Calculus - University of Calicut · School of Distance Education Calculus Page 3 CALCULUS CONTENTS...

CALCULUS

STUDY MATERIAL

B.Sc. MATHEMATICS

III SEMESTER

CORE COURSE

(2011 ADMISSION)

UNIVERSITY OF CALICUT

SCHOOL OF DISTANCE EDUCATION

CALICUT UNIVERSITY P.O. MALAPPURAM, KERALA, INDIA - 673 635

351

School of Distance Education

Calculus

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

I I I S E M E S T E R

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


Calculus

CALCULUS

CONTENTS

Module I

1 Foundations of Calculus 5

2 The Rolle’s 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


Calculus


Calculus

MODULE - I CHAPTER 1

FOUNDATIONS OF CALCULUS 1. Quick Review of Functions In this chapter we first discuss function, one of the most important concepts in mathematics. Definition Let A and B be two non-empty sets. A function (map, mapping or transformation) from A to B is a rule which assigns to each element of A a unique element of B. The set A 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 from A into B”, or “ f takes A into B”, or “ f maps A 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 of f at a. We also say that f sends or maps a into ( )f a . The set of all such

image values is called the range or image of f , 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 writing

2 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 the function. In the second

notation, the barred arrow a is read goes into. In the last notation, x is called the independent variable and y is called the dependent variable since the value of y will depend on the value of x.

• Furthermore, suppose a function is given by a formula in terms of a variable x. Then we assume, unless otherwise stated, that the domain of the function is or the largest subset 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 'A and if 'B is a subset of B , then 1( ')f B- denotes the set of elements of A 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 'A , and call 1( ')f B- the inverse image or preimage of 'B .

Example Consider the function 3( )f x x= , i.e., f assigns 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


Calculus

Consider any set A . Then there is a function from A into A which sends each element into itself. It is called the identity function on A and it is usually denoted by 1A 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 Functions Consider functions :f A Bfi and :g B Cfi , that is, where the target set B of f is the domain of g. This relationship can be pictured by the following diagram: f gA B Cfi fi Let a A˛ ; then its image ( )f a under f is in B which is the domain of g. Accordingly, we can find the image of ( )f a under the function g, 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 C or, in other words, f and g gives rise to a well defined function from A to C. The new function is called the composition of f and g, 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 a”o Here ” is used to mean equal by definition. Note that we can now add the function g fo to the above diagram of f and g as follows:

ONE TO ONE, ONTO AND INVERTIBLE FUNCTIONS Definition 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 domain A 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 a function of A onto B or that f maps A onto B. Remarks

• The term injective is used for for a one-to-one function, surjective for an onto function, 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 each element of A will correspond to a unique element of B and vice versa.

2. LIMITS OF FUNCTIONS Notation

0

lim ( )x x

f x Lfi

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

CA B

g fo

f g


Calculus

Limits of Powers and Algebraic Combinations Theorem 1 : Properties of Limits The following rules hold if lim ( )x c f x Lfi = and lim ( )x c g x Mfi = (L and M real 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 cf x g x L M

fi- = -

i.e., the limit of the difference of two functions is the difference of their limits. 3. Product Rule: ]lim ( ) ( )

fi=

x cf 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 ckf x kL

fi= (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 the limit 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 Find 3 2

24 3lim

5x c

x xxfi

+ -+

.

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

3 2

3 2

2 2

lim( 4 3)4 3lim5 lim( 5)

fi

fifi

+ -+ -

=+ +

x cx c

x c

x xx xx x

, using quotient rule

3 2

2

lim lim(4 ) lim( 3)

lim lim 5fi 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 5fi fi fi

fi fi

+ + -=

+x c x c x c

x c x c

x x

x, using rule 4

3 2

24 3

5+ -

=+

c cc

, 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.


Calculus

Example 2 Evaluate 22

lim 4 3fi -

-x

x .

Solution

22

lim 4 3fi -

-x

x ( )1/ 22

2lim 4 3fi -

= -x

x

( )1/ 2

2

2lim 4 3fi -

⎡ ⎤= -⎢ ⎥⎣ ⎦xx , using power rule with 1

2=n

( )22

lim 4 3fi -

= -x

x

22 2

4 lim lim 3fi - fi -

= -x x

x ( )2

24 lim 3

fi -= -

xx

24( 2) 3= - - 16 3= - 13= Theorem 2: Limits of Polynomials by substitution If 1

1 0( ) --= + + +Ln n

n nP x a x a x a , then

11 0lim ( ) ( ) --

fi= == + + +Ln n

n nx cP x P c a c a c a .

Theorem 3: Limits of Rational Functions by substitution (if the limit of the denominator is not zero) If ( )P x and ( )Q x are polynomials and ( ) 0„Q c , then

( ) ( )lim ( ) ( )fi=

x c

P x P cQ x Q c .

Example 3 23 2

2 20

(0) 4(0) 34 3 3lim .55 (0) 5x

x xxfi

+ -+ -= =-

+ +.

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

Identifying common factors It 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 Algebraically Theorem 3 applies only when the denominator of the rational function is not zero at the limit point c . If the denominator is zero, canceling common factors in the numerator and denominator 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 Evaluate 2

21

2limx

x xx 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


Calculus

2

21 1

2 2 1 2lim lim 31fi fi

+ - + += = =

-x x

x x xxx x

.

Example Find 0

2 2limh

hhfi

+ - .

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

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

2 2hh

+ - 2 2 2 22 2

h hh h

+ - + +=

+ +

2 2( 2 2)

hh h

+ -=

+ +

( 2 2)

hh h

=+ +

, with a common factor of h

12 2

=+ +h

, cancelling the factor h

Therefore, we obtain

0

2 2limh

hhfi

+ - 0

1lim2 2h hfi

=+ +

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

use Theorem 3 and can substitute h = 0)

12 0 2

=+ +

12 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 that lim ( ) lim ( )

x c x cg x h x L

fi fi= = .

Then lim ( )x c

f x Lfi

= .

Example Given that 2 2

1 ( ) 14 2x xu x- £ £ + for all 0x „ . Find

0lim ( )x

u xfi

.

Solution Since 0

2

lim 1 1 0 14fi

⎛ ⎞- = - =⎜ ⎟

⎝ ⎠x

x and 0

2

lim 1 1 0 12fi

⎛ ⎞+ = + =⎜ ⎟

⎝ ⎠x

x , the Sandwich Theorem

implies that 0

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. Exercises Find the limits in Exercises 1-8. 1.

12lim(10 3 )x

xfi

- 2. 3 2

2lim ( 2 4 8)x

x x xfi -

- + +


Calculus

3. 2 / 3

lim 3 (2 1)x

s sfi

- 4. 5

4lim 7x xfi -

5. 22

2lim5 6y

yy yfi

+

+ + 6. 1984

4lim ( 3)x

xfi -

+

7. 1/ 3

0lim(2 8)z

zfi

- 8. 0

5lim5 4 2h hfi + +

Find the limits in Exercises 9-15

9. 23

3lim4 3x

xx xfi -

+- +

10. 2

2

7 10lim 2x

x xxfi

+ +-

11. 2

21

3 2lim2t

t tt tfi -

+ +- -

12. 3 2

4 20

5 8lim3 16y

y yy yfi

+

-

13. 3

42

8lim16v

vvfi

--

14. 2

4

4lim2x

x xxfi

-

-

15. 2

1

8 3lim 1x

xxfi -

+ -+

16. Suppose 4lim ( ) 0x f xfi = and 4lim ( ) 3x g xfi =- . Find a)

4lim( ( ) 3)x

g xfi

+ b) 4

lim ( )x

xf xfi

c) 2

4lim( ( ))x

g xfi

d) 4

( )lim ( ) 1x

g xf xfi -

17. Suppose that 2

lim ( ) 4fi -

=x

p x , 2

lim ( ) 0fi -

=x

r x , and 2

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 ( )x

p x r xs xfi -

- +

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

18. 2( )f x x= , 2x = - 19. 1( )f x x= , 2x = -

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

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

22. The inequalities 2

21 1 cos 12 24 2

x xx

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

zero. What if anything, does this tell you about 20

1 coslimx

xxfi

- ?

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

lim ( ) 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 ( ) 0x f xfi = ? Give reasons for your answers.

24. If 22

( )lim 1x

f xxfi -

= , find (a) 2

lim ( )x

f xfi -

and (b) 2

( )limx

f xxfi -

.

25. If 20

( )lim 1x

f xxfi

= , find (a) 0

lim ( )x

f xfi

and (b) 0

( )limx

f xxfi

.


Calculus

3. FORMAL DEFINITION OF LIMIT Definition (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< - < ⇒ - <x x f x L . Example Using the Definition, show that 1lim (5 3) 2.x xfi - = Solution Set 0 1,x = ( ) 5 3f x x= - , and 2L = in the definition of limit. For any given 0e > 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- <f x . We find d by working backwards from the e -inequality. | ( ) | | (5 3) 2 | | 5 5 | e- = - - = - <f x L x x 5 | 1| e- <x

| 1| 5e

- <x

Thus we can take 5ed = . If 0 | 1| 5

ed< - < =x , then

| (5 3) 2 | | 5 5 | 5 | 1| 5 5e e⎛ ⎞- - = - = - < =⎜ ⎟

⎝ ⎠x x x .

This proves that 1lim (5 3) 2x xfi - = .

Remark In the above Example, the value of 5ed = is not the only value that will make

0 | 1|x d< - < imply | 5 5 | e- <x . Any smaller positive d will do as well. The definition does not ask for a “best” positive ,d just one that will work.


Calculus

Example Verify the following two important limits: (a)

00lim

x xx x

fi= (b)

0

limx x

k kfi

= (k constant).

Solution a) Let 0e > be given. We must find 0d > such that for all x 00 | |x x d< - < implies 0| | e- <x x . The implication will hold if d =e or any smaller positive number. This proves that

0 0lim fi =x x x x

b) Let 0e > be given. We must find 0d > such that for all x, 00 | |x x d< - < implies | | e- <k k . Since 0k k- = , we can use any positive number for d and the implication will hold. This proves that

0limx x k k- = .

Finding Deltas Algebraically for Given Epsilons In the above examples, the interval of values about 0x for which | ( ) |f x L- was less than e was symmetric about 0x and we could take d to be half the length of the interval. When such symmetry is absent, as it usually is, we can take d to be the distance from 0x to the interval’s nearer endpoint.


Calculus

Example For the limit 5lim 1 2x xfi - = , find a 0d > that works for 1e = . Solution We have to find a 0d > such that for all x 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 all 0x x„ .

| 1 2 | 1x - - <

1 1 2 1- < - - <x

1 1 3< - <x 1 1 9x< - < 2 10x< < The inequality holds for all x in the open interval (2, 10) , so it holds for all 5x „ in this interval as well. Step 2 : We find a value of 0d > that places the centered interval 5 5xd d- < < + inside the interval (2, 10) . The distance from 5 to the nearer endpoint of (2, 10) is 3. If we take 3d = or any smaller positive number, then the inequality 0 | 5 | d< - <x will automatically place x between 2 and 10 to make | 1 2 | 1x - - < .

0 | 5 | 3x< - < ⇒ | 1 2 | 1x - - < . Example Prove that 2lim ( ) 4x f xfi = if

2 , 2

( )1, 2.x x

f xx

⎧ „= ⎨

=⎩

Solution Our task is to show that given 0e > there exists 0d > such that for all x 0 | 2 |x d< - < ⇒ | ( ) 4 | e- <f x . Step 1 : We solve the inequality | ( ) 4 | e- <f x to find on open interval about 0 2x = on which the inequality holds for all 0x x„ .

For 0 2,x x„ = we have 2( )f x x= , and the inequality to solve is 2| 4 | e- <x :

2| 4 | e- <x

2 4e e- < - <x

24 4e e- < < +x

4 | | 4e e- < < +x here we 1assume that 4e<

4 4e e- < < +x this is an open interval about 0 2x = that solves the inequality


Calculus

The inequality ( ) 4 e- <f x holds for all 2x „ in the open interval ( )4 , 4 .e e- +

Step 2 : We find a value of 0d > 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 other

words, 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< - <x will automatically place x between

4 e- and 4 e+ to make | ( ) 4 | e- <f x . For all x , 0 | 2 |x d< - < ⇒ | ( ) 4 | e- <f x . This completes the proof.

1 We can assume 4e< due to the following reason: In finding a d such that for all x , 0 | 2 |x d< - < implied | ( ) 4 | 4e- < <f x , we found a d that would work for any larger e as well.

Finally, notice the freedom we gained in letting min{2 4 , 4 2}d e e= - - + - . We did not have to spend time deciding which, if either, number was the smaller of the two. We just let d represent the smaller and went on to finish the argument. Example Given that lim ( )x c f x Lfi = and lim ( )x c g x Mfi = , prove that lim( ( ) ( ))

x cf x g x L M

fi+ = + .

Solution Let 0e > be given. We want to find a positive number d such that for all x 0 | |x c d< - < ⇒ | ( ) ( ) ( ) | e+ - + <f x g x L M . Regrouping terms, we get | ( ) ( ) ( ) | ( ( ) ) ( ( ) ) |f x g x L M f x L g x M+ - + = - + - | ( ) | | ( ) |f x L g x M£ - + - , using the triangle inequality | | | | | |a b a b+ £ + Since lim ( )

fi=

x cf x L , there exists a number 1 0d > such that for all x

10 | |x c d< - < ⇒ | ( ) | 2e

- <f x L .

Similarly, since lim ( )x c g x Mfi = , there exists a number 2 0d > such that for all x

20 | |x c d< - < ⇒ | ( ) | 2e

- <g x M .

Let 1 2min{ , }d d d= , the smaller of 1d and 2d . If 0 | | d< - <x c then 1| |x c d- < , so

| ( ) | 2e

- <f x L , and 2| |x c d- < , so | ( ) | 2e

- <g x M . Therefore

| ( ) ( ) ( ) | | ( ) | | ( ) | 2 2f x g x L M f x L g x M e e e+ - + < - + - < + = .

This shows that ( )lim ( ) ( ) .x c

f x g x L Mfi

+ = +


Calculus

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- <f x L holds. Then give a value for

0d > such that for all x satisfying 00 | | d< - <x x the inequality | ( ) | e- <f x L holds. 1. ( ) 2 2,f x x= - 6,L =- 0 2,x =- 0.02e =

2. ( ) ,f x x= 1 ,2L = 01 ,4x = 0.1e =

3. ( ) 7,f x x= - 4,L = 0 23,x = 1e =

4. 2( ) ,f x x= 3,L = 0 3,x = 0.1e =

5. 1( ) ,f x x= 1,L =- 0 1,x =- 0.1e =

6. 120( ) ,f x x= 5,L = 0 24,x = 1e =

7. ( ) , 0f x mx m= > 3 ,L m= 0 3,x = 0e = >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- <f x L . 9. ( ) 3 2,f x x=- - 0 1,x =- 0.03e =

10. 2 6 5( ) ,5

x xf x x+ +

=+ 0 5,x =- 0.05e =

11. 4( ) ,f x x= 0 2,x = 0.4e =

Prove the limit statements in Exercises 12-18 12.

3lim(3 7) 2x

xfi

- = 13. 0

lim 4 2x

xfi

- =

14. 2

lim ( ) 4x

f xfi -

= if 2 , 2

( )1, 2x x

f xx

⎧ „ -= ⎨

=-⎩

15. 23

1 1lim 3x xfi= 16.

2

1

1lim 21x

xxfi

-=

-

17. 0

lim ( ) 0x

f xfi

= if 2 , 0

( ), 02

x xf x x x

<⎧⎪= ⎨

‡⎪⎩

18. 2

0

1lim sin 0x

x xfi=

20. Let

2 , 2( ) 3, 2

2, 2

x xh x x

x

⎧ <⎪

= =⎨⎪ >⎩

Show that a)

2lim ( ) 4x

h xfi

„ b) 2

lim ( ) 3x

h xfi

„ c)2

lim ( ) 2x

h xfi

„ .


Calculus

4. EXTENSIONS OF LIMIT CONCEPTS One-Sided Limits Definition (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 af 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 limit M at a , and we write lim ( )

x af x M

-fi= .

