integral, it is denoted as follows: ∫v (x) dx.
Due to the fact that V (x) is just some primitive of the original function, the expression holds: ∫v (x) dx = V (x) + C, where C is a constant. An arbitrary constant is understood to be any constant, since its derivative is equal to zero.
The properties
The properties that an indefinite integral possesses are based on the basic definition and properties of derivatives.
Consider the key points:
- the integral of the derivative of the antiderivative is the antiderivative itself plus an arbitrary constant C <=> ∫V '(x) dx = V (x) + C;
- the derivative of the integral of the function is the original function <=> (∫v (x) dx) '= v (x);
- the constant is taken out from under the integral sign <=> ∫kv (x) dx = k∫v (x) dx, where k is arbitrary;
- the integral, which 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 we can conclude that the indefinite integral is linear. Due to this, we have: ∫ (kv (y) dy + ∫ lw (y)) dy = k∫v (y) dy + l∫w (y) dy.
To fix, we consider examples of solving 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 indefinite integrals? Just find all the antiderivatives! But the principles of the search will be considered below.
Methods and Examples
In order to solve the integral, one can resort to the following methods:
- use the finished table;
- integrate in parts;
- integrate by replacing a variable;
- bringing under the differential sign.
Tables
The easiest and most enjoyable way. At the moment, mathematical analysis boasts quite extensive tables, which spell out the basic formulas of indefinite integrals. In other words, there are templates that are displayed before you and for you, it remains only to use them. Here is a list of the main table entries, which can be deduced from almost every example that has a solution:
- ∫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 number other than unity;
- ∫ (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 a tabular form and enjoy the victory. Example: ∫cos (5x -2) dx = 1 / 5∫cos (5x - 2) d (5x - 2) = 1/5 x sin (5x - 2) + C.
The solution shows that for the table example, the integrand lacks the factor of 5. We add it, while multiplying by 1/5, so that the general expression does not change.
Part Integration
Consider two functions - z (y) and x (y). They must be continuously differentiable throughout the entire domain. According to 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 equality, 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 used 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;
∫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 two previous ones, although it is more complicated. The method is as follows: let V (x) be the integral of a certain function v (x). In the event that the integral itself in the example comes across a complex one, it is likely to get confused and go the wrong way. To avoid this, the transition from the variable x to z is practiced, in which the general expression is visually simplified while maintaining the dependence of z on x.
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 substitution. And, of course, the inverse function z = y -1 (x) fully describes the dependence and interconnection of variables. An important note - the differential dx is necessarily replaced by the new differential dz, since replacing a variable in an indefinite integral implies replacing it everywhere, and not just 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 obtain 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
To solve, we rewrite the expression in the following form:
∫2 s e s ds = ∫ (2e) s ds.
We denote by a = 2e (this step is not a replacement for the argument, it is still s), we bring our seemingly complex integral to an 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.
Differential
By and large, this method of indefinite integrals is the twin brother of the principle of variable replacement, however, there are differences in the design process. Let's consider in more detail.
If ∫v (x) dx = V (x) + C and y = z (x), then ∫v (y) dy = V (y) + C.
In this case, one must not forget the 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 it is not equal to zero;
- xdx = 1 / 2d (x 2 + b);
- sinxdx = -d (cosx);
- cosxdx = d (sinx).
If we consider the general case when we calculate the indefinite integral, examples can be summed up under 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, which can be caused either by laziness or an urgent need, you can use online tips, or rather, use the calculator of indefinite integrals. Despite all the apparent complexity and controversy of the integrals, their solution is subject to a certain algorithm, which is built on the principle of "if not ... then ...".
Of course, such a calculator will not master particularly intricate examples, since there are cases in which you have to find a solution artificially, “forcibly” introducing certain elements in the process, because you cannot achieve the result in obvious ways. Despite the controversy of this statement, it is true, since mathematics, in principle, is an abstract science, and considers the need to expand the boundaries of possibilities as its paramount task. Indeed, it is extremely difficult to move upward and develop according to smooth theory, which is why it is not worth considering that the examples of solving the indefinite integrals that we gave are the top of the possibilities. But back to the technical side of the matter. At least to verify the calculations, you can use the services in which everything was written down to us. If there was a need for automatic calculation of a complex expression, then they can not be dispensed with, you will have to resort to more serious software. It is worth paying attention primarily 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 plane of application. Indeed, they cannot be used directly directly, but they are considered a necessary intermediate element in the process of deriving decisions used in practice. So, integration is inverse to differentiation, due to which it is actively involved 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, on everything that makes up the present and shapes the future. The indefinite integral, the examples of which we have considered above, is trivial only at first glance, since it is the basis for making more and more discoveries.