c01 calculus of one variable _t1_1112

8/11/2019 C01 Calculus of One Variable _T1_1112

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Calculus of One Variable

1

Calculus of One Variable_________________________________________________________

1 Real Functions of a Real Variable

1.1 Functions or Mappings

Example: The equation 2 2 1x x= + + associates to each real number x another real

number .y We say that y is a function of x .

General notion of functions

Definition: Let X and Ybe any nonempty sets. Afunction or mappingf from X to

,Y denoted by f X Y: , is a rule that assigns to each element of X exactly oneelement of .Y We say thatX is the domain of f and Yis the codomain off.

We write y f x= ( ) or f x y: to indicate that the element y Y is the value

assigned by the function f to element X . In this case, we say thaty is the image

of x .

The set of all images { }( )f x x X is called the range or image setoff, denoted byRf .

Note that if the domain and the codomain of a function f are both ,then :f

is called a real-valued function of a real variable, or simply a real function.

RemarkIn order to specify a real-valued function completely, the domain must be stated

explicitly. Otherwise the domain is taken as the largest possible subset of for

which the real-valued function can be defined.

Example: Suppose that2 1,

( )1 1.

x if xf x

x if x

< =

+

Find ( 2),f ( 1),f (2)f and2

( ).f Find the domain fD and the range fR .

Solution: 2( 2) ( 2) 4,f = = ( 1) 1 1 0,f = + = (2) 2 1 3,f = + = and

2 2( ) 1.f = +

Clearly, ( )f x is well defined for all x , thus =fD . Note that if 1,x< then2( ) 1f x x= < , and ( ) 0f x for every 1.x We conclude that

( , 1) [0, ).fR = U

1.2 Graphs of Functions


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Let f be a function. Thegraphof f is the set consisting of all points ( , ( ))x f x

in the Cartesiancoordinate plane, for all ,fD i.e., the graph is the set

.)(and),( xfyDxyx f =

Vertical Line Test.

Not every curve is the graph of a function. A curve in the xy-plane is the graph of a

function when it satisfies the vertical line property: any vertical line (a line parallel to

they-axis) intersects the curve at most once.

Symmetry.

The graph of ( )y f x= is symmetric with respect to the y-axis if and only if

( ) ( )f x f x= for all .fx D A function with this property is called an even

function.

.

The graph of )(xfy= is symmetry with respect to the origin if and only if

( ) ( )f x f x = for all .fx D A function with this property is called an odd

function.

Example: 5)( 2 +=xxf is an even function,while 3)( xxf = is an odd function.

1.3 Operations on Functions

Sums, Differences, Products and Quotients

Let f and g be two functions. Then for every x D Df g I , we define the new

functions ,f g+ f g and f g by the formulas

( )( ) ( ) ( ),f g x f x g x+ = +

( )( ) ( ) ( ),f g x f x g x =

( )( ) ( ) ( ).f g x f x g x =

They are called the sum, difference and product of functions, respectively. Thenatural domain for each of these is the set ofx-values for which both ( )f x and ( )g x

are defined, i.e., .g

D DI Moreover, at any point of x D Df g I at which ( ) , 0

we can also define the quotientfunction f gby the formula

f

gx

f x

g x

=( )

( )

( ).

For the functions ,f g+ f g and f g the domain is defined to be the intersection

of the domains of f and , and for f gthe domain is the intersection of the


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domains of f and gwith the elements where ( ) 0g x = excluded. Note also that

functions can also be multiplied by a constant c : defined by

( ) ( ) ( ( ))c f x c f x= for all .fD

Sometimes we shall write 2f to denote the product .f f Similarly, we have3f = f f f . In general, for any positive integer n, we define

factors

( ) ( ) ( ) ( )n

n

f x f x f x f x= L144424443

for all .fx D We see some examples.

Example: Given ( )f x x= and ( ) 3 .x x x= + Find ,f g+ ,f g .f g

Solution Observe that [0, )fD = and [0, ).gD = Then

(i) ( )( ) ( ) ( ) 4f g x f x g x x x+ = + = + for all f gx D D =I [0, ).

(ii) ( )( ) ( ) ( ) 2f g x f x g x x x = = for all f gx D D =I [0, ).

(iii) ( )2

( )( ) ( ) ( ) 3f g x f x g x x x x = = + for all f gx D D =I [0, ). Hence

( )( ) 3f g x x x x = + for all [0, ).x

Note also that ( )2

x since ( )2

x is not well-defined for all 0.x< But, if we

consider [0, ),x then ( )2 .x x=

Composition of functions

Let ,X Yand Zbe nonempty sets. Given two functions :f X Y and :Y Z ,

Suppose that R Df g , then the composite functiong fo function defined by

( ) ( )( ) ( )f x g f x=o

for all elementsf

x D .

X Y Z

f g

xxx x

fgo

)(xf ))(( xfgo

))(( xfg=


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Example: Given ( ) 1f x x= and 2( ) 7.g x x= Find fgo .Solution We first note that =fD \ }0{ , =fR \ }0{ , =fD , we have

( )2( )( ) ( ( )) 1 7g f x g f x x= = o

for all x 0.

1.4 Some Properties of Functions

Monotone Functions

Let A be a subset of and let Af : be a function. We say that ( )f x is

increasing on A if ( ) ( )f x f y> for all ,x y A such that ;y>

decreasing on A if ( ) ( )f x f y< for all ,x y A such that ;y>

nondecreasing on A if ( ) ( )f x f y for all ,y A such that ;x y>

nonincreasing on A if ( ) ( )f x f y for all ,y A such that ;y>

monotone on A if f is either nondecreasing on A or nonincreasing on .A

Example: The function 1( )f x x= is an increasing function on . The function

2( )f x x= is a decreasing function on . The function 3 ( ) 0f x = is a nondecreasing

and nondecreasing function on . Hence, 1 ,f 2f and 3f are monotone functions.

Example:

The function 3 ( ) 0f x = is an even function on since 3 ( ) 0f x = 3 ( )f x= for all

x . The function 1( )f x x= is an odd function on as 1( )f x = 1 ( )f x = for

all x .

.

One-to-One or Injective Functions

Let X and Y be nonempty sets. A function :f X Y is said to be one-to-one,( ) ( )f x f y= implies .x y= Equivalently, we see that a function :f X Y is one-

to-one if ( ) ( )f x f y whenever y .

Example: The function 3( )g x x= is one-to-one.

Onto or Surjective FunctionsLet X and Ybe nonempty sets. A function

:f X Y

is said to map X onto Y(or a surjective function), if for any ,y Y there exists atleast one Xa such that ( ) .f a y=


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Example:

(a) The function 1 ( )f x x= is a surjective function on since =)(1f .

(b). The function 2 ( )f x x= is a surjective function on since =)(2f .

(c). The function 23 ( )f x x= is not surjective on , since = ),0[)(3f .

We define a function ),0[; h by 2( ) .h x x= Then the function h is surjective

since [0, ).h

R =

Inverse Function

Let Xand Ybe nonempty sets. Let :f X Y and :Y X be functions. If f

and satisfy the following two conditions

(i) ( )( ) ( ( ))f g x f g x x= =o for allxY

(ii) ( )( ) ( ( ))f x g f x x= =o for allx Xthen we say that f is an inverse of ,g and is an inverse of f . However, it can be

shown that if a function f has an inverse, then it is unique, and hence the inverse of

f is commonly denoted by 1 ,f read as f inverse. Therefore, we have the

following cancellation equations:

1 1( )( ) ( ( ))f f x f f x x = =o for allxY1 1( )( ) ( ( ))f f x f f x x = =o for allx X

Example :Show that the functions 1( ) 2f x x=

and 2 2( )

x

f x =

are the inverse pair.Solution: Observe that

1 2 1 2 1 2 2( )( ) ( ( )) ( ) 2( )x xf f x f f x f x= = = =o , = 2Dx , =2D domain of 2f

22 1 2 1 2 2

( )( ) ( ( )) (2 ) xf f x f f x f x x= = = =o , for all = 1Dx , =1D domain of 1f

Theorem: Let :f X Y be a function. Then f has an inverseif and only if it is 1:1and onto.

The following algebraic procedure will produce the formula for 1 :f

Step I: Write ( ).f x= Step II: Interchange the variables and in the equation ( )y f x= to obtainthe

equation ( );x f y= solve, if possible,the equation ( )x f y= for in terms

of .x

Step III: The resulting equation in Procedure II will be 1 ( ).y f x=

Example:Find the inverse of the function 3( ) .f x x=

Solution We see that 3( )f x x= is one-to-one and onto. Therefore, it has the inverse.

Let 3( ) .f x x= = Interchanging and in this equation and then solving for ;

this yields 3 .y= Then we have1

3 .y= Hence, we have1

31( )f x x = for all x .


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1.5 Elementary FunctionsPolynomials

Apolynomial orpolynomial functionis a function defined by

1

1 1 0( ) ,n n

n np x a x a x a x a

= + + + +L x .

where n is a nonnegative integer and 0 1 1, , , ,n na a a aK are constant real numbers. If

0na then the integer n is called the degreeof the polynomial ( ).p x The constant

na is called the leading coefficient and the constant 0a is called the constant termof

the polynomial ( ).p x The domain of ( ),p x pD is . In particular, we have

a constant function is the zero degree polynomial: ( ) ;p x a=

a linear function is the first degree polynomial: ( ) ;p x ax b= +

a quadratic function is the second degree polynomial: 2( ) ;p x ax bx c= + +

a cubic function is the third degree polynomial: 3 2( ) ;p x ax bx cx d= + + +

where , , ,a b c d and s are constant real numbers.

Rational FunctionsA rational function is a function iof the form

)(

)()(

xq

xpxr =

where p(x) and q(x) are two polynomials, defined for values of x for which its

denominator is not 0..

