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School of Distance Education

Calculus Page 2

UNIVERSITY OF CALICUT

S C H O O L O F D I S T A N C E E D U C A T I O N

STUDY MATERIAL

B.Sc. MATHEMATICS

III SEMESTER

CORE COURSE - CALCULUS

Lessons prepared by:

Sri.Nandakumar M.,

Assistant Professor

Dept. of Mathematics,

N.A.M. College, Kallikkandy.

Lay out & Settings

Computer Section, SDE

Copyright

Reserved

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CALCULUS

CONTENTS

Module I

1 Foundations of Calculus 5

2 The Rolles and Mean Value Theorems 26

Module II

3 Asymptotes 34

4 Optimizations 44

5 Integration Part One 54

Module III

6 Integration Part Two 65

Module IV

7 Integration Part Three 77

8 Moments and Work 85

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MODULE - ICHAPTER 1

FOUNDATIONS OF CALCULUS1. Quick Review of Functions

In this chapter we first discuss function, one of the most important concepts inmathematics.

Definition Let A and B be two non-empty sets. A function (map, mapping ortransformation)from A to B is a rule which assigns to each element of A a unique elementof B. The setA is called the domain of the function, and the set B is called the target set.

Notations/Definitions/Remarks Functions are ordinarily denoted by symbols.

Let f denote a function from A into B. Then we write

:f A Bfi

which is read: f is a function fromA into B, or f takesA into B, or f mapsA into B.

Suppose :f A Bfi and a A . Then ( )f a [which is read f of a] will denote the unique

element of B which f assigns to a. This element ( )f a in B is called the image of a under

f or the value offat a. We also say thatf sends or maps a into ( )f a . The set of all such

image values is called the range or image off, and it is denoted by Ran (f), Im (f) or( )f A . That is

Im( ) { :f b B= there exists a A for which ( ) }f a b=

Im( )f is a subset of the target set B.

Frequently, a function can be expressed by means of a mathematical formula:Consider the function which sends each real number into its square. We may describe this

function by writing2 2 2( ) or or = =af x x x x y x

In the first notation, x is called a variable and the letter f denotes thefunction. In the secondnotation, the barred arrow a is read goes into. In the last notation, x is called theindependent variable and y is called the dependent variable since the value of y willdepend on the value of x.

Furthermore, suppose a function is given by a formula in terms of a variable x. Then weassume, unless otherwise stated, that the domain of the function is or the largestsubset of for which the formula has meaning and that the target set is .

Suppose :f A Bfi . If 'A is a subset of A , then ( ')f A denotes the set of images of

elements in ' and if 'B is a subset of B , then 1( ')f B- denotes the set of elements ofA

each whose image belongs to '.B That is,1( ') { ( ) : '} and ( ') { : ( ) '}.-= = f A f a a A f B a A f a B

We call ( ')f A the image of ' , and call 1( ')f B- the inverse image or preimage of 'B .

Example Consider the function 3( )f x x= , i.e., fassigns to each real number its cube. Then

the image of 2 is 8, and so we may write (2) 8f = . Similarly, ( 3) 27f - = - , and (0) 0f = .

Identity Function

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Consider any set A . Then there is a function from A into A which sends eachelement into itself. It is called the identity function on A and it is usually denoted by1A or AI or simply 1. In other words, the identity function 1 : fiA A A is defined by

1 ( ) =A a a

for every element a A .

Composition of FunctionsConsider functions :f A Bfi and :g B Cfi , that is, where the target set B off is the domain

ofg. This relationship can be pictured by the following diagram:f gA B Cfi fi

Let a A ; then its image ( )f a underf is in B which is the domain ofg. Accordingly, we can

find the image of ( )f a under the functiong, that is, we can find )( )g f a . Thus we have a

rule which assigns to each element a in A an element )( )g f a in Cor, in other words,fandg gives rise to a well defined function from A to C. The new function is called thecomposition offandg, and it is denoted by

g fo

More briefly, if :f A Bfi and :g B Cfi , then we define a new function :g f A Cfio by

( )( ) ( ( ))g f a g f ao

Here is used to mean equal by definition.

Note that we can now add the function g fo to the above diagram offandg as follows:

ONE TO ONE, ONTO AND INVERTIBLE FUNCTIONSDefinition A function :f A Bfi is said to be one-to-one (written 1-1) if ( ) ( ')f a f a=

implies 'a a= .

That is, :f A Bfi is 1-1 if different elements in the domainA have distinct images.

Definition A function :f A Bfi is said to be an onto function if

," $ b B a A ( )f a b= .

That is, :f A Bfi is onto if every element of B is the image of some elements in A or, in

other words, if the image of f is the entire target set B. In such a case we say that f is afunction of A onto B or that f mapsA onto B.

Remarks

The term injective is used for for a one-to-one function, surjective for an ontofunction, and bijective for a one-to-one correspondence.

If :f A Bfi is both one-to-one and onto, then f is called a one-to-one

correspondence between A and B. This terminology comes from the fact that eachelement of A will correspond to a unique element of B and vice versa.

2. LIMITS OF FUNCTIONSNotation

0

lim ( )x x

f x Lfi

= denotes ( )f x approaches the limit L as x approaches 0x .

CA B

g fo

f g

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Limits of Powers and Algebraic CombinationsTheorem 1 : Properties of Limits

The following rules hold if lim ( )x c

f x Lfi = and lim ( )x c g x Mfi = (L andMreal numbers).

1. Sum Rule: lim[ ( ) ( )]x c

f x g x L Mfi

+ = +

i.e., the limit of the sum of two functions is the sum of their limits.

2. Difference Rule: lim[ ( ) ( )]x c

f x g x L Mfi

- = -

i.e., the limit of the difference of two functions is the difference of their limits.

3. Product Rule: ]lim ( ) ( )fi

=x c

f x g x L M

i.e., the limit of the product of two functions is the product of their limits.

4. Constant Multiple Rule: lim ( )x c

kf x kLfi

= (any number k)

i.e., the limit of a constant times a function is that constant times the limit of the function.

5. Quotient Rule: ( )lim( )x c

f x Lg x Mfi

= , 0M

i.e., the limit of the quotient of two functions is the quotient of their limits, provided thelimit of the denominator is not zero.

6. Power Rule: If m and n are integers, then

lim[ ( )]m mn n

x cf x L

fi= , provided

mnL is a real number.

i.e., the limit of any rational power of a function is that power of the limit of the function,provided the latter is a real number.

Example Find3 2

2

4 3lim

5x cx x

xfi+ -

+.

Solution We note that limx c x cfi = and limx c k kfi = , where k is a constant (we will prove thisin Example 2 of Chapter 2). Now we use various parts of Theorem 1 for the evaluation ofthe required limit.

3 23 2

2 2

lim( 4 3)4 3

lim5 lim( 5)

fi

fi

fi

+ -+ -

=+ +

x c

x c

x c

x xx x

x x, using quotient rule

3 2

2

lim lim(4 ) lim( 3)

lim lim 5

fi fi fi

fi fi

+ + -=

+

x c x c x c

x c x c

x x

x, using sum rule

3 2

2

lim 4 lim lim( 3)

lim lim 5

fi fi fi

fi fi

+ + -=

+

x c x c x c

x c x c

x x

x

, using rule 4

3 2

2

4 3

5

+ -=

+

c c

c, since

) )2 2lim lim limfi fi fi

= = =x c x c x c

x x x c c c , using product rule

and )3

3 3lim limfi fi

= =x c x c

x x c , using power rule.

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Example 2 Evaluate 22

lim 4 3fi -

-x

x .

Solution

2

2lim 4 3fi -

-x

x )1/ 222

lim 4 3fi -

= -x

x

( )1/ 2

2

2lim 4 3fi -

= - x

x , using power rule with 12

=n

)22

lim 4 3fi -

= -x

x

2

2 24 lim lim 3

fi - fi -= -

x xx ( )

2

24 lim 3

fi -= -

xx

24( 2) 3= - - 16 3= - 13=

Theorem 2: Limits of Polynomials by substitution

If 11 0( )-

-= + + +Ln n

n nP x a x a x a , then

11 0lim ( ) ( )

--

fi= == + + +Ln nn n

x cP x P c a c a c a .

Theorem 3: Limits of Rational Functions by substitution (if the limit of the denominator isnot zero)

If ( )P x and ( )Q x are polynomials and ( ) 0Q c , then

( ) ( )lim

( ) ( )fi=

x c

P x P c

Q x Q c.

Example3 23 2

2 20

(0) 4(0) 34 3 3lim .

55 (0) 5x

x x

xfi

+ -+ -= = -

+ +

.

This is the limit in Example 1 with 1= -c , now done in one step.

Identifying common factorsIt can be shown that if ( )Q x is a polynomial and ( ) 0,=Q c then ( )-x c is a factor of ( )Q x .

Thus, if the numerator and denominator of a rational function of x are both zero at =x c ,then ( )-x c is a common factor.

Eliminating Zero Denominators AlgebraicallyTheorem 3 applies only when the denominator of the rational function is not zero at the limitpoint c . If the denominator is zero, canceling common factors in the numerator anddenominator will sometimes reduce the fraction to one whose denominator is no longer zero

at c . When this happens, we can find the limit by substitution in the simplified fraction.

Example Evaluate2

21

2limx

x x

x xfi+ -

-.

Solution We cannot just substitute 1x = , because it makes the denominator zero. However,we can factor the numerator and denominator and cancel the common factor to obtain

2

2

( 1)( 2)2 2

( 1)

x xx x x

x x xx x

- ++ - += =

--, if 1x .

