ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)

7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)

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INSTITUTE TECHNOLOGY BRUNEI

FACULTY OF ENGINEERING

COURSE NOTE

COURSE NUMBER: FEG1MA1

COURSE TITLE: ENGINEERING MATHEMATICS 1

DR. MD. FAZLUL KARIM

1


2/73

Differentiation

Function:

Let us consider the following question:

To fence a rectangular region with a wire of length 100 m , what is the relation between

the length and the width of the rectangle?

Figure: A rectangle

Refer to Figure, Let the rectangular region has length x and width y . Then

2 2 100x y+ =

And so

50 .y x=

y changes according to x , and a value ofx will determine a value ofy . For example, if 1x = ,

then 49y = ; if 19.4x = , then 30.6y = .

We note that since x is the length of a rectangle, 0x . It is also clear that 50x . Hence, for

this particular case, the value ofx

is restricted to the interval [ ]0,50 . Similarly,y

is also

restricted to the interval [ ]0,50 .

The relation 50y x= exhibits the concept of a function.

Definition: If a variable y depends on a variablex in such a way that each value ofx determines

exactly one value ofy, then we say thaty is a function ofx.

Definition: A function f from set A to set B , written as :f A B , is a rule that assigns, to

each element in A , a unique element in B . A is called the domain of f, and B the co-domain of

f.

2

y

x

fA B


3/73

Figure: The function f

In the above example, the function involved is

[ ] [ ]

( )

: 0,50 0,50

50

f

f x x

=

Concept of limit:

Many are puzzled by the phrase h approaches 0 but 0h . The following example gives you

an idea of what the phrase means.

Suppose one wishes to travel from A toB , where B is 100m on the right ofA .

First he reaches the midpoint CofAB .

Then he reaches the midpoint D ofCB .

He travels in this way that he always reaches the midpoint of where he is and B . In other words,

if h represents the distance between him and B , then h approaches 0 but 0h . Can u think of

another way?

The symbol ( )limx a

f x b

= means that when x approaches abut x a , then the function value

( )f x approaches b .

Definition: If the values off(x) can be made as close as we like to L by taking values ofx

sufficiently close to a (but not equal to a), then we write

Lxfax = )(lim , which is read the limit off(x) as x approaches a isL.

3

A B

A B

C

BA

C D


4/73

Example. ( )21

lim 1 1x

x x

+ =

Intuitively, when x is close to 1, 2x approaches 1 (say 1.01x = , then 2 1.0201x = ; if

1.001x = , then 2 1.002x = ), so 2 1x x+ approaches 1.

Example: Find )34(lim2

5 + xxx

Solution: )34(lim2

5 + xxx =l 3lim4lim 552

5 + xxx xxim

= ( ) 83)5(453lim5lim4lim 2552

5 =+=+ xxx x

Example: Find3

45lim

3

2 +

x

xx

Solution:

3

45lim

3

2 +

x

xx = ( )( )

4432

42.53lim45lim 3

2

32 =

+=+

xx

x

x

Continuity: A moving object cannot vanish at some point and reappear someplace else to

continue its motion. Thus, we perceive the path of a moving object as an unbroken curve, without

gaps, breaks or holes.

Definition: A function fis said to be continuous at x = a provided the following conditions are

satisfied

(i) f(a) is defined

(ii) )(lim xfax exists

(iii) )()(lim afxfax =

Example: Determine whether the following functions are continuous atx = 2.

=

==

=

=2,4

2,2

4)(,

2,3

2,2

4)(,

2

4)(

222

x

xx

xxh

x

xx

xxg

x

xxf

Solution: In all the cases the functions are identical, except at x = 2, and hence all three have the

same limit atx = 2, namely

( ) 42lim2

4lim)(lim)(lim)(lim 2

2

2222 =+=

=== xx

xxhxgxf xxxxx

The function f is undefined at x = 2, and hence is not continuous at x = 2. The function g is

defined atx = 2, but its value there is g(2) = 3 which is not same as the limit as x approaches 2,

hence g is also not continuous at x = 2. The value of the function h at x = 2 is h(2) = 4, which is

same as the limit as x approaches 2, hence h is continuous atx = 2.

4


5/73

Derivative:

Suppose a particle travels along the y -axis. Its position as time tis given by ( ) 3 23y f t t t= = +

.

When 0t= , 0y = ; when 1t= , 4y = ; when 2t= , 20y = ; etc.

The average velocity of the particle between 0t = and 1t = is 4 while the average velocity of

the particle between 0t= and 2t= is 10.

We see that the average velocity is increasing when tincreases. This shows that the velocity of

the particle changes from time to time.

Suppose tand t h+ are two different times, then the average velocity between time t and t h+

is

( ) ( )f t h f th

+ .

If we let h approaches 0, it is reasonable that

( ) ( )0

limh

f t h f t

h

+

is the instantaneous velocity of the particle at time t.

5

0 4 20

y

position of particle when 0t=




6/73

The gradient (slope) of a curve is the gradient of the tangent:

Gradient of chord:

=2 1

2 1

y y

x x

=y

x

=2 1

2 1

( ) ( )f x f x

x x

Gradient of tangent:

=

0limx

y dy

x dx

=

Differentiation from first principles:

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7/73

y

x

=

( ) ( )f x x f x

x

+

'

0 0

( ) ) )( ) lim lim

x x

dy y f x x f xf x

dx x x

+ = = = Definition: . Let ),(,),(: baCRbaf . The derivative of f at x c= is the limit

( ) ( )0

limh

f c h f c

h

+ .

If this limit exists, f is said to be differentiable at x c= .

Remarks

1. We use the symbol ( )'f c to denote the derivative of f at x c= , i.e.

( )( ) ( )

0' lim

h

f c h f cf c

h

+ = .

7


8/73

Techniques of differentiation

** 0)( =cdx

d

Example1. Letf(x)=5, then 0)5( =dx

d

** Let nbe a positive integer and ( )n

f x x= . Then ( )1

'n

f x nx

= .

Example: 11)(,5)(045 === xx

dx

dxx

dx

d

** fcxcfdx

d =))((

Example(i):4455 205.4)(4)4( xxx

dx

dx

dx

d===

(ii)

1)(

1==

xdx

dx

dx

d

** [ ] )()()()( xgdx

dxf

dx

dxgxf

dx

d=

Example: [ ] ( ) ( ) xxxdx

dx

dx

dxx

dx

d24 32424 +=+=+

Parametric differentiation:

dydy d

dxdxd

=

Implicit differentiation:

( ) ( ) xdy

f y f ydx

d ddx dy

=

Logarithmic differentiation:

( )1

ln( )dy

y

y dx

d

dx

=

The product rule:

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9/73

( ) ( ) ( )y x u x v x uv= =dy dv du

u vdx dx dx

= +

Example: Find ,dx

dyif ( ( xxxy += 32 714

Solution:dx

d

dx

dy= ( ( xxx + 32 714

= ( ) ( ) ( ) ( )147714 2332 +++ xdx

dxxxx

dx

dx

= ( )( ) ( )( ) 191408712114 24322 =+++ xxxxxxx

The quotient rule:

)(

)()(

xv

xuxy =

2

du dvv u

dy dx dx

dx v

=

Example: Differentiate:t

tey

t

cos2

2

=

Solution:

( ) 2

22

cos2

)cos2()(cos2

t

tdt

d

tetedt

d

t

dx

dytt

=

=t

tteetet ttt

2

222

cos4

)sin2()2(cos2 +

=t

ttttte t

2

2

cos2

)sincoscos2( ++

The chain rule:

(function of a function)

y is a function ofu and u is a function ofx

dy dy du

dx du dx=

Example: Differentiate (i) ( ) 913 = xy ,(ii) ( 25cos3 2 += xy

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10/73

Solution (i) Let9,13 uyxu ==

3,98 ==dx

duu

du

dy

( )89. udx

du

du

dy

dx

dy== .3= ( )88 132727 = xu

(ii) Let uyxu cos3,25 2 =+=

xdx

duu

du

dy10,sin3 ==

( )25sin30sin30. 2 +=== xxuxdx

du

du

dy

dx

dy

Chain Rule:

Ifgis differentiable atx andfis differentiable atg(x), then the composition f go is differentiable

atx. Moreover,

( ) )())(()( xgxgfxgf =

Alternatively, if ( )( ) )(, xguxgfy == then,y =f(u) and

dx

du

du

dy

dx

dy.=

Example : If ( )3cos4)( xxh = , find ( )' .h x

Solution. We first find f and g such that h = f go .