Example For the function

( ) | |xf x x= in Fig.6, we have

0

lim ( ) 1x

f x+fi

= and 0

lim ( ) 1x

f x-fi

= - .

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

Fig. 7

x

y

-2 0 2


Calculus

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

have 2

2lim 4 0x

x+fi

- = and 2

2lim 4 0x

x-fi

- =

The function does not have a left-hand limit at 2x =- or a right-hand limit at 2.x = It does 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. The theorems for limits of polynomials and rational functions hold with one-sided limits, as does the Sandwich Theorem. The connection between one-sided and two-sided limits is stated in the following theorem. 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 cf x L

fi= lim ( )

x cf x L

-fi= and lim ( )

x cf 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 ( ) limx x

f 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 that 0

1limx x+fi

⎛ ⎞⎜ ⎟⎝ ⎠

does not exist

because 1x becomes arbitrarily large and positive as 0x +fi .


Calculus

As 0x -fi , the values of 1( )f x x= become arbitrarily large and negative. Given any

negative real number B- , the values of f eventually lie below B- . (See Fig. 10 ) We write

0 0

1lim ( ) limx x

f 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 not exist because its values become arbitrarily large and negative. Example (One-sides infinite limits)

Find 1

1lim 1x x+fi - and

1

1lim 1x x-fi -

Geometric Solution: The graph of 11y x=

- is the graph of 1y x= shifted 1 unit to the right.

Therefore, 11y 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 and 11x fi ¥

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

1x fi -¥-

.

Example (Two-sided infinite limits) Discuss the behavior of

a) 21( )f xx

= near 0x = ,

b) 21( )

( 3)g x

x=

+ near 3x = - .

Solution

a) As x approaches zero from either side, the values of 21x

are positive and become

arbitrarily large (Fig.12). Hence 20 0

1lim ( ) limx x

f xxfi fi

= =¥

b) The graph of 21( )

( 3)g x

x=

+ is the graph of 2

1( )f xx

= 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 x

g xxfi - fi -

= =¥+

.

The function 1y x= shows no consistent behavior as 0x fi . We have 1fi ¥x if 0x +fi ,

but 1fi -¥x if 0x -fi . All we can say about 0

1limx xfi⎛ ⎞⎜ ⎟⎝ ⎠

is that it does not exist. The function


Calculus

21yx

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

can say that 0 21limx 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 xx x xxfi fi fi

- - -= = =

- + +-.

b) 22 2 2

2 2 1 1lim lim lim( 2)( 2) 2 44x x x

x xx x xxfi fi fi

- -= = =

- + +-.

c) 22 2

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

x xx xx+ +fi fi

- -= =-¥

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

d) 22 2

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

x xx 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

xxx x xfi fi fi

- -- -= = =-¥

- - -

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

Precise Definitions of One-sides Limits The formal definition of two-sided limit in Chapter 2 is readily modified for one-sided limits. 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> .


Calculus

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 all x

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)

1lim ( ) 1

+fi -=

xf x

b) 2

lim ( )x

f xfi

does not exist.

c) 2

lim ( ) 2x

f xfi

=

d) 1

lim ( ) 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 -

=x

f x k) 3

lim ( )x

f x+fi

does not exist.


2. 1

1lim 2x

xx+fi

-+

3. 1

1 6 3lim 1 7x

x xx x-fi

+ -⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠⎝ ⎠

4. 2

0

6 5 11 6limh

h hh-fi

- + +

5. a) 1

2 ( 1)lim | 1|x

x xx+fi

--

b) 1

2 ( 1)lim | 1|x

x xx-fi

--

6. a) ( )4

limt

t t+fi

- ⎢ ⎥⎣ ⎦ b) ( )4

limt

t t-fi

- ⎢ ⎥⎣ ⎦


7. 0

5lim 2x x-fi 8.

3

1lim 3x x+fi - 9.

5

3lim 2 10-fi - +x

xx 10. 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

-

Find the limits in Exercises 15-17.

15. 2lim1

xx -

as

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

21 3

1

o

-1 0

2y = f x( )

y

x


Calculus

16. 2 1lim 2 4

xx-+

as

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

17. 2

33 2lim

4x 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 ? Find the limits in Exercises 18-19.

18. 35

1lim 7t

⎛ ⎞+⎜ ⎟

⎝ ⎠ 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 that 2

2lim 1| 2 |x

xx+fi

-=

-

22. Let 2 1sin , 0

( ), 0

x xxf xx x

⎧ ⎛ ⎞ <⎜ ⎟⎪ ⎝ ⎠= ⎨⎪ >⎩

Find (a) 0

lim ( )x

f x+fi and (b)

0lim ( )

xf 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 f 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 the following Exercises 25-27.

25. 0

1limx x+fi

=¥ 26. 2

1lim 2x x-fi= -¥

- 27. 21

1lim1x x+fi

= -¥-

5. CONTINUITY Continuity at a point In 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 these domains. This leaves us with only three kinds of points to consider: interior points (points


Calculus

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 if it 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 we can remove the discontinuity by setting (0)f equal to this limit.

The function 21( )f xx

= 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 cf x f c+fi

= . It is left-continuous (continuous from the left) at c if lim ( )

x cf 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 only if 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 Test A 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. fg (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 where they are defined. Theorem 7 : Continuity of Polynomials and Rational Functions


Calculus

Every polynomial is continuous at every point of the real line. Every rational function is continuous 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 function

4( ) 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 < , we have ( ) ,f x x= - another polynomial. Finally, at the origin, 0lim | | 0 | 0 | (0)x x ffi = = = . Example (Continuity of trigonometric functions) The functions sin x and cos x are continuous at every value of x . Accordingly, the quotients

sintan cosxx x= coscot sin

xx x=

1sec cosx x= 1csc sinx x=

are continuous at every point where they are defined. Theorem 8 tells us that continuity is preserved under the operation of composition. Theorem 8 : Continuity of Composites If f is continuous at c , and g is continuous at ( )f c , then g fo is continuous at c . Example The following functions are continuous every where on their respective domains.

a) 2 2 5y x x= - - Using Theorems 7 and 8 (composition with the square root)

b) 23

4cos( )1

x xyx

=+

Using Theorems 6, 7, and 8 (power, composite, product, polynomial

and quotient)

c) 222

xyx-

=-

Using Theorems 7 and 8 (composite of absolute value and a rational

function) Continuous Extension to a Point As we saw in Chapter 1, a rational function may have a limit even at a point where its denominator is zero. If ( )f c is not defined, but lim ( )x c f x Lfi = exists, we can define a new function ( )F x by the rule

( ) if isin the domain of

( )if .

f x x fF x

L x c⎧

= ⎨=⎩

The function F is continuous at x c= . It is called the continuous extension of f to x c= . For rational functions f , continuous extensions are usually found by canceling common factors.


Calculus

Example Show that

2

26( )

4x xf x

x+ -

=-

is not continuous at 2x = , but has a continuous extension to 2x = , and find that extension. Solution Since (2)f is not defined, f is not continuous at x = 2. Although (2)f is not defined, if

2x „ we have

2

2( 2)( 3)6 3( ) ( 2)( 2) 24x xx x xf x x x xx- ++ - +

= = =- + +-

.

Here note that 2

2 3 5lim ( ) .2 2 2xf x

fi

+= =+

The function

3( ) 2xF x x+

=+

is equal to ( )f x for 2x „ , but is also continuous at 2x = , having there the value of 54 . Thus

F is the continuous extension of f to 2x = , and

2

22 2

6 5lim lim ( ) 44x x

x x f xxfi fi

+ -= =

-.

Continuity on Intervals A function is called continuous if it is continuous everywhere in its domain. A function that is not continuous throughout its entire domain may still be continuous when restricted to particular intervals within the domain. A function f is said to be continuous on an interval I in its domain if lim ( ) ( )x c f x f cfi = at every interior point c and if the appropriate one-sided limits equal the function values at any endpoints I may contain. A function continuous on an interval I is automatically continuous on any interval contained in I . Polynomials are continuous on every interval, and rational functions are continuous on every interval on which they are defined. Example

• The function 24= -y x is continuous on the interval [ 2, 2]- .

• The function 1=y x is continuous on the intervals ( , 0)-¥ and (0, )¥ .

• The function cos=y x is continuous on the interval ( , )-¥ ¥ . Theorem 9 : The Intermediate Value Theorem Suppose ( )f x is continuous on an interval I , and a and b are any two points of I . Then if

0y is a number between ( )f a and ( )f b , there exists a number c between a and b such that 0( )f c y= . Exercises At what points are the functions in Exercises 1-8 continuous?


Calculus

1. 1 32y xx= --

2. 21

4 3xy

x x+

=- +

3. | 1| siny x x= - + 4. cos xy x=

5. csc2y x= 6. 2tan

1x xyx

=+

7. 2 3y x= + 8. 13(2 1)y x= -

Find the limits in Exercises 9-11 9. limsin( sin )

pfi-

xx x 10. 2 2

1limsec( sec tan 1)y

y y yfi

- -

11. 0

limcos19 3sec2t t

pfi

⎛ ⎞⎜ ⎟

-⎝ ⎠

12. Define (3)g in a way that extends 2( 9)( ) ( 3)

xg x x-

=-

to be continuous at 3x = .

13. Define (1)f in a way that extends 3

2( 1)( )( 1)sf ss-

=-

to be continuous at 1s = .

14. For what value of a is 2 1, 3

( )2 , 3x x

f xax x

⎧ - <= ⎨

‡⎩ continuous at every x ?

Answers 1. All x except 2x = 2. All x except 3, 1x x= = 3. All x 4. All x except 0x =

5. All x except ,2nx np

= any integer 6. All x except ,2n np an odd integer

7. All 32x -

> 8. All x 9. 0 10.1 11. 22

12. (3) 6g = 13. 3(1) 2f = 14. 43a =


Calculus

CHAPTER 2

THE ROLLE’S AND MEAN VALUE THEOREMS EXTREME VALUES OF FUNCTIONS Theorem 1 If f is continuous at every point of a closed interval ,I then f assumes both an absolute maximum value M and an absolute minimum value m somewhere in I . That is, there are numbers 1x and 2x in I with 1( )f x 2, ( ) ,= =m f x M and ( )£ £m f x M for every other x in I . Example 1 On 2 2,p p-⎡ ⎤⎣ ⎦ , the function ( ) cosf x x= takes on a maximum value of 1 (once) and a minimum value of 0 (twice). The function ( ) sing x x= takes on a maximum value of 1 (once) and a minimum value of –1 (once) (Fig. 2). Remark The requirements that the interval be closed and the function continuous are key ingredients of Theorem 1. On an open interval, a continuous function need not have either a maximum or a minimum value. For example, the function ( )f x x= has neither a largest nor a smallest value on (0, 1) . Even a single point of discontinuity can keep a function from having either a maximum or a minimum value on a closed interval. Definition Let f be a function with domain D . Then f has an absolute maximum value on D at a point c if ( ) ( )f x f c£ for all x in D and an absolute minimum value on D at c if ( ) ( )f x f c‡ for all x in D . Absolute maximum and minimum values are called absolute extrema (or global extrema). Definition A function f has a local maximum value at an interior point c of its domain if ( ) ( )f x f c£ for all x in some open interval containing c . A function f has a local minimum value at an interior point c of its domain if ( ) ( )f x f c‡ for all x in some open interval containing c . Remarks • We can extend the definitions of local extrema to the endpoints of intervals by defining

f to have a local maximum or local minimum value at an endpoint c if the appropriate inequality holds for all x in some half-open interval in its domain containing c . In Fig. 5, the function f has local maxima at c and d and local minima at ,a e and b .

• An absolute maximum is also a local maximum. Being the largest value overall, it is also the largest value in its immediate neighborhood. Hence, a list of all local maxima will automatically include the absolute maximum if there is one. Similarly, a list of all minima will include the absolute minimum if there is one.


Calculus

Theorem 2 If f has a local maximum or minimum value at an interior point c of its domain, and if f ¢ is defined at ,c then ( ) 0f c¢ = Definition An interior point of the domain of a function f where f ¢ is zero or undefined is a critical point of f . Remark The only domain points where a function can assume extreme values are critical points and endpoints. Example Find the absolute maximum and minimum values of 2( )f x x= on [ 2,1].- Solution The function is differentiable over its entire domain, so the only critical point is where '( ) 2 0,f x x= = namely 0x = . We need to check the function’s values at 0x = and at the endpoints 2x = - and 1:x = Critical point value: (0) 0f = ; Endpoint values: ( 2) 4f - = ; (1) 1.=f The function has an absolute maximum value of 4 at 2x =- and an absolute minimum value of 0 at 0.x = Example Find the absolute extrema values of 4( ) 8g t t t= - on [ 2,1].- Solution The function is differentiable on its entire domain, so the only critical point occur where '( ) 0g t = . Solving this equation gives

38 4 0t- = or 3 2t = or 1/ 32t = , a point not in the given domain. The function’s local extrema therefore occurs at the endpoints, where we find

( 2) 32g - =- (Abs. minimum); (1) 7g = (Abs. maximum)

Example Find the absolute extrema of 2 / 3( )h x x= on [ 2,3].-

Solution The first derivative 1/ 31/ 3

2 2( )3 3

h x xx

-¢ = =

has no zeros but is undefined at 0.x = The values of h at this one critical point and at the endpoints 2x =- and 3x = are

(0) 0h = ; 2 13 3( 2) ( 2) 4h - = - = and

2 13 3(3) (3) 9h = = .

The absolute maximum value is 139 , assumed at 3x = ; the absolute minimum is 0, assumed

at 0x = .

THE ROLLE’S AND MEAN VALUE THEOREMS Theorem 1 (Rolle’s Theorem) Suppose that ( )y f x= is continuous at every point of the closed interval [ , ]a b and differentiable at every point of its interior ( , )a b . If ( ) ( ) 0,f a f b= = then there is at least one number c in ( , )a b at which ( ) 0.f c¢ = Remark

• Instead of ( ) ( ) 0,f a f b= = if ( ) ( ) 0,f a f b l= = „ then also Rolle’s theorem holds. • The hypothesis of Rolle’s Theorem are essential. If they fail at even one point, the

graph may not have a horizontal tangent. Example Verify Rolle’s Theorem for the function f defined by


Calculus

ƒ(x) = (x - a)m (x - b)n, where m and n being positive integers and x˛ [a, b]. Solution ƒ(x) = (x - a)m (x - b)n is a polynomial function in the variable x and since every polynomial function in the variable x is continuous, ƒ is continuous on [a, b]. Now differentiating ƒ with respect to x, we get, 1 1( ) ( ) ( ) ( ) ( )- -¢ = - - + - -m n m nf x m x a x b n x a x b i.e., ƒ¢(x) = (x - a)m - 1(x - b) n - 1 [(m(x - b) + n (x - a)], … (1) again a polynomial function, so that ƒ¢(x) exists at every point in the open interval (a, b). Also ƒ(a) = 0 = ƒ(b). Next to verify whether there exists a point C(c,ƒ(c)) at which .0)( =¢cf From (1), ƒ¢(x) = 0 implies

(x - a)m - 1(x - b) n - 1 [(m(x - b) + n (x - a)] = 0, which implies m(x - b) + n (x - a) = 0, since (x- a) „ 0, (x- b)„0, as a < x < b,

which implies that nmnambc

+

+= , a point within the interval (a, b) dividing the interval

in the ratio m : n. The conclusion is that there is a point c such that a< c< b, where .0)( =¢cf Hence Rolle’s theorem is verified. Example Examine whether Rolle`s theorem can be applied to the function ƒ(x) = tan x for the interval [0, π]. Solution

There is a discontinuity at 2p=x to the function ( ) tan .=f x x

Also 2( ) sec¢ =f x x which does not exist at .2p=x

HenceRolle`s theorem cannot be applied to ( )f x in [0, ].p Theorem 2 (The Mean Value Theorem) Suppose y = ƒ(x) be a function of x subject to the conditions

(1) ƒ is continuous in the closed interval [a, b]. (2) f is differentiable in the open interval (a, b)

Then there is a point c in the open interval (a, b) at which

( ) ( ) ( ).f b f a f cb a-

¢=-

… (2)

Proof We picture the graph of f as a curve in the plane and draw a line through the point ( , ( ))A a f a and ( , ( ))B b f b The line is the graph of the function

( ) ( )( ) ( ) ( )f b f ag x f a x ab a-

= + --

(point-slope equation). The vertical difference between the graphs of f and g at x is ( ) ( ) ( )h x f x g x= -


Calculus

( ) ( )( ) ( ) ( ).f b f af x f a x ab a-

= - - --

… (3)

The function h satisfies the hypotheses of Rolle’s theorem on [ , ]a b : It is continuous on [ , ]a b and differentiable on ( , )a b because both f and g are. Also, ( ) ( ) 0h a h b= = because the graphs of f and g both pass through A and B . Therefore, ' 0h = at some point c in ( , )a b . This is the point we want for equation (2). To verify equation (2), we differentiate both sides of equation (3) with respect to x , and obtain

( ) ( )( ) ( ) -¢ ¢= -

-

f b f ah x f xb a

.

