Formulas for integration by parts with examples. Complex integrals
A function F(x) differentiable in a given interval X is called antiderivative of the function f(x), or the integral of f(x), if for every x ∈X the following equality holds:
F " (x) = f(x). (8.1)
Finding all antiderivatives for a given function is called its integration. Indefinite integral function f(x) on a given interval X is the set of all antiderivative functions for the function f(x); designation -
If F(x) is some antiderivative of the function f(x), then ∫ f(x)dx = F(x) + C, (8.2)
where C is an arbitrary constant.
Table of integrals
Directly from the definition we obtain the main properties of the indefinite integral and a list of tabular integrals:
1) d∫f(x)dx=f(x)
2)∫df(x)=f(x)+C
3) ∫af(x)dx=a∫f(x)dx (a=const)
4) ∫(f(x)+g(x))dx = ∫f(x)dx+∫g(x)dx
List of tabular integrals
1. ∫x m dx = x m+1 /(m + 1) +C; (m ≠ -1)
3.∫a x dx = a x /ln a + C (a>0, a ≠1)
4.∫e x dx = e x + C
5.∫sin x dx = cosx + C
6.∫cos x dx = - sin x + C
7. = arctan x + C
8. = arcsin x + C
10. = - ctg x + C
Variable replacement
To integrate many functions, use the variable replacement method or substitutions, allowing you to reduce integrals to tabular form.
If the function f(z) is continuous on [α,β], the function z =g(x) has a continuous derivative and α ≤ g(x) ≤ β, then
∫ f(g(x)) g " (x) dx = ∫f(z)dz, (8.3)
Moreover, after integration on the right side, the substitution z=g(x) should be made.
To prove it, it is enough to write the original integral in the form:
∫ f(g(x)) g " (x) dx = ∫ f(g(x)) dg(x).
For example:
Method of integration by parts
Let u = f(x) and v = g(x) be functions that have continuous . Then, according to the work,
d(uv))= udv + vdu or udv = d(uv) - vdu.
For the expression d(uv), the antiderivative will obviously be uv, so the formula holds:
∫ udv = uv - ∫ vdu (8.4.)
This formula expresses the rule integration by parts. It leads the integration of the expression udv=uv"dx to the integration of the expression vdu=vu"dx.
Let, for example, you want to find ∫xcosx dx. Let us put u = x, dv = cosxdx, so du=dx, v=sinx. Then
∫xcosxdx = ∫x d(sin x) = x sin x - ∫sin x dx = x sin x + cosx + C.
The rule of integration by parts has a more limited scope than substitution of variables. But there are whole classes of integrals, for example,
∫x k ln m xdx, ∫x k sinbxdx, ∫ x k cosbxdx, ∫x k e ax and others, which are calculated precisely using integration by parts.
Definite integral
The concept of a definite integral is introduced as follows. Let a function f(x) be defined on an interval. Let us split the segment [a,b] into n parts by points a= x 0< x 1 <...< x n = b. Из каждого интервала (x i-1 ,
x i) возьмем произвольную точку ξ i и составим сумму f(ξ i)
Δx i где
Δ x i =x i - x i-1. A sum of the form f(ξ i)Δ x i is called integral sum, and its limit at λ = maxΔx i → 0, if it exists and is finite, is called definite integral functions f(x) of a before b and is denoted:
F(ξ i)Δx i (8.5).
The function f(x) in this case is called integrable on a segment, the numbers a and b are called lower and upper limits of the integral.
The following properties are true for a definite integral:
4), (k = const, k∈R);
5)
6)
7) f(ξ)(b-a) (ξ∈).
The last property is called mean value theorem.
Let f(x) be continuous on . Then on this segment there is an indefinite integral
∫f(x)dx = F(x) + C
and takes place Newton-Leibniz formula, connecting the definite integral with the indefinite integral:
F(b) - F(a). (8.6)
Geometric interpretation: the definite integral is the area of a curvilinear trapezoid bounded from above by the curve y=f(x), the straight lines x = a and x = b and the axis segment Ox.
Improper integrals
Integrals with infinite limits and integrals of discontinuous (unbounded) functions are called improper. Improper integrals of the first kind - These are integrals over an infinite interval, defined as follows:
(8.7)
If this limit exists and is finite, then it is called convergent improper integral of f(x) on the interval [a,+ ∞), and the function f(x) is called integrable over an infinite interval[a,+ ∞). Otherwise, the integral is said to be does not exist or diverges.
The improper integrals on the intervals (-∞,b] and (-∞, + ∞) are defined similarly:
Let us define the concept of an integral of an unbounded function. If f(x) is continuous for all values x segment , except for the point c, at which f(x) has an infinite discontinuity, then improper integral of the second kind of f(x) ranging from a to b the amount is called:
if these limits exist and are finite. Designation:
Examples of integral calculations
Example 3.30. Calculate ∫dx/(x+2).
Solution. Let us denote t = x+2, then dx = dt, ∫dx/(x+2) = ∫dt/t = ln|t| + C = ln|x+2| +C.
Example 3.31. Find ∫ tgxdx.