If uufxxg cos4)(,)(3 == then

( ) ( )2

3

3)(,sin4)(

)(cos4))(()(

xxguuf

xhxxgfxgf

==

===

Using Chain rule, ( ) ( )322 sin123)(sin4)())(()( xxxxgxgxgfxh ===

Alternatively, let y = h(x) and let uyxu cos4,3 ==

By the form of the chain rule ( )( ) ( )322 sin123sin4.)( xxxudx

du

du

dy

dx

dyxh ====

Example: If ( ) ( )100

2 3sin 1h x x x= + + , find ( )'h x .

Solution:

( ) ( ) ( )

( ) ( )

100 2

992

' differentiate .differentiate 3sin 1

100 3sin 1 2 3cos

h x x x

x x x x

= + +

= + + +

10


11/73

Example: If ( )

1

22 1sinf x x x

x

= +

, find ( )'f x .

Solution:

( )

1

22 2

2

1 1 1

' sin 2 sin cos2f x x x x x x xx x

= + +

Example: Evaluate the derivatives of the following functions:

( ) ( )x

xcxxbxxa

sin1

cos)(,sin)(,3cossin)( 2

+

Solution:

( ) ( ) ( ) ( ) ( ) ( )xxxxxxdx

da 3sin3cos33sincos)3cos(sin)( ==+

(b) By the product and chain rules:

xxxxx

xxxxxxdx

dcos

2

1sin2

2

1.cossin2)sin( 2

3

22 +=+=

(c) By the Quotient rule:

( )

( ) ( )

( ) xx

x

x

xxx

x

xxxx

x

x

dx

d

sin1

1

sin1

1sin

sin1

cossinsin

sin1

)sin0)((cos)sin(sin1

sin1

cos

2

2

22

2

=

+=

++=

=

Implicit differentiation:

Example: Find ,dx

dyifxy = 1

Solution: Differentiating implicitly yields

xy

1= , from which it follows that

2

1

0

0)(

)1()(

xx

y

dx

dy

ydx

dyx

dx

xdy

dx

dyx

dx

d

dx

xyd

==

=+

=+

=

.

Example: Find the slope of the curve 012 =+xy at the points (2,-1) and (2,1).Solution: Differentiating implicitly yields

11


12/73

[ ]

2

1,

2

12

1

012

)0(1

)1,2()1,2(

2

==

=

=

=+

dx

dy

dx

dyydx

dy

dx

dyy

dx

dxy

dx

d

We have accumulated the following formulae:

( )f x ( )'f xk (a constant)

( )0rx r

sinx

cosx

tanx

cotx

secx

cscx

1sin x

1cos x

1tan x

1cot x

1sec x

1csc x

0

1rrx

cosx

sinx2sec x

2csc x

sec tanx x

csc cotx x

2

1

1 x

2

1

1 x

2

1

1 x+

2

1

1 x

+

2

1

1x x

2

1

1x x

12


13/73

)(log xdx

d

)(sinh xdx

d

)(cosh xdx

d

)(tanh xdx

d

x

1

Coshx

Sinhx

xh2

sec

Derivative of Logarithmic and exponential function:

xx

dx

d 1)(log =

Example: Find ( ))1(log 2 +xdx

d

Solution: ( ))1(log2

+xdxd

= ( ) 12

1.1

12

2

2 +=++ xx

xdx

d

x

Example: Find

+x

xx

dx

d

1

sinln

2

Solution:

+xxx

dx

d

1

sinln

2

= ( )

++ xxx

dx

d1ln

2

1)ln(sinln2

)1(2

1cot

2

)1(2

1

sin

cos2

xx

xxx

x

x ++=

++=

Example: Differentiatex

x

tan

)(secSolution: Let

xxy

xy x

seclogtanlog

)(sec tan

==

Differentiating both sides w. r. tox,

( ) ( ) ( )xxxxxxxydx

dy

xxxxx

xdx

dy

y

xseclogsectansecseclogsectan

seclogsectansecsec

1.tan.

1

22tan22

2

+=+=

+=

Example: Find ,dx

dyif

xx

xxy

sin1sin1

sin1sin1tan 1

++

+=

Solution: On rationalizing the denominator,

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14/73

2

1

22tantan

2cos

2sin2

2sin2

tansin

cos1tan 1

2

11

=

===

=

dx

dy

xx

xx

x

x

xy

Example: If ,11

tan2

1

x

xy

+= find ,dx

dy

Solution: Putting

2

11

22

1

1.

2

1

tan2

1

22tantan

2tan

2cos

2sin2

2sin2

sin

cos1

tan

1sec11,tan

xdx

dy

xy

x

xx

+=

===

====+=

Differentiations of parametric equations:

When x and y are given in terms of a parameter, say , then by the function of a function rule of

differentiation

d

dx

dx

dy

d

d

dx

yd

d

dx

d

dy

dx

dy

==2

2

,/

Example: Find ,dxdy if ( ) ( ) cos1(,sin +== ayax

Solution:( ) 2

cot

2sin2

2cos

2sin2

cos1

sin/

2

==

==

a

a

d

dx

d

dy

dx

dy

Higher (successive) derivatives:

( ) xxdx

d

dx

dy

dx

d

xdx

dyxy

63

3,

2

23

==

==

Now,

dx

dy

dx

dis denoted by

2

2

dx

yd(i.e. 2nd derivative ofy with respect tox)

Similarly,3

3

dx

yd,(i.e. 3nd derivative ofy with respect tox) is here 6.

If y = f (x), the successive derivatives are also denoted by

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15/73

)()()(

)()()(

2

21

xfDxfDxDf

xfxfxf

yyy

yyy

n

n

n

n

D stand for the symboldx

d.

##The thn derivatives of some special functions

( )f x )(xfn

k (a constant)

nx

axe

ax +1

)log( ax +

Sin(ax+b)

Cos(ax+b)

0

!n

axnea

( )

( ) 1!1++

n

n

ax

n

( )

( ) 1

1)!1(1

+

+n

n

ax

n

++

++

baxn

a

baxn

a

n

n

2sin

2sin

Example: If ),sin(log)cos(log xbxay += show that 0122 =++ yxyyx

Solution : Given ),sin(log)cos(log xbxay +=Differentiating

)cos(log)sin(log

1).cos(log

1).sin(log

1

1

xbxaxy

xxb

xxay

+=

+=

Differentiating again,

( )0

)sin(log)cos(log

1).sin(log

1).cos(log

122

12

2

12

=++=+=+

=+

yxyyxyxbxaxyyx

xxb

xxayxy

Example: If

,6

123

2

xxx

xxy

++

= find .ny

Solution: ( ) )2(3)6(6 223 +=+=+ xxxxxxxxx

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16/73

Let236

123

2

+

++=

++

x

C

x

B

x

A

xxx

xx

Multiplying both sides by ( ) )2(3 + xxx , we get,( ) )3()2()2(312 ++++=+ xCxxBxxxAxx

Puttingx = 0,-3,2 successively on both sides, we get

21,

31,

31 === CBA

( )

+

++

+=

++

++=

+++ 111 )2(

1.