Setting ,x c= we obtain

( ) ( )( ) ( ) -¢ ¢= -

-

f b f ah c f cb a

implies ( ) ( )0 ( ) -¢= -

-

f b f af cb a

implies ( ) ( )( ) ,-¢ =

-

f b f af cb a

and the proof is complete. Remark If ƒ satisfies the two conditions as in Theorem 2, then

ƒ( a + h) = ƒ(a) + hƒ¢(a + θh), where b - a = h and c = a+ θh, where 0< q <1. Example 6 Verify Lagrange’s mean value theorem for the function ƒ(x) = ex on [0, 1]. SolutionHere ƒ(x) = ex. Since the exponential function is continuous at every point on the real line, in particular ƒ is continuous in the closed interval [ 0,1]. Also ƒ¢(x) = ex exists in the open interval (0, 1). Hence by Lagrange’s mean value theorem , we have

01)0()1()(

-

-=¢

ffcf

i.e., ec = e - 1 so that c = log (e - 1), which lies in the open interval (0, 1). Hence Lagrange’s mean value theorem is verified for the given function.

Example In the mean value theorem ƒ(a + h) = ƒ(a) + hƒ¢(a + qh), show that q = 21 if

ƒ(x) is a quadratic expression. Solution: Suppose f be the quadratic expression ƒ(x) = a x2 + b x + c, a ≠ 0 Then ƒ¢(x) = 2 ax + b By the mean value theorem, ƒ(a + h) = ƒ(a) + h ƒ¢(a + qh), or a (a+h)2+ b(a + h) + c = aa2+ b a + c +h [2a (a+ q h) + b]


Calculus

i.e., ah2 = 2 ah2q Hence q = 21 .

Example If a car accelerating from zero takes 8 sec to go 352 ft, its average velocity for the 8-sec interval is 352/8 = 44 ft/sec. At some point during the acceleration, the Mean Value Theorem says, the speedometer must read exactly 30mph (44 ft/sec). Corollary 1 If '( ) 0f x = at each point of an interval I , then ( )f x C= for all x in I , where C is a constant. Proof We want to show that f has a constant value on I . We do so by showing that 1x and 2x are any two points in I , then 1 2( ) ( ).f x f x= Suppose that 1x and 2x are two points in I , numbered from left to right so that 1 2.x x< Then f satisfies the hypotheses of the Mean Value Theorem on 1 2[ , ]:x x It is differentiable at every point of 1 2[ , ],x x and hence continuous at every point as well. Therefore,

2 1

2 1

( ) ( ) ( )-¢=

-

f x f x f cx x

… (4)

at some point c between 1x and 2x . Since 0¢=f throughout I , (4) becomes

2 1

2 1

( ) ( ) 0,f x f xx x-

=-

or 2 1( ) ( ) 0,f x f x- =

implies 1 2( ) ( ).f x f x= Remark We know that if a function f has a constant value on an interval I , then f is differentiable on I and '( ) 0f x = for all x in I . Corollary 1 provides the converse. Corollary 2 If '( ) '( )f x g x= at each point of an interval I , then there exists a constant C such that ( ) ( )f x g x C= + for all x in I . Proof At each point x in I the derivative of the difference function h f g= - is '( ) '( ) '( ) 0.h x f x g x= - = Thus, ( )h x C= on I (Corollary 1). That is, ( ) ( )f x g x C- = on I , so ( ) ( )f x g x C= + . Remark Corollary 2 says that functions can have identical derivatives on an interval only if their values on the interval have a constant difference. We know, for instance, that the derivative of 2( )f x x= on ( , )-¥ ¥ is 2x. Any other function with derivative 2x on ( , )-¥ ¥ must have the formula 2x C+ for some value of C . Example Find the function ( )f x whose derivative is sin x and whose graph passes through the point (0, 2). Solution Since ( )f x has the same derivative as ( ) cos ,g x x=- we know that ( ) cosf x x C=- + for some constant C . The value of C can be determined from the condition that (0) 2f = (the graph of f passes through (0,2) ). (0) cos(0) 2,f C= - + = or 1 2- + =c so 3.C = The formula for f is ( ) cos 3.f x x= - + Definitions Let f be a function defined on an interval I and let 1x and 2x be any two points in I . 1. f increases on I if 1 2 1 2( ) ( ).x x f x f x< ⇒ <


Calculus

2. f decreases on I if 1 2x x< 2 1( ) ( ).f x f x⇒ < Corollary 3 Suppose that f is continuous on [ , ]a b and differentiable on ( , )a b If ' 0f > at each point of ( , )a b then f 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( ) ( ) ( )( )¢- = -f 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( ) ( )f 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 Lagrange’s 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 a f cb 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 xx

THE FIRST DERIVATIVE TEST FOR LOCAL EXTREME VALUES This chapter shows how to test a function’s 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.


Calculus

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 ,<x b then f has a local minimum (maximum) value at .b Example 1 Find the critical points of

1 4 13 3 3( ) ( 4) 4f x x x x x= - = -

Identify the intervals on which f is increasing and decreasing. Find the function’s local and absolute extreme values. Solution The function f is defined for all real numbers and is continuous. The first derivative

4 1 1 23 3 3 34 4'( ) 4 )

3 3df x x x x xdx

-⎛ ⎞= - = -⎜ ⎟

⎝ ⎠

2

323

4( 1)4 ( 1)3 3

xx xx

- -= - =

is zero at 1x = and undefined at 0x = . There are no endpoints in 'f s domain, so the critical points, 0x = and 1x = , are the only places where f might have an extreme value of any kind. These critical points divide the x - axis into intervals on which 'f is either positive or negative. The sign pattern of 'f reveals the behavior of f both between and at the critical points. We display the information in a picture like the following. To make the picture, we marked the critical points on the x - axis, noted the sign of each factors of 'f on the intervals between the points, and “multiplied” the signs of the factors to find the sign of 'f . We then applied Corollary 3 of the Mean Value Theorem to determine that f decreases ( ) on ( ,0),-¥ decreases on (0,1), and increases ( ) on (1, ).¥ Theorem 1 tells us that f has no extreme at 0x = (as 'f does not change sign) and that the f has a local minimum at 1x = 'f changes from negative to positive ).

Calculus

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Calculus

MODULE - II

CHAPTER 3

ASYMPTOTES Concavity and Convexity A curve is said to be concave upwards (or convex downwards) at or near P when 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 all points 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 f over I is concave up. 2. If " 0y < on I , the graph of f over 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

.21

2/32

2

xa

dxyd

-=

Hence 2

2

dxyd

is 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 at

1P 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.


Calculus

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

when 3, , ...2 2p p=t Hence the graph of 2 cos , 0s t t= + ‡ changes concavity at

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

2 2t p p= K

Example Find the points of inflection on the curve 22

2

axxay+

= and show that they lie on a

straight line .

Solution Here ( )( )222

222

ax

xaadxdy

+

-= .

( )( )322

222

2

2 32

ax

xaxadx

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–= ,43ay –

= .

Hence the points of inflection are (0,0), ⎟⎠⎞⎜

⎝⎛

43,3 aa and ⎟

⎠⎞⎜

⎝⎛ -- 4

3,3 aa .

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

Example Prove that for the curve ,sinaxcy = every point at which it meets the -x axis is a

point of inflection. Solution

oaxc =sin or pn

ax= or

Now ax

ac

dxdy cos= , ,

ax

ac

dxyd cos33

3

-= .

For points of inflection 2

2

dxyd

=0. 0sin =ax or panx = .

Also when panx = , .0cos33

3

„-= pnac

dxyd

panx = gives the point of inflection .These are same as the points were the curve meets the 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 bdx x b x b

\

panx =

ax

ac

dxyd sin22

2

-=

\

\

bax 43 =+


Calculus

and ( ){ }2 3 12 2

21 13 2 ( ) 3( )2 2

- -⎛ ⎞= - - - - + -⎜ ⎟⎝ ⎠

d y x b a x b x bdx

32

32

1 1 3 4( ) (3 2 ) 3( )2 2 4( )

- + -⎡ ⎤= - - - - + - =⎢ ⎥⎣ ⎦ -

x a bx b x b a x bx b

For points of inflection , 2

2

dxyd

=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 curve 21 xxy+

= has three points of inflection and they are collinear

Given curve is 21xyx

=+

. ( )

2

22

11

dy xdx x

-=+

and ( )

2 3

2 32

2 61

d y x xdx 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), ⎟⎟⎠

⎞⎜⎜⎝

⎛

43,3 and ⎟⎟

⎠

⎞⎜⎜⎝

⎛ --

43,3 .

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 curve’s general shape. Step 5: Plot specific points and sketch the curve. Example Graph the function 4 34 10.y x x= - + Solution Step 1 Find '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.

\

\

Calculus

SSchool of Disstance Educaation

Page 337

Calculus

Step 3 (bends. Step 4 (Summaryover eachStep 5 (Specific

'y and yshape in

Example Solution Step 1 :

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(Concavity) S

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points and "y are zero.

step 4 as a g

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Sketch the si

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Page 3

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Calculus

5'3

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.


Calculus

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

xf x L

fi ¥=

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 -¥=

xf 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 number M such that for all x

1 10 .x Mx x

e> ⇒ - = <

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 N such that for all x

1 10 .x Nx x

e< ⇒ - = <

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 ( )

xf x L

fi –¥= and lim ( )

xg x M

fi –¥= ( 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 ( ) (anynumber )x

kf x kL kfi –¥

=

5. Quotient Rule: ( )lim , if 0( )x

f x L Mg x Mfi –¥

= „

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


Calculus

/lim [ ( )]m n

xf x

fi – ¥

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

Example Evaluate 1lim 5x xfi ¥

⎛ ⎞+⎜ ⎟⎝ ⎠

Solution 1 1lim 5 lim5 limx x xx xfi ¥ fi ¥ fi ¥

⎛ ⎞+ = +⎜ ⎟⎝ ⎠

, using sum rule

5 0 5= + =

Example Evaluate 23lim

x xp

fi -¥

Solution 23 1 1lim lim 3

fi -¥ fi -¥

⎛ ⎞= ⎜ ⎟⎝ ⎠

p px xx x x

1 1lim 3 lim limx x xx x

pfi -¥ fi -¥ fi -¥

= , using product rule

3 0 0 0.p= = Limits of Rational Function as xfi – ¥ To determine the limit of a rational function as xfi – ¥ , we can divide the numerator and denominator by the highest power of x in the denominator. What happens then depends on the degrees of the polynomials involved.

Example Evaluate 2

25 8 3lim

3 2x

x xxfi ¥

+ -+

.

Solution

22

2 25 (8/ ) (3/ )5 8 3lim lim

3 2 3 (2 / )fi ¥ fi ¥

+ -+ - =+ +x x

x xx xx x

, by dividing

numerator and denominator by 2x

5 0 0 5.3 0 3+ -= =+

Example Evaluate 311 2lim2 1x

xxfi -¥

+-

.

Solution

2 3

3 3(11/ ) (2 / )11 2lim lim ,

2 1 2 (1/ )fi -¥ fi -¥

++ =- -x x

x xxx x

by dividing numerator and denominator by

2x

0 0 0.2 0+= =-

Example Evaluate 22 3lim

7 4x

xxfi -¥

-+

.

Solution 2 2 (3/ )2 3lim lim

7 4 7 (4 / )x x

x xxx xfi -¥ fi -¥

-- =+ +

, the numerator now approaches -¥ while the denominator

approaches 7, so the ratio approaches -¥ =-¥

Example Evaluate 3

24 7lim

2 3 10x

x xx xfi -¥

- +- -

.


Calculus

Solution

3

2 24 (7 / )4 7lim lim

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

x xx xx x x xfi -¥ fi -¥

- +- + =- - - -

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

.fi ¥ =¥

Asymptotes An asymptote of a curve of infinite extent is a line whose position is approached as a limit by 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 approaches zero as the graph moves increasingly far from the origin, we say that the graph approaches the line asymptotically and that the line is an asymptote of the graph.

Example Consider the hyperbola 12

2

2

2

=-by

ax . The equation can be written as

2

2

1xax

aby -–= . . . (1)

Therefore, as x tends to infinity, the equation (1) tends to the equations xaby –= .

Hence the lines xaby = and x

aby -= are asymptotes to the curve.

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

xaxy-

=2

32 and from this it can be seen that as x fi 2a, y fi ∞, 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 xf x b f x b

fi ¥ fi -¥= =

A line x a= is a vertical asymptote of the graph if either lim ( ) or lim ( ) .

x a x af 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

xyx+=+


Calculus

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

By actual division 32

xyx+=+

becomes

112

yx

= ++

.

Hence lim ( ) 1x

f xfi –¥

= and 2 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 of 3

21( )1

xf xx-=-

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

1x = . Solution Since

23 2

2( 1)( 1)1 1,

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

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

- - + +

1lim ( )x

f xfi -

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

asymptote as 1

lim ( )x

f xfi

= 32 , a finite limit.

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

Example Find the asymptotes of the graph of 2 3( ) .

2 4xf xx-=-

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( ) 12 4 2 2 4x xf xx x-= = + +- -123

(1)

Since 2lim ( )

xf x+fi

=¥ and 2lim ( )

xf 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 TERMS

Example Graph the function 3 1.xyx+=

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 a polynomial plus remainder:

Calculus

For | |x lEqua

remaindeStep 3 F The f

' 2y =

32 1 0x - =

Step 4 Fi

The seco

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Calculus

CHAPTER 4

OPTIMIZATION Strategy for Solving Optimization Problems We follow the following steps for solving optimization problems (Maximize or minimize problems): 1. Read the problem. Read the problem until you understand it. What is unknown? What is

given? 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 or

algebraic 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 require considerable manipulation.

5. Test the critical points and endpoints. Use what you know about the shape of the function’s graph 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 urge

you to graph it over the domain that is appropriate to the problem you are solving. The graph will provide insight before you calculate and will furnish a visual context for understanding your answer. Example Find two positive numbers whose sum is 20 and whose product is as large as possible. 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 fis 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 area the rectangle can have, and what are its dimensions? Solution


Calculus

To describe the dimensions of the rectangle, we place the semicircle and rectangle in the coordinate plane (Fig.2). The length, height, and area of the rectangle can then be expressed in 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 absolute maximum 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 and endpoints. The derivative

2

2

2

2 2 44-= + --

dA x xdx x …(1)

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

2

2

2

2 2 4 0.4- + - =-

x xx

Multiplying both sides by 24 .- x ,we get 2 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 A’s 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. Example An open-top box is to be made by cutting small congruent squares from the corners of a 12-by-12-in. sheet of tin and bending up the sides. How large should the squares cut from 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 top box 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


Calculus

Since the sides of the sheet of tin are only 12 in. long, 2x cannot exceed 12 and hence 6£x 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 contribution

to economic theory. The first has to do with the relationship between point, revenue (money received), 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

=dcdx 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 revenue equals marginal cost. Example The cost and revenue functions at a soft drink company are 3 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 will maximize company’s 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= –


Calculus

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 a local maximum for loss. Theorem The production level (if any) at which average cost is smallest is a level at which the 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 marginal cost. 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 6c xd xdx x⎛ ⎞ = -⎜ ⎟⎝ ⎠

2

2( ) 2 0.c xdxdx

⎛ ⎞ = >⎜ ⎟⎝ ⎠

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

LINEARIZATION AND APPROXIMATIONS Definitions 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


Calculus

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 2.0504 4

x+ = + » + = + = ,

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 2.62

» + = + = ,

a result that is off by more than 25%.

