Qwertyuiopasdfghjklzxcvbnmqwertyu iopasdfghjklzxcvbnmqwertyuiopasdfg hjklzxcvmqwertyuiopasdfghjklzxcvbn SEKOLAH MENENGAH TEKNIK IPOH mqwertyuiopasdfghjklzxcvbnmqwerty uiopasdfghjklzxcvbnmqwertyuiopasdf ghjklzxcvbnmqwertyuiopasdfghjklzxc vbnmqwertyuiopasdfghjklzxcvbnmqw ADDITIONAL MATHEMATICS ertyuiopasdfghjklzxcvbnmqwertyuiop PROJECT WORK 1/2010 asdfghjklzxcvbnmqwertyuiopasdfghjkl zxcvbnmqwertyuiopasdfghjklzxcvbnm qwertyuiopasfghjklzxcvbnmqwertyuio ATHIRAH NUR HANI BINTI AMINUDDIN pasdfdgdhcbbnmjlzxcvbnmqwertyuio 931119 08 5888 pasdfghjklzxcvbnmqwertyuiopasdfghj 5 PERDAGANGAN 1 PUAN LOKE SIEW HO klzxcvbnmqwertyuiopasdfghjklzxcvbn mrtyuiopasdfghjklzxcvbnmqwertyuio pasdfghjklzxcvbnmqwertyuiopasdfghj
Additional mathematics project work 1
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Contents
Appreciation Appreciation ......................................................................................... ........................................................................................... .. 3 Objectives Objectives ........................................................................................... ............................................................................................... .... 4 Introduction Introduction ........................................................................................... 5 History ........................................................................................... 6 Integration Integration by parts ....................................................................... 8 Answers to more examples ......................................................... 18 Questions..........19 Further exploration ..24 Conclusion.32 Reflection..33 References 34
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Appreciation First of all, I would like to say Alhamdulillah, for giving me the strength and health to do this project work.
Not forgotten my parents for providing ever ything, such as money, to buy anything an ything that are related to this project work and their advise, which is the most needed for this project. Internet, books, computers and all that. They also supported me and encouraged me to complete this task so that I will not procrastin pro crastinate ate in doing do ing it.
Then I would like to thank my teacher, Mdm Loke Siew Ho for guiding me and my friends throughout this project. We had some so me difficulties difficulties in doing this task, but she taught us patiently until we knew what to do. She tried and tried to teach us until we understand what we supposed to do with the project work.
Last but not least, my friends who were doing this project with me and sharing o ur ideas. They were helpful that when we combined and discussed together, we had this task done.
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Objectives The aims of carrying out this t his project work are:
i.
to apply and adapt a variety of problem-solving strategies to solve problems;
ii.
to improve thinking skills;
iii.
to promote effective mathematical communication;
iv.
to develop mathematical knowledge through problem solving in a way that increases students¶ interest and confidence;
v.
to use the language of mathematics to express mathematical ideas precisely;
vi.
to provide learning environment that stimulates stimulates and enhances effective learning;
vii.
to develop positive attitude towards mathematics.