2

1

)3(

1.

3

11.

6

1!1

2

1.

2

1

3

1.

3

11.

6

1

nnn

n

nxxx

ny

xxxy

Applications of differentiations:

Example: A particle P moves along the x-axis in such a way that its position at time t is given by

tttx 24152 23 += ft.(a) Find the velocity and acceleration of P at time t.

(b) In which direction and how fast is P moving at 2 s? Is it speeding up or slowing down at

that time?

(c) When is P instantaneously at rest? When is its speed instantaneously not changing?

Solution:(a) The velocity and acceleration of P at time t are

)4)(1(624306 2 =+== ttttdt

dsv ft/s

)52(63012 === ttdt

dva ft/s2

(b) At t = 2, we have v = -12 and a = -6. Thus, P is moving to the lef with speed 12 ft/s, and,

since the velocity and acceleration are both negative, its speed is increasing.

(c) P is at rest when v = 0, that is, when t = 1 or t = 4 s. Its speed is unchanging when a =0,that is, at t = 5/2 s.

Example: Determine the rate of change of voltage, given ttv 2sin5= volts when t= 0.2s.

Solution: Rate of change of voltage is ( ) ttttttdt

dv2sin52cos102sin52cos25 +=+=

When t= 0.2, 7892.3)2.0(2sin5)2.0(2cos)2.0(102sin52cos10 =+=+= tttdt

dv

Example: The luminous intensity I candelas of a lamp at varying voltage V is given by

.10424

Vl= Determine the voltage at which the length is increasing at a rate of 0.6 candelas

per volt.

Solution: The rate of change of light with respect to voltage is given bydV

dI.

Since

( ) VVdV

dI

Vl

44

24

1082104

104

==

=

When the light is increasing at 0.6 candelas per volt then

16


17/73

voltsVV 750,1086.0 4 ==

Example: Newtons law of cooling is given by ,0kt

e= where the excess of temperature at

zero time is .0 C

Determine the rate of change of temperature after 40 s, given that

.03.0,160 == kC

Solution: The rate of change of temperature is

( ) ktekdt

d = 0

When

( ) ( ) sCedt

d

tkC

/594.103.016

,40,03.0,16

4003.0

0

==

===

Example: The distancex metres moved by a car in a time tseconds is given by

.1423 23 += tttx Determine the velocity and acceleration when (a) t= 0 and (b) t= 1.5 s.

Solution: Velocity smttdt

dxv /449 2 +==

Acceleration 22

2

/418 smtdt

xda ==

(a) When time t= 0, ( ) ( ) ( ) 22 /44018,/440409 smasmv ===+=

(b) When time t= 1.5, ( ) ( ) ( ) 22 /2345.0118,/25.1845.145.19 smasmv ===+=

Partial Differentiation

Area of a rectangle depends upon its length and breadth, hence we can say that area is the

function of two variables, i.e., its length and breadth.

zis called a function of two variablesx andy ifzhas one definite value for every pair ofx andy.

Symbolically, it is written as

z=f(x,y).

The variables x and y are called independent variables while zis called the dependent variable.

Similarly, we can definezas a function of more than two variables.

Partial derivatives: Letz=f(x,y) be function of two independent variables x andy. If we keepy

constant andx varies then zbecomes a function ofx only. The derivative ofzwith respect ofx,keeping,y as constant is called partial derivatives ofz, w.r.tox and is denoted by the symbols

),(,, yxfx

f

x

zx

etc. Then( ) ( )

x

yxfyxxf

x

zx

,,lim 0

+=

Similarly,( ) ( )

y

yxfyyxf

y

zy

,,lim 0

+=

Notation: ty

zr

x

zs

yx

zq

y

zp

x

z=

=

=

=

=

2

2

2

22

,,,,

Example: If ,325 234 yyxxz += findy

zb

x

za

)(,)(

Solution: (a)To findx

z

,y is kept constant

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18/73

( ) ( ) 223223324 6200620)1(325 yxxyxxdx

dyx

dx

dyx

dx

d

x

z+=+=+=

(b)To findy

z

,x is kept constant

( ) ( ) ( ) 341.32.2)0(5)(32)1(5 334234 =+=+=

yxyxxy

dy

dy

dy

dx

dy

dx

y

z

Example: If xyz sin= show thaty

z

xx

z

y =

11

Solution: xyx

z

yxyy

x

zcos

1,cos =

=

, sincey is kept constant

xyy

z

xxyx

y

zcos

1,cos =

=

, sincex is kept constant

Hencey

z

xx

z

y =

11

.

Application: Error determination

)(.),(,lim 0 approxxdx

dyyapprox

dx

dy

x

y

dx

dy

x

yx ==

=

x is known as absolute error inx

x

xis known as the relative error inx

100xx

is known as percentage error inx.

Example: Pressurep of a mass of a gas is given by mRTpV = , where m andR are constants, V

is the volume and Tthe temperature. Find expressions forV

p

T

p

, .

Solution: Since mRTpV = thenV

mRTp =

Hence ( ) ,V

mRT

dT

d

V

mR

T

p ==

Vis kept constant.

,1

2V

mRT

VdV

dmRT

V

p=

=

Tis kept constant.

Example: The power dissipated in a resistor is given byR

EP

2

= . Using calculus, find the

approximate percentage change inPwhenEis increased by 3% andR is decreased by 2%.

Solution:R

EP

2

= REP loglog2log =

On differentiating we get,

18


19/73

8)2(32100

1001002

1002

==

=

=

P

P

R

R

E

E

P

P

R

R

E

E

P

P

Percentage change inP= 8.

Partial derivatives of higher order:

Let z = f (x,y), theny

z

x

z

, being the functions ofx and y can be further differentiated

partially w.r. to x and y.

Symbolically

yxxyxy

yy

xx

ffforyx

for

yx

z

y

z

x

fory

for

y

z

y

z

y

forx

for

x

z

x

z

x

=

=

=

=

,,,

,,

,,

22

2

2

2

2

2

2

2

2

Example: Prove that )()( atxgatxfy ++= satisfies

=

2

22

2

2

x

ya

t

y, where f and g are assumed to be at least twice differentiable and a is any

constant.

Solution: )()( atxgatxfy ++= --------------(1)Differentiating (1) w.r. tox partially we get

)()(

)()(

2

2

atxgatxfx

y

atxgatxfx

y

++=

++=

Differentiating (1) w.r. to tpartially we get

19


20/73

2

22222

2

2

)]()([)()(

)()(

x

yaatxgatxfaatxgaatxfa

t

y

atxgaatxfat

y

=++=++=

+=

Example: If ,

xyz

eu = find the value of zyxu

3

Solution:

( )

( )

[ ]222

23

22

31

).()()21,

)())(()(,

zyxxyze

yzxxyzexyzezyx

uor

yzxxexyxzexezy

uor

xyez

u

eu

xyz

xyzxyz

xyzxyzxyz

xyz

xyz

++=

+++=

+=+=

=

=

Extremum problem (Maxima and Minima):

One application of the derivative is to determine the maximum and minimum of a differentiable

function. We shall give examples of this application.

Local maximum or local minimum is generally known as local extremum . Look at the graph of

[ ]: ,f a b in the figure and note that

(i) ( )1f c is a local maximum,

(ii) ( )2f c is a local minimum,

(iii) ( )f a is a minimum on [ ],a b

(iv) ( )3f c is a local maximum on [ ],b and it is also maximum on [ ],a b ,(v) f(a) is not a local minimum on [a,b]

(vi) )( 1cf is a maximum on [ ],

(vii) )( 2cf is a minimum on [ ],

Theorem: Iff(c) is a local extremum andfif differentiable at c, then .0)( = cf

.