Differentials Definitions 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) sin 3y 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= - , then

2(3 6) 6 .df d x xdx= - =

Every differentiation formula like ( )d u v du dvdx dx dx+

= +

has corresponding differential form like ( ) ,d u v du dv+ = + obtained by multiplying both sides by dx . Formulas for differentials


Calculus

0dc = ( )d cu cdu= ( )d u v du dv+ = + ( )d uv udv vdu= +

2

u vdu udvdv 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 ) 2sec 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 Change Let ( )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 the increase in the circle’s area A by calculating dA . Compare this with the true change AD Solution Since 2A rp= , the estimated increase is 0 0'( ) 2 2 (10)(0.1) 2dA A r dr r drp 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 pD = - = - = +

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 change 0( )

ff xD

0( )df

f x

Percentage change 0

100( )f

f xD

· 0

100( )df

f x·


Calculus

Example Show that the relative error in computing the volume of a sphere due to an error in 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 the equation

343p=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 drV r

since the differential of the constant log ( )43p is 0.

\ the error relation is

3 ,D D=V rV r . . . (2)

where DVV is the relative error in computing the volume and D r

r is the relative error in

computing the radius. The error relation (2) simply says that the relative error in computing the volume of a

sphere due to an error in measuring the radius is approximately equal to three times the relative error in the radius. Example About how accurately should we measure the radius r of a sphere to calculate the surface area 24S rp= within 1% of its true value? Solution We want any inaccuracy in our measurement to be small enough to make the corresponding increment SD in the surface area satisfy the inequality

21 4| | .

100 100rS S pD £ =

We replace SD in this inequality with

8 .dSdS dr rdrdr

p⎛ ⎞= =⎜ ⎟⎝ ⎠

This gives

24| 8 |

100rrdr pp £ , or

21 4 1| | . .8 100 2 100

r rdrr

pp

£ =

We should measure r with an error dr that is no more than 0.5% of the true value. Example The volume V of fluid flowing through a small pipe or tube in a unit of time at a fixed pressure is a constant time the fourth power of the tube’s radius r. How will a 10% increase in r affect V? Solution Here we can take 4V kr= . The differential of r and V are related by the equation


Calculus

34dVdV dr kr drdr

= = .

Hence,

3

44 4 .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 will produce 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,

343

V rp=

Taking logarithms, we get, log V = log 4

3( ) 3p + log r. Taking differentials on both sides,

3dV drV r

=

\ The error relation is 3V rV rD D

= approximately.

Given 0.02,rD = when r = 10.

\ 0.023 .00610

VVD

= · =

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 made 1100

of an inch along the edge from the correct

spot, show that the error in the angle is 29 sin

10q

p degrees approximately.

Solution Let PC be x inches. From p draw PQ perpendicular to AB. Then 3 .OQ x= -

\ 2tan .3

PQOQ x

q = =-

... (1)

Taking differentials on both sides,

\ 22

2sec(3 )

d dxx

q q =-

... (2)

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

\ 22

2sec .4 cot

d dxq qq

=

or 2sin .2

d dxqq =


Calculus

Replacing the differentials anddx dq by the increments xD and qD respectively we obtain the approximate error relation,

2sin .

2xq

qD = D

But 1100

xD = inches and since qD is measured in degrees, we take 1100

xD = 180p

·

degrees.

2sin 1 1802 100q

qp

D = · · degrees,

29 sin

10q

p= degrees.

Example The area of a triangle ABC is determined from the side a and the two angles B 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 be 2 21

2 ( ).b C c BD + D Solution The area of the triangle is given by

12 sin where .

sin sin sina b cS ab CA B C

= = =

\ 212

sin sin .sinB CS a

A= . . . (1)

Taking logarithm on both sides we obtain 21

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 dAS 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 dCS B C A

= + + +

cos cos cos cossin sin sin sin

B A C AdB dCB A C A

⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

sin cos cos sin sin cos cos sinsin sin sin sin

A B A B A C A CdB dCA B A C+ +

= +

sin ( ) sin ( )sin sin sin sin

A B A CdB dCA B A C

+ += +

sin sin .sin sin sin sin

C BdB dCA B A C

= + . . . (2)

since sin ( ) sin ( ) sinA B C Cp+ = - = and sin ( ) sin ( ) sinA C B Bp+ = - = As in (1), the other formulae for S are


Calculus

2 21 sin sin 1 sin sinand2 sin 2 sin

A C A BS b S cB C

= =

\ 2 2sin 1 sin 1and .sin sin 2 sin sin 2

C Bc bA B A C

= =

2 2

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

dS c dB b dC dS c dB b dCS S S

= + = +

\ Error is given by 2 212 ( ).S c B b CD = 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 intervals that 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 we use 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. 30( ) , 8.5f x x x= =

Answers

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

+

4. 2x 5. 5- 6. 1 412 3

x +


Calculus

CHAPTER 5 INTEGRATION - I

Sigma Notation for Finite sums We 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

as 1

( )nkk

f 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 symbol 1

nkk

a=∑ denotes the sum 1 2 na a a+ + +L . The a ’s are the terms of the sum: 1a is

the 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 of summation. 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 Evaluate 4

2

1( 3 )

kk k

=

-∑ .

Solution

4

2

1( 2 )

kk k

=

-∑ 4 4

2

1 12

k kk k

= =

= -∑ ∑

4(4 1)(8 1) 4(4 1)26 2+ + +⎛ ⎞= - ⎜ ⎟

⎝ ⎠

30 20 10= - =

Riemann Sums Given 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-< < < < <L . To make the notation consistent, we usually denote a by 0x and b by nx . The set 0 1{ , , , }nP x x x= K


Calculus

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 kk

S f c x=

= D∑ .

This sum, which depends on P and the choice of the numbers kc , is called a Riemann sum for f on 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, we need to develop a numerical way to say that partitions become finer and to determine whether the corresponding sums have a limit. We accomplish this with the following definitions. The norm of a partition P is the partition’s longest subinterval length. It is denoted by P (read “the norm of P”).

The way to say that successive partitions of an interval become finer is to say that the norms of these partitions approach zero. As the norms go to zero, the subintervals become shorter and 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 sums 1

( )nk kk

f 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 every partition P of [ , ]a b

P d< ⇒ 1

( )n

k kk

f c x I e=

D - <∑


Calculus

for any choice of the numbers kc in the subintervals 1[ , ]k kx x- . If the limit exists, we write

0 1

lim ( )n

k kP kf c x I

fi=

D =∑ .

We call I the definite integral of f over [ , ]a b , we say that f is integrable over [ , ]a b , and we say that the Riemann sums of f on [ , ]a b converge to the number I .

We usually write I as ( )b

af x dx∫ , which is read “integral of f from a to b .” Thus, if the

limit exists,

0 1

lim ( ) ( )n b

k k aP kf c x f x dx

fi=

D =∑ ∫ .

Remark The amazing fact is that despite the variety in the Riemann sums ( )k kf c xD∑ as the partitions change and the arbitrary choice of kc ’s in the intervals of each new partition, the sums always have the same limit as 0P fi as long as f is continuous. Theorem (The Existence of Definite Integrals) All continuous functions are integrable. That is, if a function f is continuous on an interval [ , ]a b , then its definite integral over [ , ]a b exists.

Functions with No Riemann Integral The function

1 when is rational

( )0 when is irrational,

xf x

x⎧

= ⎨⎩

has no Riemann integral over [0, 1] . For any partition P of [0, 1] , the upper and lower sums are

max 1 1k k k kU x x x= D = D = D =∑ ∑ ∑ , as every subinterval contains a rational number and hence kmax 1=

min 0 0k k kL x x= D = D =∑ ∑ , as every subinterval contains an irrational number and hence kmin 0= .

For the integral of f to exist over [0, 1], U and L would have to have the same limit as 0P fi . But they do not:

0

lim 0P

Lfi

= while 0

lim 1P

Ufi

= .

Therefore, f has no integral on [0,1] . No constant multiple kf has an integral either, unless k is zero. Example Express the limit of Riemann sums

2

0 1lim (3 2 5)

n

k k kP kc c x

fi=

- + D∑

as an integral if P denotes a partition of the interval [ 1, 3]- .

Solution The function being evaluated at kc in each term of the sum is ( )f x = 23 2 5x x- + . The interval being partitioned is [ 1, 3]- . The limit is therefore the integral of f from 1- to 3:


Calculus

32 2

10 1lim (3 2 5) (3 2 5)

n

k k kP kc c x x x dx

-fi=

- + D = - +∑ ∫ .

Constant Functions Result : 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 kk

f c x=

D∑ 1

n

kk

c x=

= D∑ , as ( )kf c always equals c .

1

n

kk

c x=

= D∑ ( ),c b a= - as 1

,n

kk

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 Function Definition 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. Does the definition that we have just developed for nonstandard shapes give correct results for standard 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) 22 2

xdx = - =∫ .

Notice that 2

2x is 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.


Calculus

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

length ( 0)b bxn 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 31( ) ( ) ( ) (1 )( )f c x f x x x x xD = D D = D D = D

2 2 32( ) (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

kk

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)

6b n n nn

+ += , as bx

nD =

3

2( 1)(2 1)

6b n n

n+ +

=

3 2

22 3 1

6b n n

n+ +=

3

23 12

6b

n n⎛ ⎞

= + +⎜ ⎟⎝ ⎠

. (6)

We can now use the definition of definite integral

0

1

( ) lim ( )nb

kPak

f x dx f c xfi

=

= D∑∫

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

2

0

bx dx∫ lim nn

Sfi ¥

= In this example, 0P fi is equivalent to n fi ¥ .

3

23 1lim 2

6n

bn nfi ¥

⎛ ⎞= + +⎜ ⎟

⎝ ⎠, using equation (6)

3 3

(2 0 0)6 3b b

= + + = .

With different values of b we get

31

2

0

1 13 3

x dx = =∫ , 31.5

2

0

(1.5) 3.375 1.125,3 3

x dx = = =∫ and so on.

Fig. 8


Calculus

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

1. 2

1

61k

kk= +

∑ 2. 4

1cos

kkp

=∑ 3.

31

1( 1) sink

k kp+

=

-∑

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

a) 6

1

12k

k

-

=∑ b)

5

02k

k=∑ c)

41

12k

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 depend on your choice of the lower limit of summation.

6. 1 2 3 4 5 6+ + + + + 7. 1 1 1 12 4 8 16+ + + 8. 1 1 1 11

2 3 4 5- + + +

9. Suppose that 1

5n

kk

a=

= -∑ and 1

6n

kk

b=

=∑ . Find the values of

a) 13

n

kk

a=∑ b)

1 6

nk

k

b=∑ c)

1( )

n

k kk

a b=

+∑

d) 1( )

n

k kk

a b=

-∑ e) 1( 2 )

n

k kk

b a=

-∑

Evaluate the sums in Exercises 10-14.

10. a) 10

1kk

=∑ b)

102

1kk

=∑ c)

103

1kk

=∑ 11.

7

1( 2 )

kk

=

-∑

12. 7

2

1(3 )

kk

=

-∑ 13.7

1(3 5)

kk k

=

+∑ 14. 335 5

1 1225k k

k k= =

⎛ ⎞+ ⎜ ⎟⎝ ⎠

∑ ∑

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 sum 4

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

0 1lim

n

k kP kc x

fi=

D∑ where P is a partition of [0,2]

19. 2

0 1lim ( 3 )

n

k k kP kc c x

fi=

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

20. 0 1

1lim1

n

kP k k

xcfi

=

D-

∑ , where P is a partition of [2, 3]


Calculus

21. 0 1

lim (sec )n

k kP kc x

fi=

D∑ where P is a partition of , 04p-⎡ ⎤

⎢ ⎥⎣ ⎦

Evaluate the integrals in Exercises 22-24.

22. 1

25dx

-∫ 23. 3

0( 160)dt-∫ 24.

3.4

2.10.5ds

-∫

In Exercises 25-28, graph the integrands and use areas to evaluate the integrals.

25. 4

23

2x dx

-

⎛ ⎞+⎜ ⎟

⎝ ⎠∫ 26. 3

2

39 x dx

-

-∫

27. 1

2| |x dx

-∫ 28. 1

1(2 | |)x dx

-

-∫

Use areas to evaluate the integrals in Exercises 29-30

29. 0

,b

xdx∫ 0b > 30. 2 ,b

as ds∫ 0 a b< <

Use the results of Examples 5 and 6 to evaluate the integrals in Exercises 31-36.

31. 2

1xdx∫ 32.

2

dp

pq q∫ 33.

3 72

0x dx∫

34. 12 2

0t dt∫ 35.

2a

axdx∫ 36.

32

0

b

x dx∫

In Exercises 37-38, use a definite integral to find the area of the region between the given curve and the x -axis on the interval [0, ]b , as in Example B.

37. 23y x= 38. 2y x=

39. What values of a and b maximize the value of 2( )b

ax x dx-∫ ?

(Hint: Where is the integrand positive?) 40. Upper and lower sums for increasing functions a) Suppose the graph of a continuous function ( )f x rises steadily as x moves from left

to right across an interval [ , ]a b . Let P be a partition of [ , ]a b into n subintervals of

length ( )b axn-

D = . Show by referring to the accompanying figure that the difference

between the upper and lower sums for f on this partition can be represented graphically as the area of a rectangle R whose dimensions are [ ( ) ( )]f b f a- by xD . (Hint: The difference U L- is the sum of areas of rectangles whose diagonals

0 1 1 2 1, , , n nQ Q Q Q Q Q-K lie along the curve. There is no overlapping when these rectangles are shifted horizontally onto R .)

b) Suppose that instead of being equal, the lengths kxD of the subintervals of the partition of [ , ]a b vary in size. Show that

max| ( ) ( ) |U L f b f a x- £ - D , where maxxD is the norm of P , and hence that 0lim P fi ( ) 0U L- = .


Calculus

RULES FOR INTEGRATION Rules for definite integrals

1. Zero: ( ) 0a

af x dx =∫

2. Order of Integration: ( ) ( )a b

b af x dx f x dx=-∫ ∫

3. Constant Multiples: ( ) ( )b b

a akf x dx k f x dx=∫ ∫ (Any number k )

( ) ( )b b

a af x dx f x dx- =-∫ ∫ ( 1)k =-

4. Sums and Differences: ( ( ) ( )) ( ) ( )b b b

a a af x g x dx f x dx g x dx– = –∫ ∫ ∫

5. Additivity: ( ) ( ) ( )b c c

a b af x dx f x dx f x dx+ =∫ ∫ ∫

6. Max-Min Inequality: If max f and min f are the maximum and minimum values of f on [ , ]a b , then

min ( ) ( ) max ( )b

af b a f x dx f b a- £ £ -∫

7. Domination: ( ) ( )f x g x‡ on [ , ]a b ⇒ ( ) ( )b b

a af x dx g x dx‡∫ ∫

In particular, ( ) 0f x ‡ on [ , ]a b ⇒ ( ) 0b

af x dx ‡∫

Example Suppose that

1

1( ) 5f x dx

-=∫ ,

4

1( ) 2f x dx = -∫ ,

1

1( ) 7h x dx

-=∫ .

Then

1. 1 4

4 1( ) ( ) ( 2) 2f x dx f x dx=- =- - =∫ ∫ Using Rule 2

2. - - -

+ = +∫ ∫ ∫1 1 1

1 1 1[2 ( ) 3 ( )] 2 ( ) 3 ( )f x h x dx f x dx h x dx

2(5) 3(7) 31= + = Using Rules 3 and 4

3. 4 1 4

1 1 1( ) ( ) ( ) 5 ( 2) 3f x dx f x dx f x dx

- -= + = + - =∫ ∫ ∫ Using Rule 5

We know the three general integrals:

( )b

acdx c b a= -∫ (Any constant c ) (1)


Calculus

2 2

2 2b

ab axdx = -∫ (0 )a b< < (2)

3

20 3b bx dx =∫ ( 0)b > (3)

The rules for definite integrals enable us to build on these results.

