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8/6/2019 Mathematics II 2011
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Course Title Mathematics II
Course CIC Code Math II
Credit Hours 3 Credit Hours
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1. Text BookThe main text book for the course is calculus (Early TranscendentalFunction) Third edition .
By Smith Minton
2. Course ObjectivesThis course is a transition from a course in elementary Calculus to anotheradvanced course. The main topics in this course are integration (Method of
integration-Application of integration)- Complex Analysis ² Vector andthree-dimensional Analytic Geometry- Linear algebra- Theory of Equations-Numerical Integration
The aims of this course are:
To emphasize of developing an understanding of concepts in advanced
Mathematics rather than learning by rule.To gain experience in the method of evaluating different types of integrations.
To gain an approval of importance of the complex analysis theory, Vectorand three-Dimensional, linear algebra, theory of equations and numericalintegration
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3.Course Description
The main concepts covered in this course include:�Integration Techniques and Application of Integration
�Vectors and three-dimensional Analytic Geometry
�Linear algebra
�Theory of Equations
�Complex Analysis�Numerical Integration
4. Homework ²AssignmentsProblem sets are assigned for the credit and will be graded by
the teaching Assistant (TA).Your Homework assignments should be submitted in class.
There will be four Quizzes done in the lectures
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5. Reading MaterialsThe following Chapters will covered this semester
A. Integration Techniques and Application of Integration� Trigonometric Substitution
� Integration By Parts
� Partial Fractions
� Area of a Region Between two curves
� Volume (The Disc Method ² The Shell Method)
� Improper Integrals
� Arc Length of the Curve
� Surface area
B. Vectors and three-dimensional Analytic Geometry
� Planes and lines
� Surfaces of degree two
� Vectors in three dimensions
c. Linear Algebra
� Determinates
� Matrices
� System of linear equations
�
Eigen values and Eigenvectors5
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D. Theory of Equations�Roots Properties
�Relation between coefficients and roots .
�E. Complex Analysis�Complex numbers
�Logarithmic and exponential functions
�Limits and continuity of a complex function
�
Complex integration�F. Numerical Integration�Rectangular Rule
�Trapezoidal rule
�Simpon's rule
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6. Attendance PolicyAttendance at lectures and Tutorial is a must. It will be taken during eachlecture students are not allowed to miss lectures which might contain quizzes.
7.Grading SystemThe grade will be given as follows
Description Grade Note
Final ExamParticipationQuizzesTutorials
50%
8%
32%10%
This course outline issubjects to changes ifnecessary, in which caseyou will be notified
Total 100%
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Integration ² Overview
Integration means the anti-derivative
( ) ( ) f x dx g x c! ´Why ?
Because ( ) ( )d
g x f xd x
!
Example
1
2
1tan
1dx x c
x
! ´ because 1
2
1tan
1
d x
dx x
!
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Rules of Integration
Rule 1
k dx
k x c! ´
If k is a constant
Rule 2
1
11
n
n x x d x c n
n
! {´
In the case 1n ! 1 1 x dx dx Ln x c
x
! ! ´ ´
Why ?
Why ?
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Examples
3
7
18 x x dx
x ´
4 3 / 2 6
84 3 / 2 6
x x x x c
!
3 1/ 2 7 8 x x x dx! ´
5 / 2 13 x dx
x
´
7 / 2
3
7 / 2
x Ln x x c!
21
x dx x
¨ ¸© ¹
ª º´ 2
2
12 x dx
x! ´
3 1
23 1
x x x c
!
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Rule 3
? A? A
1( )
( ) '( ) 1
1
n
n f x f x f x dx c n
n
! {
´Why ?
Examples
2 112 10 (3 7)
6 (3 7)
11
x x x d x c
!
´2 101
6 (3 7)6
x x d x! ´2 10(3 7) x x dx´ 6
6v
2 111 (3 7)
6 11
xc
!
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Rule 4
'( )( )
( )
f xd x n f x c
f x! ´
Examples
2
2
6(3 7)
3 7
xdx Ln x c
x!
´
23 7
xdx
x ´ 2
2
1 6 1(3 7)
6 3 7 6
xd x n x c
x! !
´6
6v
Why ?
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sin sintan
cos cos
cos
x x x d x d x d x
x x
n x c
! !
!
´ ´ ´
coscot
sin
sin
x x dx dx
x
Ln x c
!
!
´ ´
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Rule 5
( ) ( )'( ) f x f xe f x dx e c! ´ Why ?
Examples
tan 2 tansec
x xe x dx e c! ´2 21 12 x x x e d x e c ! ´
2 1 x x e d x´2 2
1 11 12
2 2
x x x e d x e c ! ! ´2
2v
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Rule 6
sin ( ) '( ) cos ( ) f x f x dx f x c! ´W
hy ?