Thus, by applying Theorem 3, we obtain

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2

21 1

2 2 1 2lim lim 3

1fi fi+ - + +

= = =-x x

x x x

xx x.

Example Find 02 2

limhh

hfi+ -

.

Solution We cannot find the limit by substituting 0h = , and the numerator anddenominator do not have obvious factors. However, we can create a common factor in thenumerator by multiplying it (and the denominator) by the so-called conjugate expression

2 2h+ + , obtained by changing the sign between the square roots:

2 2h

h

+ -

2 2 2 2

2 2

h h

h h

+ - + +=

+ +

2 2

( 2 2)

h

h h

+ -=

+ +

( 2 2)

h

h h=

+ +, with a common factor of h

1

2 2=

+ +h, cancelling the factor h

Therefore, we obtain

0

2 2limh

h

hfi+ -

0

1lim

2 2h hfi=

+ +(The denominator is no longer 0 at 0h = , so we can

use Theorem 3 and can substitute h = 0)

1

2 0 2=

+ +

1

2 2=

Theorem 4: The sandwich Theorem Suppose that ( ) ( ) ( )g x f x h x for all x in some open

interval containing c, except possibly at x c= itself. Suppose also thatlim ( ) lim ( )x c x c

g x h x Lfi fi

= = .

Then lim ( )x c

f x Lfi

= .

Example Given that2 2

1 ( ) 14 2

x xu x- + for all 0x . Find

0lim ( )x

u xfi

.

Solution Since0

2

lim 1 1 0 14fi

- = - =

xx

and0

2

lim 1 1 0 12fi

+ = + =

xx

, the Sandwich Theorem

implies that0

lim ( ) 1fi

=x

u x .

Example Show that if lim | ( ) | 0,fi

=x c

f x then lim ( ) 0fi

=x c

f x .

Solution By assumption lim | ( ) | 0,fi

=x c

f x so that | ( ) |f x- and | ( ) |f x both have limit 0 as x

approaches c . This combining with the fact | ( ) | ( ) | ( ) |- f x f x f x gives lim ( ) 0fi

=x c

f x by the

Sandwich Theorem.

ExercisesFind the limits in Exercises 1-8.

1.12

lim(10 3 )x

xfi

- 2. 3 22

lim ( 2 4 8)x

x x xfi -

- + +

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3.2 / 3

lim 3 ( 2 1)x

s sfi

- 4.5

4lim

7x xfi -

5. 222

lim 5 6y

y

y yfi

+

+ + 6.

1984

4lim ( 3)x xfi - +

7. 1/ 30

lim(2 8)z

zfi

- 8.0

5lim

5 4 2h hfi + +

Find the limits in Exercises 9-15

9.23

3lim

4 3xx

x xfi -+

- +10.

2

2

7 10lim

2xx x

xfi+ +

-

11.2

21

3 2lim

2tt t

t tfi -+ +

- -12.

3 2

4 20

5 8lim

3 16y

y y

y yfi

+

-

13.3

42

8lim

16vv

vfi-

-14.

2

4

4lim

2xx x

xfi-

-

15.2

1

8 3lim

1xx

xfi -+ -

+

16. Suppose4

lim ( ) 0x

f xfi = and 4lim ( ) 3x g xfi = - . Find

a)4

lim( ( ) 3)x

g xfi

+ b)4

lim ( )x

xf xfi

c) 24

lim( ( ))x

g xfi

d)4

( )lim

( ) 1x

g x

f xfi -

17. Suppose that2

lim ( ) 4fi -

=x

p x ,2

lim ( ) 0fi -

=x

r x , and2

lim ( ) 7fi -

=x

s x . Find

a)2

lim ( ( ) ( ) ( ))x

p x r x s xfi -

+ + b)2

lim ( ) ( ) ( )x

p x r x s xfi -

c)2

( 4 ( ) 5 ( ))lim

( )xp x r x

s xfi -- +

Evaluate the limit for the given value of x and function f .

18. 2( )f x x= , 2x = - 19.1

( )f xx

= , 2x = -

20. ( ) 3 1f x x= + , 0x =

21. If 22 ( ) 2cosx g x x- for all x , find 0lim ( )x g xfi .

22. The inequalities2

2

1 1 cos 1

2 24 2

x x

x

-- < < hold for all values of x close to

zero. What if anything, does this tell you about 201 cos

limx

x

xfi

-

?23. Suppose that ( ) ( ) ( )g x f x h x for all 2x and suppose that

2 2lim ( ) lim ( ) 5x x

g x h xfi fi

= = - . Can

we conclude anything about the values of , ,f g and h at 2x = ? Could (2) 0f = ? Could

2lim ( ) 0

xf xfi = ? Give reasons for your answers.

24. If22

( )lim 1

x

f x

xfi -= , find (a)

2lim ( )

xf x

fi -and (b)

2

( )lim

x

f x

xfi -.

25. If20

( )lim 1x

f x

xfi= , find (a)

0lim ( )x

f xfi

and (b)0

( )limx

f x

xfi.

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3. FORMAL DEFINITION OF LIMITDefinition

(A formal Definition of Limit)

Let ( )f x be defined on an open interval about 0x , except possibly at 0x itself. We say that

0

lim ( )x x

f x Lfi

= , if, for every number 0e> , there exists a corresponding number 0d> such that

for all x with 00 | | | ( ) |d e< - < - we have to find a suitable 0d> so

that if 1x and x is within distance d of 0 1x = , that is, if

0 | 1| ,x d< - <

then ( )f x is within distance e of 2L = , that is

| ( ) 2 | e-

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Example Verify the following two important limits:

(a)0

0lim

x xx x

fi= (b)

0

limx x

k kfi

= (k constant).

Solutiona) Let 0e> be given. We must find 0d > such that for all x

00 | |x x d< - < implies

0| | e- be given. We mustfind 0d > such that for all x,

00 | |x x d< - < implies | | e-

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Example For the limit 5lim 1 2x xfi - = , find a 0d > that works for 1e= .

Solution We have to find a0d >

such that for allx

0 | 5 | | 1 2 | 1x xd< - < - - < .

We organize the search into two steps. First we solve the inequality | 1 2 | 1x - - < to find an

interval ( , )a b about 0 5x = on which the inequality holds for all 0x x . Then we find a value

of 0d > that places the interval 5 5xd d- < < + (centered at 0 5x = ) inside the interval ( , )a b .

Step 1 : We solve the inequality | 1 2 | 1x - - < to find an interval about 0 5x = on which the

inequality holds for all0

x x .

| 1 2 | 1x - - <

1 1 2 1- < - - such that for all x

0 | 2 |x d< - < | ( ) 4 | e-

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The inequality ( ) 4 e- that places the centered interval (2 , 2 )d d- + inside the

interval ( 4 , 4 )e e- + :

Take d to be the distance from 0 2x = to the nearer endpoint of )4 , 4 .e e- + In otherwords, take

{ }min 2 4 , 4 2d e e= - - + - ,the minimum (the smaller) of the two numbers 2 4 e- - and 4 2e+ - . If d has this or any

smaller positive value, the inequality 0 | 2 | d< -

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Exercises

Each of Exercises 1-8 gives a function ( )f x and numbers L , 0x , and 0e> . In each case, find

an open interval about 0x on which the inequality | ( ) | e- such that for all x satisfying 00 | | d< - c

8. ( ) , 0f x mx b m= + > ,L m b= + 0 1,x = 0.05e=

Each of Exercises 9-11 gives a function ( )f x , a point 0x , and a positive number e. Find

0

lim ( )x x

L f xfi

= . Then find a number 0d> such that for all x

00 | |x x d< - < | ( ) | e-

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4. EXTENSIONS OF LIMIT CONCEPTS

One-Sided LimitsDefinition (Informal Definition of Right-hand and Left-hand Limits)Let ( )f x be defined on an interval ( , )a b where .a b< If ( )f x approaches arbitrarily close to

L as x approaches a from within that interval, then we say that f has right-hand limit L

at a , and we write

lim ( )x a

f x L+fi

= .

Let ( )f x be defined on an interval ( , )c a where c a< . If ( )f x approaches arbitrarily close to

M as x approaches a , from within the interval ( , )c a , then we say that f has left-hand

limitM at a , and we write

lim ( )x a

f x M-fi

= .

Example For the function

( )| |

xf x

x= in Fig.6, we have

0lim ( ) 1

xf x

+fi= and

0lim ( ) 1

xf x

-fi= - .

A function cannot have an ordinary limit at an endpoint of its domain, but it can have aone-sided limit (This is illustrated in the following Example).

Fig. 7

x

y

-2 0 2

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Example The domain of 2( ) 4f x x= - is [ 2, 2]- ; its graph is the semicircle in Fig. 7 We

have2

2lim 4 0x x+fi - = and2

2lim 4 0x x-fi - =

The function does not have a left-hand limit at 2x = - or a right-hand limit at 2.x = Itdoes not have ordinary two-sided limits at either 2- or 2.

One-sided limits have all the limit properties listed in Theorem 1, Chapter 1. The right-hand limit of the sum of two functions is the sum of their right-hand limits, and so on. Thetheorems for limits of polynomials and rational functions hold with one-sided limits, as doesthe Sandwich Theorem.

The connection between one-sided and two-sided limits is stated in the followingtheorem.

Theorem 5 (One-sided vs. Two-sided Limits)

A function ( )f x has a limit as x approaches c if and only if it has left-hand and right-hand

limits there, and these one-sided limits are equal:

lim ( )x c

f x Lfi

= lim ( )x c

f x L-fi

= and lim ( )x c

f x L+fi

=

Infinite Limits

As 0x +fi , the values of f grow without bound, eventually reaching and surpassing every

positive real number. That is, given any positive real number B , however large, the values

of f become larger still (Fig.10). Thus f has no limit as 0x +fi . It is nevertheless

convenient to describe the behavior of f by saying that ( )f x approaches as 0x +fi . We

write

0 0

1lim ( ) lim

x xf x

x+ +fi fi= = .