Turning points:

(Stationary points)

20

y

x

f

a 1c 2c 3c b


21/73

P is a maximum Q is a minimum

At a turning point: 0dy

dx= ;

Max if :

2

20

d y

dx< Min if :

2

20

d y

dx>

Inflexion point if:

2

20

d y

dx=

Definition: A number 0x is called a critical number of f if ( )0' 0f x = or f is not

differentiable at 0x x= .

21


22/73

Determination of Maxima and Minima:

Example: Find the extremum of [ ] ( ) 3 2: 0,6 , 2 7f f x x x = + .

Solution: Since f is continuous on [ ]0,6 , the Extremum Theorem tells us that f has amaximum and a minimum.

0)43(

043

0)(

2

==

=

xx

xx

xf

Or,x = 0 or 4/3

So, the critical numbers of f are4

0,3

and 6 (observe that 0 and 6 are critical numbers of f

because f is not differentiable at 0 or 6).

( )

( )

0 74 31

3 9

6 151

f

f

f

= =

=

Thus, 151 is the maximum and31

9is the minimum.

Let us note the following result which is useful in determining whether a local extremum occurs

at a critical point.

Example: Find the extremum of Rxxxxf += ,cossin)(

Solution: Since sin and cos are periodic functions with period 2 , it is enough to look at the

problem for [ ]0,2x .

( )' cos sinf x x x= and to find the critical number we let ( )' 0f x = . Then we get

tan 1x = or5

,4 4

x

=

Critical numbers of f are 50, ,4 4 and 2 . ( 0 and 2 are critical numbers because f is not

differentiable at 0 and 2 when we restrict the domain to [ ]0,2 ). f is continuous on

[ ]0,2 , thus f has a maximum and a myinimum. They occur at the critical numbers.

22


23/73

Comparing ( )0 1f = ,5

2, 24 4

f f

= =

and ( )2 1f = , we see that 24

f =

is the maximum and5

24

f =

is the minimum.

Second derivative test: Ifc is a point the interval in which the functionf(x) is defined and if

0)( = cf and 0)( cf thef(c) is a maximum if )(cf is negative and a minimum if)(cf is positive.

Example: Find for what values ofx, the following expression is maximum and minimum

respectively:

2036212 23 + xxx , find also the maximum and minimum values of the expression.

Solution: Let

36426)(

2036212)(

2

23

+=

+=

xxxf

xxxxf

Which exists for all values ofx.

Now, when f(x) is a maximum or a minimum, 0)( = xf

( )( )

6,1

0616

036426 2

==

=+

x

xx

xx

Now, whenx =1, 304212)( == xxf , which is negativeWhenx = 6, 304212)( == xxf , which is positiveHence the given expression is maximum forx = 1, and minimum forx = 6.

The maximum and minimum values of the given expression are respectively f(1), i.e. -3 and

f(6), i.e. -128.

Example: Examine whether xx1 possesses a maximum or a minimum and determine the

same.

Let

exxdx

dy

xx

xxxdx

dy

y

xx

yxy x

==

==

==

,log1,0

)1()log1(1

log11

.1

log1

log,

222

1

Differentiating (1) w. r . tox,

( )

34

2

2

22

2

log232log1

1.

.11

x

x

x

xxx

x

dx

yd

ydx

dy

y

+=

=+

Whenx = e,3

1

2

2 23.

ee

dx

yde

+= , which is negative

Therefore, forx = e, the function is a maximum, and the maximum value is ee

1

.

23


24/73

Example: A liquid form of penicillin manufactured by a pharmaceutical firm is sold in bulk at a

price of RM 200 per unit. If the total production cost forx units is2003.080000,500)( xxxC ++=

And if the production capacity of the firm is at most 30,000 units in a specified time, how many

units of penicillin must be manufactured and sold in that time to maximize the profit?

Solution: Since the total revenue for selling x units is xxR 200)( = , the profit P(x) on x unitswill be

)003.080000,500(200)()()(2

xxxxCxRxP ++== ---(1)Since the production capacity is at most 30,000 units, x must lie in the interval [0,30,000].

From (1) xxdx

dP006.0120)006.080(200 =+=

Setting 0006.0120,0 == xdx

dP

000,20=xSince the critical number lies in the interval [0,30,000], the maximum profit must occur at one of

the valuesx = 0,x = 20,000 or,x = 30,000.

000,400)(,30000

000,700)(,20000

00,500)(,0

====

==

xPx

xPx

xPx

Therefore the maximum profit P= 700,000 occurs when x = 20,000 units manufactured and sold

in the spefied times.

Example: An object is hurled upward from the roof of a building 10 m high. It rises and then falls

back; its height above ground t s after it is thrown is

1089.4 2 ++= tty m,until it strikes the ground. What is the maximum height above the ground that the object attains?

With what speed does the object strike the ground?

Solution: The vertical velocity at time t during flight is

88.98)9.4(2)( +=+== ttdt

dytv m/s.

The object is rising when v>0, that is, when ,8.9/80 8/9.8. Thus, theobject is at its maximum height at time t = 8/9.8 s, and this maximum height is

27.13108.9

88

8.9

89.4

2

max +

+

=y m.

The time tat which the object strikes the ground is the positive root of the quadratic equation

obtained by settingy = 0,

Namely,

462.28.9

196648

+

=t s.

The velocity at this time is v = -(9.8)(2.462)+8 =16.12. Thus, the object strikes the ground witha speed of about 16.12 m/s.

24


25/73

Example: A boat sails 30miles to the east from a point P, then it changes direction and sails tothe south. If this boat is sailing at a constant speed of 10miles/hr, at what rate is its distance from

the point Pincreasing

(i) 2hours after it leaves the point P(ii) 7 hours after it leaves the point P?

Solution:

(i) Since the constant speed of the boat is 10miles/hr, so 2hrs after it leaves the point P,it has traveled 20 miles and it is still sailing east. Thus the rate of its distance from

the point Pis increasing at 10miles/hr.(ii) 7hrs after it leaves P, it has sailed east 30 miles in 3hrs and south 40 miles in 4hrs.

Let its distance from P at time t after it starts sailing be s , where 3t hrs, and abethe distance traveled along the south direction.

Then2 2 230s a= +

So

2 2 .ds da

s adt dt

=

When 40. 50.a s= = Then

4010

50

8miles/hr.

ds a da

dt s dt =

=

=

Thus the rate of its distance from the point P is increasing at 8miles/hr.

Example: A d.c supply has e.m.fE = 12 V and internal resistance =1r . Prove usingcalculus that the power transferred to a load resistorR is a maximum when .1== rR

[Hints: The power,P, in the load is given by( ) 2

2

rR

REP

+= ]

Solution: The power,P, in the load is given by( ) 2

2

rR

REP

+=

SubstitutingE= 12 and r = 1 we find( ) 21

144

+=

R

RP

( ) ( )( ) ( ) 342

1

)1(144

1

)1(21441144

+=

+++=

R

R

R

RRR

dR

dP(2)

For maximum power transfer to the load we require

( )1,0

1

)1(144,0

3==

+

= RR

R

dR

dP(2)

Now,

25

P

s

a

30miles


26/73

( ) ( )

( ) ( )

016288,1

1

)2(288

1

13).1(144)144(1

2

2

46

23

2

2

==

+

=

+

++=

dR

PdR

R

R

R

RRR

dR

Pd

(4)

Which show that he turning point, 1, is a maximum. Hence maximum power is

transferred to the load whenR = r. (2)

Integration

Definition: If ( ) ( ) ( )' , ,F x f x x a b= , then f is the derivative ofF and F is called

an antiderivative (integral or indefinite integral) of f on ( ),a b .Again, IfF(x) is an integral off(x), andx = a andx = b be two given values ofx, the quantity

F(b) F(a) is defined as the definite integral off(x), denoted by the symbol

b

adxxf )( .

a b

Constant of integration: It may be noted that

),()( xfxF

dx

d= then we also have { } ),()( xfcxF

dx

d=+ where C is an arbitrary

constant. Thus, a general value of the indefinite integral += CxFdxxf )()( .

