4.1 Antiderivatives and Indefinite Integration

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.

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).

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

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).

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

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

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 )(

Notation of Antiderivatives

dxxf )( CxF )(

single function

bag of functions

Variable of IntegrationIntegrand

Constant of Integration

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

dx 3

Example

Cx 3

dxk

constant

Ckx

Example 2 Find the antiderivatives of 3

Solution

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

dxxm Cm

xm

1

1

1m

dxx5 Cx

6

6

Example

Basic Integration Rules

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

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.

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

(1) Please be aware that the indefinite integral

Note

dx

x

xx4

2 32

is NOT equal to

dxx

dxxx4

2 )32(

(2)The indefinite integration with respect to a polynomial

Note

dxxx )32( 2

can NOT be written as

dxxx 322

Practice

dxxx 386)1( 2

dx

x

xx3

3)2(

Practice

dxxx 386)1( 2 3

6 3x

2

8 2x x3 C

32x 24x x3 CCheck:

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

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

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

(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)

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

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

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

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

dxxx sin54 2

Example 9

dxxx sin54 2

Cxx

)cos(53

4 3

Check:

Cxx

cos53

4 3

Example

Solut1ion

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

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

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

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

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

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

Homework

Section 4.1 page 255 #1-41 odd, 49, 51