In writing this, we are not saying that the limit exists. Nor are we saying that there is a real

number , for there is no such number. Rather, we are saying that0

1lim

x x+

fi

does not exist

because1

xbecomes arbitrarily large and positive as 0x +fi .

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As 0x-

fi , the values of

1

( )f x x= become arbitrarily large and negative. Given anynegative real number B- , the values of f eventually lie below B- . (See Fig. 10 ) We write

0 0

1lim ( ) lim

x xf x

x- -fi fi= = - .

Again, we are not saying that the limit exists and equals the number - . There is no real

number - . We are describing the behavior of a function whose limit as 0x -fi does notexist because its values become arbitrarily large and negative.

Example (One-sides infinite limits)

Find1

1lim

1x x+fi -and

1

1lim

1x x-fi -

Geometric Solution: The graph of

1

1y x= - is the graph of

1

y x= shifted 1 unit to the right.

Therefore,1

1y

x=

-behaves near 1 exactly the way

1y

x= behaves near 0:

1

1lim

1x x+fi=

-and

1

1lim

1x x-fi= -

-.

Analytic Solution: Think about the number 1x - and its reciprocal. As 1x +fi , we have

( 1) 0x +- fi and1

1xfi

-. As 1x -fi , we have ( 1) 0x -- fi and

1

1xfi -

-.

Example (Two-sided infinite limits)

Discuss the behavior of

a) 21

( )f xx

=near

0x =,

b)2

1( )

( 3)g x

x=

+near 3x = - .

Solution

a) As x approaches zero from either side, the values of2

1

xare positive and become

arbitrarily large (Fig.12). Hence20 0

1lim ( ) limx x

f xxfi fi

= =

b) The graph of2

1( )

( 3)g x

x=

+is the graph of

2

1( )f x

x= shifted 3 units of the left.

Therefore, g behaves near 3- exactly the way f behaves near 0. Hence

23 3

1lim ( ) lim

( 3)x xg x

xfi - fi -= =

+.

The function1

yx

= shows no consistent behavior as 0x fi . We have1

fi x

if 0x +fi ,

but1

fi -x

if 0x -fi . All we can say about 01

limx xfi

is that it does not exist. The function

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2

1y

x= is different. Its values approach infinity as x approaches zero from either side, so we

can say that0 2

1lim

x xfi

=

.

Example (Rational functions can behave in various ways near zeros of their denominators)

a)2 2

22 2 2

( 2) ( 2) 2lim lim lim 0

( 2)( 2) 24x x x

x x x

x x xxfi fi fi

- - -= = =

- + +-.

b)22 2 2

2 2 1 1lim lim lim

( 2)( 2) 2 44x x xx x

x x xxfi fi fi- -

= = =- + +-

.

c)2

2 2

3 3lim lim

( 2)( 2)4x x

x x

x xx+ +fi fi

- -= = -

- +-, as the values are negative for 2,x x> near 2.

d)2

2 2

3 3lim lim

( 2)( 2)4x x

x x

x xx- -fi fi

- -= =

- +-, as the values are positive for 2,x x< near 2.

e)22 2

3 3lim lim( 2)( 2)4x x

x xx xxfi fi

- -=- +-

does not exist, as, by (c) and (d), the left hand and right hand

limits at x = 2 are not equal.

f)3 3 22 2 2

( 2)2 1lim lim lim

( 2) ( 2) ( 2)x x x

xx

x x xfi fi fi

- -- -= = = -

- - -

In parts (a) and (b) the effect of the zero in the denominator at 2x = is canceled becausethe numerator is zero there also. Thus a finite limit exists. This is not true in part (f), wherecancellation still leaves a zero in the denominator.

Precise Definitions of One-sides LimitsThe formal definition of two-sided limit in Chapter 2 is readily modified for one-sidedlimits.

Definition (Right-hand Limit)

We say that ( )f x has right-hand limit L at 0x , and write

0

lim ( )x x

f x L+fi

=

if for every number 0e> there exists a corresponding number 0d> such that for all x

0 0x x x d< < + | ( ) |f x L e- < (1)

Definition (Left-hand Limit)

We say that f has left-hand limit L at 0x , and write

0

lim ( )x x

f x Lfi

=

if for every number 0e> there exists a corresponding number 0d> such that for all x 0 0x x xd- < < | ( ) |f x L e- < (2)

Definitions (Infinite Limits)

1. We say that ( )f x approaches infinity as x approaches 0x , and write

0

lim ( )x x

f xfi

= ,

if for every positive real number B there exists a corresponding 0d > such that for all x

00 | |x x d< - < ( )f x B> .

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2. We say that ( )f x approaches minus infinity as x approaches 0x , and write

0

lim ( )x x

f xfi

= - ,

if for every negative real number B- there exists a corresponding 0d > such that for allx

00 | |x x d< - < ( )f x B< - .

Exercises

1. Which of the following statements about the function ( )y f x= graphed here are true,

and which are false?

a)1

lim ( ) 1+fi -

=x

f x

b)2

lim ( )x

f xfi

does not exist.

c)2

lim ( ) 2x

f xfi

=

d) 1lim ( ) 2x f x-fi =

e)1

lim ( ) 1x

f x+fi

= f)1

lim ( )x

f xfi

does not exist.

g)0 0

lim ( ) lim ( )x x

f x f x+ -fi fi

=

h) lim ( )x c

f xfi

exists at every c in the open interval ( 1, 1)- .

i) lim ( )x c

f xfi

exists at every c in the open interval (1, 3) .

j)1

lim ( ) 0-

fi -=

xf x k)

3lim ( )

xf x

+fi

does not exist.


2. 11

lim 2x

x

x+fi

-

+ 3. 11 6 3

lim 1 7x

x x

x x-fi

+ -

+

4.2

0

6 5 11 6limh

h h

h-fi

- + +

5. a)1

2 ( 1)lim

| 1|x

x x

x+fi

-

-b)

1

2 ( 1)lim

| 1|x

x x

x-fi

-

-

6. a) )4

limt

t t+fi

- b) )4

limt

t t-fi

-


7.0

5lim

2x x-fi8.

3

1lim

3x x+fi -9.

5

3lim

2 10-fi - +x

x

x10.

20

1lim

( 1)x x xfi-

+

11. a) 150

2limx x

+fi

b) 150

2limx x

-fi

12. 230

1limx xfi

Find the limits in Exercises 13-14.

13.2

( )lim sec

p +fi -xx 14.

0lim(2 cot )q

qfi

-


15.2

lim1

x

x -as

a) 1x +fi b) 1x -fi c) 1x +fi - d) 1x -fi -

21 3

1

o

-1 0

2

y = f x( )y

x

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16.2 1

lim2 4

x

x

-

+as

a) 2x +fi - b) 2x -fi - c) 1x +fi d) 0x -fi

17.2

3

3 2lim

4

x x

x x

- +

-as

a) 2x +fi b) 2x +fi - c) 0x -fi d) 1x +fi

e) What, if anything, can be said about the limit as 0x fi ?


18.35

1lim 7

t

+

as a) 0t +fi b) 0t -fi

19.1 43 3

1 1lim

( 1)x x

- -

as

a) 0x+

fi b) 0x-

fi c) 1x+

fi d) 1x-

fi 20. Given 0e> , find an interval (4 , 4), 0I d d= - > , such that if x lies in I , then 4 x e- < .

What limit is being verified and what is its value?

21. Use the definitions of right-hand and left-hand limits to prove that2

2lim 1

| 2 |x

x

x+fi

-=

-

22. Let

2 1sin , 0( )

, 0

x xxf x

x x

Find (a)0

lim ( )x

x+fi and (b) 0lim ( )x f x-fi ; then use limit

definitions to verify your findings. (c) Based on your conclusions in (a) and (b), can

anything be said about 0lim ( )x xfi ? Give reasons for your answer.

Use formal definitions to prove the limit statements in Exercises 23-24

23.20

1limx xfi

-= - 24.

25

1lim

( 5)x xfi -=

+

Use formal definitions from Exercise 25 in Set A to prove the limit statements in thefollowing Exercises 25-27.

25.0

1lim

x x+fi= 26.

2

1lim

2x x-fi= -

-27.

21

1lim

1x x+fi= -

-

5. CONTINUITY

Continuity at a pointIn practice, most functions of a real variable have domains that are intervals or unions of

separate intervals, and it is natural to restrict our study of continuity to functions with thesedomains. This leaves us with only three kinds of points to consider: interior points (points

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Calculus Page 22

that lie in an open interval in the domain), left endpoints, and right endpoints.

Definition A function f is continuous at an interior point x c= of its domain if

lim ( ) ( )x c

f x f cfi

= .

The function in Fig.1 is continuous at 0x = . The function in Fig.2 would be continuous ifit had (0) 1f = . The function in Fig.3 would be continuous if (0)f were 1 instead of 2. The

discontinuities in Fig.2 and Fig.3 are removable. Each function has a limit as 0x fi , and wecan remove the discontinuity by setting (0)f equal to this limit.

The function2

1( )f x

x= has an infinite discontinuity. Jumps and infinite discontinuities

are the ones most frequently encountered, but there are others.

Definition A function f is continuous at a left endpoint x a= of its domain if

lim ( ) ( )x a f x f a+fi = and continuous at a right endpoint x b= of its domain if lim ( ) ( )x b f x f b-fi = .

