2. LEARNING OBJECTIVES
• To understand the virtual loss of GM and
the calculations.
• To calculate the maximum trim allowed
to maintain a minimum stated GM.
• To understand the safe requirements for
a ship prior enter into dry dock. To
understand the critical period during dry
docking process.
3. Course Outline
• Name of Course :Chief and Second
Engineer 3000 kW or more (Unlimited
Voyage)
• Course Code/Module : ECSU , Part
A
• Subject : Mathematics and
Engineering Drawing
4. Course Outline
Module Aims
• To provide students with the familiarization to the fundamentals of calculus Mathematics
required for engineering practice and problem solving.
General Learning Objective - GLO
• Recognize that integration is the inverse process of differentiation, and apply this knowledge
to determine the area/volume/work done.
Specific Learning Objectives - SLO
Recognize that integration can be considered the reverse of differentiation
process.
Explain the integration of x, trigo. functions, 1/x, exponential functions.
Evaluate the constant of integration.
Perform the definite Integral.
Apply integration to find:
• a. Area under curves.
• Volume of solid revolution
• Work done
• Mean & root mean square (rms) values
• Centroid
5. Course outline
• Instructional Hours
• Lecture : 40 hours
• Topics Hours
• Integration as reverse of differentiation 2
• Integration of functions: x,Trig,1/x, Exponential 8
• Evaluation of constant of integration 4
• Definite integral 6
• Application of integral calculus to: 20
a. Area under curves.
b. Volume of solid revolution
c. Work done
d. Mean & root mean square (rms)
e. valuesCentroid20
6. Course Outline
Integration as the Process of Summation
Integration as the Reverse of Differentiation
Integration of functions
Applications of Integration : Areas Bounded
by Curves and Volumes of Revolution
7. • Teaching Methods -
• Combination of combination of methods as necessary -
lectures, practice
• Assessment Methods
• Lecturer Class Assessment 1 20 %
• Lecturer Class Assessment 2 20 %
• Lecturer Class Assessment 3 20 %
• Final Exam 40 %
• Recommended Texts
• K A Stroud (1992), Engineering Mathematics Programmes
And Problems
• G.S.Sharma & I.J.S.Sarna (1992), Engineering Mathematics
8. Integration : Concept and Theory
We know how to find the area of simple
geometric shapes such as the triangle below
y
2
1
x
1 2
9. Integration : Concept and Theory
But how do we find the are of geometric object
which do not have straight edges ?
y
f(x)
x
a b
10. Integration : Concept and Theory
So, how do we go about finding the area under
the curve f(x), between x=a and x=b ?
Well,
we can divide the area under the curve into
separate rectangles …
… find the area of each rectangle …
… and then sum these areas in order to find an
approximate answer to area under curve
11. Integration : Concept and Theory
y
f(x)
x
h
a b
Find area of each rectangle …
… then sum all areas between x=a and x=b
12. Process of Integration
• Integration is reverse of differentiation
• In differentiation, if f(x)= then f`(x)= 4x . Thus the integral of
• integration is the process of moving from f`(x) to f(x). By similar
reasoning, the integral of.
• Integration is a process of summation or adding parts together and an
elongated S, shown as, is used to replace the words ‘the integral of’. Hence,
from above,
‘c’ is called the arbitrary constant of integration
13. Integration
is the reverse process of differentiation.
- (PNO2)-n dPNO2 = k dt
Power series integration,
increase the exponent by one
- (PNO2)-n+1 kt
= +C
-n+1 1
Constant
and divide by the of integration
new exponent.
14. The Fundamental Theorem of Calculus
b
If F' x f x , then a f xdx Fb Fa .
If we know an anti-derivative, we can use it to find the
value of the definite integral.
15. The Fundamental Theorem of Calculus
b
If F' x f x , then a f xdx Fb Fa .
If we know an anti-derivative, we can use it to find the
value of the definite integral.
If we know the value of the definite integral, we can use it
to find the change in the value of the anti-derivative.
16. Integration of function
• The general solution of integrals of the form
• axndx, where a and n are constants is given by:
x n1
x dx n 1 C (n 1).
n
This rule is true when n is fractional, zero, or a
positive or negative integer, with the exception of n = -1.