Example:2

4 8( )

2 3

xr x

x x

=

, 1, 3.x is a rational function

The Absolute Value FunctionWe see that the absolute value function can also be expressed as a piecewise defined

function as follows.

if 0,( )

if 0.

x xf x x

x x

= =


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Real-valued nth-Root Functions

For any even n , the real-valuednth-root functionis the nonnegative real-valued

function ),0[:f defined as

1

( ) n nf x x x= = for all [0, ).x

For any odd n,the real-valuednth-root function is the function :f

defined as1

( ) n nf x x x= = for all x .

Theorem :

(i) For any even nn

n ny x y= for all , [0, );x y

(ii) For any odd nnn nxy x y= for all yx,

(iii) For any even n ( )n

n x= for all [0, );x

(iv) For any even nn nx x= for all x .

(v) For any odd n ( )n

n nn x x x= = for all x .

(vi) For any m ,n , ( ) ( )11 mnn n

mmx x= =

for all [0, ).x

Trigonometric Functions and their Inverses

Let be an anglemeasured in radians, satisfying the inequality 0 2 . < LetPbe the point on the circle of radius r.

Angles measured counterclockwise (ie, anticlockwise) form the positive x-axis

are assigned positive measures).

Angles measured clockwise are assigned negative measures.

When angles are used to describe counterclockwise or clockwise rotations, ourmeasurements can go arbitrarily far beyond 2 . In view of Figure 1.6.1, we note that

the point Pon the circle will come back to the same coordinate when 2k = + for all integer k.


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An angle measured in radians in standard position .

We define the six trigonometric functions of the angle in terms of thecoordinates of the point ),( yxP as follows:

Sine function: ( ) sinfr

= = ;

Cosine function: ( ) cosx

fr

= = ;

Tangent function: ( )f =x

y=tan ;

Cotangent function: ( )f =

y

x=cot ;

Secant function: ( )f =r

=sec ;

Cosecant function: ( )f =y

r=csc .

These functions are defined for all real numbers , which use radians as the unit

of measurement, except for those values of at which the denominators of the

fractions vanish.

As we can see, sin and cos are well defined for all . But, tan and sec

are not defined for value of . When ,0=x this means2

,k = + k an integer.

Similarly, cot and csc are not defined for values of for which ,0=y namely,

,k = k an integer. Therefore, we have

( ) sinf = for all .

( ) cosf = for all .

sin( ) tan

cosf

= = for all . with ,

2k

+ k an integer.;

1( ) cot

tanf

= = for all . with ,k k an integer.;

1( ) seccos

f = =

for all . with ,2

k + k an integer.;


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1( ) csc

sinf

= = for all . with ,k k an integer..

When we graph trigonometric functions in the coordinate plane, we usually

denote the independent variable (radians) by instead of . In what follows, we willuse the variable to denote the angle measured in radians.

It is also useful to know the values of the trigonometric functions for particular

angles and their signs in the four quadrants.

0 6 4 3 2 3 2 2

sinx 0 1 2 1 2 3 2 1 0 -1 0

cosx 1 3 2 1 2 1 2 0 -1 0 1

tanx 0 1 3 1 3 0 0

cot 3 1 1 3 0 0

Theorem (Periodicity):

For any k Z, we have

(i) sin( 2 ) sink x+ = ,for all x and k intege,.

(ii) , cos( 2 ) cosx k x+ = for all x and k integer;

(iii) tan( ) tanx k x+ = , for all x and k integer;

(iv) cot( ) cotk x+ = ,for all x and k integer;

(v) sec( 2 ) secx k x+ = ,for all x and k integer;

(vi) csc( 2 ) csck x+ = ,for all x and k integer.We summarize:

sin cos tan cot sec cscx

period 2 2 2 2

Theorem (Odd and Even Properties):

(i) The sine function is odd, i.e., sin( ) sinx = for all for all x .(ii) The cosine function is even, i.e., cos( ) cosx = for all for all x .

(iii) The tangent function is odd, i.e., tan( ) tanx x = ,for all

x \ }int,2

{ egerkk

+ . .

Theorem:

(i) 2 2sin cos 1x+ = ,for all x ;

(ii) 2 2tan 1 sec x+ = ,for all x \ }int,

2

{ egerkk

+

(iii) 2 2cot 1 cscx x+ = ,for all x \ }int,{ egerkk+ .


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Theorem(Addition Formulas):

(i) sin( ) sin cos cos siny x y x y+ = + ,for all yx,

(ii) cos( ) cos cos sin siny x y x y+ = ,for all yx,

(iii)

tan tan

tan( ) 1 tan tan

x y

x y y

+

+ = , for all yx, with 2 ,y k

+ + k integer.

(vii) sin 2 2sin cosx x= , for all x (viii) 2 2 2 2cos 2 cos sin 2cos 1 1 2sinx x x x x= = = , for all x .

Theorem (Product to Sum Formulas):

(i) ( )12sin cos sin( ) sin( )x y x y x y= + for all yx, .

(ii) ( ))cos()(cos(2

1sinsin yxyxyx += for all yx,

(iii) ( ))cos()cos(2

1coscos yxyxyx ++= for all yx, .

Theorem 1.6.7(Sum to Product Formulas)

(i) ( ) ( )( )2 2sin sin 2 sin cosx y x yx y + + = , for all yx, .(ii) ( ) ( )( )2 2sin sin 2 cos sinx y x yx y + = , for all yx, .(iii) ( ) ( )( )2 2cos cos 2 cos cosx y x yy + + = , for all yx, .(iv)

( ) ( )( )2 2

cos cos 2 sin sinx y x y

x y + = , for all yx, .

Trigonometric Functions are periodic, and hence, in general, they are not one-to-

one functions. However, we can restrict their domain setsto makethem 1:1, and so,

we can define their inverses The following table shows the restricted domains that

will make the functions one-to-one.

Function Restricted Domain Range

sinx 2 2

[- , ] [ ]1,1 cosx [0, ] [ ]1,1

tanx 2 2(- , )

Rcotx (0, ) R

secx 2 2

[0, ) ( , ] U ( , 1] [1, ) U

csc 2 2

[ , 0) (0, ] U ( , 1] [1, ) U

Definition (Inverse Trigonometric Functions):

(i) For any 1 1,x and22 y , the arc-sinefunction, denoted as 1sin ,

is defined by1siny x= sin .x=


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(ii) For any 1 1,x and ,0 y the arc-cosine function, denoted as1cos ,x is defined by

1cosy x= cos .y x=

(iii) For any x and 2 2y < < , the arc-tangentfunction, denoted as 1tan , is

defined by1tan x= tan .y x=

(iv) For any x and ,0 0, then q px aa = .Suppose x is not a


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rational number , then how do we definexa ? By a theorem which will not be

proved here, there is a unque number, denoted byxa , such that if with 1,a has the inverse.

We called this inverse function the logarithm of x to the base ,a or simply, base-a

logarithmor logarithm if the base is understood, and it is denoted by

( ) log ( )af x x= for all (0, ).x

Evidently, the natural domain and the range of logarithm are (0, ) and

respectively because the range and the domain of exponential are respectively(0, ) and .

Theorem: Let a be a positive real number such that 1.a Then we have thefollowing results.

(i) The base-a logarithm function is an injective function on (0, )x ;

(ii) for any , (0, ),x y y= log loga ax y= ;

(iii) logay x= ya x= for all (0, )x ;

(iv)log

a

x

a x= for all (0, )x ;


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(v) logx

a a x= for all x .(vi) log 1 0.a =

There are two frequently used bases for logarithms. If the base 10a= is used,then the logarithm 10log ( ), is called a common logarithm, and denoted by log( ). If

the base a e= is used, then the logarithm log ( ),e is called a naturallogarithm, and is denoted by ln( ).x

.

Theorem:Let a be positive real numbers such that 1.a Then the followingassertions hold true.

(i) log ( ) log loga a ay x y= + for all , (0, ).x y

(ii) log ( ) log loga a ay x y= for all , (0, ).x y (iii) log logy

a ax y x= for all (0, )x and x .

Theorem: Let a be positive real numbers such that 1.a Then every exponentialfunction is a power of the natural exponential function

lnx x aa e= for all x

Theorem: Let a be positive real numbers such that 1.a Then every base-alogarithm is a constant multiple of the natural logarithm

lnlog

lna

xx

a= for all (0, )x .

Example: Solve the following equation3 2 34 5 20.x x =

Solution3 2 34 5 20x x = 4 4 4 4( 3) log 4 (2 3 ) log 5 log 4 log 5x x + = +

4 4 43 2log 5 (3log 5) 1 log 5x x + = +

4 4(1 3log 5) 4 log 5x =

4 4(4 log 5) (1 3log 5).x=

Hyperbolic Functions and their Inverses

Definition: The six hyperbolicfunctions are defined as follows:

The hyperbolic sine function, denoted by sinh , is defined by

sinh2

x xe ex

= for all x

The hyperbolic cosinefunction, denoted by cosh ,x is defined by

cosh2

x xe ex

+= for all x


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The remaining hyperbolic functions are defined in terms of sinhx and coshx as

follows:

The hyperbolic tangent function, denoted by tanh , is defined by

sinhtanh cosh

x x

x x

e ex x e e

= = + for all for all x

The hyperbolic cotangent function, denoted by cothx , is defined by

coshcoth

sinh

x x

x x

x e ex

e e

+= =

for all for all x \ { }0 .

The hyperbolic secantfunction, denoted by sec hx , is defined by

1 2sech

cosh x xx

x e e= =

+, for all x

The hyperbolic cosecantfunction, denoted by csch , is defined by

1 2csch

sinh x xx

x e e= =

, for all }0{\x ..

Theorem: (Odd and Even Properties)

(i) The hyperbolic sine function is odd, i.e.,sinh( ) sinhx = for all x .

(ii) The hyperbolic cosine function is even, i.e., cosh( ) coshx x = for all x .

(iii) The hyperbolic tangent function is odd, i.e., tanh( ) tanhx = for all x

We note that the hyperbolic functions satisfy various identities similar to the identitiesfor the trigonometric functions.