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Introduction Integration is an important concept in mathematics and, together with differentiation, is one of the two main operations in calculus. Given a function of a real variable x variable x and and an interval [a [ a, b] of the real line, the definite integral
is defined informally to be the net signed area of the region in the xy -plane bounded by the graph of of , the x the x -axis, and the vertical lines x lines x = = a and x and x = = b. The term integral may integral may also refer to the notion of antiderivative, a function F whose F whose derivative is the given function . In this case it is called an indefinite integral , while the integrals discussed in this article are termed definite integrals. integrals . Some authors maintain a distinction between antiderivatives and indefinite integrals. The principles of integration were formulated independently by Isaac Newton and Gottfried Leibniz in the late 17th century. Through the fundamental theorem of calculus, which they independently developed, integration is connected with differentiation: if is a continuous real valued function defined on a closed interval [ a, b], then, once an antiderivative F of F of is known, the definite integral of over that interval is given by
Integrals and derivatives became the basic tools of calculus, with numerous applications in science and engineering. A rigorous mathematical definition of the integral was given by Bernhard Riemann. It is based on a limiting procedure which approximates the area of a curvilinear region by breaking the region into thin vertical slabs. Beginning in the nineteenth century, more sophisticated notions of integrals began to appear, where the type of the function as well as the domain over which the integration is performed has been generalised. A line integral is defined for functions of two or three variables, and the interval of integration [ a, b] is replaced by a certain curve connecting two points on the plane or in the space. In a surface integral, the curve is replaced by a piece of a surface in the three -dimensional space. Integrals of differential forms play a fundamental role in modern differential geometry. These generalizations generalizations of integral first arose f rom the needs of physics, and they play an important role in the formulation of many physical laws, notably those of electrodynamics. There are many modern concepts of integration, among these, the most common is based on the abstract mathematical theory known as Lebesgue integration, developed by Henri Lebesgue.
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History Pre-calculus
integration
Integration can be traced as far back as ancient Egypt c a. a. 1800 BC, with the Moscow Mathematical Papyrus demonstrating knowledge of a formula for the volume of a pyramidal frustum. The first documented systematic technique capable of determining integrals is the method of exhaustion of Eudoxus ( c a. a. 370 BC), which sought to find areas and volumes by breaking them up into an infinite number of shapes for which the area or volume was known. This method was further developed and employed by Archimedes and used to calculate areas for parabolas and an approximation to the area of a circle. Similar methods were independently independently developed in China around the 3rd century AD by Liu Hui, who used it to find the area of the circle. This method was later used in the 5th century by Chinese father -and-son mathematicians mathematicians [1] Zu Chongzhi and Zu Geng to find the volume of a sphere. That same century, the Indian mathematician Aryabhata used a similar method in order to find the volume of a cube. [2] The next major step in integral calculus came in Iraq when the 11th century mathematician Ibn al-Haytham (known as Alhazen in Europe) devised what is now known as "Alhazen's problem", which leads to an equation of the fourth degree, in his Book of Opti c cs . While solving this problem, he performed an integration in order to find the volume of a paraboloid. Using mathematical induction, he was able to generalize his result for the integrals of polynomials up to the fourth degree. He thus came close to finding a general formula for the integrals of polynomials, but he was not concerned with any polynomials higher than the fourth degree. [3] Some ideas of integral calculus are also found in the Siddhanta Shiromani , a 12th century astronomy text by Indian mathematician Bhskara II. The next significant advances in integral calculus did not begin to appear until the 16th century. At this time the work of Cavalieri with his method of indivisibles, indivisibles, and work by Fermat, began to lay the foundations of modern calculus. Further steps were made in the early 17th century by Barrow and Torricelli, who provided the first hints of a connection between integration and differentiation. At around the same time, there was also a great deal of work being done by Japanese mathematicians, particularly by Seki Kwa. [4] He made a number of contributions, namely in methods of determining areas of figures using integrals, extending the method of exhaustion. Newton
and Leibniz
The major advance in integration came in the 17th century with the independent discovery of the fundamental theorem of calculus by Newton and Leibniz. The theorem demonstrates a connection between integration and differentiation. This connection, combined with the comparative ease of differentiation, can be exploited to calculate integrals. In particular, the fundamental theorem of calculus allows one to solve a much broader class of problems. Equal in importance is the comprehensive mathematical framework that both Newton and Leibniz developed. Given the name infinitesimal calculus, it allowed for precise analysis of functions within continuous domains. This framework eventually became modern calculus, whose notation for integrals is drawn directly from the work of Leibniz.