In general, a function f is right-continuous (continuous from the right) at a point x c=

in its domain if lim ( ) ( )x c

f x f c+fi = . It is left-continuous (continuous from the left) at c if

lim ( )x c

x-fi = ( )f c . Thus, a function is continuous at a left endpoint a of its domain if it is

left-continuous at b . A function is continuous at an interior point c of its domain if and onlyif it is both right-continuous and left-continuous at c .

Example 1 The function 2( ) 4f x x= - is continuous at every point of its domain [ 2, 2]- .

This includes 2x = - , where f is right-continuous, and 2x = , where f is left-continuous.

Continuity TestA function ( )f x is continuous at x c= if and only if it meets the following three conditions.

1. ( )f c exists ( c lies in the domain of f )

2. lim ( )x c f xfi exists ( f has a limit as x cfi )

3. lim ( ) ( )x c f x f cfi = (the limit equals the function value)

Theorem 6 : Continuity of Algebraic Combinations

If functions f and g are continuous at x c= , then the following functions are continuous at

x c= :

1. f g+ and f g-

2. fg

3. kf, where k is any number

4.f

g(provided ( ) 0)g c

5. ( )( )mnf x (provided ( ( ))

mnf x is defined on an interval containing c , and m and n are

integers)

As a consequence, polynomials and rational functions are continuous at every point wherethey are defined.

Theorem 7 : Continuity of Polynomials and Rational Functions

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Every polynomial is continuous at every point of the real line. Every rational function iscontinuous at every point where its denominator is different from zero.

Example The following functions are continuous everywhere on their respective domains.

a) y x= Using Theorems 6 and 7 (rational power of a

polynomial)

b) 2 2 5y x x= - - Using Theorems 6 and 7, (power of a polynomial)

Example The functions 4( ) 20f x x= + and ( ) 5 ( 2)g x x x= - are continuous at every value of

x . The function4( ) 20

( )( ) 5 ( 2)

f x xr x

g x x x

+= =

-

is continuous at every value of x except 0x = and 2x = , where the denominator is 0.

Example The function ( ) | |f x x= is continuous at every value of x . If 0x > , we have

( )f x x=, a polynomial. If

0x

at each point of( , )a b

thenf

increases on[ , ]a b

.If ' 0f < at each point of ( , )a b , then f decreases on [ , ]a b .

Proof

Let 1x and 2x be two points in [ , ]a b with 1 2.x x< The Mean value Theorem applied to f on

1 2[ , ]x x says that

2 1 2 1( ) ( ) ( )( )- = -x f x f c x x (5)

for some c between 1x and 2.x The sign of the right- hand side of equation (5) is the same as

the sign of '( )f c because 2 1x x- is positive. Therefore, 2 1( ) ( )x f x> if f is positive on

( , )a b , and 2 1( ) ( )f x f x< if f is negative on ( , ) .a b

Example 11 The function 2( )f x x= decreases on ( ,0)- , where '( ) 2 0.f x x= < It increases

on (0, ) , where '( ) 2 0f x x= > .

Exercises

1. Verify Rolle`s Theorem for the following functions:

(i) f(x) = x2 , x [ 2, 3]

(ii) f(x) = 2 + (x- 1) 2/3, x [0,1]

(iii) f(x) = x (x- 1) on the interval [0, 1].

2. Verify Lagranges mean value theorem for the function

(i) f(x) = 2x3- 3x2- x , x [1,2].

(ii) f(x) = x2- 2x + 3, x (a, b).

(iii) f(x) = x2

+ 2x + 9, x [1,5].

In Exercise 3-4, find the value or values of c that satisfy the equation

( ) ( )( )

-=

-

f b f af c

b a

in the conclusion of the Mean Value Theorem for the functions and intervals

3. 2( ) 2 1, [0, 1]= + -f x x x 4. 1 1( ) , , 22

= + f x x

x

THE FIRST DERIVATIVE TEST FOR LOCAL EXTREME VALUES

This chapter shows how to test a functions critical points for the presence of

local extreme values.Theorem 1 (The First Derivative Test for Local Extreme Values)

The following test applies to a continuous function ( ).f x

At a critical point c:

1. If 'f changes from positive to negative at c ( ' 0f > for x c< and ' 0f < for x c> ), then

f has a local maximum value at c.

2. If 'f changes from negative to positive at c ( ' 0f < for x c< and ' 0f > for x c> ), then

f has a local minimum value at c.

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Calculus Page 32

3. If 'f does not change sign at c ( 'f has the same sign on both sides of c), then f has no

local extreme value at c .

At a left endpoint a :

If ' 0 ( ' 0)f f< > for ,x a> then f has a local maximum (minimum) value at .a

At a right endpoint :b

If ' 0( ' 0)f f< > for ,

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The

because

Example

is increathese val

Solution

The func

defined

domain

behavior

We conc

3x = . T

Loca

LocaSince g

and g(2)

alue of the

he function

2 Find the3( )g x x= - +

ing and deues?

tion f is co

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at all points

of g into i

of g by pi

lude that g

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maxima:

minima:is defined

is the absol

local minim

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creasing. W

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chool of Di

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axima at x

of ( )g x = -

3) 4,- = -

2) 11,- = -

interval, w

shows th

tance Educ

1

31 (1 4)- = -

from the lef

e function

, [ 3, 3]- . T

)

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is either p

of :g

3= - and3 12 5x+ +

(2g

(3) 14g = .also kno

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Calculus Page 34

MODULE - II

CHAPTER 3

ASYMPTOTES

Concavity and ConvexityA curve is said to be concave upwards (or convex downwards) at or near Pwhen at all

points near P on it lies above the tangent at P .

A curve is said to be concave downwards (or convex upwards) at or near P when at allpoints near P it lies below the tangent at P (Fig. 3 and Fig. 4) .

The Second Derivative Test for Concavity

Let ( )y f x= be twice differentiable on an interval I .

1. If " 0y > on I , the graph of fover I is concave up.

2. If " 0y < on I, the graph of fover I is concave down.

Example The curve 3y x= is concave down on ( , 0)- where " 6 0y x= < and concave

up on (0, ) where " 6 0= >y x

Example The parabola 2y x= is concave up on every interval because 2 0.y= >

Example For what values of x is the curve axy 2= concave or convex to the foot of the

ordinate.

Solution Here by successive differentiation we obtain

.2

12/32

2

x

a

dx

yd-=

Hence2

2

dx

ydis negative for all positive values x (and negative values of x are not admissible)

so that the curve in the neighbourhood of any specified point is concave downward (i.e.,concave to the foot of the ordinate of that point).

Points of Inflection

A point of a curve at which the curve changes its direction from concave upwards to

concave downwards as at P in Fig.7 or from concave downwards to concave upwards as at1P in Fig.8, is called a point of inflection. The tangent at the point is called an inflectional

tangent.

Remark The curve crosses its tangent at a point of inflection.

On the graph of a twice-differentiable function, " 0y = at a point of inflection.

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Calculus Page 35

Example For the function 2 cos , 0s t t= + , " coss t= - and " 0s = when cos 0t = i.e.,

when 3, , ...2 2

p=t Hence the graph of 2 cos , 0s t t= + changes concavity at

3, ,2 2

p= Kt i.e., the points of inflection are 3, ,2 2

t p= K

Example Find the points of inflection on the curve22

2

ax

xay

+= and show that they lie on a

straight line .

Solution Here)

( )222222

ax

xaa

dx

dy

+

-= .

)( )322

222

2

232

ax

xaxa

dx

yd

+

--=

At the point of inflection, we must have 2

2

dxyd =0..

) 03 22 =-xax . 0=x or ax 3= .

When 0=x , 0=y ;when ax 3= ,4

3ay

= .

Hence the points of inflection are (0,0),

43,3 aa and

--

43,3 aa .

These three points obviously lie on the straight line yx 4= .

Example Prove that for the curve ,sin

a

xcy = every point at which it meets the -x axis is a

point of inflection.

Solution

oa

xc =sin or pn

a

x= or

Nowa

x

a

c

dx

dycos= , ,

a

x

a

c

dx

ydcos

33

3

-= .

For points of inflection2

2

dx

yd=0. 0sin =

a

xor anx = .

Also when anx = , .0cos33

3

-= pna

c

dx

yd

anx = gives the point of inflection .These are same as the points were the curve meetsthe x -axis.

Example Show that the points of inflection of the curve )()(22 bxaxy --= lie on the line

.

Given curve is )()( 22 bxaxy --= .

Hence ( ) ( )= - -y x a x b and( ) 2( )1

( ) ( )2 ( ) 2

- + -= - + - =

- -

dy x a x bx a x b

dx x b x b

anx =

a

x

a

c

dx

ydsin

22

2

-=

bax 43 =+

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Calculus Page 36

and ( ){ }2

3 12 2

2

1 13 2 ( ) 3( )

2 2

- - = - - - - + -

d yx b a x b x b

dx

32

3 2

1 1 3 4

( ) (3 2 ) 3( )2 2 4( )

- + -

= - - - - + - = -

x a b

x b x b a x b x b

For points of inflection ,2

2

dx

yd=0 ; hence

3 4 0+ - =x a b or bax 43 =+ .

Hence the point of inflection lie on the line bax 43 =+ .

Example Prove that the curve21 x

xy

+= has three points of inflection and they are collinear

Given curve is21

xy

x=

+.

( )

2

22

1

1

dy x

dx x

-=

+and

( )

2 3

2 32

2 6

1

d y x x

dx x

-=

+.

At the point of inflection 02

2

=dx

yd. i.e., 0)3(2 3 =-xx .

0=x or 3=x ,

when 0=x , 0=y and when 3=x ,4

3=y .