Theorem: (Hyperbolic Identities)

(i) 2 2cosh sinh 1x = for all x .(ii) 2 21 tanh sechx x = for all x .(iii) 2 2coth 1 csch x = for all }0{\x .

Theorem:(Hyperbolic Addition Formulas)

(i) sinh( ) sinh cosh cosh sinhy x y x y+ = + for all yx, .

(ii) cosh( ) cosh cosh sinh sinhy x y x y+ = + for all yx, .

(v)tanh tanh

tanh( )1 tanh tanh

x yx y

x y

++ =

+for all yx, .

2 Limits and Continuity

2.1 Limits of Real Functions


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The concept of limits is the cornerstone on which the development of calculus rests

and it is the basic for all calculus problems. A good understanding of limit concept

will help explain many theorems in calculus. Before we try to give a definition of a

limit of a real function, let us look at more examples.

Example: Guess the value of the function sin( ) ,xx

f x = where x is in radians, near

0.x= Solution Evidently, the function sin( ) x

xf x = is defined for all real number x except

0.x= Although ( )f x is not well-defined at 0,x= it still makes sense to ask what

happens to the value of ( )f x as x approaches 0 without actually taking on the value

0.x= We construct the following table of values correct to seven decimal places.

sin( ) xx

f x = sin( ) xx

f x = 1.0 0.8414710 0.1 0.9983342

0.5 0.9588511 0.05 0.9995834

0.4 0.9735459 0.01 0.9999834

0.3 0.9850674 0.001 0.9999999

0.2 0.9933467 0.0000001 0.9999999999999

From the table we guess that sin( ) xx

f x = approaches 1 as x approaches arbitrarily

close to 0.In fact, this is true.

We first suggest the following informal definition of limit.

Definition: Let ( )f x be a real function defined on an open interval about a, exceptpossibly at a itself. We say that ( )f x approaches the limit L as x approaches a if

( )f x can be made arbitrarily close to L for allx sufficiently close to a (xapproach a

from either direction) but not equal to a , and we write

lim ( )x a

f x L

=

This definition is informal since the phrases such as arbitrarily close and all x

sufficiently close are imprecise. However, this definition should be clear enough to

enable us to recognize, evaluate and compute limits of specific functions.


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It should be noticed that the point ),( La is not assumed to belong to the graph ofy f x= ( ). This emphasizes the fact that lim ( )x a f x L = depends only on the

behavior of ( )f x near x a= , but not on the number ( )f a itself. An alternative

notation for lim ( )x a f x L = is ( )f x L as .a

Theorem (Uniqueness of Limits): Let ( )f x be a real function defined on an open

interval about a,except possibly at a itself. If lim ( ) ,x a f x L = then the number L is

unique.

2.2 One-Sided Limits

However, there is certain information to be conveyed the two-sided limit fails to exist.

For this reason, we define the left-hand limit and right-hand limit of ( )f x at a= asfollows.

Definition (Right-hand Limit): Let ( )f x be a real function defined on some open

interval ( , )a b We write

lim ( )x a

f x L+

=

and say that ( )f x has right-hand limit at a= if we can make the values of ( )f x

arbitrarily close to L by taking x to be sufficiently close to the right (or positive side)

of .a

Definition:(Left-hand Limit): Let ( )f x be a real function defined on some open

interval ( , )c a . We write

lim ( )x a

f x M

=

and say that ( )f x has left-hand limit at x a= if we can make the values of ( )f x

arbitrarily close to M by taking to be sufficiently close to the left (or negative

side) of .a


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Example:Evaluate2

lim ( )x

f x+ and 2lim ( ),x f x if 22

6( ) .

x

x xf x

+ =

Solution: First note that

2 if 2,2

2 if 2.

x xx

x x

=


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lim ( )x a f x L = and lim ( ) ,x a x M = where L and M are real numbers. Then the

following results hold true.

(i) ( )lim ( ) ( ) lim ( ) lim ( ) .x a x a x a

f x g x f x g x L M

+ = + = +

(ii) ( )lim ( ) ( ) lim ( ) lim ( ) .x a x a x af x g x f x g x L M = =

(iii) ( )lim ( ) lim ( ) .x a x a

cf x c f x cL

= =

(iv) ( )lim ( ) ( ) lim ( ) lim ( ) .x a x a x a

f x g x f x g x LM

=

=

(v)lim ( )( )

lim ,( ) lim ( )

x a

x a

x a

f xf x L

x g x M

= =

provided 0M

(vi) If ( ) ( )f x g x on an open interval containing ,a then lim ( ) lim ( ).x a x a

f x g x

Example:Let a .Evaluate 2lim 3 7.x ax x + + Show that lim ( ) ( ),x a p x p a = where

( )p x is a polynomial function.

Solution: 2lim (lim )x a x ax x = 2(lim )x a x a = , lim 3 3(lim ) 3x a x ax x a = = and

lim 7 7.x a = Hence2lim 3 7x ax x + + =

2 3 7.a a+ +

Let 20 1 2( ) ,n

np x a a x a x a x= + + + +L Since lim (lim )n n n

x a n n x a na x a x a a = = for

all positive integers, and0 0limx a a a = , it follows that

( )2

0 1 2 0 1lim ( ) lim lim (lim ) (lim )

n n

x a x a n x a x a n x ap x a a x a x a x a a x a x = + + + + = + + +L L 0 1 ( ).

n

na a a a a p a= + + + =L

Example:Evaluate2

3

11 1

lim .xx x

Solution Note that 3 21 ( 1)( 1).x x x x = + + Since 1x , and so 1,x it follows

that 2 3 2 2( 1) ( 1) ( 1)( 1) ( 1)( 1) ( 1) ( 1).x x x x x x x x x x = + + + = + + + Therefore,

we obtain2

1

3 2 21

lim ( 1)( 1)1 21 1 31 ( 1) lim ( 1)

lim lim .xx

xxxx xx x x x x

++ + + + +

= = =

Theorem: (Sandwich or Squeeze Theorem): Let ( ),f x ( )x and ( )h x be realfunctions, and let a . Suppose that ( ) ( ) ( )f x g x h x for all x in some open

interval containing ,a except possibly at x a= itself. If lim ( ) lim ( )x a x af x h x = =

,L then lim ( ) .x a x L =


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Example:Let a and ( )f x be a real function defined on an open interval about a,

except possibly at a itself.

Remark: lim ( ) 0x a f x = if and only if lim ( ) 0.x a f x =

Theorem:

(i)0

sinlim 1.x

x

=

(ii) limcos

.x

x

x

=

0

10

2.4 Limits at Infinity and Infinite Limits

In this section we will extend the concept of limit to allow for two situations not

covered by the definition of two-sided limit or one-sided limit in the previous section:

limits at infinity, where x becomes arbitrarily large, either positive or negative; infinite limits, which are not really limits at all (and we are not saying that the

limit exists) but provide us a useful symbolism for describing the behavior of

real functions whose values become arbitrary large positive or become negative

and arbitrarily large in absolute values.

Consider the functio 1( )x

f x = . It is clear that, =fD \ }0{ and the values of

( )f x approach 0 as takes on larger and larger positive values. Likewise,f(x)

approaches 0 as x takes on large and larger negative values.


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20

The horizontal lines 0y= is the so-called horizontal asymptote of the curve.Similarly, x=0 is the vertical asymptote of the curve.

Definition:Let ( )f x be a real function defined on an open interval ( , ).a We say

that ( )f x approaches the limit L as x approaches infinity and we write

lim ( )x

f x L

=

if ( )f x gets arbitrarily close to L as x gets arbitrary large through positive values.

Definition:Let ( )f x be a real function defined on an open interval ( , ).b We say

that ( )f x approaches the limit as approaches negative infinity and we write

lim ( )x

f x M

=

if ( )f x gets arbitrarily close to as gets negative and arbitrary large in absolute

values.

Recall that the symbol , called infinity, does not represent a real number. Wecannot use or in arithmetic in the usual way. However, we can use the

phrase approaches to mean becomes arbitrarily large positive, and the phraseapproaches to mean becomes negative and arbitrary large in absolute values.

Example:Evaluate2

2

2 2

3 5lim .x xx x

+ +

Solution Since , it follows that 21 0,

x and so 2

1 0.x

Thus,2

2

2 2

3 5lim x xx x

+ +

= 2 22

2 2 2

1 2 (1 ) (1 )2 2 233 5 1 3 5(1 )

lim lim .x x xx x

x xx x x

+ + + +

= =

Example:Evaluate lim ( )x f x and lim ( )x f x where2( ) 1.f x x x= +

Solution Observe that


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21

2 2 2 2 2( )

1 (1 (1 )) 1 (1 )

x x xf x

x x x x x= = =

+ + +

as , 0a b ab a b> = . Since 2x x= , we have

2 2

1( )

1 (1 ) 1 (1 )

x xf x

xx x x

= = + +

.

Since 21 (1 ) 1x+ and 1x x x x= as x , it follows that lim ( ) 1.x f x =

Similarly, 21 (1 ) 1x+ and ( ) 1x x x x= as ,x thus lim ( ) 1.x f x =

We next turn our attention to the notion of infinite limits. Sometimes, a function

whose values grow arbitrary large can be said that to have an infinite limit. Sinceinfinity is not a real number, it follows that infinite limits are not really limits at all,however they provide a method of describing the behavior of functions which grow

arbitrarily large positive, or become negative and arbitrarily large in absolute values.

Definition: Let a and ( )f x be a real function defined on an open interval about

,a except possibly at a itself. Then

lim ( )x a

f x+

=

means that the values of ( )f x can take arbitrarily large positive values by taking

sufficiently close to a from the right, but not equal to .a

Definition: Let a and let ( )f x a real function defined on an open interval about

,a except possibly at a itself. Then

lim ( )x a

f x+

=

means that the values of ( )f x can take arbitrarily large negative valuesby taking

sufficiently close to a from the right, but not equal to .a

Definition: Let Let a and ( )f x be a real function defined on an open intervalabout ,a except possibly at a itself. Then lim ( )x af x = if lim ( )x a f x+ = and

lim ( )x a

f x = .