Example Evaluate 22

07 5

4t t dt⎛ ⎞- +⎜ ⎟

⎝ ⎠∫ .

Solution

22

07 5

4t t dt⎛ ⎞- +⎜ ⎟

⎝ ⎠∫ 2 2 220 0 0

1 7 54

t dt tdt dt= - +∫ ∫ ∫ Rules 3 and 4

3 2 2(2) (2) (0)1 7 5(2 0)

4 3 2 2⎛ ⎞ ⎛ ⎞

= - - + -⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

, using equations (1) - (3)

2 1014 103 3

= - + = -

Example Evaluate 3 22

x dx∫ .

Solution We cannot apply equation (3) directly because the lower limit of integration is

different from 0. We can, however, use the Additivity Rule to express 3 22

x dx∫ as a difference

of two integrals that can be evaluated with equation (3):

2 3 32 2 20 2 0

x dx x dx x dx+ =∫ ∫ ∫ Rule 5

3 3 22 2 22 0 0

x dx x dx x dx= -∫ ∫ ∫ Solve for 3 22

x dx∫

3 3(3) (2)

3 3= - Eq. (3) now applies

27 8 193 3 3

= - = .

In the next chapter, we will see how to evaluate 3 22

x dx∫ in a more direct way.

The Max-Min Inequality for definite integrals (Rule 6) says that min ( )f b a- is a lower

bound for the value of ( )b

af x dx∫ and that max ( )f b a- is an upper bound.

Example Show that the value of 1

01 cos xdx+∫ cannot possibly be 2.

Solution The maximum value of 1 cos x+ on [0, 1] is 1 1 2+ = , so

1

01 cos max 1 cos (1 0)xdx x+ £ + -∫ , using max-min inequality

2 1 2£ = . The integral cannot exceed 2 , so it cannot possibly equal 2.


Calculus

Example Use the inequality 2

cos 12xx ⎛ ⎞

‡ -⎜ ⎟⎝ ⎠

, which holds for all x , to find a lower bound

for the value of 1

0cos xdx∫ .

Solution

21 1

0 0cos 1

2xxdx dx⎛ ⎞

‡ -⎜ ⎟⎝ ⎠∫ ∫ Rule 7

1 1 20 0

112

dx x dx‡ -∫ ∫ Rules 3 and 4

3(1)1 51 (1 0) 0.83

2 3 6‡ - - = » . The value of the integral is at least 5

6.

The Average Value of an Arbitrary continuous Function Definition If f is integrable on [ , ]a b , its average (mean) value on [ , ]a b is

1av( ) ( )b

af f x dx

b a=

- ∫ .

Example Find the average value of 2( ) 4f x x= - on [0, 3] . Does f actually take on this value at some point in the given domain? Solution

1av( ) ( )b

af f x dx

b a=

- ∫3 3 32 20 0 0

1 1(4 ) 43 0 3

x dx dx x dx⎛ ⎞= - = -⎜ ⎟- ⎝ ⎠∫ ∫ ∫

3(3)1 14(3 0) (12 9) 1

3 3 3⎛ ⎞

= - - = - =⎜ ⎟⎝ ⎠

The average value of 2( ) 4f x x= - over the interval [0, 3] is 1. The function assumes this value when 24 1x- = or 3x =– . Since one of these points, 3x = lies in [0, 3] , the function does assume its average value in the given domain (Fig.11 ) The Mean Value Theorem for Definite Integrals Theorem (The Mean value theorem for Definite Integrals) If f is continuous on [ , ]a b , then at some point c in [ , ]a b ,

1( ) ( )b

af c f x dx

b a=

- ∫

The continuity of f is important here. A discontinuous function can step over its average value.

Example Show that if f is continuous on [ , ],a b a b„ , and if ( ) 0b

af x dx =∫ , then ( ) 0f x = at

least once in [ , ]a b . Solution The average value of f on [ , ]a b is

1 1av( ) ( ) 0 0b

af f x dx

b a b a= = =

- -∫ .

By Theorem, f assumes this value at some point c in [ , ]a b . Exercises


Calculus

1. Suppose that f and g are continuous and that

2

1( ) 4f x dx = -∫ ,

5

1( ) 6f x dx =∫ ,

3

1( ) 8g x dx =∫

Use the rules at the beginning of this Chapter to find

a) 2

2( )g x dx∫ b)

1

5( )g x dx∫

c) 2

13 ( )f x dx∫ d)

5

2( )f x dx∫

e) 5

1[ ( ) ( )]f x g x dx-∫ f)

5

1[4 ( ) ( )]f x g x dx-∫

2. Suppose that 2

1( ) 5f x dx =∫ . Find

a) 2

1( )f u du∫ b)

2

13 ( )f z dz∫

c) 1

2( )f t dt∫ d)

2

1[ ( )]f x dx-∫

3. Suppose that f is continuous and that 3

0( ) 3f z dz =∫ and

4

0( ) 7f z dz =∫ . Find

a) 4

3( )f z dz∫ b)

3

4( )f t dt∫

Evaluate the integrals in Exercises 4-9

4. 1

37dx∫ 5.

2

05x dx∫ 6.

2

0(2 3)t dt-∫

7. ( )1

21

2z dz+∫ 8.

2 21

3u du∫ 9. 2 20

(3 5)x x dx+ -∫

Answers 1. a) 0 b) 8- c) 12- d) 10 e) 2- f) 16 2. a) 5 b) 5 3 c) 5- d) 5- 3. a) 4 b) 4-

4. 14- 5. 10 6. 2- 7. 74-


Calculus

MODULE - III

CHAPTER 6 INTEGRATION - II

FUNDAMENTAL THEOREMS OF CALCULUS In this chapter we discuss two parts of fundamental theorem of calculus. Theorem (The Fundamental Theorem of Calculus, Part 1)

If f is continuous on [ , ]a b , then ( ) ( )= ∫x

aF x f t dt has a derivative at every point of [ , ]a b

and ( ) ( )= =∫x

adF d f t dt f xdx dx , .£ £a x b (1)

Example

cos cosp-

=∫xd t dt xdx Eq. (1) with ( ) cos=f t t

2 20

1 11 1

=+ +∫

xd dtdx t x Eq. (1) with 2

1( )1

=+

f tt

Example Find dydx if

2

1cos= ∫

xy t dt .

Solution Notice that the upper limit of integration is not x but 2x . To find dydx we must

therefore treat y as the composite of

1

cosu

y t dt= ∫ and 2=u x

and apply the Chain Rule:

=dy dy dudx du dx Chain Rule

1

cosud dut dtdu dx= ∫ Substitute the formula for y .


Calculus

cos= duu dx Eq. (1) with ( ) cos=f t t

2cos 2= x x 2=u x 22 cos= x x Example Express the solution of the following initial value problem as an integral.

Differential equation: tan=dy xdx

Initial condition : (1) 5=y Solution

The function 1

( ) tan= ∫x

F x t dt

is an antiderivative of tan x . Hence the general solution of the equation is

1

tan= +∫x

y t dt C

As always, the initial condition determines the value of C :

1

15 tan= +∫ t dt C (1) 5=y

5 0= +C (2) 5.=C

The solution of the initial value problem is 1

tan 5= +∫x

y t dt .

How did we know where to start integrating when we constructed ( )F x ? We could have started anywhere, but the best value to start with is the initial value of x (in this case 1=x ). Then the integral will be zero when we apply the initial condition (as it was in Eq. 2) and Cwill automatically be the initial value of y .

The Evaluation of Definite Integrals using FTC2 Theorem (The Fundamental Theorem of Calculus, Part 2) If f is continuous at every point of [ , ]a b and F is any antiderivative of f on [ , ]a b , then

( ) ( ) ( )b

af x dx F b F a= -∫ . (3)

Example

a) [ ]00cos sin sin sin 0 0 0 0x dx x

p pp= = - = - =∫

b) [ ] ( )44

0 0sec tan sec sec0 sec 1 24x x dx x pp

p- -

= = - - = -∫

c) 4

21

3 42

⎛ ⎞-⎜ ⎟

⎝ ⎠∫ x dxx

32

4

1

4⎡ ⎤= +⎢ ⎥⎣ ⎦x x

3 32 24 4(4) (1)4 1

⎡ ⎤ ⎡ ⎤= - +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

[8 1] [5] 4= + - = . We can now see that without any restriction on the signs of a and b ,


Calculus

2 2 2

2 2 2

bb

a a

x b ax dx ⎡ ⎤= = -⎢ ⎥⎣ ⎦∫ Because 2

2x is an antiderivative of x

3 3 3

23 3 3

bb

a a

x b ax dx ⎡ ⎤= = -⎢ ⎥⎣ ⎦∫ Because 3

3x is an antiderivative of 2x

Example Find the area of the region between the x -axis and the graph of 3 2( ) 2 , 1 2= - - - £ £f x x x x x .

Solution First find the zeros of f . Since

2 2 2( ) 2 ( 2) ( 1)( 2)= - - = - - = + -f x x x x x x x x x x , the zeros are 0, 1,= -x and 2 The zeros partition [ 1, 2]- into two subintervals: [ 1, 0]- , on which 0‡f and [0, 2] , on which 0£f . We integrate f over each subinterval and add the absolute values of the calculated values.

Integral over [ 1, 0]- : 0 3 21( 2 )

-- -∫ x x x dx

04 32

14 3

-

⎡ ⎤= - -⎢ ⎥⎣ ⎦

x x x

1 1 50 14 3 12⎡ ⎤= - + - =⎢ ⎥⎣ ⎦

Integral over [0, 2] : 24 32 3 2 2

0 0( 2 ) 4 3

⎡ ⎤- - = - -⎢ ⎥

⎣ ⎦∫ x xx x x dx x 8 84 4 03 3⎡ ⎤= - - - =-⎢ ⎥⎣ ⎦

Enclosed area: Total enclosed area 5 8 3712 3 12= + - =

Example We model the voltage in our home wiring with the sine function max sin120p=V V t , which expresses the voltage V in volts as a function of time t in seconds. The function runs through 60 cycles each second (its frequency is 60 hertz, or 60 Hz). The positive constant maxV is the peak voltage.

The average value of V over a half-cycle (duration 1120

sec)

avV ( )

1120

max0

1 sin1201 0120

p=-∫ V t dt

1

120

max0

1120 cos120120 pp

⎡ ⎤= -⎢ ⎥⎣ ⎦V t

max [ cos cos0]pp

= - +V

max2p

=V .

The average value of the voltage over a full cycle is zero. If we measured the voltage with a standard moving-coil galvanometer, the meter would read zero. To measure the voltage effectively, we use an instrument that measures the square root of the average value of the square of the voltage, namely.

2rms av( )=V V


Calculus

The subscript “rms” (read the letters separately) stands for “root mean square.” Since the average value of 2 2 2

max( ) sin 120p=V V t over a cycle is

( )

160

22 2 2 max

av max1 060

( )1( ) ( ) sin 120 20VV V t dtp= =

- ∫ , (4)

the rms voltage is

2

max maxrms

( )2 2

= =V VV (5)

The values given for household currents and voltages are always rms values. Thus, “115 volts ac” means that the rms voltage is 115. The peak voltage, max rms2 2 115 163= = »V V volts, Obtained from Eq. (5), is considerably higher. Exercises Evaluate the integrals in Exercises 1-13.

1. 0

2(2 5)

-+∫ x dx 2.

34

03 4

xx dx⎛ ⎞-⎜ ⎟

⎝ ⎠∫ 3. ( )1 20

x x dx+∫ 4. 65

32

1x dx-∫

5. 0

sinp

∫ x dx 6. 3 20

2secp

∫ x dx 7. 34

4

csc cotp

pq q q∫ d 8.

2

01 cos 22p

+∫ t dt

9. 2

2

2(8 sin )p

p-+∫ y y dy 10.

1 21

( 1)-

+∫ r dr 11. 71

52

12

⎛ ⎞-⎜ ⎟

⎝ ⎠∫ u duu

12.22

21

+∫ s s dss

13. 4

4| |

-∫ x dx

In Exercises 14-17, use a substitution to find an antiderivative and then apply the Fundamental Theorem to evaluate the integral.

14. 1 30(1 2 )-∫ x dx 15.

1 20

1+∫ t t dt

16. ( )20

sin 1 2p q q+∫ d 17. 2

0sin cos4 4

p

∫ x x dx

In Exercises 18-20, find the total area between the region and the x -axis. 18. 2 2 ,=- -y x x 3 2- £ £x

19. 3 23 2= - +y x x x , 0 2£ £x

20. 13=y x , 1 8- £ £x

Answers

1. 6 2. 8 3. 1 4. 52

5. 2

6. 2 3 7. 0 8. 4p- 9.

323p 10. 8

3-

11. 34-

12. 42 8 1- + 13. 16 14. 0

15. 1 (2 2 1)3 - 16. sin 22p + 17. 2

3


Calculus

18. 283

19. 12

20. 514

APPLICATIONS OF INTEGRATION AREAS BETWEEN CURVES Definition If f and g are continuous with ( ) ( )f x g x‡ throughout [ , ],a b then the area of the region between the curves ( )y f x= and ( )y g x= from a to b is the integral of [ ]f g- from a to :b

[ ( ) ( )] .b

aA f x g x dx= -∫ (1)

To apply Eq. (1) we take the following steps. Algorithm : Finding the area between two curves Step 1. Graph the curves and draw a representative rectangle. This reveals which curve is f

(upper curve) and which is g (lower curve). It also helps find the limits of integration if we do not already know them.

Step 2. Find the limits of integration. Step 3. Write a formula for ( ) ( ).f x g x- Simplify it if we can. Step 4. Integrate [ ( ) ( )]f x g x- from a to .b The number we get is the area. Example 1 Find the area between 2secy x= and siny x= from 0 to / 4.p Solution Step 1: We sketch the curves and a vertical rectangle (Fig.1). The upper curve is the graph of

2( ) sec ;f x x= the lower is the graph of ( ) sin .g x x= Step 2: The limits of integration are already given 0, / 4.a b p= = Step 3: 2( ) ( ) sec sin .f x g x x x- = - Step 4:

/ 4

2 / 40

0(sec sin ) [tan cos ]A x x dx x x

pp= - = +∫ 2 21 [0 1] .

2 2⎡ ⎤

= + - + =⎢ ⎥⎣ ⎦

Example 2 Find the area of the region enclosed by the parabola 22y x= = and the line .y x=-

Solution In this example the limits of integration is not explicitly given. But, since the region is determined by curves that intersect, the intersection points give the limits of integration.

Step 1 Sketch the curves and a vertical rectangle. Identifying the upper and the lower curves, we take 2( ) 2f x x= - and ( ) .g x x=- The x - coordinates of the intersection points are the limits of integration.