Examples
2 28 sin(4 1) cos(4 1) x x dx x c ! ´
cos ( ) '( ) sin ( ) f x f x dx f x c! ´
2 21cos(3 1) sin(4 1)
6 x x dx x c ! ´
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Examples
(1) 7
sin cos x x dx! ´8sin
8
xc!
7sin cos x x dx´
Another solution
Let sinu x! cosdu x d x!
then7 7
sin cos x x dx u d u!´ ´8 8sin
8 8
u xc c! !
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(2) 2sin x dx´ 11 cos 2
2 x d x! ´
1 1sin 22 2 x x c
¨ ¸! © ¹ª º
Try to calculate 2cos x dx
´ ??
1 1 11 cos 2 sin 2
2 2 2 x dx x x c
¨ ¸! ! © ¹ª º
´
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(3) 1 x x dx´Let 2 1u x!
2u du d x!2
1u x !2 2
1 ( 1) 2 x x dx u u u d u ! ´ ´2 2( 1)2u u d u!
´5 3
2 25 3
u uc!
4 2(2 2 )u u d u! ´
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2 1u x! But 1/ 21u x!
Then
5 3
5 / 2 3 / 2
1 2 25 3
2 21 1
5 3
u u x x d x c
x x c
!
!
´
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(4)2
1
9d x
x´
Let
3
xu !
2
1
9 19
dx x
!¨ ¸
© ¹ª º
´
1
3du d x!
2 2
1 13
9 9 1dx d u
x u! ´ ´1
2
1 1 1tan
3 1 3d u u c
u
! ! ´ 11
tan3 3
xc ¨ ¸! © ¹
ª º
3du d x!
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In general we have
1
2 2
1 1tan
xdx c
a x a a
¨ ¸! © ¹ ª º´
(5)2
1
16
d x
x
´ 2
1
16 116
dx
x
!¨ ¸
© ¹ª º
´
Let
4
xu ! 1
4d u dx! 4 du d x!
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2 2
1 14
16 16 1dx dx
x u
!
´ ´
2
1
1du
u
!
´
In general we have
1
2 2
1sin
xdx c
aa x
¨ ¸! © ¹ª º
´
1 1sin sin4
xu c c ¨ ¸! ! © ¹
ª º
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Also we have
1
2 2
1
2 2
1
2 2
1 1
tanh
1sinh
1 cosh
xd x ca x a a
xd x c
a x a
xd x c
a x a
¨ ¸!
© ¹ ª º
¨ ¸! © ¹
ª º
¨ ¸! © ¹ª º
´´
´
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(6)
2
1
25 1dx
x ´ 2
1
1
25 1 25
d x x
!¨ ¸
© ¹© ¹ª º
´
Let1
5
xu
!
1
5d u dx!
2 2
1 15
25 125 1d x du
u x!
´ ´
5 du d x!
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2
1 1
5 1
d uu
!´
11tanh
5u c!
11 1tanh
5 5
xc ¨ ¸! © ¹
ª º
1
5
xu
!
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Definite integrals
´!
!
!
b x
a x
dx x f )(
If f is integr able and f ( x)>0 f or every x in [a,b], then
area under the gr aph of
f ( x) f r om x=a to x=b
dx
f(x)
This thin stri p is going to swee p the gr aph f r om
x=a to x=b (remember dx= ( x0)
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Note
If ( ) ( ) f x dx g x c! ´
Then
? A( ) ( ) ( ) ( )
b
b
a
a
f x d x g x g b g a! ! ´
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Problem
Evaluate the f ollowing integr al: ´
2
1
2)1( d x x x
833.4
833.1667.6
)1(2
)1(
3
)1()2(
2
)2(
3
)2(
23)1(
2323
2
1
2
1
23
2
!
!¼½
»
¬
«
¼½
»
¬
«!
¼½
»¬«
!´ x x x
dx x x
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Example
Calculate
/ 2
0
sin 2 x d x
T
´
Solution/ 2/ 2
00
1sin 2 cos 2
2
1 1cos cos 02 2
1 1( 1) (1) 1
2 2
x d x x
T T
T
« »! ¬ ¼½
« »! ¬ ¼½
« »! !¬ ¼
½
´
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Example
Calculate the area between the curves & 2 y x y x! !
and the x-axis
2 y x! y x!
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Area
1 2
0 1
2 x dx x dx! ´ ´
1 22 2
0 1
22 2
1 4 1 1 30 (4 ) (2 ) 2 1
2 2 2 2 2
x x x« » « »! ¬ ¼ ¬ ¼
½ ½
« » « »! ! !¬ ¼ ¬ ¼½ ½
Solution
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