One of the fundamental sections of mathematical analysis is integral calculus. It covers the widest field of objects, where the first is the indefinite integral. It is worth positioning it as a key, which even in high school reveals an increasing number of perspectives and opportunities that higher mathematics describes.
Appearance
At first glance, the integral seems utterly modern, relevant, but in practice it turns out that it appeared as early as 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 dating back to the 4th century BC were found. They described a method where an indefinite integral was used, the essence of which was to find the volume or area of a curvilinear figure (three-dimensionaland two-dimensional planes, respectively). The calculation principle was based on dividing the original figure into infinitesimal components, provided that their volume (area) is already known. Over time, the method has grown, Archimedes used it to find the area of a parabola. Similar calculations were carried out at the same time by scientists in ancient China, and 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 scientist-"universal" Abu Ali al-Basri, who pushed the boundaries of what was already known, deriving formulas based on the integral for calculating the sums of rows and the sums of powers from the first to the fourth, applying for this the method of mathematical induction known to us.
The minds of modern times admire how the ancient Egyptians created amazing architectural monuments without any special devices, except perhaps their hands, but is not the power of the mind of scientists of that time no less a miracle? Compared to today, their life seems almost primitive, but the solution of indefinite integrals was derived everywhere and used in practice for further development.
The next step took place in the 16th century, when the Italian mathematician Cavalieri developed the method of indivisibles, which was picked up by Pierre Fermat. It was these two personalities that laid the foundation for the modern integral calculus, which is known at the moment. They connected the concepts of differentiation and integration, which were previouslytreated 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 of unification and search for common ground was the only true one at that time, thanks to which modern mathematical analysis got the opportunity to grow and develop.
Everything has changed over time, including the notation of the integral. By and large, scientists denoted it by all means, for example, Newton used a square icon in which he placed an integrable function or simply put it next to it.
This inconsistency continued until the 17th century, when the scientist Gottfried Leibniz, a landmark for the entire theory of mathematical analysis, introduced the symbol so familiar to us. The elongated "S" is indeed based on this letter of the Latin alphabet, as it denotes the sum of antiderivatives. The integral got its name thanks to Jacob Bernoulli 15 years later.
Formal definition
The indefinite integral directly depends on the definition of the antiderivative, so let's consider it first.
An antiderivative is a function that is the inverse of a derivative, in practice it is also called primitive. Otherwise: the antiderivative of a function d is 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:
Function s(y)=y3, and its antiderivative S(y)=(y4/4).
The set of all antiderivatives of the function under consideration is the indefinite integral, it is denoted as follows: ∫v(x)dx.
Due to the fact that V(x) is only some antiderivative of the original function, the expression takes place: ∫v(x)dx=V(x) + C, where C is a constant. An arbitrary constant is any constant, since its derivative is equal to zero.
Properties
The properties that the indefinite integral has are based on the main definition and the properties of derivatives.
Let's look at the key points:
- the integral from the derivative of the antiderivative is the antiderivative itself plus an arbitrary constant С ∫V'(x)dx=V(x) + C;
- the derivative of the function integral is the original function (∫v(x)dx)'=v(x);
- constant is taken out from under the integral sign ∫kv(x)dx=k∫v(x)dx, where k is arbitrary;
- the integral 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. Thanks to this, we have: ∫(kv(y)dy +∫ lw(y))dy=k∫v(y)dy + l∫w(y)dy.
To consolidate, 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:don't know how to solve indefinite integrals? Just find all the primitives! But the principles of the search will be considered below.
Methods and examples
To solve the integral, you can resort to the following methods:
- use the prepared table;
- integrate by parts;
- integrate by changing the variable;
- bringing under the differential sign.
Tables
The easiest and most enjoyable way. At the moment, mathematical analysis boasts quite extensive tables in which the basic formulas of indefinite integrals are written. In other words, there are templates that have been developed before you and for you, it remains only to use them. Here is a list of the main table positions to which you can derive almost every example that has a solution:
- ∫0dy=C, where C is a constant;
- ∫dy=y + C, where C is a constant;
- ∫y dy=(yn+1) / (n + 1) + C, where C is a constant and n - non-one number;
- ∫(1/y)dy=ln|y| + C, where C is a constant;
- ∫eydy=ey + C, where C is a constant;
- ∫kydy=(ky/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/cos2y=tgy + C, where C is a constant;
- ∫dy/sin2y=-ctgy + C, where C is a constant;
- ∫dy/(1 + y2)=arctgy + C, where C is a constant;
- ∫chydy=shy + C, where C -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.
According to the solution, it is clear that for the tabular example, the integrand lacks a factor of 5. We add it, multiplying it 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 over the entire domain of definition. According to one of the differentiation properties, we have: d(xz)=xdz + zdx. Integrating both parts of the equation, we get: ∫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 point is that some examples can be simplified, conditionally speaking, reduce ∫zdx to ∫xdz if the latter is close to tabular form. Also, this formula can be applied more than once, achieving optimal results.
How to solve indefinite integrals this way:
need to calculate ∫(s + 1)e2sds
∫(x + 1)e2sds={z=s+1, dz=ds, y=1/2e2s, dy=e2xds}=((s+1)e2s) / 2-1/2∫e2s dx=((s+1)e2s) / 2-e2s/4+ C;
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 substitution
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 some function v(x). In the event that the integral itself in the example comes across as complex, there is a high probability of getting confused and taking the wrong path of solution. 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.
Mathematically 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 relationship of variables. Important note - the differential dx is necessarily replaced by a new differential dz, since the replacement of a variable in the indefinite integral implies its replacement everywhere, and not just in the integrand.
Example:
need to find ∫(s + 1) / (s2 + 2s - 5)ds
Apply the substitution z=(s+1)/(s2+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)/(s2+2s-5)ds=∫(dz/2)/z=1/2ln|z|+C=1/2ln|s2+2s-5|+C;
need to find the integral∫2sesdx
To solve, we rewrite the expression in the following form:
∫2sesds=∫(2e)sds.
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)sds=∫asds=as / lna + C=(2e)s / ln(2e) + C=2ses / ln(2 + lne) + C=2ses / (ln2 + 1) + C.
Bringing under the differential sign
By and large, this method of indefinite integrals is a twin brother of the variable change principle, but there are differences in the design process. Let's take a closer look.
If ∫v(x)dx=V(x) + C and y=z(x), then ∫v(y)dy=V(y) + C.
In this case, one should 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 not equal to zero;
- xdx=1/2d(x2 + 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:
need to find ∫(2s + 3)2ds, ds=1/2d(2s + 3)
∫(2s + 3)2ds=1/2∫(2s + 3)2d(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 fault of which may be either laziness or urgent need, you can use online tips, or rather, use the indefinite integral calculator. Despite all the apparent complexity and disputability of integrals, their solution is subject to a certain algorithm, which is built on the principle "if not …, then …".
Of course, such a calculator will not master especially intricate examples, since there are cases in which the solution has to be found artificially, "forcibly" introducing certain elements in the process, because the result cannot be achieved in obvious ways. Despite all 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 primary task. Indeed, it is extremely difficult to move up and develop according to smooth, run-in theories, so you should not assume that the examples of solving indefinite integrals that we have given are the height of possibilities. But back to the technical side of things. 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, then they cannot 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 out of touch with reality, as it is difficult to see the obvious areas of application. Indeed, they cannot be used directly anywhere, but they are considered a necessary intermediate element in the process of deriving solutions used in practice. So, integration is inverse to differentiation, 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, examples of which we examined above, is trivial only at first glance, since it is the basis for making more and more new discoveries.