Definition: Let Let a and ( )f x be a real function defined on an open interval

about ,a except possibly at a itself. Then lim ( )x af x = if lim ( )x a f x+ = and

lim ( )x a

f x = .

Example:Discuss about the behavior of 133

lim ,xx +

133

limxx

and 13 3lim .x x

Solution Observe that when 3 ,x + then 3 0 .x + Thus, 13x (sometimeswe denote 1

3x + ). So, we denote1

33lim .

xx +

= Clearly, the limit 133

limxx

+


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22

does not exist, but we are describing the behavior of ( )f x near positive side of 3 by

saying that the values of ( )f x become arbitrarily large positive as approaches

sufficiently close to 3. Similarly, since 3 ,x then 3 0 .x Thus, 13x .

Hence we denote1

33lim .xx + = Hence1

3 3limx x does not exist.

2.5 Continuity

Definition: Let ( )f x be a real function defined on an open interval containing .a We

say that the function ( )f x is continuousat a= if lim ( ) ( ).x a f x f a =

Note that Definition 2.5.1 implicitly requires three things if ( )f x is continuous at

:a= ( )f a is defined;

lim ( )x a f x exists;

lim ( ) ( ).x a f x f a =

If ( )f x is not continuous at ,x a= then we will say that the function ( )f x is

discontinuous at ,x a= or ( )f x has a discontinuity at .a= This is the case either

( )f a is not well defined (undefined), or l im ( )x a f x fails to exist, or ( )f a is well

defined and lim ( )x a f x exists but their values are not equal.

In particular, a function ( )f x is said to have a removeable discontinuity at x a= if

the limit lim ( )x a f x exists, but lim ( ) ( )x a f x f a either because ( )f a is

undefined or the value of ( )f a differs from the value of the limit. This refers to the

cases (i) and (iii).

The discontinuity at x a= in case (ii) is called jump discontinuity because thefunction jump from one value to another. Nevertheless, there is another type of

discontinuity illustrated in the following figures, is called infinite discontinuity

whenever either one of the following four cases lim ( ) ,x a f x = or

lim ( )x a f x = , or lim ( )x a f x+ = and lim ( )x a f x = , or lim ( )x a f x+ =

and lim ( )x a

f x = hold true.


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Example: Where are the following functions discontinuous ?

(a)2

( ) .xx

f x =

(b)21 if 0,

( )2 if 0.

x xf x

x

=

=

(c)

242

if 2,( )

2 if 2.

xx

xf x

x

=

=

Solution: (a). Note that (2)f is undefined, so ( )f x is discontinuous at 2.x=

(b) Evidently, (0) 2f = , but 0lim ( )x f x does not exist. So, ( )f x has a

discontinuity at 0.x= (c) Observe that 2 2lim ( ) lim 2 4x xf x x = + = and (2) 2.f = Thus, 2lim ( )x f x

(2),f hence ( )f x is discontinuous at 2.x=

Definition: Let ( )f x be a real function defined on an open interval containing .a

We say that the function ( )f x is continuousfrom the right at the point a= if

lim ( ) ( ).x a

f x f a+ =

Definition: Let ( )f x be a real function defined on an open interval containing .a Wesay that the function ( )f x is continuous from the left at the point x a= if

lim ( ) ( ).x a

f x f a =

Definition: Let ba, with ,a b< and let ( )f x be a real function defined on

( , ).a b We say that the function ( )f x is continuous on ( , )a b if ( )f x is continuous at

every ( , ).a b

Definition: Let ba, . with ,a b< and let ( )f x be a real function defined on

[ , ].a b We say that the function ( )f x is continuous on [ , ]a b if the following

conditions are satisfied: (i) ( )f x is continuous on ( , )a b ; (ii) ( )f x is continuous

from the right at x a= ; and (iii) ( )f x is continuous from the left at .x b=


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A real funciton that is continuous on ( , ) + is said to be continuous

everywhere or simply continuous. For example, the function 21

1( )

x

xf x

= is

continuous on the intervals ( , 1)

and (1, )+

.

Theorem: Let c be any real number, and let ( )f x and ( )g x be real functions

defined on an open interval containing .a If ( )f x and ( )x are continuous at ,x a= then

(i) ( )( ) ( ) ( )f g x f x g x+ = + is continuous at ;x a=

(ii) ( )( ) ( ) ( )f g x f x g x = is continuous at ;x a=

(iii) ( )( ) ( )cf x cf x= is continuous at ;a=

(iv) ( )( ) ( ) ( )fg x f x g x= is continuous at ;a=

(v) ( )( ) ( ) ( )f g x f x g x= is continuous at a= provided ( ) 0,g a and is

discontinuous at a= if ( ) 0.g a =

Theorem: Let ba, . Let ( )f x be a real function defined on an open interval

containing ,b let ( )x be a real function defined on an open interval containing .a If

lim ( ) ,x ag x b = and ( )f x is continuous at ,x b= then

lim ( ( )) (lim ( )) ( )x a x a

f g x f g x f b

= = .

Corollary:Let a and ( )x be a real function defined on an open interval

containing ,a and let ( )f x be a real function defined on an open interval containing

( ).a If ( )x is continuous at ,x a= and ( )f x is continuous at ( ),g a= then the

composition function ( )( )f g xo is continuous at .a=

Theorem: The following type of real functions are continuous at every point in their

domains: (i) polynomial functions; (ii) rational functions; (iii) real n-th root

functions; (iv). trigonometric functions; (v) exponential functions; (vi) logarithm

functions; and (vii) hyperbolic functions.

Theorem (The Intermediate Value Theorem): Let ba, with .a b< Suppose that

( )f x is continous on the closed interval [ , ]a b and that k is any real number between

( )f a and ( ).f b en there exists at least one real number ( , )c a b such that ( ) .f c k=


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Example: Show that the function 3( ) 3 4 9f x x x x= + + has at least one real zero in

the interval [ 3, 3].

Solution: Observe that ( 3) 15 0,f = < and (3) 33 0.f = > Since ( 3) 0 (3),f f < <

by the Intermediate Value Theorem, there exists ( 3, 3)c

such that ( ) 0.f c =

3 Differentiation

3.1 The Derivative

Suppose that )(xfy= is a function. We want to computer the slope of the graph

of )(xf at )(,( 00 xfx . We take a point near by ))(,( 00 xxfxx ++ . Then the slope of

the secant from )(,( 00 xfx to the point ))(,( 00 xxfxx ++ is

x

xfxxf

x

y

+=

)()( 00

Next we let approach 0. We define the slope of the graph at ))(,( 00 xfx to be

x

xfxxf

x

y

x

+=

)()( 00lim

0

Definition: Let ( )f x be a function defined on an open interval about a point .a= The derivative of ( )f x at ,a= denoted by '( ),f a is

0

( ) ( )limh

f a h f a

h

+

if this limit exists.

We say that ( )f x is differentiableat x a= or ( )f x has a derivativeat x = a if f(a)exists.

Example: By using definition, find the derivative of the polynomial 2( ) 8p x x x= + at.a=

Solution:


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26

2 2

0 0

0

0

( ) ( ) (( ) 8( )) ( 8 )lim lim

( 2 8)

lim

lim( 2 8) 2 8.

h h

h

h

p a h p a a h a h a a

h h

h h a

h

h a a

+ + + + +=

+ +

== + + = +

Note also that the value of the derivative of a real function ( )f x at a particular

point a= in its domain can also be expressed in several ways:

( ) ( )'( ) ( ) ( )

x x a

x a

d df a df xf a f a D f x

dx dx dx =

=

= = = = .

Let ( )f x be a differentiable real function of ,x and ( )y f x= . Then in view ofthe notation of the derivative of ( ),f x we have

( )'( ).

dy df xf x

dx dx= =

It is sometimes useful to be able to refer, or view, the quantities dy and dx in such a

way that their quotient is the derivative .dy dx We can justify this by regarding dx =

x as a new independent variable, called thedifferential of , and defining a new

dependent variable dy , called the differential of ,y as a function of and dx by

( )'( ) ,

df xdy dx f x dx

dx= =

or simply, ( ) '( ) .df x f x dx=

For example, if 2 ,y x= then we have 2 ,dy dx x= or equivalently, we obtain2 .dy x dx=

Example:Determine whether the following functions are differentiable at 0.x= (i)( ) sinf x x= . (ii) ( ) .f x x=

Solution:(i) 0 0 0lim ( (0 ) (0)) lim (sin sin 0) lim sin 1h h hf h f h h h h h + = = = . So, the

function ( ) sinf x x= is differentiable at 0x= , and '(0) 1.f =

(ii) Firstly, we consider the right-hand limit,

0 0 0

(0 ) (0) ( ) (0)lim lim lim 1,h h h

f h f f h f h

h h h+ + +

+ = = =

and the left-hand limit,

0 0 0

(0 ) (0) 0

lim lim lim 1.h h h

f h f h h

h h h

+

= = =


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27

Hence 0lim ( (0 ) (0))h f h f h + does not exist. We conclude that the function

( )f x x= is not differentiable at 0.x=

Let ba, with ,b a> and let ( )f x be a function defined on ( , ).a b We say that( )f x is differentiable on ( , )a b if ( )f x is differentiable at every point ( , ),a b and

we say that ( )f x is differentiable everywhere, or simply, differentiable if ( )f x is

differentiable on ( , ).

For instance, ( )f x x= is a differentiable function. Let be an arbitrary real number.Then we have

0 0 0 0

( ) ( )lim lim lim lim1 1.h h h h

f x h f x x h x h

h h h

+ + = = = =

Thus, ( )f x is differentiable throughout ( , ) with '( ) 1f x = for all ( , ).x

Example:

Consider the function ( )f x x= . Although it is differentiable from the right at 0x= ,

and is differentiable from the left at 0x= , it is not differentiable at 0. Other than that,it is differentiable everywhere. (x 0 ).