Calculus

Step 2 We find the limits of integration by solving 22y x= - and y x=- simultaneously for :x

22 x x- =- implies 2 2 0x x- - = implies ( 1)( 2) 0x x+ - = implies 1, 2.x x= - = The region runs from 1x = - to 2.x = The limits of integration are 1, 2.a b= - = Step 3 2 2( ) ( ) (2 ) ( ) 2f x g x x x x x- = - - - = - +

22 .x x= + - Step 4

22 32

2

11

[ ( ) ( )] (2 ) 22 3

b

a

x xA f x g x dx x x dx x-

-

⎡ ⎤= - = = - = + -⎢ ⎥

⎣ ⎦∫ ∫

4 8 1 14 22 3 2 3

⎛ ⎞ ⎛ ⎞= + - - - + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3 9 96 .2 3 2

= + - =

Example 3 Find the area of the region in the first quadrant that is bounded above by y x= and below by the x - axis and the line 2.y x= - Solution Step 1 The region’s upper boundary is the graph of ( ) .f x x= The lower boundary changes from

( ) 0g x = for 0 2x£ £ to ( ) 2g x x= - for 2 4x£ £ (there is agreement at 2).x = We partition the region at 2x = into subregions A and B and sketch a representative rectangle for each subregion. Step 2 The limits of integration for region A are 0a = and 2.b = The left-hand limit for region B is 2.a = To find the right-hand limit, we solve the equations y x= and 2y x= - simultaneously for :x 2x x= - implies 2 2( 2) 4 4x x x x= - = - + .

implies 2 5 4 0x x- + = implies ( 1)( 4) 0x x- - = implies 1, 4.x x= =

Only the value 4x = satisfies the equation 2.x x= - The value 1x = is an extraneous root introduced by squaring. The right-hand limit is 4.b = Step 3 For 0 2 :x£ £ ( ) ( ) 0f x g x x x- = - =

For 2 4 :x£ £ ( ) ( ) ( 2) 2.f x g x x x x x- = - - = - +


Calculus

Step 4 We add the area of subregions A and B to find the total area:

Total area 2 4

0 2

area ofarea of

( 2)

BA

x dx x x dx= + - +∫ ∫144244314243

42 2

3/ 2 3/ 2

0 2

2 2 23 3 2

xx x x⎡ ⎤⎡ ⎤

= + - +⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

3/ 2 3/ 2 3/ 22 2 2(2) 0 (4) 8 8 (2) 2 43 3 3

⎛ ⎞ ⎛ ⎞= - + - + - - +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2 10(8) 2 .3 3

= - =

Integration with Respect to y If a region’s bounding curves are described by ,y functions of the approximating rectangles are horizontal instead of vertical and the basic formula has y in place of .x Example By integrating with respect to y , find the area of the region in the first quadrant that is bounded above by y x= and below by the x - axis and the line 2.y x= - Solution Step 1 We sketch the region and a typical horizontal rectangle based on a partition of an interval of y -values (Fig.4). The region’s right-hand boundary is the line 2,x y= + so ( ) 2.f y y= + The

left-hand boundary is the curve 2 ,x y= so 2( ) .g y y= Step 2 The lower limit of integration is 0.y = We find the upper limit by solving 2x y= + and

2x y= simultaneously for :y

22y y+ =

implies 2 2 0y y- - = implies ( 1)( 2) 0y y+ - = implies 1, 2y y=- = The upper limit of integration is 2.b = (The value 1y =- gives a point of intersection below the x - axis and we do not consider). Step 3 2 2( ) ( ) 2 2f y g y y y y y- = + - = + - Step 4

2

2

0[ ( ) ( )] [2 ]

b

aA f y g y dy y y dy= - = + -∫ ∫

22 3

0

22 3y yy

⎡ ⎤= + -⎢ ⎥⎣ ⎦

4 8 104 .2 3 3

= + - =


Calculus

Example Combining integrals with formulas from geometry, find the area of the region in the first quadrant that is bounded above by y x= and below by the x - axis and the line

2.y x= -

Solution The area we want is the area between the curve , 0 4,y x x= £ £ and the x - axis, minus the area of a triangle with base 2 and height 2:

Area 4

0

1 (2)(2)2

x dx= -∫

4

3/ 2

0

2 23⎡ ⎤

= -⎢ ⎥⎣ ⎦x

2 10(8) 0 2 .3 3

= - - =

Exercises Find the areas of the regions enclosed by the lines and curves in Exercises 1-5. 1. 2 2y x= - and 2y = 2. 4y x= and 8y x=

3. 2y x= and 2 4y x x=- + 4. 4 24 4y x x= - + and 2y x=

5. | |y x= and 5 6y x= + (How many intersection points are there?)

Find the areas of the regions enclosed by the lines and curves in Exercises 6-9. 6. 22 , 0,x y x= = and 3y = 7. 2 4 4y x- = and 4 16x y- =

8. 2 0x y+ = and 23 2x y+ = 9. 2 1x y= - and 2| | 1x y y= -

Find the areas of the regions enclosed by the curves in Exercises10-11. 10. 24 4x y+ = and 4 1x y- =

11. 24 4x y+ = and 4 1,x y+ = for 0x ‡ Find the areas of the regions enclosed by the lines and curves in Exercises 12-15. 12. 2siny x= and sin 2 , 0y x x p= £ £

13. cos( / 2)y xp= and 21y x= -

14. 2sec ,y x= 2tan ,y x= / 4,x p=- and / 4x p=

15. 3sin cosx y y= and 0, 0 / 2x y p= £ £ Answers 1. 32 / 3 2. 48/5 3. 8/3 4. 8 5. 5/3 (There are three intersection points). 6. 18 7. 243/8 8. 8. 8/3 9. 2

10. 104 /15 11. 56 /15 12. 4 13. 4 43 p-

14. / 2p 15. 2

Volume of solid of revolution


Calculus

Solids of revolution are solids whose shapes can be generated by revolving the plane regions about axes. Volume of solid of revolution – revolution about the x-axis

The volume of the solid generated by revolving about the x-axis the region between the x-axis and the graph of the continuous function

( ), ,y R x a x b= £ £ is

.)]([]radius[ 22 ∫∫ ==b

a

b

a

dxxRdxV pp .

Example The region between the curve , 0 4,y x x= £ £ and the x - axis is revolved about the x - axis to generate a solid. Find its volume. Solution The volume is

2[ ( )]b

aV R x dxp= ∫

4

2

0[ ]x dxp= ∫

42 24

00

(4) 8 .2 2xxdx xp p p

⎡ ⎤= = = =⎢ ⎥

⎣ ⎦∫

Example 4 Find the volume of the solid generated by revolving the region bounded by y x= and the lines 1, 4y x= = about the line 1.y =

Solution The volume is

4

2

1[ ( )]V R x dxp= ∫

4

2

1[ 1]x dxp= -∫ ( ) 1R x x= -

4

1[ 2 1]x x dxp= - +∫

42

3/ 2

1

2 72 .2 3 6x x x p

p⎡ ⎤

= - + =⎢ ⎥⎣ ⎦

Example Find the volume of the solid formed by the revolution about the major axis of an ellipse with axes 2a and 2b (a > b). Solution

The equation of the ellipse is 12

2

2

2=+

by

ax


Calculus

Since a > b, the major axis is x-axis. Hence, the required volume is obtained by the revolution about the x-axis of the area enclosed by the portion of the ellipse and the x- axis.

Since the ellipse is symmetrical about the axes, the required volume is twice that of the volume obtained by revolving the part of the ellipse in the first quadrant about the x-axis.

In the portion of the ellipse in the first quadrant x varies from x = 0 to x = a. \ The required volume is given by

( ) .34

3222 2

0

32

2

2

0

22

22 abxxa

abdxxa

abdxyV

aab

a

pppp =⎥⎦

⎤⎢⎣

⎡-=-=·= ∫∫ 2

Example Find the volume of the solid generated by revolving the part of the curve x2 (y - x2) =3 between x = 1 and x = 2 about the x-axis. Solution: The given equation of the curve can be written as

22 3

xxy +=

and here x varies from x = 1 to x = 2. Hence the volume of revolution is given by

pppp40

593133

3 2

1

2

1

32

12

22 =⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦⎤

⎢⎣⎡

-⎥⎦

⎤⎢⎣

⎡=⎟

⎠⎞

⎜⎝⎛

+== ∫∫ xxdx

xxdxyV

b

a

.

Example One arch of the sine curve y = sin x revolved round the x-axis. Find the volume of the solid so generated. Solution An arch of the sine curve is the portion between two consecutive contacts of the curve with its base (i.e. x-axis). Since y = 0 at the contact of the curve with the x axis, we have

0 = y = sin x implies x = np, for any integer n. This implies in particular (substituting n = 0 and n = 1) that an arch of the curve lies between x = 0 and x= p. Hence the volume of revolution of this arch is given by

[ ] .22

2sin22

2cos1sin2

000

22 ppppp ppp

=⎭⎬⎫

⎩⎨⎧

-=-

== ∫∫∫xxdxxdxxdxy

b

a

Example Show that the volume of the segment of a solid sphere of height h and base radius

a is ( )hah -331 2p . Deduce the volume of a hemisphere.

Solution

( ) [ ]⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡-=-==

-

-

--∫∫

a

ha

aha

a

ha

a

ha

xxadxxadxyV3

32222 ppp

( )hah -= 331 2p .

When h = c, the segment becomes a solid hemisphere and its volume is given by

substituting h = c in the above and that is .32 3ap


Calculus

Example 9 Find the volume of the solid generated by the revolution about the x-axis of the

loop of the curve xaxaxy

+

-=

322 .

Solution The loop is formed between x = 0 and x = 3a.

\ ∫∫ +

-==

aa

dxxaxaxdxyV

3

0

23

0

2 3pp

∫ +

-=

a

dxxaxax

3

0

323p

∫ +

+-=

a

dxxaaxx

3

0

23 3p

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛

++-+-=

a

dxxa

aaaxx3

0

322 444p , by the actual division of 23 3axx +- by xa +

( )a

xaaxaaxx3

0

3223

log4423 ⎥

⎥⎦

⎤

⎢⎢⎣

⎡++-+

-=p

( )3

3 3 3 327 18 12 4 log 3 4 log3a a a a a a a ap ⎛ ⎞-= + - + + -⎜ ⎟

⎝ ⎠

⎟⎠⎞

⎜⎝⎛

+-+-=aaaaaa 4log412189 3333p

( )4log433 +-= ap

( )2log833 +-= ap

Volume of solid of revolution – revolution about the y-axis

The volume of the solid generated by the revolution of area between the curve x = R(y), the y axis and two abscissas at the points y = c and y = d is given by the formula

.)]([]radius[ 22 ∫∫ ==d

c

d

c

dyyRdyV pp

Example Find the volume of the solid generated by revolving the region between the y-axis

and the curve 2 , 1 4,x yy

= £ £ about the y-axis.

Solution We draw figures showing the region, a typical radius, and the generated solid (Fig.8a and 8b). The volume is

42

1[ ( )]V R y dyp= ∫

24

1

2 dyy

p⎛ ⎞

= ⎜ ⎟⎝ ⎠∫ 2( )R y

y=


Calculus

44

211

4 14p p⎡ ⎤

= = -⎢ ⎥⎣ ⎦∫ dy

y y

344

p ⎛ ⎞= ⎜ ⎟

⎝ ⎠3 .p=

Example Find the volume of the solid generated by revolving the region between the parabola 2 1x y= + and the line 3x = about the line 3.x = Solution We draw figures showing the region, a typical radius, and the generated solid (Fig.9). The volume is

2

2

2[ ( )]V R y dyp

-= ∫

2

2 2

2[2 ]y dyp

-= -∫ 2 2( ) 3 ( 1) 2R y y y= - + = -

2

2 4

2[4 4 ]y y dyp

-= - +∫

25

3

2

443 5

yy yp-

⎡ ⎤= - +⎢ ⎥

⎣ ⎦

64 2 .15p

=

Example Find the volume of the solid formed by the revolution about the minor axis of an ellipse with axes 2a and 2b. (a > b) Solution

The equation of the ellipse is 12

2

2

2=+

by

ax

Since a > b, the minor axis is the y-axis. Hence, the required volume is obtained by the revolution about the y-axis of the area enclosed by the portion of the ellipse and the y- axis.

Since the ellipse is symmetrical about the axes, the required volume is twice that of the volume obtained by revolving the part of the ellipse in the first quadrant about the y-axis In the portion of the ellipse in the first quadrant y varies from y = 0 to y = b. \ The required volume is given by

( ) .34

3222 2

0

32

2

2

0

22

22 bayyb

badyyb

badyxV

bbd

c

pppp =⎥⎦

⎤⎢⎣

⎡-=-=·= ∫∫ 2

Remarks We have seen that the volumes obtained by revolving the ellipse about the major

and minor axes are different.


Calculus

Remark 1 The volume of the solid formed by the revolution about the major axis of an

ellipse with axes 2a and 2b. (a > b) is .34 2abp

Remark 2 The volume of the solid formed by the revolution about the minor axis of an ellipse

with axes 2a and 2b ( a > b) is .34 2bap .

Special Case We know that the volume of the sphere is obtained by the revolution of the circle about one of its diameter. When a = b, the ellipse in example becomes a circle with radius a. Hence using the result obtained in example, we get the volume of the sphere

having radius a as .34 3ap

MODULE - IV

CHAPTER 7 INTEGRATION - III

LENGTHS OF PLANE CURVES Definition A function with a continuous first derivative is said to be smooth and its graph is called a smooth curve.

Definition If f is smooth on ],,[ ba the length of the curve )(xfy = from a to b is

( )∫∫ ¢+=⎟⎠⎞

⎜⎝⎛

+=b

a

b

adxxfdx

dxdyL 2

2)(11 …(1)


Calculus

Example Find the length of the curve 3 / 24 2 1, 0 1 .

3y x x= - £ £

Solution

We use equation (1) with 0, 1a b= = , and 3/ 24 2 13

y x= - . Then

1/ 2 1/ 24 2 3 2 23 2

dy x xdx

= = ( )2 21/ 22 2 8 .dy x x

dx⎛ ⎞

= =⎜ ⎟⎝ ⎠

The length of the curve from 0x = to 1x = is

2

1 1

0 01 1 8dyL dx x dx

dx⎛ ⎞= + = +⎜ ⎟⎝ ⎠∫ ∫

8

duu= ∫ , by taking 1 8 , 8 .u x du dx= + = 3 / 21

8 3 / 2u⎡ ⎤= ⎢ ⎥⎣ ⎦

13/ 2

0

2 1 (1 8 )3 8

x⎡ ⎤= +⎣ ⎦ , by replacing 1 8 , 8 .u x du dx= + = 1 3 .6

=

Example 2 Find the length of the arc of the semi cubical parabola 32 xy = extending from the origin to the point (1, 1). Solution Differentiating the given equation with respect to x we get

.2332

22

yx

dxdyandx

dxdyy ==\

\ Length of the arc from (0, 0) to (1, 1) is

∫∫∫==

+=⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟

⎠⎞

⎜⎝⎛

+=1

02

41

0

222

491

2311

xx

b

adx

yxdx

yxdx

dxdys

∫∫ +=+==

1

0

1

03

4

491

491 dxxdx

xx

x

( ) ( )[ ] ( ) .8131327194

27194

10

2/31

0

2/1 -=+=+= ∫ xdxx

Example 3 Find the length of the curve y = log sec x between the points given by the

intersection of the curve with the lines x = 0 and 3x p= .

Solution Differentiating the given equation with respect to x, we get .tantansec.

sec1 xxx

xdxdy

==

\ Length of the curve from x = 0 to x = p/3 is given by

∫∫∫==

=+=⎟⎠⎞

⎜⎝⎛

+=3/

0

3/

0

22

sectan11pp

xx

b

axdxdxxdx

dxdys


Calculus

( )[ ] 3/0tanseclog pxx +=

( )0tan0seclog3

tan3

seclog +-⎟⎠⎞

⎜⎝⎛

+=pp

( ) ( )01log32log +-+=

( )32log += .

Example 4 For the catenary cxcy cosh= , show that y2 = c2 + s2, where s is the length of the

arc measured from its vertex to the point (x, y).

Solution The point of intersection of the curve and its axis is called vertex of the

curve. Since replacing x by -x in cxc cosh does not alter its value, the

given curve is symmetrical about the y-axis. Hence the vertex lies on the y-axis. Hence the x coordinate of the vertex is 0.

Differentiating the given curve with respect to x, we obtain

cx

ccxc

dxdy sinh1sinh ==

\ ∫∫ +=⎟⎠⎞

⎜⎝⎛

+=xx

dxcxdx

dxdys

02

0

2sinh11

x

x

cxcdx

cx

00

sinhcosh ⎥⎦⎤

⎢⎣⎡

== ∫

cxcc

cxc sinh0sinhsinh =-=

\ .coshsinh1sinh 2222222222 ycxc

cxc

cxccsc ==⎟

⎠⎞

⎜⎝⎛+=+=+

Example 5 Prove that the length s of the curve 3/23/23/2 ayx =+ measured from (0, a) to

the point (x, y) is given by 3 2

23 axs = . Also find the entire length.

