CHAPTER
1
CHAPTER
4
CHAPTER
2
CHAPTER
5
CHAPTER
3
Editor
Assoc.Prof.Dr. Barış ERBAŞ
Authors
Assoc.Prof.Dr. Derya ÇELİK Assoc.Prof.Dr. Nihal EGE Prof.Dr. Julius KAPLUNOV Assoc.Prof.Dr. Danila PRIKAZCHIKOV Prof.Dr. Nülifer ÖZDEMİR
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MATHEMATICS II E-ISBN 978-975-06-2942-6
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Eskişehir, Republic of Turkey, December 2018
CHAPTER 1 Indefinite Integral
Introduction ... 71
Ordinary Differential Equations (Odes) of The First Order ... 71
General Definitions ... 71
Existence and Uniqueness of Solution of an İnitial Value Problem ... 72
Geometrical Interpretation ... 73
Separation of Variables ... 74
First Order Homogeneous ODEs with Constant Coefficients ... 75
Linear ODEs of the First Order ... 75
Applications of The First-Order ODes ... 77
Newton’s Law of Cooling ... 77
Radioactive Decay... 78
Population Growth ... 78
Logistic Growth ... 79
CHAPTER 4 Differential Equations Introduction ... 47
Derivation of Total Function from Marginal Function ... 47
Consumer and Producer Surplus ... 48
Lorenz Curve and Gini Index ... 52
Present and Future Value of Income ... 54
CHAPTER 3 Applications of Integral Introduction ... 23
Area Problem ... 23
Definite Integral ... 27
Properties of the Definite Integral ... 28
Fundamental Theorem of Calculus ... 31
Definite Integral Revisited ... 32
Area bounded by two curves ... 32
Average Value of a Function ... 33
CHAPTER 2 Definite Integral Introduction ... 3
Indefinite Integral ... 3
Fundamental Rules of Integration ... 4
Techniques of Integration ... 7
Change of Variable Method (Substitution Method) ... 7
Integration by Parts ... 11
Introduction ... 91
Matrices ... 91
Systems of Linear Equations ... 96
Introduction ... 117
Structure of a Linear Programming Problem ... 118
System of Linear Inequalities ... 118
Constraint Set ... 121
Goal Function. General Linear Programming Problem ... 122
Graphical Method of Solution ... 123
Polygonal Sets and Corner Point ... 123
The Corner Point Theorem ... 126
Examples ... 127
CHAPTER 6 Introduction to Linear Programming Introduction ... 141
What is a Graph ... 141
Konigsberg Bridge Problem ... 143
Planar Graphs ... 146
Colouring Vertices ... 149
Trees ... 151
Introduction ... 163
Normal Form of Games: Strategies and Payoff Functions ... 163
Game of k Players in Normal Form ... 163
Equilibrium Pair ... 165
Zero Sum Games ... 167
Maximin - Minimax Strategies. Lower and Upper Values ... 169 CHAPTER 8 Game Theory
Dear Students,
In the second volume, we introduce more advanced concepts of mathematical knowledge including differential equations, linear programming, graph theory and game theory. All this new material is widely used in a variety of diverse real-world problems. In particular, graph and game theories as well as linear programming find exciting applications in modern economics.
The style of this volume is not much different to that of the first one. It is oriented to a broad student community interested in modern mathematical methods and their implementations.
We trust that as you go along the book you will find it easier and easier to grasp the undeservedly called “difficult” notions of mathematics and will realise that you are able to solve the problems you encounter in your lectures.
I am grateful to all the authors who had put an intense effort to complete the book in time. As in every book, the mistakes are inevitable and naturally I will be grateful for letting me know about any spotted typos and mistakes. Unfortunately, these are virtually unavoidable for the first edition of such a textbook.
I do hope that you will have at least a little bit of joy and fun from reading this volume.
Editor
Assoc.Prof.Dr. Barış ERBAŞ
Learning Outcomes
What kind of rules are there about indefinite integral?
How can we apply the partial integration method to the integral
What is an antiderivative for the function
What is the indefinite integral of
e
2x?Can you find the integral by using the change of variable method?
3 5
1 2
4
Chapter Outline
Introduction Indefinite Integral Techniques of Integration
Key Terms
Antiderivative Indefinite Integral Change of Variables
Integration by Parts
After reading this chapter, you will be able to answer the following
questions
We studied the methods of finding out how quickly a given quantity changes with respect to another one, that is, finding the derivative of a given function. Now, let us consider the opposite of this problem. That is, let us try to find the function whose derivative is already known.
When someone asks you what is the function F whose derivative is F'(x) = 2x for all x in an open interval (a, b), then you immediately give the following answer: F(x) = x2. We call the function x2 an antiderivative of the function 2x. But the function with the derivative 2x is not just F(x)
= x2. For example, the derivative of the function x2 + 1 is also 2x. The function x2 + 1 is another antiderivative of the function 2x. In the most general case, the derivative of the function x2 + c for any constant c is 2x because the derivative of a constant is zero. Thus, in an open interval, there are an infinite number of antiderivatives of the function 2x. We call these family of antiderivatives the indefinite integral of the function 2x and denote the family by the following notation using the symbol called the integral sign:
Since
then F(x) = x is an antiderivative of the function 1.
Thus one can write,
Similarly, F(x) = x3 is an antiderivative of 3x2, since
In the general case, a function F is called an antiderivative of f on an interval (a, b) if
F' (x) = f (x) for all x in (a, b).
If F is an antiderivative of f on an interval (a, b), then the most general antiderivative of f on (a, b) is F(x)+ c where c is an arbitrary constant. This property tells us, in order to find all antiderivatives of a given function f , all we need to do is to find a single function F whose derivative is f , the rest of the antiderivatives of f can then be found by adding arbitrary constants to the function F.