Integration is differentiation in reverse:

26

)(xfy =


27/73

differentiating.2y x= gives 2

dyx

dx=

integrating

2xreverses this and we get back 2

x c+22xdx x c= +

c can be any number c is an arbitrary constant

2xdx is called an Indefinite Integral.Example: Find cos .xdx

Solution: Let ( ) sinF x x= . Then ( )' cosF x x= . Thus ( ) sinF x x= is an antiderivativeof cosx . Therefore

{ } += Rccxxdx sincos

(You will see later that1

ln ,dx x k x

= + where ln x is the logarithmic function with basee .)

Example: Determine: + dxx )73cos(Solution: Let 3x+7=u

Then dudxdxdu

31,3 ==

Therefore cxcudu

udxx ++=+==+ )73sin(31

sin3

1

3cos)73cos(

Example: Find sin cosrx xdx where r is a rational number and 1r .

Solution: Let sinu x= , then cosdu

xdx

= and cosdu xdx= .

1

1

sin cos

1

sin,

1

r r

r

r

x xdx u duu

Cr

xC C

r

+

+

== +

+

= + +

27


28/73

Fundamental Integrals:

+=+=

+=+=+=

=+=+=+

=

+=+=+=++

=+

,cotcos)(,tansec)(

sectansec)(,tansec,tan

sec)(

sincos,sin

cos)(,cossin,cos

sin)(

)(,)(,log1

)(,1

)(

22

22

1

cxxdxecxcxxdxix

cxxdaxxviiicxxdxcm

mxmxdxvii

xxdxc

m

mxmxdxvicxxdxc

m

mxmxdxv

cedxeivcm

edxeiiicxdx

xiic

n

xdxxi

xxmx

mxn

n

Techniques and applications of Integration:

Example: Find Nnxdxxn ,sectan 2Solution: Let tanu x= , then 2sec .du xdx=

++=++==++

Rccn

xc

n

udxuxdxx

nnnn ,

1

tan

1sectan

112

Example: Find sin , , 0.ax dx a a

Solution: Let u ax= , then du a dx= and

( )

sin sin

1sin

1 cos

1cos ,

duax dx u

a

udua

u Ca

ax C C a

=

=

= +

= +

Example: Determine + dxxx )673

4( 2

Solution: + dxxx )673

4( 2 can be written as + dxxxdxdx 2673

4

Hence + dxxx )673

4( 2 =4x+ cxx

++

+

++

126

11.

7

3 1211

=4x+ cxx +32

214

3

Example: Determine (a) ( )

dttbdxx

xx 23

1)(,4

32

Solution: Rearranging into standard integral form gives:

cxxdxdxxdxx

xdx

x

xdx

x

xx+===

4

3

6

1

4

3

2

1

4

3

4

2

4

32 3233

28


29/73

Example: Find the indefinite integrals:

dxeecdxx

xbdx

x

xa

xx ++ 1)(,)ln3sin(

)(,1

)(2

Solution:

(a)dx

x

x +12 , Let duxdxxdxduxu 2

1,2,12 ==+=

( ) cxcxcuu

du++=++=+== 1ln1ln2

1ln

2

1

2

1 22

(b)

dxx

x

)ln3sin(, Let dx

x

duxu3

,ln3 ==

cxcuudu +=+== )ln3cos(31

cos3

1sin

3

1

(c)

dxee xx +1 , Let dxedueuxx =+= ,1

( ) cecuduu x ++=+== 23

2

3

2

1

13

2

3

2

Example: Find tan .x dxSolution:

( )

1

sintan

cos

1Let cos , then sin

ln

ln cos

ln cos

ln sec , is an arbitrary constant

xx dx dx

x

du u x du x dxu

u C

x C

x C

x C C

=

= = =

= +

= +

= +

= +

Example: Find2

.xx e dxSolution: Let 2u x= , then 2du x dx= .

29


30/73

2

2

1

2

1

2

1, is an arbitrary constant2

x u

u

x

xe dx e du

e C

e C C

=

= +

= +

Example: Findsin

2 cos

xdx

x+ .

Solution: Let 2 cos , sinu x du x dx= + = .sin 1

2 cos

ln ,

ln 2 cos

xdx du

x u

u C C

x C

= +

= +

= + +

Example: Findsin

.x

dxx

Solution: Let1

,2

u x du dxx

= =

sinsin 2

2cos ,

2cos

xdx u du

x

u C C

x C

=

= +

= +

Example: Find2

2 .xx dx

Solution: Let 2u x= , then 2 .du x dx=2

2

12 2

2

12

2

1 2,

2 ln 22

2ln2

x u

u

u

x

x dx du

du

C C

C

=

=

= +

= +

Integration by parts: ( ) =+=+=

gffggfgfgffggfgffg ,,

Sometimes it is written as = vduuvudvThis is known as the formula of integration by parts.

30


31/73

Example: Find ln .x dxSolution:

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

( ) ( )

( )

( )

ln 1.ln

' ,

' ,

1ln ,

ln 1

ln ln ,

ln

x dx x dx

f x g x dx

f x g x f x g x dx

x C x x C dx Cx

Cx C x dx

x

x C x x C x K K

x x x K

=

==

= + +

= + +

= + +

= +

Example: Find1sin .xdx

Solution: We let ( )' 1f x = and ( ) 1sing x x= , then ( )f x x= and ( )2

1'

1g x

x=

.

( ) ( )

( ) ( ) ( ) ( )

1 1

1

2

sin 1.sin

'

'

sin1

xdx xdx

f x g x dx

f x g x f x g x

xx x dxx

=

=

=

=

To find

21

xdx

x , let 21u x= , then 2 .du xdx=

2

1

2

1

2

2

1

21

1

2

112

2

1 ,

x dudx

ux

u du

uC

x C C

=

=

= +

= +

Putting this into (1), we get

31


32/73

1 1 2sin sin 1xdx x x x C = + + .

Example: Find sin .xe xdx

Solution:( ) ( )

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

( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( )

sin '

' ' , sin

sin cos , ' cos

s cos '

' ' ,

x

x

x x x

x

x

e xdx f x g x dx

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

e x e xdx f x e g x x

e xdx f x g x dx

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

=

= = =

= = =

=

= =

( )

( ) ( ) ( )

cos

cos sin , ' sin

cos sin

x x x

x x

g x x

e x e x dx f x e g x x

e x e xdx

=

= = == +

Put this into (1), then

( )sin sin cos sin

sin cos sin

x x x x

x x x

e xdx e x e x e xdx

e x e x e xdx

= +

=

So 2 sin sin cos ,x x xe xdx e x e x= thus

( )

1sin sin cos ,

2

x x xe xdx e x e x C C = +

Example: Evaluate: 20 sin2

d

Solution: dsin2 = ++= cd sin2cos22)cos()cos(2

20 sin2

d = [ ] 2sin2cos2 20 =+

Example: Find2

1

1dx

x .

Solution: Let secx u= , then sec tandx u udu= and

2 2

1 1sec tan

1 sec 1

sec

ln sec tan ,

dx u udux u

udu

u u C C

=

=

= + +

32

x

2 1x


33/73

Refer to the figure, since secu x= ,

2tan 1u x= .

Thus2

21 ln 1 .

1dx x x C

x= + +

Example: Find2

1.