Solution.∫ tgxdx = ∫sinx/cosxdx = - ∫dcosx/cosx. Let t=cosx, then ∫ tgxdx = -∫ dt/t = - ln|t| + C = -ln|cosx|+C.
Example3.32 . Find ∫dx/sinxSolution.
Example3.33. Find .
Solution. = .
Example3.34 . Find ∫arctgxdx.
Solution. Let's integrate by parts. Let us denote u=arctgx, dv=dx. Then du = dx/(x 2 +1), v=x, whence ∫arctgxdx = xarctgx - ∫ xdx/(x 2 +1) = xarctgx + 1/2 ln(x 2 +1) +C; because
∫xdx/(x 2 +1) = 1/2 ∫d(x 2 +1)/(x 2 +1) = 1/2 ln(x 2 +1) +C.
Example3.35 . Calculate ∫lnxdx.
Solution. Applying the integration-by-parts formula, we get:
u=lnx, dv=dx, du=1/x dx, v=x. Then ∫lnxdx = xlnx - ∫x 1/x dx =
= xlnx - ∫dx + C= xlnx - x + C.
Example3.36 . Calculate ∫e x sinxdx.
Solution. Denote u = e x , dv = sinxdx, then du = e x dx, v =∫sinxdx= - cosx → ∫ e x sinxdx = - e x cosx + ∫ e x cosxdx. The integral ∫e x cosxdx is also integrable by parts: u = e x , dv = cosxdx, du=e x dx, v=sinx. We have:
∫ e x cosxdx = e x sinx - ∫ e x sinxdx. We got the relation ∫e x sinxdx = - e x cosx + e x sinx - ∫ e x sinxdx, whence 2∫e x sinx dx = - e x cosx + e x sinx + C.
Example 3.37. Calculate J = ∫cos(lnx)dx/x.
Solution. Since dx/x = dlnx, then J= ∫cos(lnx)d(lnx). Replacing lnx through t, we arrive at the table integral J = ∫ costdt = sint + C = sin(lnx) + C.
Example 3.38 . Calculate J = .
Solution. Taking into account that = d(lnx), we make the substitution lnx = t. Then J = 
.
Example 3.39
. Calculate the integral J = 
.
Solution. We have: 
. Therefore =
=
=. entered as sqrt(tan(x/2)).
And if you click on Show steps in the upper right corner in the result window, you will get a detailed solution.
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Integration by parts- a method used to solve definite and indefinite integrals, when one of the integrands is easily integrable and the other is differentiable. A fairly common method for finding integrals, both indefinite and definite. The main sign when you need to use it is some function consisting of a product of two functions that cannot be integrated point-blank.
Formula
In order to successfully use this method, it is necessary to disassemble and learn the formulas.
Formula for integration by parts in the indefinite integral:
$$ \int udv = uv - \int vdu $$
Formula for integration by parts in a definite integral:
$$ \int \limits_(a)^(b) udv = uv \bigg |_(a)^(b) - \int \limits_(a)^(b) vdu $$
Solution examples
Let's consider in practice examples of solutions of integration by parts, which are often offered by teachers in tests. Please note that under the integral symbol there is a product of two functions. This is a sign that this method is suitable for the solution.
| Example 1 |
| Find the integral $ \int xe^xdx $ |
| Solution |
|
We see that the integrand consists of two functions, one of which, upon differentiation, instantly turns into unity, and the other is easily integrated. To solve the integral, we use the method of integration by parts. Let us assume that $ u = x \rightarrow du=dx $ and $ dv = e^x dx \rightarrow v=e^x $ We substitute the found values into the first integration formula and get: $$ \int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C $$ If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner! |
| Answer |
|
$$ \int xe^x dx = xe^x - e^x + C $$ |
| Example 4 |
| Calculate the integral $ \int \limits_0 ^1 (x+5) 3^x dx $ |
| Solution |
|
By analogy with the previous solved examples, we will figure out which function to integrate without problems, which to differentiate. Please note that if you differentiate $ (x + 5) $, then this expression will be automatically converted to unity, which will be "on hand" for us. So we do this: $$ u=x+5 \rightarrow du=dx, dv=3^x dx \rightarrow v=\frac(3^x)(ln3) $$ Now all unknown functions have been found and can be put into the second formula for integration by parts for a definite integral. $$ \int \limits_0 ^1 (x+5) 3^x dx = (x+5) \frac(3^x)(\ln 3) \bigg |_0 ^1 - \int \limits_0 ^1 \frac (3^x dx)(\ln 3) = $$ $$ = \frac(18)(\ln 3) - \frac(5)(\ln 3) - \frac(3^x)(\ln^2 3)\bigg| _0 ^1 = \frac(13)(\ln 3) - \frac(3)(\ln^2 3)+\frac(1)(\ln^2 3) = \frac(13)(\ln 3 )-\frac(4)(\ln^2 3) $$ |
| Answer |
| $$ \int\limits_0 ^1 (x+5)3^x dx = \frac(13)(\ln 3)-\frac(4)(\ln^2 3) $$ |
By a definite integral from a continuous function f(x) on the final segment [ a, b] (where ) is the increment of some of its antiderivatives on this segment. (In general, understanding will be noticeably easier if you repeat the topic of the indefinite integral) In this case, the notation is used
As can be seen in the graphs below (the increment of the antiderivative function is indicated by ), a definite integral can be either a positive or a negative number(It is calculated as the difference between the value of the antiderivative in the upper limit and its value in the lower limit, i.e. as F(b) - F(a)).