The family of all antiderivatives of a function f , F(x)+ c is called the indefinite integral of f and denoted by
where F' (x) = f (x) for all x in (a, b). Here, the symbol ∫ is called the indefinite integral sign, f (x) is the integrand, c is the constant of integration, and dx denotes the independent variable we are integrating with respect to. The process of finding antiderivatives of a given function is called integration.
Now reconsider the indefinite integral of 3x2. We know that for an arbitrary constant c, x3 + c is an antiderivative of 3x2. If we plot the graphs of every member of this family, we obtain an infinite
Let F and G be two functions such that F' = f and G' = f for all x from an open interval. Then F – G is constant, that is
F(x) = G(x) + c for some .
For this example, if the range is selected then these graphs fill the plane. When a point is selected in the plane, there is a single function that passes through that point and belongs to that family. For example, in order to find the function that belongs to the family which passes through the point (1,2), we write 1 instead of x and 2 instead of y in the equation of y = x3 + c:
2 = 13 + c c = 1.
Then the function we are looking for is y = x3 + 1.
Example: Find the function F(x), for which F'(x) = x+ 1 and the graph of the function passes through the point (2, 3).
Solution: We may easily obtain
If we write 2 instead of x and 3 instead of y in the following equation
4 + c = 3 c = –1.
Thus, the function is
Fundamental Rules of Integration
We may derive integration rules for certain functions with the help of differentiation.
• (Power Rule) For
is an antiderivative of the function xr because
Then,
where
• (Logarithm Rule) For the particular case is an antiderivative of the function ln x for x > 0 since,
Then,
for x > 0.
• (Exponential Rule) Figure 1.1 Members of the family of antiderivatives of
f(x)=3x2 Find the function F(x), for
which F'(x) = 4x3 + 6x2 and the graph of the function passes through the point (0,1).
1
1 x
1 c=0 c=1
c=-1 -1
-1
the difference) of any two functions is the sum (or the difference) of the derivatives of these functions. The derivative of a function multiplied by a constant k is equal to the derivative of that function times the constant k. These properties are also valid for integration. That is, given the arbitrary functions f and g and a real constant k, we have
Find the following integrals
2
Some basic rules:
Example:
Solution:
where c = c1 + c2 + c3. Thus, we obtain the integral as
where c = c1 + c2 + c3 + c4. Example:
Solution:
We know that the derivative of the product of two functions is not equal to the product of the derivatives of each of the functions. The same applies to the indefinite integral. That is, the indefinite integral of the product of two functions is not equal to the product of the indefinite integrals of each of the functions.
Examine the following example to observe that this equality cannot be achieved.
Example: We will show that
To this end, it is enough to show that the derivative of the right-hand side of the expression is not equal to x3. Let us substitute the integrals
in the expression
If we take the derivative of the function
we obtain
which, apparently, is not equal to x3.
This difficulty forces us to investigate certain approaches to be used in the evaluation of integrals containing complicated functions where the rules presented above may not be employed.
TECHNIQUES OF INTEGRATION
Change of Variable Method (Substitution Method)
Up until now, we have learned how to find simple integrals such as
However, we do not have any idea how to calculate the following integrals yet:
We now introduce a method to find the integrals of such functions, which is one of the most useful techniques for evaluating integrals. This technique is called the substitution method in which we change the variable of the integrand to something more convenient, or to put it more correctly, to a form where the previous rules may be applied directly.
This method is usually applied if a certain part of the integrand can be viewed as the derivative of remaining part of the integrand.
For example, in the integral,
u, we have
Now, the integral that looked very difficult, with this substitution, becomes very straightforward to evaluate by using the power rule. On doing so we obtain
You can check the result with a quick differentiation if you’d like.
We now give a simple proof of this method which is obtained from the chain rule for the derivatives.
This method says that
From the chain rule, we know that
Therefore, we obtain the equality
Now, using this rule, let us obtain the method of change of variables, or sometimes called as the substitution method, which considerably simplifies the calculations in applications.
Changing the variable
u = g(x) in the last equality, we get
du = g' (x) dx We note that after integrating with respect
to the new variable, do not forget to “back substitute” and get the integral back in terms of the original variable.
3
which is the statement of the change of variables method.
Example:
Solution: If
u = In x then
If we rewrite the integral with respect to u and integrate, we obtain
Example: Find the integral
Solution: Writing u = x2 + 2, we get
and the integral may easily be calculated as where u = g(x).
For the differentiable function u = g(x), the term du = g' (x) dx is called the differential of u.
Solution:
1. In the integral dx if we set u = x3 + 4 then,
Therefore, the integral becomes
Finally, writing back x3+4 instead of u, we obtain
2. As for the integral we choose u = 2x + 1 which gives
Hence, we obtain
3. For let us choose This gives
The integral is thus obtained as
4. In the case of we take u = x3 + 5 for which
Inserting the new variable in the integral and evaluating, we find returning back in the end to the original variable x,
5. How about Let us choose Then
Hence the result is found as
Evaluate the following integral
5
Employing the substitution method, we know how to find the integral
Now let us consider the following integral
At this point, we do not know how to find this integral since there is no substitution that we can use for the functions inside this integral.
Here, we will need to use integration by parts which is derived from the differentiation rule of the product of any two functions. Let us recall the product rule for the derivatives:
(f (x) g (x))' = f' (x) g (x) + f (x) g' (x).