Hence the point of inflection are (0, 0),

4

3,3 and

--

4

3,3 .

Obviously, the above points lie on the line yx 4= .

Hence the points of inflection are collinear.

Graphing with y' and y" Algorithm for Graphing y = f(x )

Step 1: Find 'y and "y .

Step 2: Find the rise and fall of the curve.

Step 3: Determine the concavity of the curve.

Step 4: Make a summary and show the curves general shape.

Step 5: Plot specific points and sketch the curve.

Example Graph the function 4 34 10.y x x= - +

Solution

Step 1Find 'y and "y .

4 34 10y x x= - + 3 2 24 12 4 ( 3)y x x x x= - = -

Critical points: 0x = , 3x = 2" 12 24 12 ( 2)y x x x x= - = - Possible inflection points 0, 2.x x= =

Step 2 (Rise and fall) Sketch the sign pattern for 'y and use it to describe the behavior of y.

\

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chool of Di tance Education

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Step 3 (

bends.

Step 4

(Summarover eac

Step 5

(Specific'y and

shape in

Example

Solution

Step 1 :

5

3y x=

Concavity)

y and generainterval. T

points and" are zero.

step 4 as a

15 Graph

Find

2 2

3 35 (x x x- =

ketch the s

l shape) Suhen combin

curve) PlotIndicate an

uide to ske

5 2

3 35y x x= -

' ".y and y

5)- T

chool of Di

gn pattern

marize thee the shapes

the curveslocal extre

ch the curv

.

e x - interc

tance Educ

for "y and

informatioto show th

interceptsme values

. (Plot addi

pts are at x

ation

use it to de

from stepscurves ge

if conveniend inflectio

tional point

0 and 5.x= =

cribe the

2 and 3. Sheral form.

nt) and then points. U

s as needed.

ay the gra

ow the sha

points whee the gener

)

h

e

real

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5

'3

y =

10

" 9y =

Step 2

(Rise and

Sketch t

the sign

Fromsign pat

Howeve

1. the f

2. 'y fi

3. the c

2 1

3 310

3x

-

- =

1 4

3 310

9x x

- -

+

fall)

e sign patt

pattern for

tern for "y

, knowing

unction y =

as 0x fi

ncavity do

1

35 ( 2)3x x

-

-

4

310(9 x x

-

= +

rn for 'y a

" and use i

, we see t

hat5 2

3 35x x- is c

- and 'y fi

es not chan

chool of Di

ritical poin

) Possible i

d use it to

to describe

at there is

ontinuous,

- as x fi

e at 0x = (s

tance Educ

ts: 0 andx x=

flection poi

escribe the

the way th

an inflecti

0+ (See the f

tep3) tells

ation

2.=

nt: 0 andx x=

behavior of

Sgraph ben

n points a

ormula for

s that the g

1.= -

y.

ep 3(Concas.

t 1x = - , b

' in step 2)

aph has a c

vity) Sket

tt not x =

and

sp at 0x =

h

e0.

.

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Calculus Page 40

ASYMPTOTES

Definitions

1. We say that ( )f x has the limit L as x approaches infinity and write

lim ( )x

f x Lfi

=

if, for every number 0,e> there exists a corresponding number M such that for all x

| ( ) | .x M f x L e> - <

2. We say that ( )f x has the limit L as x approaches minus infinity and write

lim ( )fi -

=x

f x L

if, for every number 0,e> there exists a corresponding number N such that for all x

| ( ) | .x N f x L e< - <

Example Show that 1lim 0x xfi

= .

Solution Let 0e> be given. We must find a numberM such that for all x

1 10 .x M

x xe> - = <

The implication will hold if 1/M e= or any larger positive number. This proves

1lim 0.x xfi

=

Example Show that 1lim 0.x xfi -

=

Solution (Ref. Fig. 2)Let 0e> be given. We must find a number Nsuch that for all x

1 10 .x N

x xe< - = <

The implication will hold if 1/N e= - or any number less than 1/ .e- This proves

1lim 0.

x xfi - =

Theorem 1 (Properties of Limit as x )

The following rules hold if lim ( )x

f x Lfi

= and lim ( )x

g x Mfi

= (L and M real numbers).

1. Sum rule lim [ ( ) ( )]x

f x g x L Mfi

+ = +

2. Difference rule lim [ ( ) ( )]x

f x g x L Mfi

- = -

3. Product Rule lim ( ) ( )]x

f x g x L Mfi

=

4. Constant Multiple Rule lim ( ) (any number )x

kf x kL k fi

=

5. Quotient Rule:( )

lim , if 0( )x

f x LM

g x Mfi =

6. Power Rule: If andm n are integers, then

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Calculus Page 41

/lim [ ( ) ]m nx

f xfi

/m nL= provided /m nL is a real number.

Example Evaluate 1lim 5x xfi

+

Solution 1 1lim 5 lim 5 limx x xx xfi fi fi

+ = +

, using sum rule

5 0 5= + =

Example Evaluate2

3lim

x x

p

fi -

Solution2

3 1 1lim lim 3fi - fi -

=

pp

x xx x x

1 1lim 3 lim lim

x x xx xp

fi - fi - fi -= , using product rule

3 0 0 0.p= = Limits of Rational Function asxfi To determine the limit of a rational function as x fi , we can divide the numerator and

denominator by the highest power of x in the denominator. What happens then depends onthe degrees of the polynomials involved.

ExampleEvaluate2

2

5 8 3lim

3 2xx x

xfi + -

+.

Solution22

2 2

5 (8/ ) (3/ )5 8 3lim lim

3 2 3 (2 / )fi fi

+ -+ -=

+ +x x

x xx x

x x, by dividing

numerator and denominator by 2x

5 0 0 5.

3 0 3

+ -= =

+

Example Evaluate3

11 2lim

2 1xx

xfi -+

-.

Solution2 3

3 3

(11/ ) (2 / )11 2lim lim ,

2 1 2 (1/ )fi - fi -

++=

- -x x

x xx

x xby dividing numerator and denominator by

2x

0 00.

2 0

+= =

-

Example Evaluate 22 3lim7 4xxxfi -

-+

.

Solution2 2 (3 / )2 3

lim lim7 4 7 (4 / )x x

x xx

x xfi - fi -

--=

+ +, the numerator now approaches - while the denominator

approaches 7, so the ratio approaches - = -

Example Evaluate3

2

4 7lim

2 3 10xx x

x xfi -- +

- -.

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Calculus Page 42

Solution3

2 2

4 (7 / )4 7lim lim

2 3 10 2 (3 / ) (10 / )x x

x xx x

x x x xfi - fi -

- +- +=

- - - -, numerator ,fi - denominator 2,fi so ratio

.fi

=

AsymptotesAn asymptote of a curve of infinite extent is a line whose position is approached as a limitby a tangent to the curve as the point of contact recedes indefinitely along the curve.

Definition If the distance between the graph of a function and some fixed line approacheszero as the graph moves increasingly far from the origin, we say that the graph approachesthe line asymptotically and that the line is an asymptote of the graph.

Example Consider the hyperbola 12

2

2

2

=-b

y

a

x. The equation can be written as

2

2

1x

ax

a

by -= . . . (1)

Therefore, as x tends to infinity, the equation (1) tends to the equations xa

by = .

Hence the lines xa

by = and x

a

by -= are asymptotes to the curve.

Example Consider the equation ) 32 2 xxay =- . The given equation can be written as

xa

xy

-=

2

32 and from this it can be seen that as xfi 2a, yfi , so that the line x = 2a is an

asymptote to the given curve (Fig. 1). In Fig. 1, dashed line denotes the asymptote of the

curve 2 3(2 ) .y a x x- =

Definitions A line y b= is a horizontal asymptote of the graph of a function ( )y f x= if

either

lim ( ) or lim ( ) .x x

f x b f x bfi fi -

= =

A line x a= is a vertical asymptote of the graph if either

lim ( ) or lim ( ) .x a x a

f x f x+ -fi fi

= =

Example The coordinate axes are asymptotes of the curve 1yx

=

The x - axis is asymptote of the curve on the right because 1lim 0x xfi

= and on the

left because 1lim 0.x xfi -

=

The y-axis is an asymptote of the curve both above and below because

0 0

1 1lim and lim .

+ -fi fi= = -

x xx x

Notice that the denominator is zero at 0x = and the function is undefined.

Example Find the asymptotes of the curve

3.

2

xy

x

+=

+

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Solution We are interested in the behavior as x fi and as 2x fi - , where thedenominator is zero.

By actual division 3

2

xy

x

+=

+becomes

11

2y

x= +

+.

Hence lim ( ) 1x

f xfi

= and2 2

lim ( ) , lim ( )+ -fi - fi -

= = x x

f x f x and hence, by the Definition, the

asymptotes are the lines 1y = and 2.x = -

Example Verify that the graph of3

2

1( )

1

xf x

x

-=

-has a vertical asymptote at 1x = - but not at

1x = .

Solution

Since2

3 2

2( 1)( 1)1 1,

1 ( 1)( 1) 1x x xx x x

x x x x- + +- + += =

- - + +

1lim ( )

xf x

fi -= and hence the graph has vertical asymptote 1x = - . But 1x = is not an

asymptote as1

lim ( )x

f xfi

= 32

, a finite limit.

Oblique AsymptotesIf the degree of the numerator of a rational function is one greater than the degree ofthe denominator, the graph has an oblique asymptotes, that is a linear asymptotethat is neither vertical nor horizontal.

Example Find the asymptotes of the graph of2

3( ) .

