Presentation on the topic of the indefinite integral. Presentation for the lesson "Indefinite integral. Calculation methods". The largest and smallest values of the function
Anoshina O.V.
workshop for bachelors [Certificate of the Ministry of Education of the Russian Federation] / V. S.
Shipachev; ed. A. N. Tikhonova. - 8th ed., revised. and additional Moscow: Yurayt, 2015. - 447 p.
2. V. S. Shipachev, Higher Mathematics. Full course: textbook
for acad. Bachelor's degree [Certificate of UMO] / V. S. Shipachev; ed. BUT.
N. Tikhonova. - 4th ed., Rev. and additional - Moscow: Yurayt, 2015. - 608
with
3. Danko P.E., Popov A.G., Kozhevnikova T..Ya. higher mathematics
in exercises and tasks. [Text] / P.E. Danko, A.G. Popov, T.Ya.
Kozhevnikov. At 2 o'clock - M .: Higher School, 2007. - 304 + 415c.
Test. Performed in accordance with:
Tasks and guidelines for the performance of examinations
in the discipline "APPLIED MATHEMATICS", Yekaterinburg, FGAOU
VO "Russian State Vocational Pedagogical
University", 2016 - 30s.
Choose the option of control work by the last digit of the number
record book.
2.
Exam
antiderivative function f x defined on
some interval if F x f x for
each x from this interval.
For example, the cos x function is
antiderivative function sin x , since
cos x sin x . Obviously, if F x is an antiderivative
functions f x , then F x C , where C is some constant, is also
antiderivative function f x .
If F x is some antiderivative
function f x , then any function of the form
F x F x C is also
antiderivative function f x and any
primitive can be represented in this form. Definition. The totality of all
antiderivatives of the function f x ,
defined on some
in between is called
indefinite integral of
functions f x on this interval and
denoted by f x dx . If F x is some antiderivative of the function
f x , then they write f x dx F x C , although
it would be more correct to write f x dx F x C .
We, according to the established tradition, will write
f x dx F x C .
Thus the same symbol
f x dx will denote as the whole
set of antiderivatives of the function f x ,
and any element of this set.
integrand, and its differential to the integrand. Really:
1.(f (x)dx) (F (x) C) F (x) f (x);
2.d f (x)dx (f (x)dx) dx f (x)dx.
differential continuously (x)
differentiable function is equal to itself
this function up to a constant:
d (x) (x) dx (x) C,
since (x) is an antiderivative of (x).
antiderivatives, then the function f1 x f 2 x
also has an antiderivative, and
f1 x f 2 x dx f1 x dx f 2 x dx ;
5. Kf x dx Kf x dx ;
6. f x dx f x C ;
7. f x x dx F x C .
a 1
x
2. x a dx
C, (a 1) .
a 1
dx
3. ln x C .
x
x
a
4.a x dx
C.
ln a
5. e x dx e x C .
6. sin xdx cos x C .
7. cos xdx sin x C .
dx
8.2 ctgx C .
sin x
dx
9. 2tgx C .
cos x
dx
arctgx C .
10.
2
1 x
dx
arcsin x C .
1x2
dx
1
x
12. 2 2 arctan C .
a
a
a x
13.
14.
15.
dx
a2x2
x
arcsin C ..
a
dx
1
x a
ln
C
2
2
2a x a
x a
dx
1
a x
a 2 x 2 2a log a x C .
dx
16.
x2 a
log x x 2 a C .
17. shxdx chx C .
18. chxdx shx C .
19.
20.
dx
ch 2 x thx C .
dx
cthx C .
2
sh x
properties: 1
1. dx d (ax)
a
1
2. dx d (ax b),
a
1 2
3.xdxdx,
2
1 3
2
4. x dx dx .
3
Decision. In the table of integrals we find
cos xdx sin x C .
Let us transform this integral to a tabular one,
taking advantage of the fact that d ax adx .
Then:
d5 x 1
= cos 5 xd 5 x =
cos 5xdx cos 5x
5
5
1
= sin 5 x C .
5
3x x 1 dx .
Decision. Since under the integral sign
is the sum of four terms, then
expand the integral as a sum of four
integrals:
2
3
2
3
2
3
x
3
x
x
1
dx
x
dx
3
x
dx xdx dx .
x3
x4 x2
3
x C
3
4
2
use the following properties
integrals:
If f x dx F x C , then
f x b dx F x b C .
If f x dx F x C , then
1
f ax b dx F ax b C .
a
1
6
2
3
x
dx
2
3
x
C
.
3 6
5
The following integrals are taken by the method of integration by parts:
a) x n sin xdx, where n 1.2...k;
b) x n e x dx , where n 1,2...k ;
c) x n arctgxdx , where n 0, 1, 2,... k . ;
d) x n ln xdx , where n 0, 1, 2,... k .