If we integrate both sides of this equality, we obtain the method of integration by parts in terms of the following equality:
We can transform this formula into a form that is more memorable in practice by changing the variable. So, let us perform a couple of substitutions:
u = f (x), du = f' (x)dx, v = g (x), dv = g' (x)dx.
These substitutions give us the following for- mula that most people know as the integration by parts formula:
In the method of integration by parts, firstly
If we choose
u = x and dv = ex dx then
Now substituting all the results in the equation
we obtain the result of the integral as
We note that once we have taken the last integral in the problem, we will add in the constant of integration c to get our final answer.
In this example we could have also chosen u = ex and dv = xdx.
But in this case, we would get
and
Since we face a more difficult integral as a result of these choices, i.e.,
these substitutions do not work. However, when we choose the first substitutions, the second term
Solution: If we use the following choices
u = ln x and dv = dx = 1.dx we get
With these results inserted in the integration by parts rule we obtain the integral,
Example: Find the following integrals.
Solution:
1. Let u = (ln x)2 and dv = dx. Then
And the integral becomes
2. Let u = ln x and dv = x dx. Then
The integral is evaluated as,
we will evaluate
Let u = 2x + 3 and dv = ex dx. Then du = 2dx and v = ex. Thus we obtain the result of the integral as follows
Find the following integral
7 6
RECTILINEAR MOTION
A physicist who knows the velocity of a particle might wish to know its position at a given time. An engineer who can measure the variable rate at which water is leaking from a tank wants to know the amount leaked over a certain time period. A biologist who knows the rate at which a bacteria population is increasing might want to deduce what the size of the population will be at some future time. In each case, the problem is to find a function F whose derivative is a known function f. If such a function F exists, it is called an antiderivative of f.
Antidifferentiation is particularly useful in analyzing the motion of an object moving in a straight line. Recall that if the object has position function s = f (t), then the velocity function is v(t) = s'(t). This means that the position function is an antiderivative of the velocity function. Likewise, the acceleration
Example: A particle moves in a straight line and has acceleration given by a(t) = 6t + 4. Its initial velocity is v(0) = –6 cm/s and its initial displacement is s(0) = 9 cm. Find its position function (t).
Solution: Since v' (t) = a(t) = 6t + 4, antidifferentiation gives
Note that v(0) = C. But we are given that v(0) = –6, so C = –6 and
v(t) = 3t2 + 4t – 6
Since v(t) = s' (t), s is the antiderivative of v:
Reading Text
Summary
LO 1
functionA function F is called an antiderivative of f on an interval (a, b) if F' (x) = f (x)
for all x in (a, b). Thus, an antiderivative of is a function such that its derivative is
We can take since
Observe that the derivative of is equal to As a result, G(x) is another antiderivative of
LO 2
What is the indefinite integral of e2x ?The indefinite integral of is the family of functions with derivative e2x. That is, it is the family of antiderivatives of e2x. Then
LO 3
What kind of properties are there about indefinite integral?There are certain rules when integrating functions, which are direct consequences of the differentiation process. We list some of them below:
Besides, for any real number k and any functions f and g we have the following equalities:
Summary
LO 4
the change of variable method?This integral can be calculated by different ways.
The first is if we set u = x2 + 3 then
Hence the integral is evaluated as
The second way is if we take then
Thus, the integral is obtained as follows
LO 5
How can we apply the integration by parts method to the integralLet u = ln x and dv = x2 dx then
If we substitute what we find in the following equation into the formula
then we obtain
1
Which of the following functions is an antiderivative of e2x+1?a. ex2 b. 2e2x + 1 c.
d. x2ex e. (2x+1)ex
2
Which of the following functions is an antiderivative ofa.
b. ln(5x3+x2+1) c.
d. (5x3+x2+1) ln x e.
3
a. x3+2 ln x + 3x + c b. 3x2 + 5x + ln x + c c. (x2 + x) (ln x + 3x) + c d. x2 + 3 ln x + c e. 2 ln x + 3x2 + c
4
a.
b.
c.
d.
e.
5
a. x8 + 5x4 + c b.
c. (x4 + 5)11 + 3x2 + c d.
e. 30x2 (x4 + 5)9 + c
6
a. (x2 + x) e5x + c b.
c. 2e5x + 10e2x + c d.
e.
7
Which of the following functions is the antiderivative of with F (1) = 3 ?a. F (x) = x2 + x + 1 b.
c.
d.
e.
8
a.
b.
c.
d.
e.
9
a. (x2 + 3x) e–x + c b. –(3x + 4) e–x + c c. 3e–x + x2 + c d. 3x2 – ex + c e.
10
a. x2 ln (2x + 1) + c b. 2 ln (3x2 – 3x + 5) + c c.
d. ln (3x3 + x2) + c e.
Test Y ourself
1. c Since
2. a Since
and
for an arbitrary constant c.
3. b
4. c Let u = x + 2 then du = dx and x = u – 2.
Answer K ey for “T est Y ourself ”
7. d Since
then for every c ∈ !, + c is an antiderivative of . If we use F (1) = 3 then we obtain
8. a Let u = ln (7x) and dv = x3 dx. Then
9. b
Let u = 3x + 1 and dv = e–x dx then du = 3dx and v = –e–x.
10. c Let u = 3x2 + 3x + 5 then du = (6x + 3) dx = 3(2x + 1) dx.
Answer K ey for “T est Y ourself ”
Suggested Answers for “Y our T urn”
Find the following integrals
your turn 1
your turn 2
graph of the function passes through the point (0,1).
For a constant c,
Since F (0) = 1 holds
F (0) = c = 1.
As a result we obtain
F (x) = x4 + 2x3 + 1.