2 4

xf x

x

-=

-

Solution We are interested in the behavior as x fi and also as 2,x fi where the

denominator is zero. By actual division, we obtain

{

2

linear remainder

3 1( ) 1

2 4 2 2 4

x xf x

x x

-= = + +

- -123(1)

Since2

lim ( )x

f x+fi = and 2lim ( )x f x-fi = - , the line 2x = is a two-sided asymptote. As

,x fi the remainder approaches 0 and ( ) ( / 2) 1.f x xfi + The line ( / 2) 1y x= + is an

asymptote both to the right and to the left.

GRAPHING WITH ASYMPTOTES AND DOMINANT TERMSExample Graph the function

3

1.xy x+=

Solution We investigate symmetry, dominant terms, asymptotes, rise, fall, extreme values,and concavity.

Step 1

Symmetry. There is none.

Step 2 Find any dominant terms and asymptotes. We write the rational function as apolynomial plus remainder:

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For | |x l

Equremaind

Step 3

The

'y =

3

2 1x - =

Step 4 F

The sec

i.e., whe

arge, 2y x

tion (2) reer is zero.

ind 'y and

first

3

2

1 2xx

x x- =

0 i.e., when

nd "y and

nd derivati

32 2x + =

y =

. For x near

eals a ver

nalyze the fu

2

1- is u

3 1

2x = i.e.,

etermine the

ve " 2y = +

i.e., when

chool of Di

2 1 .x

+

zero, 1/y

tical asym

nctions crit

defined at

when 3x =

curves conc

3

3 3

2 2 2x

x

+=

3 1x = - i.e.,

tance Educ

x .

tote at x

cal points.

0= and ze

10.82

avity.

is undefine

when x = -

ation

(2)

0 , where

here does th

ro when

d at 0x =

1.

the denom

curve rise a

2

10x

x- =

nd zero w

inator of t

d fall?

derivati

i.e., wh

en3

22

x+ =

e

e

n

0

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CHAPTER 4

OPTIMIZATIONStrategy for Solving Optimization Problems

We follow the following steps for solving optimization problems (Maximize orminimize problems):

1. Read the problem. Read the problem until you understand it. What is unknown? What isgiven? What is sought?

2. Draw a picture. Label any part that may be important to the problem.

3. Introduce variables. List every relation in the picture and in the problem as an equation oralgebraic expression.

4. Identify the unknown. Write an equation for it. If you can, express the unknown as a

function of a single variable or in two equations in two unknowns. This may requireconsiderable manipulation.

5. Test the critical points and endpoints. Use what you know about the shape of the functionsgraph and the physics of the problem. Use the first and second derivatives to identify

and classify critical points (where 0=f or does not exist).

Whenever you are maximizing or minimizing a function of a single variable, we urgeyou to graph it over the domain that is appropriate to the problem you are solving. Thegraph will provide insight before you calculate and will furnish a visual context forunderstanding your answer.

Example Find two positive numbers whose sum is 20 and whose product is as large aspossible.

Solution

If one number is x, then the other is (20 ).-x Their product is

2( ) (20 ) 20 .= - = -f x x x x x

We want the value or values of x that make ( )f x as large as possible. The domain of f

is the closed interval 0 20. x

We evaluate f at the critical points and endpoints. The first derivative,

( ) 20 2 , = -f x x

is defined at every point of the interval 0 20 x and is =( ) 0f x only at 10.=x Listing the

values of f at this one critical point and the endpoints gives

Critical-point value : 2(10) 20(10) (10) 100= - =f

Endpoint values : (0) 0, (20) 0.= =f f

We conclude that the maximum value is (10) 100.=f The corresponding numbers are

10 and (20 10) 10= - =x (Fig.1).

Example 2 A rectangle is to be inscribed in a semicircle of radius 2. What is the largest areathe rectangle can have, and what are its dimensions?

Solution

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To describe the dimensions of the rectangle, we place the semicircle and rectangle in thecoordinate plane (Fig.2). The length, height, and area of the rectangle can then be expressedin terms of the position x of the lower right hand corner:

Length: 2x 2Height: 4 -x 2Area: 2 4 .-x x

Notice that the values of x are to be found in the interval 0 2. x

Now, our practical problem reduces to the mathematical problem of finding absolutemaximum value of the continuous function

2( ) 2 4= -A x x x

on the domain [0, 2]. We do this by examining the values of A at the critical points andendpoints. The derivative

22

2

22 4

4

-= + -

-

dA xx

dx x(1)

is not defined when 2=x . The derivative (1) is equal to zero when2

2

2

22 4 0.

4

-+ - =

-

xx

x

Multiplying both sides by 24 .-x ,we get2 22 2(4 ) 0- + - =x x

28 4 0- =x 2 2=x

2.= x

Of the two zeros, 2=x and 2,= -x only 2=x lies in the interior of As domain and

makes the critical-point list. The values of A at the endpoints and at this one critical point

are:Critical-point value: )2 2 2 4 2 4= - =A Endpoint values : (0) 0,=A (2) 0.=A

The area has a maximum value of 4 when the rectangle is 24 2- =x units high and

2 2 2=x units long.

ExampleAn open-top box is to be made by cutting small congruent squares from the cornersof a 12-by-12-in. sheet of tin and bending up the sides. How large should the squares cutfrom the corners be to make the box hold as much as possible?

Solution

We take the corner squares are x inches on a side and hence the height of the open topbox is x. Note that the volume of the box is =V hlw ,

where h is the height, l is the length and w is the width. Here

;=h x

12 12 2 ;= - - = -l x x x

12 12 2 .= - - = -w x x x

Hence the volume of the box is a function of the variable x:2 2 3( ) (12 2 ) 144 48 4 .= - = - +V x x x x x x

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Since the sides of the sheet of tin are only 12 in. long, 2x cannot exceed 12 and hence6x and the domain of ( )V x is the interval 0 6. x

A minimum value of 0 at 0=x and 6=x and a maximum near 2.=x To locate the exact

point, we examine the first derivative of with respect to :x

2 2144 96 12 12 (12 8 ) 12 (2 ) (6 ).= - + = - + = - -dV

x x x x x xdx

At the critical point 0=dVdx

, which gives 2=x and 6=x . Only 2=x lies in the interior of

the function domain and is the required critical- point.

Critical-point value: (2) 128=V .

Endpoint values: (0) 0, (6) 0= =V V

The maximum volume is 128 in3. The cut-out squares should be 2 in. on a side.

COST AND REVENUE IN ECONOMICS

Here we want to point out two of the many places where calculus makes a contributionto economic theory. The first has to do with the relationship between point, revenue (moneyreceived), and cost.

Suppose that

( ) =r x the revenue from selling x items

( ) =c x the cost of producing the x items

( ) ( ) ( )= - =p x r x c x the profit from selling x items.

The marginal revenue and marginal cost at this production level (items) are

=drdx

marginal revenue

=dc

dx

marginal cost.

Next theorem is about the relationship of the profit to these derivatives.

Theorem Maximum profit (if any) occurs at a production level at which marginal revenueequals marginal cost.

Example The cost and revenue functions at a soft drink company are3 2( ) 6 15 ,= - +c x x x x and ( ) 9=r x x

where x represents thousands of soft drink bottles. Is there a production level that willmaximize companys profit? If so, what is it?

Solution

( ) 9 ,=r x x 3 2( ) 6 15= - +c x x x x

Differentiating with respect to x, we obtain

( ) 9, =r x 2( ) 3 12 15 = - +c x x x

By the Theorem, maximum profit occurs when ( ) ( ) =r x c x , which implies

23 12 15 9- + =x x 23 12 6 0- + =x x

implies 23 12 6 0- + =x x

implies 2 4 2 0- + =x x

Solving the above quadratic equation, we obtain 4 16 4 22

-=x 2 2=

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The possible production levels for maximum profit are 2 2= +x thousand units and

2 2= -x thousand units. A quick glance at the graphs in (Fig.) or at the corresponding

values of r and c shows 2 2= +x to be a point of maximum profit and 2 2= -x to be alocal maximum for loss.

Theorem The production level (if any) at which average cost is smallest is a level at whichthe average cost equals the marginal cost.

Example The cost function at a soft drink company is 3 2( ) 6 15= - +c x x x x (x in thousands

of units). Is there a production level that minimizes average cost? If so, what is it?

Solution

Cost : 3 2( ) 6 15= - +c x x x x

Marginal cost : 2( ) 3 12 15 = - +c x x x

Average cost : 2( )

6 15= - +c x

x xx

By the Theorem, the production level is minimum at which average cost equals marginalcost. Hence

2 26 15 3 12 15x x x x- + = - +

implies 22 6 0- =x x

implies 2 ( 3) 0- =x x

implies 0=x or 3=x

Since 0,>x the only production level that might minimize average cost is three thousand

units.

We check the derivatives:

2( ) 6 15= - +c x

x xx

Average cost

( )2 6

c xdx

dx x

= -

2

2

( )2 0.

c xdxdx

= >

The second derivative is positive, so 3000x = units gives an absolute minimum.

LINEARIZATION AND APPROXIMATIONSDefinitions If f is differentiable at ,x a= then the approximating function

( ) ( ) ( )( )= + -L x f a f a x a (1)

is the linearization of f at a . The approximation( ) ( )f x L x

of f by L is the standard linear approximation of f at a . The point x a= is the

center of the approximation.

Example Find the linearization of ( ) 1f x x= + at 3x = .

Solution

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We evaluate Equation (1) for f at 3a = . With 1/ 21( ) (1 )2

- = +f x x , we have

1(3) 2, '(3) ,

4

f f= = and the linearization is

1 5( ) 2 ( 3) .