When calculating the integrals a) and b) enter
n 1
notation: x n u , then du nx dx , and, for example
sin xdx dv , then v cos x .
When calculating the integrals c), d) denote for u the function
arctgx , ln x , and for dv they take x n dx .
Decision.
u x, du dx
=
x cos xdx
dv cos xdx, v sin x
x sin x sin xdx x sin x cos x C .
x ln xdx
dx
u ln x, du
x
x2
dv xdx, v
2
x2
x 2 dx
ln x
=
2
2 x
x2
1
x2
1x2
ln x xdx
ln x
C.
=
2
2
2
2 2
directly pick up the primitive
for f x we cannot, but we know that
she exists. Often found
antiderivative by introducing a new variable,
according to the formula
f x dx f t t dt , where x t and t is the new
variable
axb
dx ,
x px q
containing a square trinomial in
the denominator of the integrand
expressions. Such an integral is also taken
change of variables method,
previously identified in
the denominator is a full square.
2
dx
.
x4x5
Decision. Let's transform x 2 4 x 5 ,
2
selecting a full square according to the formula a b 2 a 2 2ab b 2 .
Then we get:
x2 4x5 x2 2x2 4 4 5
x 2 2 2 x 4 1 x 2 2 1
x 2 t
dx
dx
dt
x t 2
2
2
2
x 2 1 dx dt
x4x5
t1
arctgt C arctg x 2 C.
1 x
1 x
2
dx
tdt
1 t
2
x t, x t 2 ,
dx2tdt
2
t2
1 t
2
dt
1 t
1 t
d (t 2 1)
t
2
1
2
2tdt
2
dt
log(t 1) 2 dt 2
2
1 t
ln(t 2 1) 2t 2arctgt C
2
ln(x 1) 2 x 2arctg x C.
1 t 2 1
1 t
2
dt
the problem of finding the area of a curvilinear
trapezoid.
Let on some interval be given
continuous function y f (x) 0
Task:
Plot its graph and find F area of the figure,
bounded by this curve, two straight lines x = a and x
= b, and from below - a segment of the abscissa axis between the points
x = a and x = b. The figure aABb is called
curvilinear trapezoid
f(x)dx
Under a definite integral
a
from a given continuous function f(x) on
this segment is understood
the corresponding increment
primitive, that is
F (b) F (a) F (x) /
b
a
The numbers a and b are the limits of integration,
is the interval of integration.
values of the antiderivative integrand
functions for upper and lower limits
integration.
Introducing the notation for the difference
b
F (b) F (a) F (x) / a
b
f (x)dx F (b) F (a)
a
Newton-Leibniz formula.
integration variable notation, i.e.
b
b
a
a
f (x)dx f (t)dt
where x and t are any letters.
2) A definite integral with the same
outside
integration is zero
a
f (x)dx F (a) F (a) 0
a 3) When rearranging the limits of integration
the definite integral reverses its sign
b
a
f (x)dx F (b) F (a) F (a) F (b) f (x)dx
a
b
(additivity property)
4) If the interval is divided into a finite number
partial intervals, then the definite integral,
taken over the interval is equal to the sum of the defined
integrals taken over all its partial intervals.
b
c
b
f(x)dx f(x)dx
c
a
a
f(x)dx 5) A constant multiplier can be taken out
for the sign of a definite integral.
6) A definite integral of the algebraic
sums of a finite number of continuous
functions is equal to the same algebraic
the sum of definite integrals of these
functions.
integral.
b
f (x)dx f (t) (t)dt
a
a(), b(), (t)
where
for t[; ] , the functions (t) and (t) are continuous on;
5
Example:
1
=
x 1dx
=
x 1 5
t04
x 1 t
dt dx
4
0
3
2
t dt t 2
3
4
0
2
2
16
1
t t 40 4 2 0
5
3
3
3
3
Definition. Let the function f(x) be defined on
infinite interval , where b< + . Если
exist
b
lim
f(x)dx,
b
a
then this limit is called improper
integral of the function f(x) on the interval
}
Main literature
1. V. S. Shipachev, Higher Mathematics. Basic course: textbook andworkshop for bachelors [Certificate of the Ministry of Education of the Russian Federation] / V. S.
Shipachev; ed. A. N. Tikhonova. - 8th ed., revised. and additional Moscow: Yurayt, 2015. - 447 p.
2. V. S. Shipachev, Higher Mathematics. Full course: textbook
for acad. Bachelor's degree [Certificate of UMO] / V. S. Shipachev; ed. BUT.