Let u = x2 + 1 then du = 2x dx a)
b)
your turn 3
Suggested Answers for “Y our T urn”
a.Let u = 2 + e3x then
We now write the integral in the variable u as
If we return back to the variable x again we obtain,
b. Let us consider the following integral
Let u = x3 + 2x. Then
Writing the integral with respect to u, evaluating it and returning back to the original variable x, we find
your turn 4
Evaluate the following integral
Let u = –3x2 + 1 then du = –6x dx, in other words, Thus the integral is obtained as follows
your turn 5
Suggested Answers for “Y our T urn”
Let u = 3x + 1 and dv = e–2xdx. Then
If we use the equality
we obtain the integral as follows
your turn 6
Find the following integral
For u = ln 3x and dv = 5x4 + 1 then and v = x5 + x.
your turn 7
Learning Outcomes
How will you employ the Fundamental theorem to calculate the definite integral?
How would you interpret the definite integral related to the area of a planar region R
geometrically? How do you calculate definite integral of a
simple function by geometrical means?
In what type of problems do you encounter definite integrals?
3 1 2
4
Chapter Outline
Introduction Area Problem Definite Integral
Key Terms
Area Problem
Upper and Lower Riemann Sums Definite Integral
Fundamental Theorem Average Value
After reading this chapter, you will be able to answer the following
questions
The study of calculus introduces certain notions to define the motion of objects, and just like the operations of addition and multiplication it introduces operations such as differentiation and integration. The concept of integral, just as the concept of derivative introduced in the first volume Mathematics I, is at the centre of mathematical analysis. It was mentioned already that the notion of derivative is encountered in the problems of determining the instantaneous velocity of a moving object, or the geometrical problem of finding the tangent line drawn to the graph of a function at a given point on the graph. Although these problems originated about 300 years ago, the problem of finding the area of a planar region bounded by some given curves dates back nearly 2500 years to the times of ancient Greeks.
It is subject of this chapter to introduce the concept of definite integral which emerged from determining the area of planar regions. Although the concepts of derivative and the definite integral may seem to be based on quite independent geometrical problems, we will present a relation between these two concepts which establishes the foundation of mathematical analysis, namely the Fundamental Theorem of Calculus.
Let us state the area problem from which the notion of integral originates.
AREA PROBLEM
The calculation of areas of planar regions bounded by given curves, which dates back as early as the ancient Greeks (Eudoxus, BC 390-337), were achieved through a method called the method of exhaustion by
inscribing and circumscribing the region in question by polygons the areas of which
to calculate the area of a circle (disk) by inscribing and circumscribing the circle with polygons. Over the years, this method was improved and the notion of definite integral was founded.
In this chapter, we will find the area of a bounded region determined by the graph of the function y=f (x). We assume that the function f (x) is defined and continuous on a closed interval [a,b], and it is also nonnegative, i.e. f (x) ≥ 0. The region will be such that it will be below the graph of f, above the x-axis, to the right of the line x=a and to the left of the line x=b. We simply state the mentioned area as the area under the graph of the function f defined on the interval [a,b] (see Figure 2.3).
Figure 2.2 Polygons inscribing and circumscribing the circle.
y
y= f (x)
using the known areas of certain polygons to calculate the area under the graph of the function f (x)=x2 defined on the interval [0,1].
In order to apply the method, we choose rectangles that enclose the region both from inside and outside and approach the region with these rectangles in the following way: First, using the points 0,1/2, and 1 let us divide the interval [0,1] into two subintervals [0,1/2] and [1/2,1].
The set {0,½,1} of the points so chosen is called a partition of the interval [0,1]. Making use of the subintervals, we construct rectangles covering the region both from inside and outside. Since the function f is an increasing function, it takes its smallest values at the left end and its largest values at the right end of each interval. Therefore, by choosing the smallest value of the function as the height of the rectangle and the length of the subinterval as the length of the base, we approach the area of the region by two rectangles constructed inside the region. Similarly, if we take the right end points for the heights of the rectangles, i.e.
if we choose the largest values of the function at each subinterval as the height of the rectangle, we obtain two rectangles approaching the area of the region from outside. We thus constructed, as seen in Figure 2.5, a rectangle on the subinterval [0,1/2] with height f (0)=0, and another one on the subinterval [1/2,1] with height f (1/2)=1/4. Let us denote the sum of the areas of these rectangles by L2 (f ) and call it the lower sum of the function f.
Since the length of each subinterval is ½ this sum is found as
The value of this lower sum is an approximate value for the area of region R and is smaller than the actual area (see Figure 2.5). In a similar manner, we construct two rectangles on the subintervals [0,1/2] and [1/2,1] with the heights
, respectively. We denote the sum of these rectangles by U2 (f ), and call it the upper sum for f, and since each subinterval has length ½, the upper sum becomes
As observed in Figure 2.5 (b), this is the sum of the area of the rectangles covering the region R and gives a larger value than the actual area.
x y
0 1
y= x2
R 1
Figure 2.4 The area under the graph of y=x2 on the interval [0,1].
y
x
12 1
14 1
y =x2
f (12) R
Figure 2.5 (a) The lower sum corresponding to the partition
Denoting the area of R by A, it is clear that L2 (f ) ≤ A ≤ U2 (f ).
The difference between the actual area A and the areas of the rectangles inside and outside the region is considerably different. To reduce this difference and approach to the real value of the actual area we re-divide each subinterval obtained in the previous step into two subintervals, therefore obtaining As before, we construct rectangles with bases on the subintervals and approach to the area of the region R from inside and outside respectively. To this end, first, we approach the area from inside by constructing rectangles with bases on the subintervals and heights as the smallest values of the functions assumed at the left end points of each subinterval. Then we repeat the same process by approaching the area from outside and, in this case, we take the heights to be the right end points of the subintervals (see Figure 2.6 (a)).