Numbers a And b are called the lower and upper limits of integration, respectively, and the segment [ a, b] – segment of integration.
Thus, if F(x) – some antiderivative function for f(x), then, according to the definition,
(38)
Equality (38) is called Newton-Leibniz formula . Difference F(b) – F(a) is briefly written as follows:
Therefore, we will write the Newton-Leibniz formula like this:
(39)
Let us prove that the definite integral does not depend on which antiderivative of the integrand is taken when calculating it. Let F(x) and F( X) are arbitrary antiderivatives of the integrand. Since these are antiderivatives of the same function, they differ by a constant term: Ф( X) = F(x) + C. That's why
This establishes that on the segment [ a, b] increments of all antiderivatives of the function f(x) match up.
Thus, to calculate a definite integral, it is necessary to find any antiderivative of the integrand, i.e. First you need to find the indefinite integral. Constant WITH excluded from subsequent calculations. Then the Newton-Leibniz formula is applied: the value of the upper limit is substituted into the antiderivative function b , further - the value of the lower limit a and the difference is calculated F(b) - F(a) . The resulting number will be a definite integral..
At a = b by definition accepted
Example 1.
Solution. First, let's find the indefinite integral:
Applying the Newton-Leibniz formula to the antiderivative
(at WITH= 0), we get
However, when calculating a definite integral, it is better not to find the antiderivative separately, but to immediately write the integral in the form (39).
Example 2. Calculate definite integral
Solution. Using formula
Properties of the definite integral
Theorem 2.The value of the definite integral does not depend on the designation of the integration variable, i.e.
(40)
Let F(x) – antiderivative for f(x). For f(t) the antiderivative is the same function F(t), in which the independent variable is only designated differently. Hence,
Based on formula (39), the last equality means the equality of the integrals
Theorem 3.The constant factor can be taken out of the sign of the definite integral, i.e.
(41)
Theorem 4.The definite integral of an algebraic sum of a finite number of functions is equal to the algebraic sum of definite integrals of these functions, i.e.
(42)
Theorem 5.If a segment of integration is divided into parts, then the definite integral over the entire segment is equal to the sum of definite integrals over its parts, i.e. If
(43)
Theorem 6.When rearranging the limits of integration, the absolute value of the definite integral does not change, but only its sign changes, i.e.
(44)
Theorem 7(mean value theorem). A definite integral is equal to the product of the length of the integration segment and the value of the integrand at some point inside it, i.e.
(45)
Theorem 8.If the upper limit of integration is greater than the lower one and the integrand is non-negative (positive), then the definite integral is also non-negative (positive), i.e. If
Theorem 9.If the upper limit of integration is greater than the lower one and the functions and are continuous, then the inequality
can be integrated term by term, i.e.
(46)
The properties of the definite integral make it possible to simplify the direct calculation of integrals.
Example 5 Calculate definite integral
Using Theorems 4 and 3, and when finding antiderivatives - table integrals (7) and (6), we obtain
Definite integral with variable upper limit
Let f(x) – continuous on the segment [ a, b] function, and F(x) is its antiderivative. Consider the definite integral
(47)
and through t the integration variable is designated so as not to confuse it with the upper bound. When it changes X the definite integral (47) also changes, i.e. it is a function of the upper limit of integration X, which we denote by F(X), i.e.
(48)
Let us prove that the function F(X) is an antiderivative for f(x) = f(t). Indeed, differentiating F(X), we get
because F(x) – antiderivative for f(x), A F(a) is a constant value.
Function F(X) – one of the infinite number of antiderivatives for f(x), namely the one that x = a goes to zero. This statement is obtained if in equality (48) we put x = a and use Theorem 1 of the previous paragraph.
Calculation of definite integrals by the method of integration by parts and the method of change of variable
where, by definition, F(x) – antiderivative for f(x). If in the integrand we make the change of variable
then, in accordance with formula (16), we can write
In this expression
antiderivative function for
In fact, its derivative, according to the rule of differentiation of a complex function, is equal
Let α and β be the values of the variable t, for which the function
takes respectively the values a And b, i.e.
But, according to the Newton-Leibniz formula, the difference F(b) – F(a) There is
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We study the concept of "integral"
Integration was known in ancient Egypt. Of course, not in a modern form, but still. Since then, mathematicians have written a great many books on the subject. 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 a basic knowledge of the basics of mathematical analysis. Information about , which is also necessary for understanding integrals, is already in 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 a figure, the mass of a non-uniform body, the distance traveled during uneven movement, and much more. It should be remembered that an integral is the sum of an infinitely large number of infinitesimal terms.
As an example, imagine a graph of some function. How to find the area of a figure bounded by the graph of a function?
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.
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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 suggest you figure out the intricacies of the solution yourself, and if something is unclear, ask questions in the comments.
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