N. Tikhonova. - 4th ed., Rev. and additional - Moscow: Yurayt, 2015. - 608
with
3. Danko P.E., Popov A.G., Kozhevnikova T..Ya. higher mathematics
in exercises and tasks. [Text] / P.E. Danko, A.G. Popov, T.Ya.
Kozhevnikov. At 2 o'clock - M .: Higher School, 2007. - 304 + 415c.
Reporting
1.Test. Performed in accordance with:
Tasks and guidelines for the performance of examinations
in the discipline "APPLIED MATHEMATICS", Yekaterinburg, FGAOU
VO "Russian State Vocational Pedagogical
University", 2016 - 30s.
Choose the option of control work by the last digit of the number
record book.
2.
Exam
Indefinite integral, its properties and calculation Antiderivative and indefinite integral
Definition. The function F x is calledantiderivative function f x defined on
some interval if F x f x for
each x from this interval.
For example, the cos x function is
antiderivative function sin x , since
cos x sin x . Obviously, if F x is an antiderivative
functions f x , then F x C , where C is some constant, is also
antiderivative function f x .
If F x is some antiderivative
function f x , then any function of the form
F x F x C is also
antiderivative function f x and any
primitive can be represented in this form. Definition. The totality of all
antiderivatives of the function f x ,
defined on some
in between is called
indefinite integral of
functions f x on this interval and
denoted by f x dx . If F x is some antiderivative of the function
f x , then they write f x dx F x C , although
it would be more correct to write f x dx F x C .
We, according to the established tradition, will write
f x dx F x C .
Thus the same symbol
f x dx will denote as the whole
set of antiderivatives of the function f x ,
and any element of this set.
Integral Properties
The derivative of the indefinite integral isintegrand, and its differential to the integrand. Really:
1.(f (x)dx) (F (x) C) F (x) f (x);
2.d f (x)dx (f (x)dx) dx f (x)dx.
Integral Properties
3. Indefinite integral ofdifferential continuously (x)
differentiable function is equal to itself
this function up to a constant:
d (x) (x) dx (x) C,
since (x) is an antiderivative of (x).
Integral Properties
4. If the functions f1 x and f 2 x haveantiderivatives, then the function f1 x f 2 x
also has an antiderivative, and
f1 x f 2 x dx f1 x dx f 2 x dx ;
5. Kf x dx Kf x dx ;
6. f x dx f x C ;
7. f x x dx F x C .
1. dx x C .
a 1
x
2. x a dx
C, (a 1) .
a 1
dx
3. ln x C .
x
x
a
4.a x dx
C.
ln a
5. e x dx e x C .
6. sin xdx cos x C .
7. cos xdx sin x C .
dx
8.2 ctgx C .
sin x
dx
9. 2tgx C .
cos x
dx
arctgx C .
10.
2
1 x
Table of indefinite integrals
11.dx
arcsin x C .
1x2
dx
1
x
12. 2 2 arctan C .
a
a
a x
13.
14.
15.
dx
a2x2
x
arcsin C ..
a
dx
1
x a
ln
C
2
2
2a x a
x a
dx
1
a x
a 2 x 2 2a log a x C .
dx
16.
x2 a
log x x 2 a C .
17. shxdx chx C .
18. chxdx shx C .
19.
20.
dx
ch 2 x thx C .
dx
cthx C .
2
sh x
Properties of differentials
When integrating, it is convenient to useproperties: 1
1. dx d (ax)
a
1
2. dx d (ax b),
a
1 2
3.xdxdx,
2
1 3
2
4. x dx dx .
3
Examples
Example. Calculate cos 5xdx .Decision. In the table of integrals we find
cos xdx sin x C .
Let us transform this integral to a tabular one,
taking advantage of the fact that d ax adx .
Then:
d5 x 1
= cos 5 xd 5 x =
cos 5xdx cos 5x
5
5
1
= sin 5 x C .
5
Examples
Example. Calculate x3x x 1 dx .
Decision. Since under the integral sign
is the sum of four terms, then
expand the integral as a sum of four
integrals:
2
3
2
3
2
3
x
3
x
x
1
dx
x
dx
3
x
dx xdx dx .
x3
x4 x2
3
x C
3
4
2
Independence of the type of variable
When calculating integrals, it is convenientuse the following properties
integrals:
If f x dx F x C , then
f x b dx F x b C .
If f x dx F x C , then
1
f ax b dx F ax b C .
a
Example
Compute1
6
2
3
x
dx
2
3
x
C
.
3 6
5
Integration methods Integration by parts
This method is based on the formula udv uv vdu .The following integrals are taken by the method of integration by parts:
a) x n sin xdx, where n 1.2...k;
b) x n e x dx , where n 1,2...k ;
c) x n arctgxdx , where n 0, 1, 2,... k . ;
d) x n ln xdx , where n 0, 1, 2,... k .