1dx

x+

Solution: Let tanx u= , then 2secdx udu= .

2 2

2

1 1

1 1 tan

1

secsec

sec

ln sec tan ,

dx dux u

uduu

udu

u u C C

=+ +

==

= + +

Since tan u x= , we have 2sec 1u x= + ,

thus2

2

1ln 1 .

1dx x x C

x= + + +

+

Example: Find 2

1.

2 2dx

x x+ +Solution: ( )

22 2 2 1 1.x x x+ + = + + Let 1 tanx u+ = , then 2secdx udu= .

( )2

2

2

2

1 1

1 12 2

1sec

tan 1

sec

ln sec tan ,

ln 2 2 1

dx dxxx x

uduu

udu

u u C C

x x x C

=+ ++ +

=+

=

= + +

= + + + + +

Partial fraction:

Find + dxxxx

322

33

1

x

2

1 x+

1

u


34/73

Solution:( )( )13322 +

=+ xx

x

xx

x, writing it in the partial fraction, we have

1

1.

4

1

3

1.

4

3

322 +

+=

+ xxxxx

Rccxx

dxxdxxdxxxdxxx

x

+++=

++=

++=+ ,1ln

4

13ln

4

3

1

1

4

1

3

1

4

3

1

1

.4

1

3

1

.4

3

322

Example: Find( ) ( )

1.

1 2dx

x x +

Solution: First we shall express( ) ( )

1

1 2x x + as a partial fraction.

( ) ( )1 1 1 1 1

1 2 3 1 3 2x x x x

= + +

Then( ) ( )

1 1 1 1 11 2 3 1 3 2

dx dx dxx x x x

= + + +

To find1

1dx

x , we let 1u x= , then .du dx=

1 1

1

1 1

1

ln ,

ln 1

dx dux u

u C C

x C

=

= +

= +

Similarly, RCCxdxx

++=+

22 ,2ln2

1

( )( )RCCxxdx

xx++=

+ ,2ln3

11ln

3

1

21

1

THEOREM (Fundamental Theorem of Calculus)

Suppose

(i) f is integrable on [ ],a b

(ii) F is continuous on [ ],a b and 'F f= on ( ),a b

Then ( ) ( )b

af F b F a=

34


35/73

Definite integral: Definite integrals are those in which limits are applied.

Example: Evaluate (a) a

dxx0

2and (b) ( )

+2

1

2 23 dxxx

Solution: (a)

33

1 3

0

3

0

2 axdxx

aa

==(b)

( )2

9)2()1(

2

3)1(

3

14)4(

2

3)8(

3

12

2

3

3

123

2

1

232

1

2 =

++=+=+

xxxdxxx

Example: Evaluate ( )dxx

3

2

24

Solution: ( )( )

3

25

3

2)2.(4

3

33.4

3

44

333

2

33

2

2 =

=

=

xxdxx

Example: Evaluate: 20 2sin3

xdx

Solution:

3)11(2

3)0cos(cos

2

3

)0(2cos2

2cos2

32cos

2

132sin3

2

0

2

0

===

=

=

xxdx

Example: Find1

0

sin .x dx

Solution: We know that ( ) cosF x x= is an antiderivative of ( ) sin .f x x=

Now F is continuous on [ ]0,1 and 'F f= on ( )0,1 ,

f is continuous on [ ]0,1 and f is integrable on [ ]0,1 .

By the Fundamental Theorem of Calculus,

( ) ( )

1 1

0 0

1

0

sin

|

1 0

cos1 cos 0

1 cos1.

x dx f

F

F F

=

=

= = +=

Remarkb

af is often written as ( )

b

af x dx to emphasize on the variable x .

35


36/73

Example: Find ( )132

11 .x dx+

Solution: ( ) ( )1

21f x x= + is continuous on [ ]1,3 , so it is integrable on [ ]1,3 . We need to

find one antiderivative of f .

( ) ( )

( )

11 22

1

2

1

2

1 let 1 , then

2

3

21 , is a constant

3

x dx u du u x du dx

u c

x c c

+ = = + =

= +

= + +

We need only one antiderivative of f , we choose

( ) ( )3

22

1

3

F x x= +

Thus

( ) ( )

( ) ( )

( ) ( )

( )

133

21

1

3 3

2 2

1 |

3 1

2 21 3 1 1

3 3

28 2 2 .

3

x dx F x

F F

+ =

=

= + +

=

Example: Evaluate: +2

0 2 123xxdx

Solution: Let 12 2 += xu

1,0

,9,2

4,4

====

==

ux

uwhenx

x

dudxx

dx

du

Thus [ ] 3192

3

2

14

3

4

3

4

3

12

3

9

1

2

1

9

1

2

19

1

2

0 2=====

+

uduu

u

du

x

xdx

Example: 202

cos

xdx

Solution:

36


37/73

( )

4sin

4

1

4

2sin4

1

22cos1

2

1cos2

2

1cos

2

0

2

0

2

0

2

0

22

=+=

+=+== xx

dxxxdxxdx

Example:

1

0 2

1

1

sindx

x

x

Solution: Putting dxx

dx2

1

1

1,sin

==

Whenx =0,2

,1;0

=== x

821

sin 22

0

2

2

0

1

0 2

1

=

==

=

ddxx

xI

Example: a

dxxa0

22

Solution:Put dadxax cos,sin ==

Whenx =0,2

,;0 === ax

( )42

2sin

22cos1

2

1.cos

22

0

2

2

0

22

0

22 aadadaI

=

+=+==

General properties of definite Integral:

)()(,,)()(.,)()(.

,)()()(.,)()(.

0000

xafxfifdxxfndxxfivdxxafdxxfiii

bcadxxfdxxfdxxfiidxxfdxxfi

anaaa

a

c

c

a

b

a

a

b

b

a

+===


38/73

Example: Show that =+2

0 4cossin

sin

dx

xx

x

Solution:

++

+=

+=

+

=+=

2

0

2

0

20

20

20

sincos

cos

cossin

sin2

sincos

cos

2cos

2sin

2sin

cossin

sin

dxxx

xdx

xx

xI

dxxx

xdx

xx

x

dxxx

xI

[ ]4

,2cossin

cossin2

0

20

2

0

====++

= Ixdxdxxx

xx

Example: A car is travelling at 72 km/h. At a certain instant its brakes are applied to produce aconstant deceleration of 0.8 m/s2. How far does the car travel before coming to a stop?

Solution: Let s(t) be the distance the car travels in the t seconds after the breaks are applied. Then

),/(8.0)( 2smts = so the velocity at time t is given by

+== 18.08.0)( ctdtts m/s.Since 72)0( =s km/h = 20 m/s, we have 201 =c . Thus,

tts 8.020)( =

38


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And

( ) +== .4.0208.020)( 22 cttdtttsSince s(0) = 0, we have 02 =c and .4.020)(

2ttts = When the car has stopped, its velocitywill be 0. Hence, the stopping time is the solution tof the equation

tts 8.020)(0 ==

That is, t = 25 s. The distance travelled during deceleration iss(25) = 250 m.

Applications of integration:

Let us look at a few applications of integration. We shall see how the integral can be used to find

the length of a curve, the area of a surface of revolution and the volume of a solid of a revolution.

Length of a curve:

The length of the curve ( ) [ ], ,y f x x a b= is defined as 10

1

lim .n

k kP

k

T T=

The length kl of the line segment joining 1kT and kT is given by

( ) ( ) ( )( )

22

1 1 .k k k k k l x x f x f x = + If fis a function with continuous first order derivative on the interval [ ],a b , then the length of

the curve ( ) [ ], ,y f x x a b= is defined by the integral

( )( )2

1 ' .b

af x dx+

Example: Find the length of the curve 23

xy = from (1,1) to ( 22,2 .