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integrals
While Newton and Leibniz provided a systematic approach to integration, their work lacked a degree of rigour. Bishop Berkeley memorably attacked infinitesimals as "the ghosts of departed quantities". Calculus acquired a firmer footing with the development of limits and was given a suitable foundation foundation by Cauchy in the first half of the 19th century. Integration was first rigorously formalized, using limits, by Riemann. Although all bounded piecewise continuous functions are Riemann integrable on a bounded interval, subsequently more general functions were considered, to which Riemann's definition does not apply, and Lebesgue formulated a different definition of integral, founded in measure theory (a subfield of real analysis). Other definitions of integral, extending Riemann's and Lebesgue's approaches, were proposed. Notation
Isaac Newton used a small vertical bar above a variable to indicate integration, or placed the variable inside a box. The vertical bar was easily confused with or , which Newton used to indicate differentiation, and the box notation was difficult for printers to reproduce, so these notations were not widely adopted. The modern notation for the indefinite integral was introduced by Gottfried Leibniz in 1675 (Burton 1988, p. 359; Leibniz Leib niz 1899, p. 154). He adapted the integral i ntegral symbol, , from an elongated letter s, standing for summa (Latin for "sum" or "total"). The modern notation for the definite integral, with limits above and below the integral sign, was first used by Joseph Fourier in Mémoires of the French Academy around 1819±20, reprinted in his book of 1822 (Cajori 1929, pp. 249±250; Fourier 1822, §231).
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INTEGRATION BY PARTS One of very common mistake mista ke students usually do is
To convince yourself that it is a wrong formula, take f take f ( x) = x and g and g ( x)=1. Therefore, one may wonder what to do in this case. A partial answer is given by what is called Integration by Parts. In order to understand this technique, recall the formula
which implies
Therefore if one of the two integrals and is easy to evaluate, we can use it to get the other one. This is the main idea behind Integration by Parts. Let us give the practical steps how to perform perf orm this technique: 1
Write the g the g iven iven inte g inte g ral ral
where you identi f identi f y the two f unctions f(x) unctions f(x) and g(x). g(x). Note that i f i f you you are g are g iven iven only one f one f unction, then set the second one to be the constant f constant f unction g(x)=1. unction g(x)=1. 2
Introduce the intermediary f intermediary f unctions u(x) u(x) and v(x) v(x) as:
Then you need to make one derivative ( o f f(x)) and one inte g inte g ration ration ( o f g(x)) to g et et
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Note that at this step, you have the choice whether to di ff di ff erentiate f(x) erentiate f(x) or g(x). g(x). We will discuss this in little more details later . 3
Use the f the f ormula
4
Take care o f o f the the new inte g inte g ral ral
.
The first problem one faces when dealing with this technique is the choice that we encountered in Step 2. There is no general rule to follow. It is truly a matter of experience. But we do suggest not to waste time thinking about the best choice, just go for any choice and do the calculations. In order to appreciate whether your choice was the best one, go to Step 3: if the new integral (you will be handling ) is easier than the initial one, then your choice was a good one, otherwise go back to Step 2 and make the switch. It is after many integrals that you will start to have a feeling for the right choice. In the above discussion, we only considered indefinite integrals. For the definite integral
, we have two ways to go:
1
Evaluate the indefinite integral
which gives
2
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Use
the above steps describing Integration by Parts directly on the given definite integral. This is how it goes: (i) Write down the given definite integral
where you identify the two t wo functions f functions f ( x) and g and g ( x). (ii) Introduce the intermediary functions u( x) and v( x) as:
Then you need to make one derivative (of f f ( x)) and one integration (of g g ( x)) to get
(iii) Use the formula
(iv)
Take care of the new integral
.
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The following examples illustrate the most common cases in which you will be required to use Integration by Parts:
EXAMPLE 1 Evaluate
Let us follow the steps 1
This is an indefinite integral involving one function. fu nction. The second needed function is g is g ( x) = 1. Since the derivative of this function is 0, the only choice left is to differentiate the other function
.