Example:By using definition, evaluate the derivative at any x for the following

functions. (i). ( )f x c= , where cis a real constant. (ii) ( ) sin .f x x= Solution

(i). Given ( ) ,f x c= then we have0 0 0

( ) ( )lim lim lim 0 0.h h h

f x h f x c c

h h

+ = = =

(ii). Let ( ) sin .f x x= Then we have

0 0

0

( ) ( ) sin( ) sinlim lim

(sin cos cos sin ) sinlim

h h

h

f x h f x x h x

h h

x h x h x

h

+ + =

+ =

= )h

sinh(xcos

h

1cosxsin(lim

0h+

hx

hx

hh

sinhlimcos

1coshlimsin

00 +

=

x

xx

cos

1)(cos0)(sin

=+=

3.2 Differentiability Theorems

Theorem: Let c be an arbitrary real number, and let x Then

(i) 0;d

cdx

=

(ii) 1;

d

xdx =


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28

(iii) 1n nd

nxdx

= for any positive integer .n

(iv) 1n nd

nxdx

= for any positive integer ,n and any x \ }0{ .

Theorem: Let ,a b be any real numbers, and let ( )f x and ( )x be real functions

defined on an open interval containing x. If ( )f x and ( )g x are differentiable at ,x

then the following rules hold true.

(i) The function ( )( ) ( ) ( )af bg x af x bg x+ = + is differentiable at ,x and

( )d

dxaf x bg x a

d

dxf x b

d

dxg x( ) ( ) ( ) ( ).+ = +

(ii) The function ( )( ) ( ) ( )fg x f x g x= is differentiable at ,x and

( )ddx

f x g x ddx

f x g x f x ddx

g x( ) ( ) ( ) ( ) ( ) ( ) .=

+

(iii) The function ( )( ) ( ) ( )f g x f x g x= is differentiable at , provided

( ) 0g x , and

2

( ) ( ) ( ) ( )( )

.( ) ( )

d df x g x f x g x

d f x dx dx

dx g x g x

=

In more convenient notations ,we have:

'')'( bgafbgaf +=+

'')'( fggffg +=

2g

'gfg'f)'

g

f(

=

Corollary:Let naaa ...,, ,21 and 1( ), ( ), , ( )nf x f x f xK be real functions defined on

an open interval containing x. If 1( ), ( ), , ( )nf x f x f xK are differentiable at ,x thenthe following hold true:

(i) The function ( )( ) ( )af x af x= is differentiable at , and

( )( ) ( ).d d

af x a f xdx dx

=

(ii) The function (1 )( ) 1 ( )f x f x= is differentiable at , provided ( ) 0f x , and

2

( )1

.( ) ( )

df x

d dx

dx f x f x

=

(iii) The function1

( )n

i iia f x

= is differentiable at , and


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29

1 11

( ) ( ).n n

i i i i

i i

d da f x a f x

dx dx= =

=

(iv). The function 1 2( ) ( ) ( )nf x f x f xL is differentiable at ,x and

( )1 2 1 2 1 2 3

1 1

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

n n n

n n

d d df x f x f x f x f x f x f x f x f x f x

dx dx dx

df x f x f x

dx

= +

+ +

L L L

L L

------------------------------------------------------------------------

(f g h)=f g h+f g h+f g h .

-------------------------------------------------------------------------

Theorem: Let ( )f x be a real function defined on an open interval containing x. If

( )f x is differentiable at ,x then ( )f x is continuous at .x

In other words, if ( )f x is not continuous at ,x then it is not differentiable at .x

Next, we list the derivatives for some elementary real functions.

1 Derivatives of Polynomials and n-th Root Functions

(a) ( )1 1 21 1 0 1 2 1( 1) 2 ,n n n n

n n n n

da x a x a x a na x n a x a x a

dx

+ + + + = + + + +L L

for any real number , where n is any positive integer, and n

aaa ,...,,21

(b) ( ) ( ) 1 ,m

n m n mnd d nx x xdx dx m

= = for any nonnegative real number x , where

,m n are any positive integers.

2 Derivatives of Trigonometric Functions and theirInverses

sin cos ,d

x xdx

= x cos sin ,d

x xdx

= x

2tan sec ,d

x

dx

=

x \ }2

{

k+ , k .

2cot csc ,d

x

dx

=

x \ }{ k , k .

sec sec tan ,d

x x xdx

=

x \ }2

{

k+ , k .

csc csc cot ,d

x x xdx

=

x \ }{ k , k .

d

dxx

xxsin , . =


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30

1

2

1tan ,

1

dx

dx x

=+

x d

dxx

x xxcsc , . =

>1

2

1

11

3 Derivatives of Logarithmic and Exponential Functions.

(a) 1ln log ,ed dx xdx dx x

= = for any positive real number .x

(b)1

log ,ln

a

dx

dx x a= for any positive real number ,x where a is any positive

real number with 1.a

(c) ,x xd

e edx

= for any real number .x

(d)

(ln ) ,

x xd

a a adx = for any real number .

4 Derivatives ofHyperbolic Functions.

sinh cosh ,d

xdx

= x 2coth csc h ,d

xdx

= x

cosh sinh ,d

x xdx

= x sec h sec h tanh ,d

x xdx

= x

2tanh sec h ,d

xdx

= x csc h csc h coth ,d

x x xdx

=

x 0x

3.3 The Chain Rule

Theorem: (The Chain Rule): Let ( )f x and ( )g x be real functions. If ( )g x is

differentiable at , and ( )f x is differentiable at ( ),g x then ( )( )f g xo is

differentiable at x with

( ) '( ) '( ( )) '( )f g x f g x g x=o .

In terms of Leibniz notation, if ( )y f u= and ( ),u g x= then

dy dy du

dx du dx= ,

r

Example: Find the derivatives of the following function. Simplify your answers

whenever possible. (i). Let 43 )73()( +== xxfy . (ii) xexxfy +== 2sec)( .

Solution:

(i). Let 73 3 += xu . Then 4)( uufy == . So


31/51


31

)9(4 23 xudx

du

du

dy

dx

dy== )9())73(4( 233 xx += 332 )74(36 += xx

(ii). Letxexu += 2sec . Then uy= .By the chain rule, we have

dx

du

du

dy

dx

dy= xexx

u+= tansec2(

2

1 2 )x

x

ex

exx

+

+=

2

2

sec2

tansec2.

3.4 Higher Order Derivatives

Let ( )f x be a differentiable function, then we obtain '( );f x i.e., the first

derivative of ( )f x with respect to .x This derivative may itself be a differentiable

function; if so, we obtain the second derivativeof ( )f x with respect tox , and wewrite

2(2 )

2

( ) ( )''( ) ( ) .

d df x d f xf x f x

dx dx dx

= = =

In general, the n-th derivative of ( )f x with respect to ,x for any positive integer n, is

defined by induction as follows:

))x(f(dx

d)x(f )1n()n( =

where )x(f

)1n(

denotes the (n-1)th derivative of f and is denoted by

( )f xd f x

dx

d

dxf xn

n

n

n

n

( ) ( )( )

( ) .= =

Example:Let 4( )p x x= and let ( ) ,nq x ax= a Find ( ) ( )kp x for 1,2,3,4,5,k=

and find ( 1) ( ),nq x ( ) ( )nq x and ( 1) ( ).nq x+

Solution We see that (1) 3( ) '( ) 4 ,p x p x x= = (2) 3 2( ) (4 ) 13 ,p x d dx x x= = (3) ( )p x = 2(12 ) 24 ,d dx x x= (4) ( ) (24 ) 24,p x d dx x= = and (5) ( ) (24) 0.p x d dx= =

Similarly, we have (1) 1( ) ,nq x anx = (2) 2( ) ( 1) ,nq x an n x = (3) 3( ) ( 1)( 2) ,nq x an n n x = and hence, we have

( 1) ( ) ( 1)( 2) 3 2 ,nq x an n n x = L ( ) ( ) ( 1)( 2) 3 2 1 !,nq x an n n an= =L

( 1) ( ) 0.nq x+ =

3.5

Implicit Differentiation


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32

When a function is specified by an equation of the form ( )f x= , we say that is

determined as an explicit function of .x However, some equations involving the two

variables x and determine interesting relation ships between these variables even

though the equations cannot be brought into the explicit form ( )y f x= . For instance,2 2 25x y+ = and sin( ) 1xy = .

However, we can still regard it as defining as one or more functions of

implicitly, even if we cannot solve for these functions explicitly.

However we can find the derivative dy dx by a technique called implicit

differentiation. The main idea is to differentiate the given equation with respect to ,x

noting that is a function of .x

We summarize the procedures as follows:

(i) impose a condition on the variable y so that the given equation implicitlydefines y as a differentiable function of ;x

(ii) differentiate both sides of the given equation, remembering that the factor

dy dx will occur in any differentiation of a term involving the function ,

according to the chain rule; and

(iii) solve the resulting equation for dy dx .

Example: Find dy dx if 3sin cos .x y y= +

Solution By differentiate the equation 3sin cosx y y= + with respect to , weobtain

3( sin ) ( cos ),d d

y x y ydx dx

= +

2sin cos 3 sindy dy dy

x y x y ydx dx dx

+ =

2(3 sin sin ) cos .dy

y x y xdx

=

Hence we conclude that

2

cos.