Thus, we obtain the following lower sum for the
x
12 1
14
1 y=x
f (12)
f (1)
R
Figure 2.5 (b) The upper sum corresponding to the partition
y
x 14 1
2 3
4 1
161 14 169
1
y=x2
f(14) f(12)
f(34) R
Figure 2.6 (a) The lower sum corresponding to the partition
y
14 169
1 y=x2
f(1) f(34)
f(1)
it is seen in Figure 2.6, these values are better estimations for the value A of the area. It is also clear that L4 (f ) ≤ A ≤ U4 (f ).
If the lower and upper sums corresponding to
the partition sets and are
compared we arrive at the following inequality:
L2 (f ) ≤ L4 (f ) ≤ A ≤ U4 (f ) ≤ U2 (f )
It means that the lower sum corresponding to the partition approaches A from below with increasing values, whereas the upper sum approaches A
from above with decreasing values. We observe, then, that by adding new points to the partition set, in other words, by increasing the number of rectangles, we obtain better approximations to the sought for area A.
We observe that as we increase the number of points in the partition set, we obtain at every step a better estimation to the value of the area A than the previous one. Let us, then, choose the partition of the interval [0,1] which forms n subintervals of equal lengths These subintervals are As before, in order to calculate the lower sum, we construct rectangles with bases as the subintervals and the heights with the values of the function on the left end point of each subinterval and approach the area from inside the region. By summing the areas of the rectangles so obtained we find the lower sum as
Similarly, the upper sum maybe calculated by summing the areas of rectangles with bases on the subintervals and heights with values of the function on the right-end points of the subintervals covering the region from outside. The upper sum is the total area of the rectangles given as
The lower and upper sums obtained above contain sums of squares of the numbers from 1 to n–1, and from 1 to n respectively. Using the formula presented for the sum of squares and rearranging the terms, we represent the lower and upper sums as a function of n as Calculate the lower and upper sums L5 (f ) and U5 (f ) for the partition
Compare your results with the values L2 (f ), L4 (f ), U2 (f ), U4 (f ).
1
The sum of squares of the natural numbers from 1 to n is given by the formula
subintervals, i.e. the number of rectangles, the second and third rows correspond to the lower and upper sums respectively.
Table 2.1
n 1 2 3 4 5 6 7 … 18 19 20 21 …
Ln (f ) 0 0,125 0,18 0,22 0,24 0,25 0,27 … 0,31 0,307 0,308 0,309 …
Un (f ) 1 0,625 0,52 0,47 0,44 0,42 0,41 … 0,36 0,360 0,358 0,357 …
It is evident from Table 2.1 that as n increases, i.e. the number of rectangles grow, the lower and upper sums approach to the area with increasing and decreasing values respectively.
To express this result with mathematical certainty, we need to show that by increasing n without bound, that is letting n tend to infinity, the values of the lower and upper sums approach the area A as close as desired. Remember that approaching procedure is defined in terms of the limit concept. Therefore, as n→∞
In the limit, we find that the lower and upper sums are equal. In other words, the rectangles enclosing the region both from inside and outside approach each other, and therefore to the region R, so that the area A of the region R is obtained
DEFINITE INTEGRAL
The functions considered in the area problem discussed in the previous section were specifically chosen nonnegative. We may remove this constraint and carry out similar calculations for a function f defined and continuous on an interval [a,b].
It is the purpose of this section to extend the “area” problem examined in the previous section to an arbitrary function defined and continuous on the interval [a,b] and introduce a number which will be called the “definite integral” of the function f.
Let f be a function defined and continuous on the interval [a,b]. First, let us choose the partition {x0, x1, x2, … ,xn–1 , xn} composed of the points x0, x1, x2, … ,xn–1, xn such that a=x0, b=xn and a=x0 < x1 < x2 < … < xn–1 < xn=b. With this partition the interval [a,b] is divided into n subintervals [x0,x1], [x1,x2], … , [xn–1, xn].
1,xi], i=1,2,…,n. The length of each interval is Δxi=xi–xi-1. Since the function f is continuous on [a,b] it assumes a minimum and a maximum value on each of the subintervals denoted by mi and Mi, i=1,2,…,n, respectively. For each x ∈ [xi-1,xi ], i=1,2,…,n we have the inequality
mi Δxi ≤ f (x)Δxi ≤ M Δxi
Adding the minimum values on the left-hand side of this inequality for each interval and denote this sum by Ln (f ) we obtain
Ln (f )=m1 Δx1+m2 Δx2+…+mn-1 Δxn-1+mn Δxn. Similarly, adding the maximum values on the right-hand side and denoting the sum by ve Un (f ) we get
Un (f )=M1 Δx1+M2 Δx2+… +Mn-1 Δxn-1+Mn Δxn These sums are exactly the same as the ones we obtained before in the case of positive functions, but differ slightly when the function is not positive.
In the general case, these sums are called lower and upper Riemann sums respectively. It is clear from the inequality above that the lower sums are always smaller than the upper sums, i.e.
Ln (f ) ≤ Un (f )
All the results obtained for the area problem are also valid in this case in that as the number of points of the partition increases, the lower Riemann sum increases and the upper Riemann sum decreases as illustrated in Figure 2.8.
unboundedly the lower and upper Riemann sums corresponding to each partition approache the same number. If we denote this number by I, as n gets large enough we write
the number I is denoted by the symbol
This number is called the definite integral of the function f on the interval [a,b].
Properties of the Definite Integral
Let us list below some of the essential properties that the integral satisfies.