When calculating the integrals a) and b) enter
n 1
notation: x n u , then du nx dx , and, for example
sin xdx dv , then v cos x .
When calculating the integrals c), d) denote for u the function
arctgx , ln x , and for dv they take x n dx .
Examples
Example. Calculate x cos xdx .Decision.
u x, du dx
=
x cos xdx
dv cos xdx, v sin x
x sin x sin xdx x sin x cos x C .
Examples
Example. Calculatex ln xdx
dx
u ln x, du
x
x2
dv xdx, v
2
x2
x 2 dx
ln x
=
2
2 x
x2
1
x2
1x2
ln x xdx
ln x
C.
=
2
2
2
2 2
Variable replacement method
Let it be required to find f x dx , anddirectly pick up the primitive
for f x we cannot, but we know that
she exists. Often found
antiderivative by introducing a new variable,
according to the formula
f x dx f t t dt , where x t and t is the new
variable
Integration of functions containing a square trinomial
Consider the integralaxb
dx ,
x px q
containing a square trinomial in
the denominator of the integrand
expressions. Such an integral is also taken
change of variables method,
previously identified in
the denominator is a full square.
2
Example
Calculatedx
.
x4x5
Decision. Let's transform x 2 4 x 5 ,
2
selecting a full square according to the formula a b 2 a 2 2ab b 2 .
Then we get:
x2 4x5 x2 2x2 4 4 5
x 2 2 2 x 4 1 x 2 2 1
x 2 t
dx
dx
dt
x t 2
2
2
2
x 2 1 dx dt
x4x5
t1
arctgt C arctg x 2 C.
Example
To find1 x
1 x
2
dx
tdt
1 t
2
x t, x t 2 ,
dx2tdt
2
t2
1 t
2
dt
1 t
1 t
d (t 2 1)
t
2
1
2
2tdt
2
dt
log(t 1) 2 dt 2
2
1 t
ln(t 2 1) 2t 2arctgt C
2
ln(x 1) 2 x 2arctg x C.
1 t 2 1
1 t
2
dt
Definite integral, its main properties. Newton-Leibniz formula. Applications of a definite integral.
The concept of a definite integral leads tothe problem of finding the area of a curvilinear
trapezoid.
Let on some interval be given
continuous function y f (x) 0
Task:
Plot its graph and find F area of the figure,
bounded by this curve, two straight lines x = a and x
= b, and from below - a segment of the abscissa axis between the points
x = a and x = b. The figure aABb is called
curvilinear trapezoid
Definition
bf(x)dx
Under a definite integral
a
from a given continuous function f(x) on
this segment is understood
the corresponding increment
primitive, that is
F (b) F (a) F (x) /
b
a
The numbers a and b are the limits of integration,
is the interval of integration.
Rule:
The definite integral is equal to the differencevalues of the antiderivative integrand
functions for upper and lower limits
integration.
Introducing the notation for the difference
b
F (b) F (a) F (x) / a
b
f (x)dx F (b) F (a)
a
Newton-Leibniz formula.
Basic properties of a definite integral.
1) The value of a definite integral does not depend onintegration variable notation, i.e.
b
b
a
a
f (x)dx f (t)dt
where x and t are any letters.
2) A definite integral with the same
outside
integration is zero
a
f (x)dx F (a) F (a) 0
a 3) When rearranging the limits of integration
the definite integral reverses its sign
b
a
f (x)dx F (b) F (a) F (a) F (b) f (x)dx
a
b
(additivity property)
4) If the interval is divided into a finite number
partial intervals, then the definite integral,
taken over the interval is equal to the sum of the defined
integrals taken over all its partial intervals.
b
c
b
f(x)dx f(x)dx
c
a
a
f(x)dx 5) A constant multiplier can be taken out
for the sign of a definite integral.
6) A definite integral of the algebraic
sums of a finite number of continuous
functions is equal to the same algebraic
the sum of definite integrals of these
functions.
3. Change of variable in a definite integral.
3. Replacing a variable in a certainintegral.
b
f (x)dx f (t) (t)dt
a
a(), b(), (t)
where
for t[; ] , the functions (t) and (t) are continuous on;
5
Example:
1
=
x 1dx
=
x 1 5
t04
x 1 t
dt dx
4
0
3
2
t dt t 2
3
4
0
2
2
16
1
t t 40 4 2 0
5
3
3
3
3
Improper integrals.
Improper integrals.Definition. Let the function f(x) be defined on
infinite interval , where b< + . Если
exist
b
lim
f(x)dx,
b
a
then this limit is called improper
integral of the function f(x) on the interval
}