Solution Differentiating the equation of the curve with respect to x, we get

032 3/13/1 =⎟

⎠⎞

⎜⎝⎛

+ --

dxdyyx implies

3

3

3/1

3/1

xy

yx

dxdy

-=-=-

-

Hence the length of the arc from (0, a) to the point (x, y) (i.e. abscissa from x = 0 to x = x ) is given by

∫∫ ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛-+=⎟

⎠⎞

⎜⎝⎛

+=xx

dxxy

dxdxdys

0

2

3

3

0

211

∫+

=x

dxx

yx0 3/2

3/23/2


Calculus

∫=x

dxxa

0 3/2

3/2

, since x2/3 + y2/3 = a2/3

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛=

xdx

xa

0

21

3/2

3/2

3 2

0

3/23/1

03/13/1

23

3/2axxadxxa

xx

=⎥⎥⎦

⎤

⎢⎢⎣

⎡== ∫ - . . . (2)

Finding the entire length of the curve Since replacing x by -x does not alter the equation 3/23/23/2 ayx =+ , the given curve is symmetrical about the y-axis. Similarly, the given curve is symmetrical about the x-axis. That is, the given curve is symmetric about both the axes, and hence there are 4 symmetrical portions. \ The total length of the curve is 4 times the length of the arc in the first quadrant. Also, the given curve x2/3 + y2/3 = a2/3 touches the x-axis at the point ( )0,a and the y-axis at ( )a,0 . Hence in the first quadrant arc varies from (0, 0) to (a, 0) i.e. abscissa varies from x = 0 to x = a. Hence, substituting x = a in equation (2), we get the length of the arc in

first quadrant as .23

23 3 2 aaa ==

Hence the entire length of the curve is .6234 aa =·

Formula for the length of a smooth curve x = g(y), c y d£ £

2

21 1 ( ( )) .d d

c c

dxL dy g y dydy

⎛ ⎞¢= + = +⎜ ⎟

⎝ ⎠∫ ∫ …(2)

Example Find the length of the curve 2/ 3( / 2)y x= from 0x = to 2.x = Solution The derivative

1/ 31/ 32 1 1 2

3 2 2 3dy xdx x

-⎛ ⎞⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

is not defined at 0x = , so we cannot find the curve’s length with equation (1) We therefore rewrite the equation to express x in term of y :

2 / 3

2xy ⎛ ⎞= ⎜ ⎟

⎝ ⎠

implies 3/ 2

2xy =

implies 3/ 22x y=

From this we see that the curve whose length we want is also the graph of 3/ 22x y= from 0y = to 1y =

The derivative


Calculus

1/ 2 1/ 232 32

dx y ydy

⎛ ⎞= =⎜ ⎟⎝ ⎠

is continuous on [0,1]. We may therefore use equation (2) to find the curve’s length:

2

1

01 1 9

d

c

dxL dy y dydy

⎛ ⎞= + = +⎜ ⎟

⎝ ⎠∫ ∫

9

duu= ∫ , by taking 1 9 , 9 .u y du dy= + =

3/ 21

9 3/ 2u⎡ ⎤= ⎢ ⎥⎣ ⎦

13/ 2

0

1 2 (1 9 )9 3

y⎡ ⎤= +⎣ ⎦

2 (10 10 1).27

= - .

Exercises 1. Find the length of the curve y = log sec x between the points given by x = 0 and x =

p/4 . 2. Find the length of the arc of the parabola y2 = 4x from the vertex up to the point (4, 4).

3. Find the length of the curve 27y2 = 4x3 from x = 0 to x = 3. 4. Find the length of the arc of the parabola y2 = 4ax measured from the vertex to one

extremity of the latus rectum. 5. Find the length of the arc of the parabola x2 = 4y from the vertex to an extremity of the

latus rectum. 6. Show that the length of the arc of a curve 27y2 = 4x3 measured between the origin and

the point (6, 4 2) is equal to 2(3 3 – 1). 7. Prove that the length of the arc of the parabola y2 = 4ax cut off by the latus rectum is

2a[ +log(1+ 2)]. 8. Find the length of the arc of the following curves:

(i) 3y = (x-3) x between x = 0 and x = 9;

(ii) 11log

+

-= x

x

eey from x = 1 to x = 2.

9. Find the length of the loop of the curve 3 ay2 = x (x - a)2 . 10. Find the entire length of the curve 27 y3 = 4 x2 from x = 0 to x = 4. Answers 1. log( 2+1) 2. 2 5+log(2+ 5) 3. 4 2-2 4. a [a +1/2 log(3+2 2)]

5. log( 2+1)+ 2 8.(i) 12; (ii) log (e2 + 1 - e) 9. 4a/ 3 ( )649881458

1.10 23 -

AREAS OF SURFACES OF REVOLUTION If a plane curve revolves about an axis lying in its own plane, then its perimeter generates a surface of revolution. For example, a sphere is obtained by revolution of a


Calculus

semicircle about one of its diameter and a cylinder is obtained by the revolution of a straight line about another parallel line.

Consider the surface obtained by revolving about the x-axis the portion of the arc of the curve y = ƒ(x) that is intercepted between the points whose x- coordinates are a and b. Then its surface area A is given by

( )∫∫ ¢+=⎟⎠⎞

⎜⎝⎛

+=b

a

b

a

dxxfxfdxdxdyyA 2

2)(1)(212 pp

Example Find the area of the surface generated by revolving about the x-axis, the arc of the parabola y2 = 4ax from (0, 0) to the point where x = a. Solution Here y2 = 4ax and x varies from 0 to a. Differentiating,

adxdyy 42 = or

xa

axa

ya

dxdy

===2

22

Therefore, required surface area A is given by

∫ ⎟⎠⎞

⎜⎝⎛

+=

a

dxdxdyyA

0

212p ∫ +=

a

dxxaax

0

122p

∫+

=a

dxx

axax0

22p ∫ +=a

dxaxa0

4p

( ) [ ]12238

324 2

0

2/3 -=⎥⎦⎤

⎢⎣⎡

+= axaaa

pp

Example Find the area of the surface generated by revolving the arc of the catenary

cxcy cosh= from x = 0 to x = c about the x-axis.

Solution

Here cx

ccxc

dxdy sinh1sinh =·=

Hence the required area is given by

∫∫ +=⎟⎠⎞

⎜⎝⎛

+=cc

dxcx

cxcdx

dxdyyA

0

2

0

2

sinh1cosh212 pp

∫=c

dxcxc

0

2cosh2p

* ∫+

=c

dxcx

c0

2

2cosh12p

[ ]⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦⎤

⎢⎣⎡

+=c

c

cxcxc

00 2sinh

2p

( )2sinh22

2sinh2

2

+=⎟⎠⎞

⎜⎝⎛+=

cccc pp


Calculus

Aliter:(after the step *)

c

cx

cx

c cx

cx

xeecdxeecA

0

22

0

22

22224

22⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+-=

++=

--

∫p

p

( )2sinh22

021

212

222

222

+=⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛

+--+-=- cceec pp

Example Find the area of the surface generated by revolving about the x-axis, the portion in the first and second quadrants of the circle 222 ayx =+ . Solution Differentiating the given equation with respect to x, we get

022 =+dxdyyx

implies yx

dxdy

-=

In the first and second quadrants of the portion of the circle, x varies from x = -a to x = a. \ the required area is given by

∫∫--

+=⎟⎠⎞

⎜⎝⎛

+=a

a

a

a

dxyxydx

dxdyyA 2

22

1212 pp

.4222 222

22

adxadxadxy

xyya

a

a

a

a

a

pppp ===+

= ∫∫∫---

Example Find the area of the surface generated by revolving the curve 2 , 1 2y x x= £ £ , about the x -axis.

Solution We evaluate the formula

2

2 1b

a

dyS y dxdx

p ⎛ ⎞= + ⎜ ⎟⎝ ⎠∫

with

11, 2, 2 , dya b y xdx x

= = = =

2

2

1

12 2 1S x dxx

p⎛ ⎞

= + ⎜ ⎟⎝ ⎠∫

2

1

12 2 1x dxx

p= +∫

2

1

12 2 xx dxx

p += ∫


Calculus

2

1

12 2 xx dxx

p += ∫ 2

14 1x dxp= +∫

2

3/ 2

1

2 84 . ( 1) (3 3 2 2)3 3

x pp ⎤= + = -⎥⎦.

Revolution About the y - axis If the arc rotates about the y axis, the area thus formed is obtained by taking x instead of y in the formula. Surface Area Formula for Revolution About the y-axis If ( ) 0x g y= ‡ is smooth on [ , ]c d , the area of the surface generated by revolving the curve

( )x g y= about the y -axis is

2

22 1 2 ( ) 1 ( '( )) .d d

c c

dxS x dy g y g y dydy

p p⎛ ⎞

= + = +⎜ ⎟⎝ ⎠

∫ ∫

Example The line segment 1 , 0 1x y y= - £ £ , is revolved about the y -axis to generate a cone. Find its lateral surface area. Solution

0, 1, 1 , 1,dxc d x ydy

= = = - =-

221 1 ( 1) 2dx

dy⎛ ⎞

+ = + - =⎜ ⎟⎝ ⎠

and

2

1

02 1 2 (1 ) 2

d

c

dxS x dy y dydy

p p⎛ ⎞

= + = -⎜ ⎟⎝ ⎠

∫ ∫

12

0

12 2 2 2 12 2yyp p

⎡ ⎤ ⎛ ⎞= - = -⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦

2.p= Aliter: The above calculation agrees with the formula from geometry;

Lateral surface area = base circumference slant height 2.2

p· =

The Short Differential Form The equations

2

2 1b

a

dyS y dxdx

p ⎛ ⎞= + ⎜ ⎟⎝ ⎠∫ and

2

2 1d

c

dxS x dydy

p⎛ ⎞

= + ⎜ ⎟⎝ ⎠

∫

are often written in terms of the arc length differential 2 2ds dx dy= + as

2b

aS y dsp= ∫ and 2

d

cS x dsp= ∫

In the first of these, y is the distance from the x -axis to an element of arc length ds . In the second, x is the distance from the y-axis to an element of are length ds. Both integrals have the short differential form

2 (radius)(band width) 2 ,S dsp pr= =∫ ∫


Calculus

where r is the radius from the axis of revolution to an element of arc length ds . In any particular problem, we would then express the radius function r and the arc length differential ds in term of a common variable and supply limits of integration for that variable. Example 6 Find the area of the surface generated by revolving the curve 3 ,0 1/ 2,y x x= £ £ about the x - axis . Solution We start with the short differential form: 2S dspr= ∫

2 ydsp= ∫ For revolution about the x-axis, the radius function is .yr =

2 22 y dx dyp= +∫ , as 2 2ds dx dy= +

We then decide whether to express dy in terms of dx or dx in terms of dy . The original form of the equation, 3y x= , makes it easier to express dy in terms of dx , so we continue the calculation with

3 2 2 2 2 2 2, 3 , (3 )y x dy x dx and dx dy dx x dx= = + = +

41 9x dx= + . With the substitutions, x becomes the variable of integration and

1/ 2 2 2

02

x

xS y dx dyp

=

== +∫

1/ 2 3 4

02 1 9x x dxp= +∫ Now substitute 4 31 9 , / 36 .u x du x dx= + =

1/ 2

4 3/ 2

0

1 22 (1 9 )36 3

xp ⎤⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟ ⎥⎝ ⎠⎝ ⎠ ⎦

3/ 291 1

27 16p ⎡ ⎤⎛ ⎞= + -⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

3/ 225 1251 1

27 16 27 64p p⎡ ⎤⎛ ⎞ ⎛ ⎞= - = -⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

6 11 7 2 8

p= .

Exercises 1. Show that the surface area of the curved surface of a hemisphere of radius a is 2pa2. 2. Show that the area of the surface of the solid obtained by revolving the arc of the curve y

= sin x from x = 0 to x = p about the x-axis is 2p[ 2+log(1+ 2)]. 3. Find the area of the surface formed by the rotation of the curve y2 = 8x about the axis of x

from x = 2 to x = 7. 4. Find the area of the curved surface obtained by revolving about the x-axis the portion of

the parabola axy 82 = included between the origin and the ordinate x = 3a. 5. Show that the area of the surface generated by revolving about the x-axis the arc of the

parabola axy 42 = from the vertex to the end of a latus rectum is .3

12282ap

⎟⎠⎞⎜

⎝⎛ -


Calculus

6. Show that the area of the surface formed by revolving the ellipse x2 + 4y2 = 16 about the x-axis is

⎟⎟⎠

⎞⎜⎜⎝

⎛+

33418p

7. Find the area of the curved surface obtained by revolving about the x-axis the curve 6xy = x4 + 3 from x = 2 to x = 7.

8. Find the area of the curved surface obtained by revolving about the x-axis loop of the curve ( )223 axxay -= .

Answers 3. 152 2 p /3 4. 56pa2/3 7. 7615p/64 8. pa2/2

CHAPTER 8 MOMENTS AND WORK

Moments and Centers of Mass Definition If an object of mass m is placed in a location where the acceleration of gravity is

,g the object’s weight there is =F mg (as in Newton’s second law). Masses Along a Line

Consider masses 1 2, ,m m and 3m on a rigid -x axis supported by a fulcrum at the origin.

The resulting system may or may not balance depending on how large the masses are and how they are arranged.

Each mass km exerts a downward force km g equal to the magnitude of the mass times the acceleration of gravity. Each of these forces has a tendency to turn the axis about the origin, the way you turn a seesaw. This turning effect, called a torque, is measured by multiplying the force km g by the signed distance kx from the point of application to the origin. Masses to the left of the origin exert negative (counterclockwise) torque. Masses to the right of the origin exert positive (clockwise) torque.

The sum of the torques measures the tendency of a system to rotate about the origin. This sum is called the system torque.

System torque 1 1 2 2 3 3= + +m gx m gx m gx . …(1) The system will balance if and only if its torque is zero. If we factor out the g in Eq. (1), we obtain System torque = 1 1 2 2 3 3( ).+ +g m x m x m x …(2) Thus the torque is the product of the gravitational acceleration g, which is a feature of

the environment in which the system happens to reside, and the number 1 1 2 2 3 3(m x + m x + m x ), which is a feature of the system itself, a constant that stays the same

no matter where the system is placed. The number 1 1 2 2 3 3( )+ +m x m x m x is called the moment of the system about the origin. It

is the sum of the moments 1 1 2 2 3 3, ,m x m x m x of the individual masses.

=OM Moment of system about origin =∑ k km x

We usually want to know where to place the fulcrum to make the system balance, that is, at what point x to place it to make the torque zero.


Calculus

The torque of each mass about the fulcrum in this special location is given by: Torque of km about =x (signed distance of km from x ) ·

(downward force) ( ) .= -k kx x m g When we write the equation that says that the sum of these torques is zero, we get an

equation we can solve for :x ( ) 0,- =∑ k kx x m g as sum of the torques equals zero

( ) 0,- =∑ k kg x x m using Constant Multiple Rule for Sums

( ) 0,- =∑ k k km x xm as g divided out, km distributed

0,- =∑ ∑k k km x xm using difference rule for sums

,=∑ ∑k k km x x m by rearrangement and applying constant Multiple rule again

.=∑∑

k k

k

m xx

m

This last equation tells us to find x by dividing the system’s moment about the origin by the system’s total mass:

system moment about origin .system mass= =∑∑

k k

k

x mx

m

The point x is called the system’s center of mass (c.m.). Formula for center of mass of One Dimensional Objects–c.m. of Wires and Thin Rods

Imagine a long, thin strip lying along the x - axis from x a= to x b= and cut into small pieces of mass kmD by a partition of the interval [ , ].a b

The thk piece is kxD units long and lies approximately kx units from the origin. Now observe three things.

First, the strip’s center of mass x is nearly the same as that of the system of point masses we would get by attaching each mass kmD to the point .kx

system moment .system massx »

Second, the moment of each piece of the strip about the origin is approximately ,k kx mD so the system moment is approximately the sum of the :k kx mD

System moment .k kx m» D∑

Third, if the density of the strip at kx is ( ),kxd expressed in terms of mass per unit length, and d is continuous, then kmD is approximately equal to ( )k kx xd D (mass per unit length times length): ( ) .k k km x xdD » D

Combining these three observations gives

system momentsystem massx »

( ).