2
We have
which gives
3
We have the formula
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4
Since
we get
Since the derivative of
is 1/ x 1/ x,, it is very common that whenever an integral
involves a function which is a product of differentiate
with another function, to
and integrate the other function.
EXAMPLE 2 Evaluate
First let us point out that we have a definite integral. Therefore the final answer will be a number not a function function of x! Since Since the derivative derivative or or the integral of lead to the same function, it will not matter whether we do one operation or the other. Therefore, we concentrate on the other function . Clearly, if we integrate we will increase the power. This suggests that we should differentiate and integrate . Hence
After integration and differentiation, we get
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The integration by parts formula gives
It is clear that the new integral is not easily obtainable. Due to its similarity with the initial integral, we will use integration by parts for a second time. The same discussion as before leads to
which implies
The integration by parts formula gives
Since
, we get
which finally implies
Easy calculations give
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From this example, try to remember that most of the time the integration inte gration by parts will not be enough to give you the answer after one shot. You may need to do some extra work: another integration by parts or use other techniques,....
EXAMPLE 3 Find
The two functions involved in this example do not exhibit any special behavior when it comes to differentiating or integrating. Therefore, we choose one function to be differentiated and the other one to be integrated. We have
which implies
The integration by parts formula gives
The new integral is similar in nature to the initial one. One of the common mistake is to do another integration by parts in which we integrate and differentiate . This will simply take you back to your original integral with nothing done. In fact, what you would have done is simply the reverse path of the integration by parts (Do the calculations to convince yourself ). Therefore we continue doing another integration by parts as
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which implies
Hence
Combining both formulas we get
Easy calculations give
After two integration by parts, we get an integral identical to the initial one. You may wonder why and simply because the derivative and integration of are the same while you need two derivatives of the t he cosine function to generate the same function. Finally easy algebraic manipulation gives
Try to find out how did we get the constant
C ?
In fact we have two general formulas for these kind of integrals
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and
EXAMPLE 4 Evaluate
Looking at the two functions involved in this example, we see that the function x function x is easy to integrate or differentiate. The real problem is how to handle it is not clear how to integrate this one but its derivative is the rational
function
. First
. Therefore this suggests the following
which implies
The integration by parts formula gives
The new integral will be handled hand led by using the technique of integrating of integrating rational functions.. But we can also do the following (which comes up doing the same ideas functions used in partial in partial fraction decomposition )
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Using
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this we get
The main idea behind this example is valid for many other functions f unctions such as: , , etc.. In fact, this is how the integration by parts should be carried whenever the integral is given as a product of f of f ( x) and one the previous inverse-functions, try to integrate f integrate f ( x) and differentiate the inverse-function. The same remark holds for the function
.
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Answers to More Examples
y
Problem 1: Find
.
Answer: y
Problem 2: Find
.
Answer: y
Problem 3: Find
Answer:
.
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Questions: The diagram below shows shows the gate of an art gallery. A concrete structure structure is built at the upper part of the gate and the words µART GALLERY¶ is written on it. The top of the concrete structure is flat whereas the the bottom is parabolic in shape. The concrete structure structure is supported by two vertical pillars at both ends. The distance between the two pillars is 4 metres and the height of the pillar is 5 metres. The height of the concrete structure is 1 metre. The shortest distance from point A of the concrete structure to point B, that is the highest point on the parabolic shape, is 0.5 metres.
A
0.5 m 1m
B
4
m
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(a) The parabolic shape of the concrete structure can be represented by various functions depending on the point of reference. reference. Based on different different points of reference, reference, obtain at least three different functions which can be used t o represent the curve of this concrete co ncrete structure. ( b) b) The front surface of this concrete structure will be painted before the words µART GALLERY¶ is written on it. Find the area to be painted.