4 4 4

xL x x= + - = +

At 3.2x = , the linearization in Example 2 gives

5 3.21 1 3.2 1.250 0.800 2.050

4 4x+ = + + = + = ,

which differs from the true value 4.2 2.04939 by less than one thousand. The

linearization in Example 1 gives

1 1 3.2+ = +x

3.21 1 1.6 2.6

2

+ = + = ,

a result that is off by more than 25%.

DifferentialsDefinitions Let ( )y f x= be a differentiable function. The differential dx is an independent

variable. The differential dy is

'( ) .dy f x dx=

Example Find dy if

a) 5 37y x x= + b) sin3y x=

Solution

a) 4(5 37)dy x dx= + b) (3cos3 )dy x dx=

Remark If 0dx and we divide both sides of the equation '( )dy f x dx= by dx , we

obtain the familiar equation

'( ).dy

f xdx

=

This equation says that when 0dx , we can regard the derivative /dy dx as a quotient of

differentials.

We sometimes write

'( )df f x dx= in place of '( )dy f x dx= , and call df the differential of f . For instance,

if 2( ) 3 6f x x= - , then2(3 6) 6 .df d x xdx= - =

Every differentiation formula like( )d u v du dv

dx dx dx

+= +

has corresponding differential form like

( ) ,d u v du dv+ = +

obtained by multiplying both sides by dx .

Formulas for differentials

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0dc =

( )d cu cdu=

( )d u v du dv+ = +

( )d uv udv vdu= +

2

u vdu udvd

v v

- =

1( )n nd u nu du--

(sin ) cosd u u du=

(cos ) sind u u du= -

2(tan ) secd u u du=

2(cot ) cscd u u du= -

(sec ) sec tand u u u du=

(csc ) csc cotd u u u du= -

Example

a) 2 2(tan 2 ) sec (2 ) (2 ) 2 sec 2d x x d x xdx= =

b)2 2 2

( 1) ( 1).

1 ( 1) ( 1) ( 1)

x dx xd xx xdx dx xdx dxd

x x x x

+ - + + - = = =

+ + + +

The Differential Estimate of ChangeLet ( )f x be differentiable at 0x x= . The approximate change in the value of f when x

changes from 0x to 0x dx+ is

0'( ) .df f x dx=

Example The radius r of a circle increases from 0 10r = m to 10.1m. Estimate theincrease in the circles area A by calculating dA . Compare this with the true change AD

Solution

Since 2A r= , the estimated increase is

0 0'( ) 2 2 (10)(0.1) 2dA A r dr r dr p p= = = = 2m .

The true change is

{ {

2 2(10.1) (10) (102.01 100) 2 0.01 .dA error

A p p p p D = - = - = +

Absolute, Relative, and Percentage Change

As we move from 0x to a nearby point 0x dx+ , we can describe the change is f in three

ways:True Estimate

Absolute change 0 0( ) ( )f f x dx f xD = + - 0'( )df f x dx=

Relative change0( )

f

f x

D

0( )

df

f x

Percentage change0

100( )

f

f x

D

0

100( )

df

f x

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Example Show that the relative error in computing the volume of a sphere due to an errorin measuring the radius is approximately equal to three times the relative error in the radius.

Solution

Let V be the volume and r be the radius of the sphere. Then they are related by theequation

343

=V r . . . . (1)

Taking logarithms on both sides of (1), we obtain

)4log log 3 log3p= +V r Now taking differentials, we get

3 ,=dV dr V r

since the differential of the constant log

)

4

3is 0.

\ the error relation is

3 ,D D

=V r

V r. . . (2)

whereDV

Vis the relative error in computing the volume and

D r

ris the relative error in

computing the radius.

The error relation (2) simply says that the relative error in computing the volume of asphere due to an error in measuring the radius is approximately equal to three times therelative error in the radius.

Example About how accurately should we measure the radius r of a sphere to

calculate the surface area2

4S r= within 1% of its true value?Solution We want any inaccuracy in our measurement to be small enough to make thecorresponding increment SD in the surface area satisfy the inequality

21 4| | .

100 100

rS S

pD =

We replace SD in this inequality with

8 .dS

dS dr rdr dr

p

= =

This gives24

| 8 |100

rrdrp , or

21 4 1| | . .

8 100 2 100

r rdr

r

p =

We should measure r with an error dr that is no more than 0.5% of the true value.

Example The volume Vof fluid flowing through a small pipe or tube in a unit oftime at a fixed pressure is a constant time the fourth power of the tubes radius r.How will a 10% increase in raffect V?

Solution

Here we can take 4V kr= .

The differential of r and V are related by the equation

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34dV

dV dr kr dr dr

= = .

Hence,3

4

44 .

dV kr dr dr

V kr r = =

The relative change in V is 4 times relative change in r, so a 10% increase in r willproduce a 40% increase in the flow.

Example The radius of a sphere is found to be 10 cms, with a possible error of 0.02 cms.What is the relative error in the computed volume?

Solution

If V is the volume and r is the radius of the sphere,

34

3V rp=

Taking logarithms, we get,

log V= log 43

( ) 3p + log r.

Taking differentials on both sides,

3dV dr

V r=

\ The error relation is 3V r

V r

D D= approximately.

Given 0.02,rD = when r= 10.

\ 0.02

3 .00610

V

V

D= =

Hence, relative error in the volume = 0.006.

Example ABCD is a regular protractor in which 6AB = inches, 2BC = inches and O is

the mid-point of .AB An angle BOP is indicated by a mark P on the edge CD . If in

setting an angle q degrees, a mark is made1

100of an inch along the edge from the correct

spot, show that the error in the angle is29 sin

10

qdegrees approximately.

Solution

Let PC be x inches. From p draw PQ perpendicular to AB.

Then 3 .OQ x= -

\ 2

tan .3

PQ

OQ xq = =

-... (1)

Taking differentials on both sides,

\ 22

2sec

(3 )d dx

xq q =

-... (2)

From (1) not that 3 2 cot .x q- =

\ 22

2sec .

4 cotd dxq q

q=

or2sin

.2

d dxq

q =

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Replacing the differentials anddx dq by the increments xD and qD respectively we

obtain the approximate error relation,2sin

.2 x

q

qD = D

But1

100xD = inches and since qD is measured in degrees, we take

1

100xD =

180

degrees.2sin 1 180

2 100

qqD = degrees,

29 sin

10

q= degrees.

Example The area of a triangle ABC is determined from the side a and the two anglesB and .C If there are small errors in the values of B and ,C show that the resulting error

in the calculated value of the area will be2 21

2 ( ).b C c BD + D Solution

The area of the triangle is given by

12

sin where .sin sin sin

a b cS ab C

A B C= = =

\ 212

sin sin.

sin

B CS a

A= . . . (1)

Taking logarithm on both sides we obtain21

2log log ( ) log sin log sin log sin .S a B C A= + + -

Taking differentials on both sides, we get

cos cos cos,

sin sin sin

dS B C AdB dC dA

S B C A= + -

since the differential of a is 0 as there is no error in the measured value of a.

we have o180 , 0.A B C dA dB dC+ + = \ + + =

cos cos cos( )

sin sin sin

dS B C AdB dC dB dC

S B C A= + + +

cos cos cos cos

sin sin sin sin

B A C AdB dC

B A C A

= + + +

sin cos cos sin sin cos cos sin

sin sin sin sin

A B A B A C A CdB dC

A B A C

+ += +

sin ( ) sin ( )

sin sin sin sin

A B A CdB dC

A B A C

+ += +

sin sin.

sin sin sin sin

C BdB dC

A B A C= + . . . (2)

since sin ( ) sin ( ) sinA B C C+ = - =

and sin ( ) sin ( ) sinA C B B+ = - =

As in (1), the other formulae for S are

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2 21 sin sin 1 sin sinand2 sin 2 sin

A C A BS b S c

B C= =

\ 2 2sin 1 sin 1

and .sin sin 2 sin sin 2

C Bc bA B A C= =

2 22 21 1 1, i.e., ( ).

2 2 2

dS c dB b dC dS c dB b dC

S S S= + = +

\ Error is given by 2 212

( ).S c B b C D = D + D

Exercises

In exercises 1-3, find the linearization ( )L x of ( )f x at x a=

1. 4( ) 1f x x at x= =

2. 3( ) 1f x x x at x= - =

3. ( ) 4f x x at x= =

We want linearizations that will replace the functions in Exercise 4-6, over intervalsthat include the given points 0x . To make our subsequent work as simple as possible,

we want to center each linearization not at 0x but at a nearby integer x a= at which

the given function and its derivative are easy to evaluate. What linearization do weuse in each case?

4. 2 0( ) 2 , 0.1f x x x x= + =

5. 2 0( ) 2 4 3, 0.9f x x x x= + - = -

6. 3 0( ) , 8.5f x x x= =

Answers

1. 4 3x - 2. 2 2x - 3. 1 1xx

+

4. 2x 5. 5- 6.1 4

12 3x +

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CHAPTER 5

INTEGRATION - ISigma Notation for Finite sumsWe use the capital Greek letter S (sigma) to write an abbreviation for the sum

1 2( ) ( ) ( )nf t t f t t f t tD + D + + DL

as1

( )n

kkf t t

=D . When we write a sum this way, we say that we have written it in sigma

notation.

Definitions (Sigma Notation for Finite Sums)

The symbol1

n

kka

= denotes the sum 1 2 na a a+ + +L . The a s are the terms of the sum: 1a isthe first term, 2a is the second term, ka is the k th term, and na is the n th and last term. The

variable k is the index of summation. The values of k run through the integers from 1 to n .The number 1 is the lower limit of summation; the number n is the upper limit ofsummation.

Sum Formulas for Positive Integers

The first n integers:1

( 1).