1. The definite integral is independent of the choice of the integral variable, i.e.
2. If the lower and upper limits of integration is the same, the value of the integral is zero:
.
x y
x0 x1 x2 x3... xi 1− xi ... xn−1xn
Figure 2.8 Upper and lower Riemann sums.
The definite integral of a function is a real number.
The deformed “S” in the symbol ∫ is called the integral, the numbers a and b are called the lower and upper limits of the integral, the function f (x) is called the integrand and dx is called differential element signifying according to which variable the integral is calculated.
the integration changes the sign of the integral:
.
4. Given that f and g are integrable functions on the interval [a,b] and k is a real number we have
,
i.e. the integral of the sum of the functions f and g is equal to the sum of the integrals of f and g.
Also,
.
For the two constants k and l, the two properties just given may be combined to give
which is called the “linearity of the integral”.
5. Let f be integrable on the interval [a,b] and let c ∈ [a,b]. Then, the definite integral may be written as the sum of two integrals as
(see Figure 2.9).
of f over the interval [a,b] is equal to the area A under the graph of f; that is
Similarly, if f (x)≤ 0 for all x ∈ [a,b] then the integral is equal to the negative of the area between the graph of f and the x-axis:
If the function takes both negative and positive values along the interval [a,b], the definite integral cannot immediately be interpreted as the area under the graph of the function f. However, taking advantage of the fifth property, the definite integral may be calculated by splitting the function into
y
y= f(x)
x y
a b
A
Figure 2.10
x y
a b
A
Figure 2.11
Example: For any constant k,
Solution: From the linearity of the integral we have
The constant function f (x)=1 is a positive function, so the definite integral of f is the area under its graph (see Figure 2.13) which is the area of the rectangle with base length b–a and height k.
Therefore,
Example: The graph of the function f is given below. Evaluate the value of the integral
function does not assume positive values, the value of the integral is equal to the area A1. The graph of the function is a trapezoid with base lengths 2 and 4, and height 1. This gives the area of the trapezoid as A1=3. On the other hand, the function is nonpositive in the interval [2,3] so that the value of the integral is –A2. The area A2 is the area of the triangle whose base is 1 and height is 1, and thus In the end, using properties (5) and (6) the value of the sought for integral is
From the examples considered, it is observed that if the area under the graph of the function f is not known, or not easy to calculate, then this approach of evaluating the definite integral is not very useful. Also, constructing and calculating the lower and upper Riemann sums to find the value of a definite integral is not practical and in many cases not quite suited for arbitrary functions. It is these difficulties that lead to the discovery of the following tool called the “Fundamental Theorem of Calculus” which relates the definite integral to the operation of differentiation.
x y
a A c c b
A
A
Figure 2.12
x y
a b
1
Figure 2.13
x y
2 3
−2
1 A1
A2
Figure 2.14
The graph of the function f is given below. If what is the value of the area A2?
2
x y
−
A A
A Figure 2.15
Calculus
In this section, we introduce arguably the most important theorem of mathematical analysis: the fundamental theorem of calculus. Beside presenting a very practical way of calculating definite integrals, the fundamental theorem also establishes the link between the derivative and the integral concepts.
Let f be a continuous function defined on the interval [a,b] and let there be a function F defined on the interval [a,b] such that F '(x)=f (x). The function F discussed in Chapter 1 is called the antiderivative of f and is not uniquely defined.
The fundamental theorem states that the definite integral of f on the interval [a,b] is given in terms of the antiderivative F by the following rule:
We now give an example on how to employ this theorem.
Example: The velocity of a particle moving along the horizontal axis is given by the equation v(t)=2t–2.
The particle is located at x=1 at the time t=0. Let us first determine the function giving the particle’s location on the x-axis and then calculate the distance the particle travels between the times t=2 and t=4.
In our investigation of derivative, we found that the velocity of a particle is the time derivative of the particle’s position. In other words, if we denote the position by F(t), then we have F '(t)=v(t)=2t–2.
This means that F is the indefinite integral of the velocity function v:
.
To fully specify the position function, we need to determine the constant c. Since the particle
theorem since it tells us that the distance travelled is the difference of the position function at the end points of [2,4].
As a second approach, we might have calculated the area under the graph of v(t)=2t–2 in the time interval [2,4] as illustrated in Figure 2.16 which is, again, 8 units.
Thus, the area under the graph of the velocity function determines how much the particle travels in the time interval [0,4]. Hence, since the area under a graph is the definite integral, we have calculated the sought for distance in two different ways. As a result, we obtain
We have therefore confirmed that the definite integral may be calculated through the fundamental theorem.
If we introduce the notation to denote the difference F(b)–F(a) then we may write
t y
y = u( t)
Figure 2.16 The area under the graph of the velocity function on the interval [2,4].
We may employ the integration techniques
under the graph of f (x)=x2 on the interval [0,1].
Solution: If we denote the required area by A, then
We need to find the antiderivative F such that F'(x)=x2. One such function is apparently
Thus, the area is found as
Example: Evaluate the integral
Solution: The antiderivative of the integrand is Taking the difference of this function at the lower and upper limits of integral we find
Example: What is the value of the definite integral
Solution: The antiderivative of the function is ln x. From property (5) we get
Example: Evaluate the definite integral
Solution: Let us consider the indefinite integral
Inserting back the original variable x, we obtain an antiderivative as Finally, the definite integral is found as
Definite Integral Revisited
We may come across the definite integral in many different kinds of problems. The following two subsections present only two of such circumstances where the definite integral is used.
Area bounded by two curves
The area problem considered at the beginning of this chapter dealt with the area of a plane region bounded by the graph of a function
Evaluate the area of the region bounded by the parabola y=–2x2+6x+8 and the x-axis.