Solution: 21

2

3x

dx

dy=

The required arc length

dxxL +=2

14

91 , Let dxduxu

4

9,

4

91 =+=

4

222,

4

131 ==== uxux

39

( )y f x=

0T

1T1k

T

kTnT

0a x= 1a x= 1kx kx

x

nb x=


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09.2]4

13

4

22[

27

8

27

8

9

4 23

2

34/22

4/13

2

34/22

4/13

2

1

=== uduuL

Example: Find the length of the curve ( )3

21y x= + from 0x = to 4.x =

Solution: Let ( ) ( )321f x x= + , then

( ) ( )1

23

' 12

f x x= + and ( )( ) ( )2 9

1 ' 1 1 .4

f x x+ = + +

The length of the curve

( )( )

( )

4

0

4

0

49

13

4932

13

3

2

1 ' 2

19 13

2

1let 9 13

2 9

1

27

1343 13

27

f x dx

x dx

duu u x

u

= +

= +

= = +

=

=

Example: Determine the length of an arc of the cycloid )cos1(),sin( =+= ayax ,measured from the vertex (i.e., the origin)

Solution: Here ( )2

cos2sincos1 22

22

aad

dy

d

dx

d

ds =++=

+

=

Also at the origin 0= .

Hence the required length, from 0= to any point is

2sin4

2cos2

0

adas == .

Areas of a plane curves:

The definite integral b

a

b

aydxeidxxf ,.,)( represents the area bounded by the curves y =

f(x), thex-axis and the two fixed ordinatesx = a andx =b.

Example: Find the area of the quadrant of the ellipse

12

2

2

2

=+b

y

a

xbetween the major and minor axes.

Solution: The required area is

1,2

2

0 2

222

0=+== b

y

a

xdxxa

a

bydx

aa

, for the curve

y= 20 ,cos.cos

daaa

bputting sinax =

40


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= ( ) abab

dab

4

1

2

2sin

22cos1

2

2

0

2

0=

+=+

Example: Find the area A of the plane region lying above the x-axis and under the curve

.3 2xxy =Solution:

We need to find the points where the curve 23 xxy = meets the x-axis. These are solutions ofthe equation

).3(30 2 xxxx ==The only roots are x = 0 and 3. Hence, the area of the region is given by

( )2

9)00(

3

27

2

27

3

1

2

33

3

0

323

0

2 ==== xxdxxxA square units.Example: Find the area under the curve ,sin xy = abovey = 0 fromx = 0 to .=xSolution: The required area is

2)11(cossin0

0

====

xdxxA square units.

**Area between two curves

If f and g are continuous functions on the interval [a,b] and if )()( xgxf for all x in [a,b],

then the area of the region bounded above by y = f(x), below by y = g(x), on the left by x = a and

on the right by the line x = b is

dxxgxfA

b

a

)]()([ =

Example: Find the area of the bounded, plane region R lying between the curves

xxy 22 = and .4 2xy =

Solution: First, we must find the intersections of the curves, so we solve the equations

simultaneously:

0)1)(2(2

0422

42

2

22

=+

=

===

xx

xx

xyxxy

So x = 2 or x = -1

Since ,21,24 22 xxxx The area A of R is given by

41


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( ) ( )

)13

24(4)8(

3

2)2(4

3

24)224(]24[

2

1

23

2

1

2

2

1

22

+++=

+=+==

xxxdxxxdxxxxA

= 9 square units.

Example: Find the area of the region bounded above by y =x+6, bounded below by 2xy = , and

bounded on the sides by the linesx = 0 andx = 2.

Solution: The required area

( )3

34

36

2]6[

2

0

322

0

2 =

+=+=

xx

xdxxxA

Example . Find the area of the region bounded by the curves2y x= and 3.y x=

Solution: The two curves2y x= and 3y x= meet when 2 3x x= , i.e. ( )2 1 0x x = which

gives 0x = or 1x = . Note that on the interval [ ]0,1 , the graph of 2y x= lies above that of3y x= . Thus the area of the region between 2y x= and 3y x= is

( )1

3 41

2 3

003 4

1

12

x xx x dx

=

=

Surface of Revolution:

If fis a continuously differentiable non-negative function on [ ],a b ,then the area of a

surface generated by revolving the curve ( ) [ ], ,y f x x a b= , about the x -axis is

42

y

x

1

1

2y x=

3y x=


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( ) ( )( )2

2 1 'b

af x f x dx +

Suppose that a parametric curve in the upper half of the xy -plane defined by a pair ofcontinuously differentiable functions

( ) ( ) [ ], , ,x x t y y t t a b= =

( ) ( )( ) ( )( )2 22 ' ' .b

ay t x t y t dt +

Example: Find the surface area of a sphere with radius .r

Solution: The surface of a sphere can be generated by revolving a semicircle 2 2y r x=

, [ ],x r r about the x -axis.

Let ( ) 2 2f x r x= , [ ],x r r . Then ( ) 2 2' .x

f xr x

=

( ) ( )( )2

2

2 2

2 2

The surface area of a sphere 2 1 '

2 1

2

r

r

r

r

r

r

f x f x dx

xr x dx

r x

rdx

= +

= +

=

24 r=

Volume of Revolution:

The Disc Method:

The volume of the solid of revolution generated by revolving about the x -axis the region

bounded by the curve ( ) [ ], ,y f x x a b= , where f is continuous, and the x -axisfrom x a= to x b= is given by

( )( )2b

af x dx .

Example: Find the volume of a sold ball having radius a.

Solution: The ball can be generated by rotating the half-disk, axaxay ,0 22

about the x-axis. Therefore its volume is

( ) ( ) 30

32

0

222

22

3

4

3

122 axxadxxadxxaV

aa

a

a

====

cubic units.

Example: Find the volume of the solid of revolution when the region which is bounded by

the curves 2y x= and 2y x= is revolved about(i) The x -axis, (ii) The y -axis

43


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Solution: Let us first find the points of intersection between the two given curves. Solving

the two equations 2y x= and 2y x= , we have2 2 ,x x= i.e. ( )2 0,x x =

So, 0x = or 2x = , and the curves meet at ( )0,0 and ( )2,4 .

(i) The region T bounded by the two curves (refer to the figure above) is revolved about thex -axis. Note that a small vertical slice of the solid which is perpendicular to the axis of

revolution is in the form of a washer with cross sectional area

( ) ( )22 22 .x x

The volume obtained by revolving T about the x -axis is

( ) ( )2

3 52 22 2

00

42

3 5

64

15

x xx x dx

=

=

(ii) The region T is revolved about the y -axis. A small horizontal slice of the solid which is

perpendicular to the axis of revolution is in the form of a washer with cross sectional area

( )2

2

.2

yy

The volume obtained by revolving T about the y -axis is

( )42 2 324

00

8

2 2 12 3

y y yy dy

= =

44

y

42

x

( )2,4

2y x=2y x=

T


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The Shell Method:

The volume of the solid of revolution is

( )2 .b

axf x dx

Example: Find the volume of the solid generated by revolving the region bounded by the

curve2

1y

x= and the x -axis between 1x = and 2x = about

(i) The y -axis

(ii) The x -axis

(iii) The line 1x =

Solution:

(i) The given region is revolved about the y -axis. Subdivide the region into small strips

that are parallel to the axis of revolution, then each strip when revolved generates a

cylindrical shell. The volume of the solid of revolution about the y -axis is

2 2

2 11

12 2 ln

2 ln 2

x dx xx

=

=

(ii)When the given region is revolved about the x -axis, we subdivide the region into

small strips that are perpendicular to the axis of revolution. Each strip when revolvedgenerates a disc. The volume of the solid of revolution is

22

2 311

1 1 7

3 24dx

x x

= =

45

x

y

2

1y

x=

1 2


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(iii)When the given region is revolved about the line 1x = , a vertical strip at coordinate x

is at distance ( )1 1x x = + from the axis of revolution; and when revolved it generates acylindrical shell. The volume of solid of revolution is

( )2

2

211

1 1 12 1 2 ln 2 ln 2

2x dx x

x x

+ = = +

Multiple Integrals: Multiple integral is a natural extension of a definite integral to a function of

two variables (double integral) or three variables (triple integral) or more variables. Double andtriple integrals are useful in evaluating are, volume, mass, centroid and moments of inertia of

plane and solid regions.

