4.1 antiderivatives and indefinite integration. after this lesson, you should be able to: write the...
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4.1 Antiderivatives and Indefinite Integration
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After this lesson, you should be able to:
Write the general solution of a differential equation.Use indefinite integral notation for antiderivatives.Use basic integration rules to find antiderivatives.Find a particular solution of a differential equation.
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Definition of an AntiderivativeIn many cases, we would like to know what is the function F(x) whose derivative is a given function f(x). Or,
F’(x) = f(x)
For instance, if a given function f(x) = –sinx, then the derivative of F(x) = cosx + 4 is the given function f(x).
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Definition of an AntiderivativeNotice that even we could find the antiderivative of a given function, the answer is not unique! If f(x) = sinx + cosx
This is why F(x) is called an antiderivative of f(x), rather then the antiderivative of f(x). In fact, all antideritives of f(x) differs at a constant.
then F1(x) = sinx – cosx – 1
F2(x) = sinx – cosx + 4
F3(x) = sinx – cosx – 9
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Antiderivatives and Indefinite Integration
Differentiable Functions
ALL functions
62
2
x
x
Dderivative operator
x2
antiderivative operator/ indefinite integral operator
dxxf )( is read as “the antiderivative of f with respect to x” or “the indefinite integral of f with respect to x”.
This expression denotes (all) the antiderivatives of f(x).
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Theorem 4.1 Representation of Antiderivatives
This Theorem tells us that we can represent the entire family of antideritives of a given function by adding a constant to a known antiderivative.If f(x) = sinx + cosx
then F(x) = sinx – cosx + C
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Theorem 4.1 Representation of Antiderivatives
The constant C is called the constant of integration. The family of functions represented by G is the general antiderivative of f.If f(x) = sinx + cosx
then F(x) = sinx – cosx + C is the general solution of the differential equation F ’(x) = sinx + cosx
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Notation of Antiderivatives
dxxF )(' CxF )(
)()(' xfxFdx
dy
dxxfdxxFdy )()('
Let F(x) be an antiderivative of f(x), and y be all antiderivatives f(x), then
CxFy )(
We differentiate (*), then
(*)
dyy
The operation of finding all solutions of this equation is called antidifferentiation (or indefinite integration) denoted by the sign dxxf )(
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Notation of Antiderivatives
dxxf )( CxF )(
single function
bag of functions
Variable of IntegrationIntegrand
Constant of Integration
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xdx5
single function
bag of functions
dxxf )( CxF )( iff F’(x) = f(x)
Cx 2
2
5
Example 1 Find the antiderivatives of 5x
Indefinite Integration
Solution
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dx 3
Example
Cx 3
dxk
constant
Ckx
Example 2 Find the antiderivatives of 3
Solution
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dxx Cx
2
2
dxx2 Cx
3
3
dxxm Cm
xm
1
1
1m
Example 3 Find the antiderivatives of the following
Solution
dxx dxx2 and dxx5
m is not necessary to be an integer. mR
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dxxm Cm
xm
1
1
1m
dxx5 Cx
6
6
Example
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Basic Integration Rules
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dxx
Cx
23
23
dxx 21
Cx 23
3
2
dxx2
1
Cx
1
1
dxx 2 Cx 1
Example 4 Find the antiderivatives of the following
Solution
Example
and
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dxxx )(
Cx
2
2
dxxx )( 21
Cx 23
3
2
Example 5 Find the antiderivatives of the following
Solution
Example
dxxxdx 21
Cxx
23
2
22
32
1
2
2C
x 22
3
3
2Cx 21
22
3
3
2
2CCx
x
Cxx
23
3
2
2
2Also refer to middle of P. 251 for another example.
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Example 6 Find the antiderivatives of the following
Solut1ion
ExampleTip: Sometimes, simplifying or rewriting before integration
dx
x
xx4
2 32
dxx
dxx
dxx 432
321
dx
x
xx4
2 32
dxxdxxdxx 432 32
Cxxx
32
111
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(1) Please be aware that the indefinite integral
Note
dx
x
xx4
2 32
is NOT equal to
dxx
dxxx4
2 )32(
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(2)The indefinite integration with respect to a polynomial
Note
dxxx )32( 2
can NOT be written as
dxxx 322
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Practice
dxxx 386)1( 2
dx
x
xx3
3)2(
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Practice
dxxx 386)1( 2 3
6 3x
2
8 2x x3 C
32x 24x x3 CCheck:
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dx
x
xx3
3)2(
dxx
x
x
x33
21
3
dxxx 25
32
1
1
x
23
23
3
xC
x
1
23
3
32
x
C
x
1
xx
2 C
check:
Practice
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Initial Conditions and Particular SolutionsIn the beginning of this section, we have already known that all antideritives of f(x) differs at a constant. This means that the graphs of any two antiderivatives of f are VERTICAL translation of each other.