3
x y
−
y = − x + x +
R
Figure 2.17
generalise this situation and try to formulize the area bounded by two curves (the graphs of two functions) with the help of the definite integral.
Let us be given two continuous functions on the interval [a,b] such that f (x)≤g (x) for all x ∈ [a,b].
It is clear that the graph of f (x) is always below the graph of g (x). Then, denoting the area bounded by the graphs of f and g by A, this area is written in terms of a definite integral as
(see Figure 2.18).
Example: Evaluate the area bounded by the graphs of the functions f (x)=x and g (x)=x2 on the interval [0,1].
Figure 2.19 depicts that for 0≤x≤1, g (x)≤f (x).
bounded by the graphs of h(x)=x and k(x)=x+1 on the interval [0,1].
Solution: According to Figure 2.20, h(x)≤k(x) over the interval [0,1]. Therefore, the area in question may easily be obtained through the definite integral as follows:
x y
a b
y = g( x)
y = f ( x) A
Figure 2.18
x y
y=x y= x
A
Figure 2.19 The area bounded by the graphs of y=x and y=x2 on the interval [0,1].
Calculate the area of the plane region bounded by the curves
4
x
y x
=−
−
y =x y =x +
Figure 2.20 The area bounded by the graphs of y=x+1 and y=x2 on the interval [0,1].
y1,y2,…,yn which is defined as
and is called the arithmetic mean or the average of the given numbers. It should be noted that the average is taken over a set of discrete numbers. It is interesting to ask at this point that how would we calculate the average of continuous quantities such as the average of all numbers in an interval, or the average temperature T(t) measured continuously throughout a day? It is then our purpose in this section to introduce the concept of the mean of a continuous function f on an interval [a,b].
To this end, since we already know how to calculate the average of a finite number of points, let us choose a finite number of equally distant points x1,x2,…,xn from the interval [a,b] such that a=x0<x1<x2<…<xn-1<xn=b. These points will divide [a,b] into n subintervals [x0,x1],[x1,x2],…
,[xn-1,xn] all of which have equal lengths. Let us know calculate the arithmetic mean of the values f (x1),f (x2),…,f (xn). Since subintervals have equal
lengths we write which
gives Consequently, the average value of f at these selected points is
This value is just an estimation of the average value of f over the interval [a,b]. Then, by increasing the number of points, i.e. by increasing the value of n we can get better approximation to the average value. In the case of n→∞, we arrive at the definite integral since the obtained sum will correspond to the definition of the definite integral. Therefore, denoting the average value of f by favg we find
Assume that the minimum value of f on the interval [a,b] is m, and its maximum value on this interval is M. As depicted in Figure 2.21, the area under the graph of f is between the area of the small and large rectangles. This may be put mathematically as
Dividing both sides by (b–a), we obtain
namely, the average value favg is between the smallest and largest value of the function f. Remember that continuous functions assume every value between their minimum and maximum values on a given interval [a,b], then there always exists at least one particular point x- ∈ [a,b] such that
Also, if we rewrite
then we observe from Figure 2.22 that the left- hand side of the equality corresponds to the area of the rectangle with height f(x- ) and base b–a, and the right-hand side is just the area under the graph of f, which are equal to each other.
x M
m
a b
m
M
Figure 2.21
Example: Evaluate the average value of the function f (x)=2+x2 on the interval [0,2] and find the point which gives this mean value.
Solution: According to the foregoing discussion we write
The point that gives the average value may be calculated by solving the equation
i.e. by solving The quadratic equation has two roots which are and Since is not in the interval [0,2] the point we are looking for is
revenue through the sale of x units of a commodity is given by R(x)=100+4x+2x2. What is the average revenue of this producer if he sells between 2 units and 10 units of his product?
Solution: The revenue is a continuous function of the product sold in x units. If the sale is between 2 to 10 units then we are asked to find the average value of the function R(x)=100+4x+2x2 on the interval [2,10]. It is now a straightforward matter to find the average value which is
a ¯x b x
f ( ¯x)
Figure 2.22 Average value of f over the interval [a,b].
Evaluate the average value of the function on the interval [1,8].
5
A French nobleman, Georges Louis Leclerc, Comte de Buffon, posed the following problem in 1777:
“Suppose that you drop a short needle on ruled paper-what is then the probability
Reading Text
If a short needle, of length ℓ, is dropped on paper that is ruled with equally spaced lines of ℓ≤d, then the probability that the
needle comes to lie in a position where it crosses one of the lines is exactly
If you drop any needle, short or long, then the expected number of crossing will be
E=p1+2p2+3p3+…
where p1 is the probability that the needle will come to lie exactly one crossing, p2 is the probability that we get exactly two crossings, p3 is the probability for three crossings, etc. The probability that we get at least one crossing, which Buffon’s problem asks for, is thus
p=p1+p2+p3+…
(Events where the needle comes to lie exactly on a line, or with an end- point on one of the lines, have probability zero - so they can be ignored throughout our discussion.)
On the other hand, if the needle is short then the probability of more than one crossing is zero, p2=p3=…=0, and thus we get E=p. The probability that we are looking for is just the expected number of crossings. This reformulation is extremely useful, because now we can use linearity of expectation. Indeed, let us write E(t) for the expected number of crossings that will be produced by dropping a straight needle of length l. If this length is l = x +y, and we consider the
“front part” of length x and the “back part” of length y of the needle separately, then we get
E(x+y)=E(x)+E(y)
since the crossings produced are always just those produced by the front part, plus those of the back part.