Double integral over regionA may be evaluated by two successive integrations. IfA is described

as [ ]bxayyyxfyxf ,),()( 2121

Then =b

a

y

y

dxdyyxfdxdyyxf ]),([),(2

1

)(: 22 xyyc =

a b

46

1 1 2

x

2

1y

x=

)(: 11 xyyc =R


47/73

Partial definite integrals: b

a

b

a

dyyxfdxyxf ),(,),(

Example:22

21

0

221

0

2

1

0

2 yxyxdxydxxy ===

33

1

0

31

0

2

1

0

2 xxydyyxdyxy ===

This two-stage of integration process is called iterated integration.

We introduce the following notation:

=

d

c

b

a

d

c

b

a

dydxyxfdxdyyxf ),(),(

=

b

a

d

c

b

a

d

c

dxdyyxfdydxyxf ),(),(

These integrals are called iterated integrals.

Example: Evaluate ( ) ( ) 4

2

3

1

3

1

4

2

240)(,240)( dxdyxybdydxxya

Solution: (a)

( ) ( ) [ ]

[ ] 112680

)1280(40240240

3

1

2

3

1

3

1

3

1

4

2

2

4

2

3

1

4

2

==

==

=

xx

dxxdxxyydxdyxydydxxy

Similarly,(b) 112

Example: Evaluate the double integral

R

xdAy 2

over the rectangle { }.10,23:),( = yxyxRSolution:

6

5

6

5

2

5

2

1

0

3

1

0

2

2

3

1

0

221

0

2

3

22 =

==

==

ydyydy

xyxdxdyyxdAy

R

Example: Evaluate +20 0 )cos(

dxdyyx

Solution: We have ( )[ ]

02

0

2

0 0sin)cos( yxdydxyxdyI +=+=

= [ ]dyyy ++20 )0sin()sin(

= [ ] [ ] 2cos2sinsin 2020

==

ydyyy

47


48/73

Example: Evaluate dyedxx

x

y

1

0 0

Solution:

( ) ( ) ( )12

1

211)(

1

0

21

0

1

0

0

1

0

1

00=

===

= e

xexdxedxxxexedxdyedx

x

x

yx

x

y

Example: Evaluate xydxdy over the region in the positive quadrant for which 1+yx .Solution:

B(0,1)

x+y=1

O A(1,0) x

Solution:x+y=1 represents a straight lineAB in the figure. The limits fory are 1 x and 0.

Required integral = ( )( ) 21

0

1

0

21

0

1

0

1

01

21

2xxdxyxdxydyxdx

x

x

=

=

= ( )24

1

432

22

12

2

11

0

43232

1

0=

+=+

xxxdxxxx

Area in polar coordinate: Area = rdrd

Evaluate, =

2

0 sin

a

ardrdI

Solution: ( ) +==

==

2

0

22

0

22

sin

2

0

22

0 sin2cos1

4cos

22d

ad

ad

rrdrdI

a

a

a

a

22

2sin

4

22

0

2 aa

=

=

Triple integral: Let V be a given three-dimensional domain in space, bounded by a closed

surface S. Letf(x,y,z) be a continuous function in Vof the rectangular coordinates x,y,z. Then a

triple integral of f over the domain Vis defined as

== VViiv dxdydzzyxfdvpfvpfi ),,()()(lim 0 ------------(1)Volume: Volume of a solid contained in the domain V is given by the triple integral (1) with

f(x,y,z) = 1,

i.e. volume = V dxdydz

Evaluate: ( ) 32,21,10:, ++ zyxRdxdydzzyxR

Solution:( )

3

2

22

1

3

2

1

0

2

1

1

0 2)(

++=++

zyxdydxdzzyxdydx

48


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( ) ( )[ ] ( )

( )( ) ( )[ ] ( )

2

94

2

45225428

1

4

522

2

1

1.5222

123

2

1

1

0

2

1

0

221

0

2

1

21

0

2

1

1

0

222

1

1

0

=

+=

+=++++=

++=

++=++++=

xx

dxxxxdxyx

dx

dyyxdxyxyxdydx

Example: Find the total mass of the region in the cube 10,10,10 zyx with densityat any point given byxyz.

Solution: Mass =

8

1

24

1

4

1

22

1

2

1

0

21

0

1

0

1

0

21

0

1

0

1

0

21

0

1

0

1

0=

==

=

=

zzdzdz

zydydzyz

xxyzdxdydz

Complex numbers

For any real numberx,x2 0 and thereforex2 + 1 1>0 forxR. Thus the equationx2 + 1 = 0 has no solution in R. This equation has a solution in the complex number field.

Note that i, -i are the solutions of the equationx2 + 1 = 0. i = 1 .

Definition: A number of the form a + ib is called a complex number when a and b are real

numbers and i = 1 . We denote the set of all complex numbers by C. Two complex numbers z1= (a1,b1) andz2 = (a2,b2) are equal ifa1=a2 and b1=b2.A pair of complex numbers a + ib and a ib are said to be conjugate of each other.

Fundamental operations with complex numbers:

1. Addition:z1 +z2 = ( ) ( ) ( ) ( )21212211 bbiaaibaiba +++=+++ = (a1+a2, b1+b2)

2. Subtraction:z1 -z2 = ( ) ( ) ( ) ( )21212211 bbiaaibaiba +=++ = (a1-a2, b1-b2)

3. Multiplication:z1z2 = ( )( ) ( ) ( )211221212211 babaibbaaibaiba ++=++ = (a1a2 b1b2,a1b2+a2b1)

4. Division: 22

2

2

2112

2

2

2

2

2121

22

22

22

11

22

11 .

ba

babai

ba

bbaa

iba

iba

iba

iba

iba

iba

+

+

+

+=

+

+=

+

+

Graphical representation of a complex number:

y

49

x

(a,b)


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Ifz= (a,b), then a and b are the real and imaginary parts of the complex numberz. The complexnumbers can be represented as points in R2 and we call the x-axis and y-axis to be the real and

imaginary axis respectively. The plane itself is called the Argand plane. Also one can think of acomplex number as a vector from the origin to the point (a,b).

Example: Let 1 3z i= + and 2 1 2z i= + . Then their sum 1 2z z+ is given by

1 2 (3 ) ( 1 2 ) 2 3z z i i i+ = + + + = +

And their difference 1 2z z is given by

1 2 (3 ) ( 1 2 ) 4z z i i i = + + =

Their product 1 2z z is given by

2

1 2 (3 )( 1 2 ) 3 6 2 3 5 2 5 5z z i i i i i i i= + + = + + = + = + ,

Example: We find the real and imaginary part of1/ z. Since

2 2( )( )x iy x iy x y+ = + ,

We have

2 2 2 2 2 2

1 1 1

.

x iy

z x iy x iy x iy

x iy x iy

x y x y x y

= =+ +

= =

+ + +

And therefore

2 2 2 2 2 2 2 2

1 1,

x Rz y IzR I

z x y x y z x y x y

= = = =

+ + + +From these equation, we conclude that

1

0 0Rz Rz> >And

10 0Iz I

z>

ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)

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