3 sin sin

dy y x

dx y y x=

3.6 The Mean Value Theorem

Theorem:(Rolles Theorem): Let ( )f x be a continuous function on the closed

interval [ , ]a b and differentiable on ( , ).a b If ( ) ( ),f a f b= then there exists ( , )c a b

such that '( ) 0.f c =

In other worlds,Rolles Theorem asserts that if function is continuous on a closed

interval and differentiable everywhere except possibly at endpoints and if the values

of the function are equal at the endpoints, then the graph of the function must have at

least one horizontal tangent.:


33/51


33

The hypotheses of the Rolles Theorem are all necessary for the conclusion: if

( )f x fails to be continuous at even one point of [ , ],a b or fails to be differentiable at

even one point of ( , ),a b then there may be no horizontal.tangent line

TheRolles Theorem gives no indication of how many points c there may be on

the curve between a and b where the tangent is horizontal. If the graph of ( )f x is astraight horizontal line, then every point on the line between a and b has the required

property. In general, there may be one or more than one point.

Theorem (Mean Value Theorem): Let ( )f x be a continuous function on [ , ]a b and

differentiable on ( , ).a b Then there exists ( , )c a b such that

( ) ( )'( ) .

f b f af c

b a

=

Example: Show that sin x< for all 0.x> Solution Since 1 sin 1x for all x ,it follows that

sinx x< for all x >1.

Next we consider (0,1].x Clearly, ( ) sinf x x= is differentiable and continuous on

(0,1].x For any (0,1],x we see that there exists (0, )c x such that

( ) (0) sin sin 0'( ) cos( ) 1 sin .

0 0

f x f xf c c x x

x

= = < for any , ( , )x y a b with x y> (if

( ) ( )f x f y< for any , ( , )x y a b withx y< ).However, if ( ) ( )f x f y for any , ( , )y a b with y> we say ( )f x is non-

decreasing on ( , )a b ; and if ( ) ( )f x f y for any , ( , )y a b with ,y> then we say

( )f x is non-increasing on ( , ).a b

Theorem: Let ( )f x be differentiable on an open interval ( , ),a b and continuous on

[ , ].a b Then

(i) if '( ) 0f x > for all ( , ),a b then ( )f x is increasing on ( , );a b

(ii) if '( ) 0f x < for all ( , ),a b then ( )f x is decreasing on ( , );a b

(iii) if '( ) 0f x for all ( , ),a b then ( )f x is non-decreasing on ( , );a b

(iv) if '( ) 0f x for all ( , ),a b then ( )f x is non-increasing on ( , ).a b

Example: Show that 3( )f x x= is increasing on .

Solution Firstly, we note 2'( ) 3 .f x x= Clearly, '( ) 0f x > for all ( ,0) (0, ),x U

this implies ( )f x is increasing throughout ( ,0) and (0, ). Moreover, for any real

numbers yx, withy


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Theorem: Let ( )f x and ( )x be differentiable for all Mx> for some M . If

'( ) 0g x for all Mx> and lim ( ) lim ( ) 0,x xf x g x = = then

( ) '( )lim lim ,( ) '( )x x

f x f x

x g x =

provided the second limit exists.

Theorem:Let ( )f x and ( )g x be differentiable for all Mx> for some M .

If '( ) 0g x for all Mx> and lim ( )x f x = or lim ( )x f x = and

lim ( )x g x = or lim ( ) ,x g x = then

( ) '( )lim lim ,

( ) '( )x x

f x f x

g x g x

=

provided the second limit exists.

Remarks

(i) We can sometimes handle the indeterminate forms of type 0 and byusing algebraic manipulations to convert them into the indeterminate form of type

0 0 or . Here again, we do not mean to suggest that there is a number 0

or any more than we mean to suggest that there is a number 0 0 or .These forms are not numbers but descriptions of functions behavior.

(ii) Limits that lead to the indeterminate forms of type 0 00 , , and 1

can besometimes be handled by taking logarithms first. We use lHspitals rule to find

the limit of the logarithm and the exponential to find the original function

behavior; i.e., if lim(ln ( )) ,x a

f x L

= < then

ln ( )lim ( ) lim .f x Lx a x a

f x e e

= =

Here, amay be eitherfiniteor infinite.

Example:Evaluate each of the following limits ( if it exits)::

(i)0

sinlim .x x

(ii).10

lnlim .

xx

+ (iii)

ln(2 )lim .

x

x

e

x+

(vi) ( )1lim 1 .x

xx

+ (v)0

lim .xx

+

Solution

(i).0 0 0

sin coslim lim lim cos 1.

1x x x

x xx

= = =

(ii).2

1

1 10 0 0

lnlim lim lim ( ) 0.x

x x xx x

xx

+ + + = = =

(iii). (2 )ln(2 )lim lim lim lim lim1 1.1 2

x xx x x

x xx x x x x

e ee e ex e e

++ = = = =+


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36

(iv). Find ( )1lim 1 .x

xx

+

Let xy )1

1( += . Then )1

1ln()ln(x

xy += .

=

)ln(lim yx

( )( ) ( )( ) ( ) ( )

( )

2

2

1 11

1

1 1 1

1ln 1 1lim ln 1 lim lim lim 1.

1

xx x

xx x x x

x xx

x

+++ = = = =

+

Hence, eeeeyy

y

xx

x

x

x =====+

1)ln(lim

lnlimlim)1

1(lim

(v). Consider0

lim .xx

x+

Firstl let xxy= Then xxy lnln = .

=

yx

lnlim

2

1

1 10 0 0 0 0

lnlim ln( ) lim ln lim lim lim ( ) 0.x x

x x x x x

x x

xx x x x

+ + + + + = = = = =

Therefore,

=

x

xxlim 1limlim

0lnlim

ln ====

eeeyy

y

xx

x

3.7 Differentiability of Inverses

Now let us look at inverse function from the point of view of differentiability. We

recall from Theorem 1.4.6 that a real function ( )f x is one-to-one on an interval if andonly if the function ( )f x has the inverse on that interval, denoted by 1( ).f x We first

have the following interesting result.

Theorem: Suppose that a function f has an inverse g. Assume that f is differentiable

on an open interval I and f(a) = b. Then g(b) =)a('f

1.

or,

)x('f

1))x(f()'f( 1 =

or, if we write 1 ( ),y f x= then ( ) ,f x y= and hence we have

1 .dy dx

dx dy=

Example:If ( ) 2 sin ,f x x x= + find 1( ) '( 1).f +

Solution Firstly, we see that ( )f x is one-to-one since '( ) 2 cos 0f x x= + > for all

all x and hence the inverse exists. Furthermore, we notice that2

( ) 1,f = +

hence, by the above theorem we have


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1( ) '( 1).f + =)

2('f

1

=

2cos2

1

+

=2

1

3.8 Some Applications of Derivatives

Definition: (Local Extreme Value) A real valued function ( )f x has a local maximum

value(or relative maximum value) at 0x= in its domain Df if there exists an open

interval Icontaining 0 such that

0( ) ( )f x f x for all x I Df .

Similarly, ( )f x has a local minimum value (or relative minimum value) at 0x x= inits domain Df if there exists an open interval Icontaining 0 such that

0( ) ( )f x f x for all x I Df .

Local maximum and minimum values are called local extreme values.

We can extend the definition of local extreme value to the endpoint of intervals

by defining ( )f x to have a local maximum or local minimum value at the endpoint

0x= if the appropriate inequality holds for all x in some half-open interval in its

domain containing 0x .

The graph of a real function ( )f x with local maximum values at ,x b= ,d= k=

and ,q= and with local minimum values at ,a= ,c= x h= and .p=

Definition: (Absolute Extreme Values): Let 0x be a real number, and let ( )f x be a

real-valued function with domain fD containing 0 . Then f has an absolute

maximum value(orglobal maximum value) at 0 D if

0( ) ( )f x f x for all x Df ,


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38

and the real value0

( )f x is called the maximum value of ( )f x on .f

D Similarly,

( )f x has an absolute minimum value(global minimum value) at 0 fx D if

0

( ) ( )f x f x for all ,f

x D

and the real value 0( )f x is called the minimum value of ( )f x on .fD

The maximum and minimum values of ( )f x are called extreme valuesof ( ).f x

The above figure shows the graph of a real function ( )f x with absolute maximum

and minimum values at b= and ,x a= respectively.

Example: Determinethe extreme values of 2( )f x x= on ( 1,1).

Solution: Note that ( ) 0f x for all ( 1,1) since 2 0x for all x ( 1,1).

Hence ( )f x has the absolute minimum value at 0.x= However, since2

1x < for all ( 1,1), and 2 2a b< for any 0 1,a b < < and any 1 0,b a < < it follows that( )f x has no absolute maximum values on ( 1,1).

However,f has an absolute maximum f(1) =1 on [-1,1].

Definition: Letf(x) be a real-valued function. A critical pointof the function ( )f x is

an element fc D such that either '( ) 0f c = or '( )f c fails to exist.

Example:Find the critical points of 3 5( ) (4 ).f x x x= Solution First we have

2 5 3 535 2 5

12 8'( ) (4 ) .

5

xf x x x x

x

= =

Therefore '( ) 0f x = when 32,x= and '( )f x does not exist when 0.x= Thus, the

critical points of ( )f x are 32

x= and 0.x=

Theorem: (The Closed Interval Method) .Let ( )f x be a continuous function on

[ , ].a b The extreme values (either local or absolute) of ( )f x on [ , ]a b must occur

either

(i). at an endpoint oftheclosed interval [ , ];a b or

(ii).at point [ , ],x a b where f x' ( ) ;= 0 or(iii).at point [ , ],x a b where '( )f x fails to exist.


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39

Example:Find the maximum and minimum values of x2x3)x(f 32

= on [ 1,1]. Solution: Firstly, we see that ( )f x is a continuous function on [ 1,1].

Furthermore, we have1 3 1 3

'( ) 2 2 2( 1)f x x x

= = .