2

n

k

n nk

=

+=

The first n squares: 2

1

( 1)(2 1).

6

n

k

n n nk

=

+ +=

The first n cubes:2

3

1

( 1).

2

n

k

n nk

=

+ =

Example Evaluate4

2

1

( 3 )k

k k=

- .

Solution4

2

1

( 2 )k

k k=

- 4 4

2

1 1

2k k

k k= =

= -

4(4 1)(8 1) 4(4 1)2

6 2

+ + + = -

30 20 10= - =

Riemann SumsGiven an arbitrary continuous function ( )y f x= on an interval [ , ]a b , we partition the

interval into n subintervals by choosing 1n - points, say 1 2 1, , , ,nx x x -K between a and b

subject only to the condition that

1 2 1na x x x b-< < < <

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is called a partition of [ , ]a b .

The partition P defines n closed subintervals

0 1 1 2 1[ , ], [ , ], , [ , ].n nx x x x x x-K

The typical closed subinterval 1[ , ]k kx x- is called the k th subinterval of P.

The length of the k th subinterval is 1k k kx x x -D = - .

In each subinterval 1[ , ]k kx x- , we select a point kc and construct a vertical rectangle from

the subinterval to the point ( , ( ))k kc f c on the curve ( )y f x= . The choice of kc does not

matter as long as it lies in 1[ , ]k kx x- .

If ( )kf c is positive, the number ( )k kf c xD =height base is the area of the rectangle. If

( )kf c is negative, then ( )k kf c xD is the negative of the area. In any case, we add the n

products ( )k kf c xD to form the sum

1

( )n

p k k

k

S f c x=

= D .

This sum, which depends on P and the choice of the numbers kc , is called a Riemann

sum forfon the interval [a, b].As the partitions of [ , ]a b become finer, the rectangles defined by the partition

approximate the region between the x -axis and the graph of f with increasing accuracy. So

we expect the associated Riemann sums to have a limiting value. To test this expectation, weneed to develop a numerical way to say that partitions become finer and to determinewhether the corresponding sums have a limit. We accomplish this with the followingdefinitions.

The norm of a partition P is the partitions longest subinterval length. It is denoted byP (read the norm of P).

The way to say that successive partitions of an interval become finer is to say that the normsof these partitions approach zero. As the norms go to zero, the subintervals become shorterand their number approaches infinity.

Example The set {0, 0.2, 0.6, 1, 1.5, 2}P= is a partition of [0, 2] . There are five subintervals of

:[0, 0.2], [0.2, 0.6], [0.6, 1], [1, 1.5]P , and [1.5, 2].

The lengths of the subintervals are 1 0.2,xD = 2 0.4,xD = 3 0.4,xD = 4 0.5,xD = and 5 0.5.xD =

The longest subinterval length is 0.5, so the norm of the partition is 0.5P = . In this

example, there are two subintervals of this length.

Definition (The Definite integral as a Limit of Riemann Sums)Let ( )f x be a function defined on a closed interval [ , ]a b . We say that the limit of the

Riemann sums1

( )n

k kkf c x

=D on [ , ]a b as 0P fi is the number I if the following condition

is satisfied:

Given any number 0e> , there exists a corresponding number 0d> such that for everypartition P of [ , ]a b

P d< 1

( )n

k k

k

f c x I e=

D -

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Calculus Page 58

32 2

101

lim (3 2 5) (3 2 5)n

k k kP

k

c c x x x dx-fi

=

- + D = - + .

Constant FunctionsResult : If ( )f x has the constant value c on [ , ]a b , then

( ) ( )b b

a af x dx c dx c b a= = - .

Verification: Suppose that f has the constant value ( )f x c= over [ , ]a b . Then, no matter

how the kc s are chosen,

1

( )n

k k

k

f c x=

D 1

n

k

k

c x=

= D , as ( )kf c always equals c .

1

n

k

k

c x=

= D ( ),c b a= - as1

,n

k

k

x b a=

D = - length of interval [ , ]a b

Since the sums all have the value ( )c b a-

, their limit is ( )c b a-

and the integral value is( )c b a- .

Example

a)4

13 3(4 ( 1)) (3)(5) 15dx

-= - - = =

b)4

1( 3) 3(4 ( 1)) ( 3)(5) 15dx

-- = - - - = - = -

The Area Under the Graph of a Nonnegative FunctionDefinition Let ( ) 0f x be continuous on [ , ]a b . The area of the region between the graph

of f and the x -axis is

( )

b

aA f x dx= .Whenever we make a new definition, as we have here, consistency becomes an issue. Doesthe definition that we have just developed for nonstandard shapes give correct results forstandard shapes? The answer is yes, but the proof is complicated and we will not go into it.

Example Using an area, evaluate the definite integral ,b

ax dx 0 a b< < .

Solution We sketch the region under the curve y x= , a x b , and see that it is a trapezoid

with height ( )b a- and bases a and b . The value of the integral is the area of this trapezoid:

2 2

( )2 2 2

b

a

a b b ax dx b a

+= - = - .

In particular, 2 25

1

( 5) (1)2

2 2xdx = - = .

Notice that2

2

xis an antiderivative of x , further evidence of a connection between

antiderivatives and summation.

Example Use a definite integral to find the area of the region between the parabola 2y x=

and x -axis on the interval [0, ]b .

Solution We evaluate the integral for the area as a limit of Riemann sums.

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We sketch the region (a nonstandard shape) and partition [0, ]b into n subintervals of

length( 0)b b

xn n

-D = = . The points of the partition are

0 0,x = 1 ,x x= D 2 2 ,x x= D K , 1 ( 1) ,nx n x- = - D nx n x b= D = .

We are free to choose the kc s any way we please. We choose each kc to be the right-hand

endpoint of its subinterval, a choice that leads to manageable arithmetic. Thus,1 1,c x=

2 2 ,c x= and so on. The rectangles defined by these choices have areas

2 2 3

1( ) ( ) ( ) (1 )( )f c x f x x x x xD = D D = D D = D

2 2 3

2( ) (2 ) (2 ) (2 )( )f c x f x x x x xD = D D = D D = D

M 2 2 3( ) ( ) ( ) ( )( )nf c x f n x x n x x n xD = D D = D D = D

The sum of these areas is

nS1

( )n

k

k

f c x=

= D

2 3

1

( )n

k

k x=

= D

3 2

1

( )n

k

x k=

= D , as 3( )xD is a constant.3

3

( 1)(2 1)

6

b n n n

n

+ += , as

bx

nD =

3

2

( 1)(2 1)

6

b n n

n

+ +=

3 2

2

2 3 16

b n n

n

+ +=

3

2

3 12

6

b

n n

= + +

. (6)

We can now use the definition of definite integral

01

( ) lim ( )nb

kPa

k

f x dx f c xfi

=

= D

to find the area under the parabola from 0x = to x b= as

2

0

b

x dx lim nn Sfi = In this example, 0P fi is equivalent to n fi .3

2

3 1lim 2

6n

b

n nfi

= + +

, using equation (6)

3 3

(2 0 0)6 3

b b= + + = .

With different values of b we get31

2

0

1 1

3 3x dx = = ,

31.52

0

(1.5) 3.3751.125,

3 3x dx = = = and so on.

Fig. 8

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Exercises

Write the sums in Exercises 1-3 without sigma notation. Then evaluate them.

1.2

1

6

1k

k

k= + 2.

4

1

cosk

k=

3.3

1

1

( 1) sink

k k

+

=

-

4. Which of the following express 1 2 4 8 16 32+ + + + - in sigma notation?

a)6

1

1

2k

k

-

=

b)5

0

2k

k=

c)4

1

1

2k

k

+

=-

5. Which formula is not equivalent to the other two?

a)14

2

( 1)

1

k

k k

-

=

-

- b)

2

0

( 1)

1

k

k k=

-

+ c)

1

1

( 1)

2

k

k k=-

-

+

Express the sums in Exercises 6-8 in sigma notation. The form of your answer will dependon your choice of the lower limit of summation.

6. 1 2 3 4 5 6+ + + + + 7.1 1 1 1

2 4 8 16+ + + 8.

1 1 1 11

2 3 4 5- + + +

9. Suppose that1

5n

k

k

a=

= - and1

6n

k

k

b=

= . Find the values of

a)1

3n

k

k

a=

b)1 6

nk

k

b

=

c)1

( )n

k k

k

a b=

+

d)1

( )n

k k

k

a b=

- e)1

( 2 )n

k k

k

b a=

-

Evaluate the sums in Exercises 10-14.

10. a)

10

1kk

= b)10

2

1kk

= c)10

3

1kk

= 11.7

1( 2 )

kk

=-

12.7

2

1

(3 )k

k=

- 13.7

1

(3 5)k

k k=

+ 14.335 5

1 1225k k

kk

= =

+

In Exercises 15-16, graph each function ( )f x over the given interval. Partition the interval

into four subintervals of equal length. Then add to your sketch the rectangles associated

with the Riemann sum4

1( )k kk f c x= D , given that kc is the (a) left-hand endpoint, (b) right-

hand endpoint, (c) midpoint of the k th subinterval. (Make a separate sketch for each set ofrectangles.)

15. 2( ) 1f x x= - , [0,2] 16. ( ) sin ,f x x= [ , ]p-

17. Find the norm of the partition [0, 1.2, 1.5, 2.3, 2.6, 3]P= .Express the limits in Exercises 18-21 as definite integrals.

18. 20

1

limn

k kP

k

c xfi

=

D where P is a partition of [0,2]

19. 20

1

lim ( 3 )n

k k kP

k

c c xfi

=

- D , where P is a partition of [ 7, 5]-

20.0

1

1lim

1

Calculus 277

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