( )k k k k k

k k k

x m x x xm x x

d

d

D D» »

D D∑ ∑∑ ∑

.... (2)


Calculus

The sum in the last numerator in Eq. (2) is a Riemann sum for the continuous function ( )x xd over the closed interval [ , ].a b The sum in the denominator is a Riemann sum for the

function ( )xd over this interval. We expect the approximations in (2) to improve as the strip is partitioned more finely, and we are led to the equation

( )

.( )

b

ab

a

x x dxx

x dx

d

d=∫∫

This is the formula we use to find the center of mass .x

Moment, mass, and Center of Mass of a Thin Rod or Strip Along the x-axis with Density Function ( )xd

Moment about the origin : ( )= ∫b

O aM x x dxd … (3a)

Mass : ( )b

aM x dxd= ∫ … (3b)

Center of mass : = OMx M … (3c)

Example Show that the center of mass of a straight, thin strip or rod of constant density lies halfway between its two ends. Solution We model the strip as a portion of the x - axis from x a= to x b= . Our goal is to show that ( ) / 2,x a b= + the point halfway between a and .b

The key is the density’s having a constant value. This enables us to regard the function ( )xd in the integrals in Eqs. (3) as a constant (call it d ), with the result that

2 2 21 ( )2 2⎡ ⎤= = = = -⎢ ⎥⎣ ⎦∫ ∫

bb b

O a a aM x dx xdx x b add d d

[ ] ( )b b b

aa aM dx dx x b ad d d d= = = = -∫ ∫

2 2

0( )2( )

b aMx M b a

d

d

-= =

- 2+ -

=-

(b a)(b a)(b a)

.2a b+=

Exercises 1. An 80 lb- child and a 100 lb- child are balancing on a seesaw. The 80 lb- child is 5 ft

from the fulcrum. How far from the fulcrum is the 100 lb- child? 2. The ends of two thin steel rods of equal length are welded together to make a right-

angled frame. Locate the frame’s center of mass. (Hint : Where is the center of mass of each rod?)

Exercises 3-6 give density functions of thin rods lying along various intervals of the x - axis. Use Eqs. (3a-c) to find each rod’s moment about the origin, mass, and center of mass. 3. ( ) 4, 0 2x xd = £ £ 4. ( ) 1 ( / 3), 0 3= + £ £x x xd

9. ( ) 1 (1/ ), 1 4= + £ £x x xd 6. 2 , 0 1

( ), 1 2- £ £⎧

= ⎨£ £⎩

x xx

x xd

Answers 1. 4 ft 2. ( / 4, / 4)L L


Calculus

3. 0 8, 8, 1M M x= = = 4. 0 15 / 2, 9 / 2, 5 / 3M M x= = =

5. 0 73 / 6, 5, 73 / 30M M x= = = 6. 0 3, 3, 1M M x= = = Masses Distributed over a Plane Region –

c.m. of Two Dimensional Objects

Suppose we have a finite collection of masses located in the plane, with mass km at the point ( , )k kx y . The mass of the system is given by

System mass : .kM m=∑

Each mass km has a moment about each axis. Its moment about the x - axis is ,k km y and its moment about the y -axis is .k km x The moments of the entire system about the two axes are Moment about x - axis : ,x k kM m y=∑

Moment about y -axis : .y k kM m x=∑

The x - coordinate of the system’s center of mass is defined to be

.y k k

k

M m xx M m= =∑

∑ … (4)

With this choice of ,x as in the one-dimensional case, the system balances about the line x x= .

The y -coordinate of the system’s center of mass is defined to be

.k kx

k

m yMy M m= =∑

∑ … (5)

With this choice of ,y the system balances about the line y y= as well. The torques exerted by the masses about the line y y= cancel out. Thus, as far as balance is concerned, the system behaves as if all its mass were at the single point ( ), .x y We call this point the system’s center of mass.

Thin, Flat Plates

In many applications, we need to find the center of mass of a thin, flat plate: a disk of aluminum, say, or a triangular sheet of steel. In such cases we assume the distribution of mass to be continuous, and the formulas we use to calculate x and y contain integrals instead of finite sums. The integrals arise in the following way.

Imagine the plate occupying a region in the xy -plane, cut into thin strips parallel to one of the axes (in Fig.8, the y -axis). The center of mass of a typical strip is ( , ).x y% % We treat the strip’s mass mD as if it were concentrated at ( , ).x y% % The moment of the strip about the y -axis is then .x mD% The moment of the strip about the x - axis is .y mD% Equations (4) and (5) then become


Calculus

D

= =D

∑∑%yM x m

x M m and .

D= ==

D∑∑%

x y mMy M m

As in the one-dimensional case, the sums are Riemann sums for integrals and approach these integrals as limiting values as the strips into which the plate is cut become narrower and narrower. We write these integrals symbolically as

x dm

xdm

= ∫∫%

and .y dm

ydm

= ∫∫%

Moments, Mass, and Center of Mass of a Thin Plate Covering a Region in the plane. Moment about the x –axis xM y dm= ∫ % …(6a)

Moment about the y –axis yM x dm= ∫ % … (6b)

Mass : M dm= ∫ … (6c)

Center of Mass : ,= =y xM Mx yM M … (6d)

Example 4 Find the center of mass of a thin plate of constant density covering the region bounded above by the parabola 24y x= - and below by the x - axis. Solution

Since the plate is symmetric about the y -axis and its density is constant, the distribution of mass is symmetric about the y -axis and the center of mass lies on the y -axis. This means that 0.x = It remains to find / .xy M M=

A trial calculation with horizontal strips leads to an inconvenient integration

4

02 4 .xM y y dyd= -∫

We therefore model the distribution of mass with vertical strips instead . The typical vertical strip has

center of mass (c.m.) : 24( , ) , ,2

xx y x⎛ ⎞-= ⎜ ⎟⎝ ⎠

% %

length : 24 ;- x width : ;dx area : 2(4 ) ;= -dA x dx

mass : 2(4 ) ,dm dA x dxd d= = -

distance from c.m. to x - axis: 24 .2

xy -=%

The moment of the strip about the x - axis is

2

2 2 24 (4 ) (4 ) .2 2xy dm x dx x dxdd-= - = -%

The moment of the plate about the x - axis is

2 2 2

2(4 )2xM y dm x dxd

-= = -∫ ∫% … (7)

2 2 4

2

256(16 8 ) .2 15x x dxd d-

= - + =∫

The mass of the plate is


Calculus

2 2

2

32(4 ) .3M dm x dxd d-

= = - =∫ ∫ … (8)

Therefore,

(256 /15) 8 .(32 / 3) 5xMy M

dd

= = =

The plate’s center of mass is the point ( )8( , ) 0, .5=x y

Example Find the center of mass of the plate in Example 4 if the density at the point ( , )x y is 22 ,xd = twice the square of the distance from the point to the y -axis.

Solution The mass distribution is still symmetric about the y -axis, so 0.x = With 22 ,xd = Eqs.

(7) and (8) become

2 22 2 2 2 2

2 2(4 ) (4 )2xM y dm x dx x x dxd

- -= = - = -∫ ∫ ∫%

2 2 4 6

2

2048(16 8 ) ,105x x x dx-

= - + =∫

2 22 2 2

2 2(4 ) 2 (4 )

- -= = - = -∫ ∫ ∫M dm x dx x x dxd

2 2 4

2

356(8 2 ) .15x x dx-

= - =∫

2048 5 8 .105 256 7xMy M

1= = =

The plate’s new center of mass is

( )8( , ) 0, .7=x y

Example 6 Find the center of mass of a wire of constant density d shaped like a semicircle of radius .a Solution

We model the wire with the semicircle 2 2y a x= - . The distribution of mass is symmetric about the y -axis, so 0.x = To find ,y we imagine the wire divided into short segments. The typical segment has

length : ,ds a dq= mass : ,dm ds a dd d q= = (mass per unit length times length) distance of c.m. to x - axis : sin .y a q=%

Hence,

2

0 0

0

sin [ cos ] 2 .-= = = =

∫∫∫ ∫

% a a dy dm ay aadm a d

pp

p

q d q d qd p pd q

The center of mass lies on the axis of symmetry at the point (0, 2 / ),a p about two-thirds of the way up from the origin (Fig.15). Exercises In Exercises 1-6, find the center of mass of a thin plate of constant density covering the given region.


Calculus

1. The region bounded by the parabola 2y x= and the line 4.y =

2. The region bounded by the parabola 2y x x= - and the line y x=-

3. The region bounded by the y -axis and the curve 3= -x y y 0 1.y£ £ 4. The region bounded by the axis and the curve cosy x= , / 2 / 2xp p- £ £ .

5. The region bounded by the parabolas 22 4y x x= - and 22y x x= - .

6. The “triangular” region in the first quadrant between the circle 2 2 9x y+ = and the lines 3x = and 3.y = (Hint : Use geometry to find the area.) (0, 0), ( , 0), (0, )a b

Answers 1. 0, 12 / 5x y= = 2. 1, 3/ 5= = -x y 3. 16 /105, 8 /15x y= = 4. 0, /8x y p= =

5. 1, 2 / 5x y= = - 6. 24

x yp

= =-

WORK When a body moves a distance d along a straight line as a result of being acted on

by a force of constant magnitude F in the direction of motion, we calculate the work W done by the force on the body with the formula (constant-force formula for work)

.W Fd= . . . (1) Right away we can see a considerable difference between what we are used to calling

work and what this formula says work is. If you push a car down the street, you will be doing work on the car, both by your own reckoning and by Eq. (1). But if you push against the car and the car does not move, Eq. (1) says you will do no work on the car, even if you push for an hour. Example If you jack up the side of a 2000-lb car 1.25 ft to change a tire (you have to apply a constant vertical force of about 1000 lb) you will perform 1000 1.25 1250· = ft-lb of work on the car. In SI units, you have applied a force of 4448 N through a distance of 0.381m to do 4448 0.381 1695 j· » of work. Definition The work done by a variable force ( )F x directed along the x-axis from

tox a x b= = is

( ) .b

aW F x dx= ∫ . . . (2)

The units of the integral are joules if F is in newtons and x is in meters, and foot-pound if F is in pounds and x in feet. Example Find the work W done by the force of 2( ) 1/ NF x x= along the x-axis from

1m to 10mx x= = Solution The required work is given by

1010

21 1

1 1 1 1 0.9J10

⎤= =- =- + =⎥⎦∫W dxx x

Hooke’s Law for Spring: F = kx Hooke’s law says that the force it takes to stretch or compress a spring x length units

from its natural (unstressed) length is proportional to x. In symbols, F kx= . . . (3)


Calculus

The constant k, measured in force units per unit length, is a characteristic of the spring, called the force constant (or spring constant) of the spring. Hooke’s law (Eq. 3) gives good results as long as the force doesn’t distort the metal in the spring. We assume that the forces in this section are too small to do that. Example Find the work required to compress a spring from its natural length of 1 ft to a length of 0.75 ft if the force constant is 16lb/ftk = . Solution

We picture the uncompressed spring laid out along the x-axis with its movable end at the origin and its fixed end at 1ftx = (Fig.2). This enables us to describe the force required to compress the spring from 0 to x with the formula 16F x= . To compress the spring from 0 to 0.25ft , the force must increase from

(0) 16 0 0lbF = = to (0.25) 16 0.25 4lbF = = . The work done by F over this interval is

0.25

016 ,= ∫W xdx

using Eq. (2) with 0,a = 0.25,b = ( ) 16F x x=

0.252

08 0.5ft lb⎤= =⎦x .

Example A spring has a natural length of 1 m. A force of 24 N stretches the spring to a spring to a length of 1.8 m.

a) Find the force constant k. b) How much work will it take to stretch the spring 2m beyond it natural length? c) How far will a 45-N force stretch the spring?

Solution a) The force constant: We find the force constant from Eq. (3). A force of 24 N stretches the

spring 0.8 m, so 24 (0.8)k= , Eq.(3) with 24, 0.8F x= = 24 / 0.8 30 N / mk = =

b) The work to stretch the spring 2m.: We imagine the unstressed spring hanging along the x-axis with its free end at 0x = (Fig.3). The force required to stretch the spring mx beyond its natural length is the force required to pull the free end of the spring x units from the origin. Hooke’s law with k = 30 says that this force is

( ) 30 .F x x= The work done by F on the spring from 0mx = to 2 mx = is

22 2

0 030 15 60J.W xdx x ⎤= = =⎥⎦∫

c) How far will a 45-N force stretch the spring? We substitute 45mF = equation 30F x= to find

45 30 ,x= or 1.5m.x = A 45-N force will stretch the spring 1.5m. No calculus is required to find out. Pumping Liquids from containers

How much work does it take to pump all or part of the liquid from a container? To find out, we imagine lifting the liquid out one thin horizontal slab at a time and applying the


Calculus

equation W Fd= to each slab. We then evaluate the integral that leads to as the slabs become thinner and more numerous. The integral we get each time depends on the weight of the liquid and the dimensions of the container, but the way we find the integral is always the same. We illustrate this is in the coming examples. The Weight of Water Because of variations in the earth’s gravitational field, the weight of a cubic foot of water at sea level can vary from about 62.26 lb at the equator to as much as 62.59 lb near the poles, a variation of about 0.5%. The weight of a cubic meter of water is 9800 N. Example How much work does it take to pump the water from a full upright circular cylindrical tank of radius 5m and height 10m to a level of 4m above the top of the tank? Solution

The typical slab between the planes at y and y y+D has volume of 2(radius) (thickness)V pD =

2 3(5) 25 m .y yp p= D = D The force F required to lift the slab is equal to its weight, 9800F V= D , as water weighs 9800 3N/ m . 9800(25 ) 245,000 N.y yp p= D = D The distance through which F must act is about (14 )m,y- so the work done lifting the

slab is about force × distance 245,000 (14 ) J.W y ypD = = - D

The work it takes to lift all the water is approximately 10 10

0 0245,000 (14 ) J.W W y yp» D = - D∑ ∑

This is a Riemann sum for the function 245,000 (14 )yp - over the interval 0 10.y£ £ The work of pumping the tank dry is the limit of these sums as || || 0 :P fi

10 10

0 0245,000 (14 ) 245,000 (14 )W y dy y dyp p= - = -∫ ∫

102

0

245,000 14 245,000 [90]2yyp p

⎡ ⎤= - =⎢ ⎥

⎣ ⎦

669,272,118 69.3 10 J.» » · Example The conical tank in Fig.5 is filled to within 2 ft of the top with olive oil weighing

357 lb/ft . How much work does it take to pump the oil to the rim of the tank? Solution

We imagine the oil divided into thin slabs by planes perpendicular to the y -axis at the points of a partition of the interval [0, 8].

The typical slab between the planes at y and y y+D has a volume of about

( )2

2 2 31(radius) (thickness) ft .2 4V y y y ypp pD = = D = D

The force ( )F y required to lift this slab is equal to its weight, 257( ) 57 lb,4F y V y yp= D = D


Calculus

as Weight = weight per unit volume × volume The distance through which ( )F y must act to lift this slab to the level of the rim of the

cone is about (10 )y- ft, so the work done lifting the slab is about

257 (10 ) ft lb4W y y ypD = - D

The work done lifting all the slabs from 0y = to 8y = to the rim is approximately

8

2

0

57 (10 ) ft lb.4W y y yp» - D∑

This is a Riemann sum for the function 2(57 / 4)(10 )y yp - on the interval from 0y = to 8.y = The work of pumping the oil to the rim is the limit of these sums as the norm of the

partition goes to zero:

8 2

0

57 (10 )4W y y dyp= -∫

8 2 3

0

57 (10 )4 y y dyp= -∫

83 4

0

1057 30,561 ft lb.4 3 4y yp ⎡ ⎤

= - »⎢ ⎥⎣ ⎦

Exercises 1. A mountain climber is about to haul up a 50-m length of hanging rope. How much work

will it take if the rope weights 0.624N/m? 2. An electric elevator with a motor at the top has a multistrand cable weighing 4.5 lb/ft.

When the car is at the first floor, 180 ft of cable are paid out, and effectively 0 ft are out when the car is at the top floor. How much work does the motor do just lifting the cable when it takes the car from the first floor to the top?

Calculus - University of Calicut · School of Distance Education Calculus Page 3 CALCULUS CONTENTS...

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