SOLUTIONS: (a)
Function
1
Maximum point (0,4.5) and pass through point (2,4)
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2
Maximum point (0, 0.5) and pass through point p oint (2, 0) 0)
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3
Maximum point (2, 4.5) and pass through point p oint (0, 4)
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(b)
Area to be painted = Area of rectangle - Area under the curve
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FURTHER (a)
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EXPLORATION
You are given four different shapes of of concrete structures as shown in the diagrams below. All the structures have the same thickness of 40 cm and and are symmetrical.
Structure 1
Structure 2 0.5 m
0.5 m
1m
1m
5
4
5
m
4
m
Structure 3
m
Structure 4 0.5 m
0.5 m
1m
1m
1m
2m 5
4
m
m
m
5
4
m
m
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Given that the cost to construct 1 cubic metre of concrete is RM840.00, determine which structure will cost the minimum to construct.
(ii) ii)
As the president of the Arts Club, you are given the opportunity to decide on the shape of the gate to be constructed. Which Which shape would you choose? Explain and elaborate on your reasons for choo sing the shape.
SOLUTIONS: Structure 1 0.5 m
1m
5
4
m
m
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Structure 3 0.5 m
1m 1m 5
4
m
m
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(ii) ii) As the president of the Arts Club, I will decide Structure 4 as the shape of the gate to be constructed. It is because Structure 4 will cost the minimum and it is easier to be co nstructed compared to Structure 1 which is a curve.
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The following questions refer to the concrete structure in the diagram below. If the value of k of k increases increases with a common difference of 0.25 m; (i)
complete Table 1 by finding the values of k of k and and the corresponding areas of the concrete structure to be painted.
(ii) ii)
observe the values of the area to be painted from Table 1. Do you see any pattern? Discuss.
0.5 m 1m k
k ( k (m) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
) Area to be painted( painted(
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(ii) ii) There is a pattern in the area to be painted.
The area to be painted decreases as the k increases k increases 0.25m and form a series of numbers:
3, 2.9375, 2.875, 2.8125, 2.75, 2.6875, 2.625, 2.5625, 2.5
We can see that t he difference between each term and the next term is the same.
We can deduce that t hat this series of numbers is an Arithmetic Progression (AP) AP), with a common difference, In conclusion, when k increases k increases 0.25m, the area to be painted decreases by -0.0625
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Express the area of the concrete co ncrete structure to be painted in terms of k of k . Find the area a k approaches the value of 4 and predict the shape of the concrete structure.
The shape of the concrete co ncrete structure will be a rectangle with length 4m and breadth 0.5m, which may look like this:
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Conclusion A gate is a point of entry to a space enclosed by walls, or a moderately sized opening in a fence. Gates may prevent or control entry or exit, or they may be merely decorative. Other terms for gate include yett and port. Larger gates can be used for a whole building, such as a castle or fortified town, or the actual doors that block entry through the gatehouse. Other than that, selection of gate also depends on beautifyi beaut ifying ng or some religious believe in feng sui which is bring luck in life. As I doing this project, I notice that quadratic function and integration can be so close in our daily life. There are many shape of gate outside there. Different shapes of the gate have different cost. From quadratic function and integration, we can know area of the gate. From the area we can get volume og the gate. As the result, we can know cost of the gate by times volume with RM840 ( price price for 1m³) 1m³) After we know concept of quadratic function and integration, we can apply it in our life. In order to meet the budget or saving money, we can capable to decide on which shape or design more favorable and reasonable.
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Reflection While I conducting this project, a lot of information that I found.I have learnt how to build a concrete gate structure with with good quality and proper price. Apart from that, this project encourage the student to work together and share their knowledge. It is also encourage student to gather information from the internet, improve thinking skills and promote effective mathematical communication. Last but not least, I proposed this project should be continue because it brings a lot of moral value to the student and also test the student¶s understanding in Additional Mathematics.
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References y
http://en.wikipedia.org/wiki/Integral
y
http://www.sosmath.com/
y
http://www.scribd.com/
y
http://www.facebook.com/
y
Additional Mathematics Mathematics Form 5
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