7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
1/73
INSTITUTE TECHNOLOGY BRUNEI
FACULTY OF ENGINEERING
COURSE NOTE
COURSE NUMBER: FEG1MA1
COURSE TITLE: ENGINEERING MATHEMATICS 1
DR. MD. FAZLUL KARIM
1
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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=
position of particle when 1t=
position of particle when 2t=
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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:
6
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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:
8
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
9
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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,
13
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
14
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
15
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
17
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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 =
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
+===
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
39/73
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=
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
40/73
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
41/73
= ( ) 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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
42/73
( ) ( )
)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=
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
43/73
( ) ( )( )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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
44/73
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
45/73
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
46/73
(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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
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
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
49/73
( ) ( )[ ] ( )
( )( ) ( )[ ] ( )
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)
7/27/2019 ITB - ENGINEERING MATHEMATICS 1 (FEG1MA1)
50/73
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>
Top Related