Notice that '( )f x is not well-defined at 0,x= and '( ) 0f x = when 1.x=

Hence 0x= and 1= (it also happens that 1= is an endpoint of the domain of( ))f x are the critical points of ( ).f x

Since the extreme values must occur either at 1,x= 0x= or 1.x= Thus,

from ( 1) 5,f = (0) 0f = and (1) 1.f = we know that ( )f x has a maximum value 5at 1x= and a minimum value 0 at 0.x=

Theorem (First Derivative Test): Let ( )f x be a real continuous function, and let

0 .D Then the following test applies to ( ).f x

At a critical point 0 :x

(i) if '( ) 0f x > for 0 ,x< and '( ) 0f x < for 0 ,x x> then ( )f x has a local

maximum value at0;x x=

(ii) if '( ) 0f x < for 0 ,x< and '( ) 0f x > for 0 ,x x> then ( )f x has a local

minimum value at 0 ;x x=

(iii) if '( )f x has the same sign on both sides of 0 ,x x= then ( )f x has no local

extreme value at 0 .x x=

At a left endpoint 0 :x

(i) if '( ) 0f x < for 0 ,x> then ( )f x has a local maximum value at 0 ;x x=

(ii) if '( ) 0f x > for 0 ,x> then ( )f x has a local minimum value at 0 .x x=

At a right endpoint 0 :x

(i) if '( ) 0f x < for 0 ,x< then ( )f x has a local minimum value at 0 ;x x=

(ii) if '( ) 0f x > for 0 ,x< then ( )f x has a local maximum value at 0 .x x=

Example*:Find and classify the local and absolute extreme values of the

function f(x)=x3-3x + 15 on the interval [-2,2].

Solution: f(x)=3x2-3.

Find the critical points by solving f(x)=3x2-3=0.

So x = -1 , 1 are the critical points.

-1 1

x+1 - + +

x-1 - - +

f(x) + - +


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40

Hence f has a local maximum atx = -1and a local minimum at x = 1.

f(-1)=17 ,local maximum. f(1)=13 , local maximum.

f(-2)=13, f(2)=17.

Abslute maximum is f(2)=17 and absolute minimum is f(1)=13.

(The second Derivative Testfor Local Extreme Values)

Let f x( ) be differentiable on an interval I containing 0x= and f x' ( ) also

differentiable at 0 .x x= Suppose that 0'( ) 0.f x =

(i) If 0''( ) 0,f x < then )( 0xf is a local maximum.

(ii) If 0''( ) 0,f x > then )( 0xf is a local minimum.

(iii) If 0''( ) 0,f x = then no conclusion can be drawn.

Example and exercise: By using the second derivatives test, find the LOCALmaximum and minimum of the following functions whose derivatives are given:

(i). 184272)( 23 +++= xxxxf . =)(' xf )7)(2(6 ++ xx

(ii). 196234

1)( 234 ++= xxxxxf . )8)(4)(3()(' ++= xxxxf

4.1 Riemann Sums - The Definite Integral

Definition: Let { }0 1, , , nP x x x= K be a partition of [ , ]a b , and let f x( ) be a real

function defined on [ , ].a b ARiemann sum nS of f x( ) on [ , ]a b with respect to Pis

a sum of the form

1

1

( )( ),n

n i i i

i

S f t x x =

=

where ti is an arbitrary element of 1[ , ]i ix for all , 1, 2, , .i n= K

Area of the first rectangle = f(t1)(x1-x0),etc.


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41

Definition: (Definite Integral) Let ba, with ,b a> and let f x( ) be a function

defined on closed interval [ , ]a b . We say that ( )f x is Riemann integrable or simply

integrable on [ , ]a b if the limit of theRiemann sums exists. The value of this limit is

called definite integral of ( )f x over [ , ],a b denoted as

10

1

( ) lim ( )( ),b n

i i iP

ia n

f x dx f t x x

=

= where [ ]1,i i it x x .

Remarks

1. We stress that the definite integral of ( )f x over [ , ]a b is a number; it is not a

function of .x It depends on the numbers a and b and the particular function

( ).f x

2. The definite integral does not depend on the variable ,x which is a dummy

variable, thus we have

( ) ( ) .b b

a a

f x dx f t dt=

The real numbers a and b are called the limits of integration; a is the lower

limit and b is the upper limits.

4.2 Basic Properties of The Definite Integral

In this section, we list some of the most important properties of the definite integral.

Theorem . Leta,b be real numberswith .b a> and f x( ) a function defined on the

closed interval [ ]a b, . If ( )f x is continuous on [ ], ,a b then ( )f x is integrable on

[ ], .a b

Theorem: Let a, b be real numbers with .b a> Let f x( ) and g(x) be continuousfunctions defined on the closed interval ],[ ba and , be any real constants. Then

the following properties hold true.

(i) ( ) 0.a

af x dx=

(ii) ( ) ( ) ( ) ( ) .b b ba a a

f x g x dx f x dx g x dx + = + (iii) f x( ) is integrable on each subinterval [ , ]c d of [ , ].a b Moreover

( ) ( ) ( )b c b

a a cf x dx f x dx f x dx= + for all ( , ).c a b

(iv) ( ) ( ) .a b

b af x dx f x dx=

(v) If ( ) ( )f x g x for all [ ], ,x a b then ( ) ( ) .b b

a af x dx g x dx

(vi) ( ) ( ) .b b

a af x dx f x dx


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42

(vii) The integral of an odd function over an interval symmetric about zero is

zero; i.e., if ( ) ( )f x f x = for all [ , ],x a a then ( ) 0.a

af x dx

=

(viii) The integral of an even function over an interval symmetric about zero is

twice the integral over the positive half of the interval; i.e., if ( ) ( )f x f x =

for all [ , ],a a then0

( ) 2 ( ) .a a

af x dx f x dx

=

4.3 Areas of Plane Regions

In this section we the use of definite integrals to represent plane areas. We recall that

if ( ) 0,f x the definite ( )b

af x dx measures the area between the graph of ( )f x and

the x-axis from x a= to x b= with .b a> However,if ( ) 0f x < for all [ , ],x a b

then the definite integral ( )b

af x dx is negative. Thus, in order to express the total

area bound by the curve ( ),y f x= 0,y= x a= and b= , counting all the area

positively, we should integrate the absolute value of ( );f x i.e., ( ) .b

af x dx

If A and B are the bounded areas illustrated in the diagram above, then

( ) ,b

af x dx A B=

and the area bound by the curve ( ),y f x= 0,y= x a= and b= is

( ) .b

af x dx A B= +

Example: Consider the area bounded by sin ,y x= 0,y= 0x= and 3 2.x = Since

sin 0x> for all [0, ],x and sin 0x< for all [ ,3 2],x it follows that

sin 0 ,sin

sin 3 2.

x xx

x x

=

<

Then the area bounded by sin ,y x= 0,y= 0x= and 3 2x = is

3 3

2 2

0 0sin sin ( sin ) .x dx x dx x dx

= +


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Suppose that a plane region R is bounded by two graphs of continuous functions,

( )f x= and ( ),y g x= and the vertical straight lines a= and x b= with .b a> If

)x(g)x(f for all [ , ],x a b then the area A of R is the bounded area above x -axis shown in f the following figure.

Then the area A is( ) ( ) ( ) ( ) .

b b b

a a af x g x dx f x dx g x dx =

More generally, if the restriction ( ) ( )f x g x is removed, then the total area lying

between ( )y f x= and ( ),y g x= and the vertical straight lines a= and b= with,b a> is given by

( ) ( ) .b

aA f x g x dx=

In view of figure above, we see that ( ) ( )f x g x for all [ , ] [ , ],a c d b U and

( ) ( )x f x for all [ , ].x c d Then the total bounded area lying between ( )f x=

and ( ),g x= and the vertical straight lines a= and b= with ,b a> is given by

( ) ( )

( ) ( ) ( ( ) ( )) ( ) ( ) .

b

a

c d b

a c d

A f x g x dx

f x g x dx f x g x dx f x g x dx

=

= + +

4.4 The Fundamental Theorem of Calculus

Theorem: (The FirstFundamental Theorem of Calculus)

Let ( )f x be a continuous on [ ]a b, , then the function ],[: bag defined by


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( ) ( )x

ag x f t dt=

is continuous on [ ]a b, , and differentiable on ( , ),a b and

( )'( ) ( ) ( )x

a

dx f t dt f x

dx= = for all [ , ].a b

Example:Find ( )1 sin .xd

t dtdx

Solution: Since ( ) sinf x x= is continuous throughout , hence

( )1 sin sinxd

t dt xdx

= for all x .

Class exercise: =x

2 ?tdt7cosdx

d --------- cos7x

=x

1?tdt5costsin

dx

d --------- sin x cos5x

Definition: A function ( )F x is called an antiderivative of a given function ( )f x on

an interval Iif ( )F x is differentiable on Iand

( ) ( )d

F x f xdx

= for all .I

Example: 2( )f x x= , then it is not difficult to discover an antiderivative of ( ).f x Let31

3( ) ,F x x= then '( ) ( ).F x f x= SoF(x)is an antiderivative off(x).

So is each 313

( )H x x c= + with c is any real number.

More examples:

( )f x Antiderivative of ( )f x

( ) sinf x x= ( ) cosF x x C= +

( ) xf x e= ( ) xF x e C= +

( ) 1f x x= ( ) lnF x x C= +

( ) sinhf x x= ( ) coshF x x C= +

Theorem:(The Second Fundamental Theorem of Calculus):

Let ( )f x is a continuous on [a,b] and ( )F x is an antiderivative of ( )f x on

[ ], .a b Then

( ) ( ) ( ) ( ),b b

aaf x dx F x F b F a= =


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45

Example:Evaluate dxx5

2

38 .

Solution: LetF(x)=2x4, then 3x8)x(Fdxd = , hence

dxx5

2

38 =F(x)2

5

=F(5) - F(-2)=2(625 -16)= 1018

4.5 Indefinite Integrals

Definition: Let ( )f x be a continuous function, and let ( )F x be an antiderivative

function of ( ).f x The most general anti-derivative form for a function f x( ) is

denoted by

f x dx( ) ,

and is called an indefinite integral of ( ),f x and we have

f x dx F x C( ) ( ) ,= +

where Cis an arbitrary constant.