By induction on n this “functional equation”
implies that E(nx)= nE(x) for all and then that
that E(x)=cx for all x > 0, where c=E(1) is some constant.
But what is the constant? For that we use needles of different shape. Indeed, let’s drop a “polygonal” needle of total length l, which consists of straight pieces. Then the number of crossings it produces is (with probability 1) the sum of the numbers of crossings produced by its straight pieces. Hence, the expected number of crossings is again
E=cl,
by linearity of expectation. (For that it is not even important whether the straight pieces are joined together in a rigid or in a flexible way!)
The key to Barbier’s solution of Buffon’s needle problem is to consider a needle that is a perfect circle C of diameter d, which has length x = dr. Such a needle, if dropped onto ruled paper, produces exactly two intersections, always! The circle can be approximated by polygons. Just imagine that together with the circular needle C we are dropping an inscribed polygon Pn , as well as a circumscribed polygon P n. Every line that intersects Pn, will also intersect C, and if a line intersects C then it also hits P
n. Thus, the expected numbers of intersections satisfy E(Pn)≤E(C)≤E(P n).
Now both Pn and P n are polygons, so the number of crossings that we may expect is “c times length” for both of them, while for C it is 2, whence
c l (Pn)≤2≤c l (P n).
Both Pn, and P n approximate C for n→∞.
In particular,
and thus, for n→∞ we infer from the previous inequality that
c d π≤2≤c d π, which gives
y
x
The trick to obtain an “easy” integral is to first consider the slope of the needle; let’s say it drops to lie with an angle of α away from horizontal, where α will be in the range 0≤α≤π/2. A needle that lies with angle α has height l sin α, and the probability that such a needle crosses one of the horizontal lines of distance d is Thus, we get the probability by averaging over the possible angles α, as
*Aigner, Martin, Günter M. Ziegler, and Alfio Quarteroni.
Proofs from the Book. Vol. 274. Berlin: Springer, 2010.
Summary
LO 1
How would you interpret the definite integral related to the area of a planar region R geometrically?Let y=f (x) be a non negative function defined on the interval [a,c] and R1 be the region bounded by the graph of f, the x-axis, the right of the line x=a and the left of the line x=c. Let the area of this region be denoted by A1. In this case, we have seen that the area A1 may be found on using the method exhaustion with the help of the definite integral and we denoted this area by
However, the function y=f (x) may assume negative values on the interval, say, [c,b] Let us denote the region bounded by the graph of f, the x-axis, the lines x=c and x=b by R2, and the area corresponding to the region R2 by A2. In this case the definite integral of f on the interval [c,b] is a negative number and this integral does not correspond to the area of the region R2. Nonetheless, the function –f assumes positive values on this interval and the area A2 of the region is then expressed as
If the two cases are combined, we may express the area of the region R=R1 ∪ R2 between the graph of the function f and the x-axis over the interval [a,b] as a definite integral given by
.
LO 2
How do you calculate definite integral of a simple function by geometrical means?In view of the discussion given above, the area of a planar region may be calculated using the defi- nite integral of a function f defining one of the sides of this region. Let us now state the converse:
Is it possible to calculate the definite integral of a function f with the help of the area bounded by the graph of f ? The answer is an immediate “yes” if the region is a known geometrical figure.
This implies that if the function given above has the property that its graph determines a known geometrical figure on the interval [a,b], then its area may be evaluated as
LO 3
calculate the definite integral?Let us state the fundamental theorem once again: Let f be a continuous function defined on the interval [a,b] and let there be an antiderivative function F defined on the interval [a,b] such that F ' (x)=f (x). In this case, the definite integral of f may be calculated as
Thus, we may apply the methods of evaluating indefinite integrals to the calculation of definite integrals. To find the function F satisfying the equation F' (x)=f (x) means to find the indefinite integral of the function f. Therefore, to calculate the definite integral we first find the antiderivative of f, i.e. then we only need to compute the difference F (b)–F (a).
LO 4
In what type of problems you encounter definite integrals?Since the area of a plane region may be calculated by means of the definite integral, we can extend this notion to the calculation of the area bounded by the graphs of two functions. Let f and g be two given functions defined and continuous on the interval [a,b] such that f (x)≤g(x) for all x ∈ [a,b]. If the are region below the graph of g and above the graph of f is denoted by A, then this area may be found by the definite integral
In addition to calculation of areas, the definite integral may also be employed to evaluate the mean value of a function. If the average value of a function defined and continuous over the interval [a,b]
is denoted by favg then the average value is calculated by the definite integral
Summary
1
What is the area between the x-axis and the graph of the function f (x)=2x+2 over the interval [–1,2]?a. 10 b. 9
c. 6 d. 5
e. 3
2
Which of the following is the value of the integrala. b.
c. d.
e.
3
a. 27 b. 32 c. 44 d. 52 e. 54
4
What is the valve of ? a. 40 b. 32c. -25 d. -10 e. -15
5
It is known that on the interval [0,4], the functions f and g satisfyWhat is the value
a. 32 b. 21 c. 17 d. 15 e. 10
6
If andfor the continuous function f, then which of the following is the value of the integral
a. 4 b. 3
c. –2 d. –4 e. –5
7
For x≥0, what is the area between the curves y=2x3, y=8x?a. 8 b. 9 c. 12 d. 16 e. 24
8
According to the graph of the function f given above, what is the value of the integrala. – b.
c.
d.
e.
9
Which of the following is the mean value of the function f (x)=x3–x2+1 between x=–1 and x=1?a.
b.
c.
d.
e.
10
What is the area under the graph of on the interval 1≤ x ≤ e3?a. 1 b. 2 c. e d. 3 e. 3e
Test Y ourself
x y
−
−
