Post on 25-Mar-2018
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CHAPTER 4 INTEGRATION
4.1 INTRODUCTION TO INTEGRATION
Why do we need to study Integration?
The Petronas Towers
of Kuala Lumpur
Often we know the relationship involving the rate of change of two variables, but we may need to know the direct relationship between the two variables. For example, we may know the velocity of an object at a particular time, but we may want to know the position of the object at that time.
To find this direct relationship, we need to use the process which is opposite to differentiation. This is called integration (or antidifferentiation).
The processes of integration are used in many applications.
The Petronas Towers in Kuala Lumpur experience high forces due to winds. Integration was used to design the building for strength.
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Sydney Opera House
The Sydney Opera House is a very unusual design based on slices out of a ball. Many differential equations (one type of integration) were solved in the design of this building.
Wine cask
Historically, one of the first uses of integration was in finding the volumes of wine-casks (which have a curved surface).
Other uses of integration include finding areas under curved surfaces, centres of mass, displacement and velocity, fluid flow, modelling the behaviour of objects under stress, etc.
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4.1.1 Antiderivatives and The Indefinite Integral
Integration is the inverse process of differentiation.
'dy
f xdx
'f x dx y c
Two types of integrals
Indefinite integral
Definite integral
22d
xdx
22x c 4x
4xdx
C is arbitrary constant
integrand
The term dx identifies
x as the variable of
integration
integral
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4.2 Indefinite Integral
4.2.1. Integration Of Constants nax
1
, 1 ( )1
nn ax
ax dx c n power rulen
Examples
Evaluate the following equation;
i. 24x dx
ii. 3 dp
iii. s ds
iv. 3
1dx
x
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Exercise
a) 1 dx
b) dx c) 3
4dx
d) 3
3dx
x
e) 2x dx f) 2
31
3x dx
g) 1
33x
h) 3
8dx
x
i) 10 x dx
4.2.2. Integration Of Summation & Subtraction
n m n max bx dx ax dx bx dx
1 1
1 1
n max bx
n m c
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Example Find the derivative for each of the following functions.
a) 24 4 5x x dx
b) 26 9x x x dx
c) 2
2
46t dt
t
d) 2
x dxx
e) 2
1x dx
x
f) 1
2p dpp
g) 2
4
43x dx
x
h) 2x x dx i)
4
2 4x xdx
x
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4.2.3. INTEGRATION BY SUBSTITUTION
n
ax b dx , n
Step 1 : Let “u”
Step 2 : Differentiate “u”
Step 3 : Make as a subject
Step 4: Integrate the “u” equation
1
1
1
1
1
nn
n
n
duax b dx u
a
uc
n a
uc
n a
Step 5 : Substitute back “u”
So the final answer is
1
1
nax b
n a
Examples:
Evaluate the following equation:
Find 4
3 5 2x dx
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Example
Find the following integration by substitution:
a). dxx 5)72( b). dx
x 4)85(
1
C). dxx23 d).
dxx 3
2
)76(
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4.2.4 INTEGRATION OF 1
xOR LOGARITHM FUNCTION
11dx x dx
x
, n=-1
ln xc
dx
dx
1 1
lndx ax b cax b a
Example 1:
i. 1
4dx
x
ii. 1
12 4 9x dx
iii. 1
3 2dt
t
iv. 128 6 2 1x x dx
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Exercise:
4.2.5. INTEGRATION OF EXPONENTIAL FUNCTION
x xe dx e c
1ax axe dx e ca
ax bax b e
e dx cd
ax bdx
a) 1
5dx
x
b) 1
2 1dx
x c) 2
2
3
xdx
x d) 2
31
xdx
x
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Example:
a) 5xe dx
b) 1
2x
e dx c) dxe x2
4
d) 3 13
xe dx
e) 3 2 23te t dt f)
21 x
x
edx
e
4.2.5. INTEGRATION OF TRIGONOMETRIC FUNCTION
2
cossin
sincos
tansec
ax bax b dx c
dax b
dx
ax bax b dx c
dax b
dx
ax bax b dx c
dax b
dx
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2
1sin cos
1cos sin
1sec tan
ax dx ax ca
ax dx ax ca
ax dx ax ca
Example:
a) sin 3 3x dx
b) 2 2cos 2 4 sec 2x x dx c) sin 2x dx
d) 22sec 3x dx
e) cos 2sinx x dx f) ∫
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Challenge:
1. 2
33
1
3
pdp
p p
2. 7
2 31k k dk
3. tan x dx
4.