Cxxdxxyei 32 13..
C=2
C=1
C=0
C=–1
C=–2
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Initial Conditions and Particular Solutions
Cxxdxxyei 32 13..
C=2
C=1
C=0
C=–1
C=–2
Graph must pass through point (0, –2).
Cxxy 3
So, x = 0, y = –2 is the solution of the equation
Or,,002 3 C 2C
23 xxy 23 xxy
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(To find a particular solution, we need an initial condition)
Example 7 Find f(x) if f ’(x) = 6x – 4 and f(1) = 2.
differential equation
use this to get a bag of functions
containing f(x)
initial condition
use this to reach into bag and pull
out a particular f(x)
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dxxxf 46)(
Example 7 Find f(x) if f’ (x) = 6x – 4 and f(1) = 2.
Cxx
42
6 2
Cxx 43 2
Cxxxf 43)( 2
2)1(f C )1(4)1(3 2
C 43 23C
343)( 2 xxxf
check:
Solut1ion
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Example 8 Find f(x) if7)( and sin3cos2)(' fxxxf
differential equation
initial condition
dxxxxf )sin3cos2()( Cxx cos3sin2
Cxxxf cos3sin2)(
)(7 f C cos3sin2
C )1(30)2(7C37
4C4cos3sin2)( xxxf
Answer:
Example
Solut1ion
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dx
xd sin xcos dxx cos Cx sin
dx
xd cos xsin dxx sin Cx cos
dx
xd tan x2sec dxx sec2 Cx tan
USED MOST OFTEN
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dx
xd sec xx tansec dxxx tansec Cx sec
dx
xd csc xx cotcsc dxxx cotcsc Cx csc
dx
xd cot x2csc dxx csc2 Cx cot
USED MOST OFTEN
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dxxx sin54 2
Example 9
dxxx sin54 2
Cxx
)cos(53
4 3
Check:
Cxx
cos53
4 3
Example
Solut1ion
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Example
Example 10 Find the antiderivatives of the following xdxx 23 cossin
xdxx 23 cossin xxdx coscos)sin( 22
xxdx coscos)1(cos 22 xdxx cos)cos(cos 24
xxdxxd coscoscoscos 24
duuduu 24 Cuu
35
35
Cxx
3
cos
5
cos 35
check:
Solut1ion
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s(t) position function
v(t) velocity
a(t) acceleration
v(t) = s’ (t)a(t) = v’ (t) dttatv )()(
dttvts )()(
v(t) = s’ (t)
a(t) = v’ (t)
a(t) = s’’ (t)
Motion Along a Line
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NOTE Velocity is negative when falling (or positive when thrown up)
2sec32)( ftta
acceleration due to gravity
2sec8.9)( mta
Free Falling Objects
2sec8.9 mg 2sec
32 ftg
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Example
Example 11 The Grand Canyon is 1800 meters deep at its deepest point. A rock is dropped from the rim above this point. Write the height of the rock as function of the time t in the seconds. How long will it take the rock to hit the canyon floor?Solut1ion
)()(' tvth
Let h(t), v(t) and a(t) be the height, the velocity, and the acceleration of the rock at time t. Then 8.9)()(' tatv
1800)0( h 0)0( v
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Example
Solut1ion
)()(' tvth
Let h(t), v(t) and a(t) be the height, the velocity, and the acceleration of the rock at time t. Then 8.9)()(' tatv
1800)0( h 0)0( v
18.9)8.9()()( Ctdtdttatv 0)0(8.9)0( 1 Cv 01 C
So ttv 8.9)(
229.4)8.9()()( Ctdttdttvth
1800)0(9.4)0( 2 Ch 18002 C
18009.4)( 2 tth
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Example
Solut1ionWhen h(t) = 0, the rock will hit the canyon floor. So
(The negative root will be discard)
018009.4)( 2 tth
17.19t
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Homework
Section 4.1 page 255 #1-41 odd, 49, 51