Polynomial Functions

(i). ,a dx ax C = + where a .(ii).

1

,1

nn xdx C

n

+

= ++ where { }\ 1 .n Z

(iii). ( )2 1

0 1 0 1 ,2 1

nn

n n

x xa a x a x dx a x a a C

n

+

+ + + = + + + ++ L L

where n is a positive integer and naaa ,...,, 10 .

(iv). For each 1r ,1

.1

rr xdx C

r

+

= ++

Trigonometric Functions

(i). sin cos .dx x C = + (ii). cos sin .x dx x C= + (iii). tan ln sec .x dx x C= + (iv). cot ln sin .dx x C = + (v). sec ln sec tan .dx x x C = + + (vi). csc ln csc cot .x dx x x C= + (vii). 2sec tan .x dx x C= +


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46

(viii). 2csc cot .x dx x C= + (ix). sec tan sec .x x dx x C= + (x). csc cot csc .x x dx x C= +

Exponential Functions

(i). .x xe dx e C = +

(ii). ,ln

xx aa dx C

a= + where 1.a>

Hyperbolic Functions

(i). sinh cosh .dx x C = + (ii). cosh sinh .dx x C = +

(iii). ( )tanh ln cosh .x dx x C= + (iv). coth ln sinh .dx x C = +

(v). 2sec h tanh .dx x C = + (vi). 2csc h coth .dx x C = +

Algebraic Functions

(i).1

ln ,dx x C x

= + }0{\x .

(ii). 12

1 tan .

1dx x C

x

= ++

(iii). 122

1 1 ln ,

11

xdx C

xx

+= +

x \ }1,1{

(iv). 122

1 1 ln ,

11

xdx C

xx

= +

+ x \ }1,1{

(v). 12

1 sinh .

1dx x C

x

= ++

(vi).1

2

1

sin ,1 dx x C x

= + 1 1.x < < .

Example:Find 3 6 ,x dx and evaluate3

3

06 .x x dx

Solution First we see that

3 3 4 214

6 6 3 .x dx x dx x dx x x = =

By the Second Fundamental Theorem of Calculus, we have

3 33 4 214 00

81 276 3 27 .4 4

x

xx x dx x x =

= = = =


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47

4.6 Techniques of Integration

Integration by Parts

Theorem: Let ( )f x and ( )x be differentiable functions. Let ( )G x be an

antiderivative of ( )g x ( that is, )()(' xgxG = ). Then ( ) ( )f x g x is integrable, and

( ) ( ) ( ) ( ) ( ) '( ) .f x g x dx f x G x G x f x dx=

Furthermore, if we let ( )u f x= and dx)x(gdv= , then

.u dv uv vdu=

We note that the integration by parts formula can be applied to definite integrals

as well as to indefinite integrals. The corresponding formulation is

( ) ( ) ( ) ( ) ( ) '( ) ,b bx b

x aa af x g x dx f x G x G x f x dx

=

==

and

,b bx b

x aa au dv uv v du

=

==

Example: Find ln .x dx

Solution Let )ln(xu= , dxdv= . So dx

x

1du= , v= and therefore

( )ln (ln ) 1 ln 1 ln ,dx x x x x dx x x dx x x x C = = = +


Example:Find ln .x dx Solution: Let u=ln x , dv = xdx.

Then du = dxx

1, 2

2

1xv = .

Hence dxx

1)x2

1(xlnx2

1xdxlnx

22

= = Cx41

xlnx2

1 22+ #

Example:Find 2 3 .xe dx

Solution: let dxedv,xu x32 == . Then x3e3

1v,xdx2du == .

( ) ( )2 3 2 3 3 2 3 31 1 1 23 3 3 32 .x x x x xe dx x e e x dx x e xe dx= =

ow, consider dxxex3

. Let u=x , .dxedvx3

= So du=dx ,x3

e3

1v= .


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48

Then ( ) ( )3 3 3 3 3 3 31 1 1 1 1 13 3 3 3 3 91 ,x x x x x x xe dx x e e dx xe e dx xe e B= = = + where B

is an arbitrary constant.

Thus, we have

( )2 3 2 3 3 3 2 3 3 31 2 1 1 1 2 23 3 3 9 3 9 27 ,x x x x x x xe dx x e xe e B x e xe e C = + = + +

where 23

.C B= #

Example:Evaluate 30

cos .xe x dx

Solution: x3eu= , dxxcosdv= . dxe3du x3= , sinv= . So,

3 3 3 3 3cos (sin ) sin (3 ) sin 3 sin .x x x x xe x dx e x x e dx e x e x dx= = Next, consider dxxsine

x3 .

Let x3eu= , dxxsindv= .So, v= - cos x , dxe3du x3= . So,3 3 3 3 3sin ( cos ) ( cos )(3 ) cos 3 cos .x x x x xe x dx e x x e dx e x e x dx= = + .

So,

( )3 3 3 3cos sin 3 cos 3 cosx x x xe x dx e x e x e x dx= + 3 3 3sin 3 cos 9 cos .x x xe x e x e x dx= +

Therefore,3 3 310 cos sin 3 cos ,x x xe x dx e x e x= +

so,3 3 331

10 10cos sin cos .x x xe x dx e x e x= + +C.

Hence

( ) ( )3

3 3 3 33110 10

0 0

3 3 3cos sin cos 1 .

10 10 10

xx x x

x

ee x dx e x e x e

=

=

= + = = #

Integration by Substitution

Example: Consider the indefinite integral2xxe dx .

If we let

2

,u x= then 2 ,du

dx x= and hence 2 .du xdx= Therefore, we can simplify theindefinite integral into a simple form

2 2 21 1 12 2 2

2 2 .x x x ue dx xe dx e x dx e du= = = Since 1

2,u ue du e C = + where Cis an arbitrary constant, it follows that

2 212

.x xxe dx e C= + #Theorem:Let ( )f x be a continuous function and let )(xgu= be a differentiable

function. If ( )F x is an antiderivativeof ( )f x (that. is, )()(' xfxF = ),.then

= ))x(g(Fdx)x('g))x(g(f +C.


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49

Example: Evaluate2

4 1.

2

xdx

x

+

+

Solution: Let. xx2u 2 += then dx)1x4(du += and hence,

Cxx2lnCulnduu

1dx

xx2

1x4 22

++=+==+

+


Example: Find2

1,dx

a x if 0.a>

Solution: Let sin .x a u= Then cos ,dxdu

a u= and hence cos .dx a u du= Thus,

1

2

cos1

1 sin ,cos

a u du x

dx du u C C a u aa x

= = = + = +


Example: Find dxax + 22

1 if 0.a>

Solution: Let uax tan= , then uduadx 2sec= . Therefore,

+===+Cu

adu

adu

ua

uadx

a

11

1

sec

sec122

2

22 a

x

a

1tan1 = C+


Integrating Rational Functions-Partial Fractions

In this section we are concerned with indefinite integrals of the form

( ),

( )

P xdx

Q x

where ( )P x and ( )Q x are polynomials. The main technique for solving this type of

problems is using the partial fraction decomposition (see Section 1.0.5) with the help

of the formula

'( )ln ( ) ,

( )

f xdx f x C

f x= +

where ( )f x is a differentiable function and Cis an arbitrary constant.

Example:Find3 2

2

3.

1

x xdx

x

+

+

Solution By long devision, we see that3 2

2 2

3 33

1 1

x x xx

x x

+ += +

+ +. Hence


50/51


50

3 2

2 2

2 2

2 2

22 1

3 33

1 1

33

1 11 2 1

3 32 1 1

13 ln( 1) 3tan ,

2 2

x x xdx x dx

x x

xx dx dx dx

x xx

x dx dx dxx x

xx x x C

+ += +

+ +

= +

+ += +

+ +

= + + +


An Application: Work Done by Variable Force

When a force is applied to an object and it causes a displacement to the object, then

we say that some work has been done by the force.If the force F is constant and the

object moves in a straigh line in the direction of Fwith displacement d, then we

define the work done to be W = Fd. If Fis measured in Newton and the displacement

is in meter, then the work done isFd J, whereJrepresents Joule, that is

1J = 1 ewton.Meter.

If the force is a variable force and it causes an object to move in a straight line in the

direction of F with displacement d, what would be the work done?

Suppose that the force F is a continuous force acting along the x-axis and its

magnitude isF(x) at x, where F(x) is a continuous function,and it causes an object to

make a displacement fromx=a to x=b, we will define the work done by the force.

We first partition the interval [a,b] into nsubintervals by the points

bxxxax nn == ,110 ,...,,.

We then choose an arbitary point ],[ 1 iii xxt .Then the force at it is )( itF .If nis

large and 1= iii xxx , then the work done by the force in moving the object

from1ix to ix is approximately

)( ii tFW = ix .

Hence the total weork done is approximately

==

==n

ii

n

ii tFWW

11

)( ix .

Now let n and 01= iii xxx , as n for all i, then we define the work

done by the force in moving the object from x=a to x=b to be


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=

n

i

iin

xtFW1

)(lim = dxxFb

a)(

The integral exists, sinceF (x) is continous on [a,b].

Example 4.7.5. A force of 55 N is required to hold a spring that has been stretched

from its natural length of 10 cm to a lengh of 20 cm. How much work work is done in

stretching the spring 16 cm from its natural length to 22 cm?

Solution: According to Hookes law, the force required to hold the spring stretched x

meter beyong its natural lengh is

F(x)=kx,

where kis the spring constant. (hereFis measured in Newton).

When the spring is stretched 10 cm from its natural lenght to 20 cm, the amountstretched is 10 cm = 0.1 m. Thus,

55)1.0( =F

and so k1.055= ,So,

550=k .Hence,

xxF 550)( = Therefore, the work done is

16.0

22.0

2

550550 2

22.0

16.0xxdxW == = 6.27J.

c01 calculus of one variable _t1_1112

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