3
23
5
1
x
x
edx
e
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Solve the integration below by using substitution:
a). 1
4 6dx
x b).
3
5
4 5 4dx
x c).
43 xx e dx
d). 5
4
x
xdx
e
e). sin 2 cos2x x dx f). 3 42 sinx x dx
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Integrate for 2sin mx dx and 2cos mx dx
2 1sin 1 cos 2
2mx dx mx dx
2 1cos 1 cos 2
2mx dx mx dx
Example:
a) 2sin 2x dx
b) 2cos 3x dx
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c) dxx
3sin4 2
4.3 DEFINITE INTEGRAL
4.3.1 Definite Integral
If f x dx F x c
So b
a
f x dx F b F a
Examples:
1. dx2
0
5
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2.
2
1
2 )32( dxx
3. 4
4 2
2
xe dx
4. 4
2
3
sec 3z dz
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5. 5
3
3
2 4dx
x
Exercise:
1. 4
2
0
4 3x dx
2. 4
0
3 4x dx
3. 9
1
3 xdx
x
4. 2
3
2
1 2x x dx
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4.3.2 DEFINITE INTEGRAL BY SUBSTITUTION METHOD
1. Find the value of
a) 4
3
2
4 2x dx
b) 32
6 42
0
2x x
x e dx
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c) 4
2
0
sec 3x dx
d) 2
0.2
0
3 te dt
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Exercise:
1. 1
2
2
4 1x dx
2. 2
52
1
2 3x x dx
3.
1 2
23
0 1
xdx
x
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4.4 PROPERTIES OF THE DEFINITE INTEGRAL
a) b b
a a
kf x dx k f x dx
b) 1 2 1 2
b b b
a a a
f x f x dx f x dx f x dx
c) b a
a b
f x dx f x dx
d) b c c
a b a
f x dx f x dx f x dx
Examples:
i. Given that 6
3
4f x dx , find the value of:
a) 6
3
3 f x dx
b) 6
3
2f x dx
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c) 3
6
f x dx
d) 4 6
3 4
f x dx f x dx
e) k if 6
3
12f x kx dx
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Exercise:
Given that 2
1
4f x dx and 3
2
6f x dx , evaluate each of following
a) 1
2
f x dx
b) 2
1
5 f x dx
c) 3
1
f x dx
d) 2
1
4 3f x dx
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POLITEKNIK KOTA BHARU JABATAN MATEMATIK, SAINS DAN KOMPUTER
BA 201 ENGINEERING MATHEMATICS 2
PAST YEAR FINAL EXAMINATION QUESTIONS
JULAI 2010
Integrate this following:
i. ∫
ii. ∫( )
iii. ∫
iv. ∫
( )
v. ∫
vi. ∫
( )
JANUARI 2010
a) Integrate this following:
i. ∫( ) ( )
ii. ∫
iii. ∫
b) Evaluate the integrals:
i. ∫ ( )
ii. ∫ (
)
JULAI 2009
Integrate:
a) ∫( )
b) ∫
c) ∫( )
d) ∫ ( )
e) ∫ ( )( )
JANUARI 2009
a) Solve the following of indefinite
integrals:
i. ∫
ii. ∫
iii. ∫
iv. ∫ ( )
b) Solve the following of definite
integrals:
i. ∫ ( )
ii. ∫ ( )
JULAI 2008
Integrate:
i. ∫
ii. ∫
iii. ∫ ( )
iv. ∫ ( )( )
v. ∫ ( )
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JANUARI 2006
a) Solve the following of indefinite
integrals:
i. ∫( )
ii. ∫( )
iii. ∫
iv. ∫ ( )
b) Solve the following of definite
integrals:
i. ∫ ( )
ii. ∫ ( )