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LECTURE NOTE
ON
MATHEMATICS FOR SOCIAL SCIENTISTS II
(MTS 108)
BY
ADEOSUN SAKIRU ABIODUN
E-mail: [email protected]
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Course Outline
Functions and equations: Definition and types of functions, definition and types of
equations. Matrices. Introduction to statistical distribution and density functions,
especially the binomial, Poisson and normal. Introduction to calculus functions of the
variable and their continuity. Techniques of differentiation; logarithmic, trigonometric
and exponential functions. Integral calculus; Optimization of functions, maximal,
minimal and in flexional points, and the application of these concepts in Business
and Economics.
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ยง 1.0 FUNCTIONS
Mathematical modeling is an attempt to describe some part of the real world
in mathematical terms. Our models will be functions that show the relationship
between two or more variables. These variables will represent quantities that we
wish to understand or describe. We will begin by reviewing the basic concept of
functions. In short, we call any rule that assigns or corresponds to each element in
one set precisely one element in another set a function.
Definition of a function: A function ๐ from ๐ท to ๐ is a rule that assigns to each
element ๐ฅ in ๐ท one and only one (unique) element ๐ฆ = ๐(๐ฅ) in ๐ .
The set ๐ท in the definition is called the domain of f. We might think of the
domain as the set of inputs. We then think of the values ๐(๐ฅ) as outputs. The set of
all the possible outputs (set of functional), ๐ is called the range of f. The letter
representing elements in the domain is called the independent variable and the letter
representing elements in the range is called the dependent variable. Thus, if
๐ฆ = ๐(๐ฅ), ๐ฅ is the independent variable while ๐ฆ is the dependent variable. Note that
๐ฆ = ๐(๐ฅ) is read as โ๐ฆ is a function of ๐ฅโ.
Example 1: A restaurant serves a steak special for N500. Write a function that
models the amount of revenue made from selling these specials. How much revenue
will 15 steak specials earn?
Solution: We first need to decide if the independent variable is the price of the
steak specials, the number of specials sold, or the amount of revenue earned. Since
the price is fixed at N500 per special and revenue depends on the number of special
sold, we choose the independent variable, ๐ฅ, to be the number of specials sold and
the dependent variable, ๐ = ๐(๐ฅ) to be the amount of revenue.
โด ๐ = ๐ ๐ฅ = 500๐ฅ. We note that ๐ฅ must be a whole number, so the domain is
= 0,1,2, โฆ . Hence, when selling 15 steak specials, ๐ = ๐ 15 = 500 15 = 7500. So
the revenue is N7500.
Example 2: If ๐ ๐ฅ = ๐ฅ3 โ 3๐ฅ + 4, find the values of : ๐(0), ๐(โ1), ๐(๐), ๐(3๐ฅ).
Solution: ๐ 0 = 03 โ 3 0 + 4 = 4
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๐ โ1 = โ1 3 โ 3 โ1 + 4 = 6
๐ ๐ = ๐3 โ 3๐ + 4
๐ 3๐ฅ = 3๐ฅ 3 โ 3 3๐ฅ + 4 = 27๐ฅ3 โ 9๐ฅ + 4
Example 3: If ๐ ๐ฅ =๐ฅ+1
๐ฅ2+๐ฅโ2, determine
(a) the image of โ1
(b) ๐(๐ฅ โ 1)
(c) What values of ๐ฅ will have no images for this function?
Solution: (a) The image of ๐ฅ = โ1 is ๐(โ1):
๐ โ1 =โ1+1
โ1 2+ โ1 โ2=
0
โ2= 0. (Note ๐ฅ = โ1 is the zero for this function)
(b) ๐ ๐ฅ โ 1 =๐ฅโ1+1
๐ฅโ1 2+ ๐ฅโ1 โ2=
๐ฅ
๐ฅ2โ๐ฅโ2
(c) ๐ ๐ฅ =๐ฅ+1
๐ฅ2+๐ฅโ2=
๐ฅ+1
๐ฅ+2 ๐ฅโ1 .
When ๐ฅ = โ2, ๐ โ2 =โ1
0 and when ๐ฅ = 1, ๐ 1 =
2
0. And since division by 0 is
meaningless (undefined), therefore ๐(โ2) and ๐(1) do not exist. Hence the values
๐ฅ = โ2 and ๐ฅ = 1 have no images for this function.
Example 4: Given the functions ๐ ๐ฅ = ๐ฅ2 โ 1, ๐ ๐ฅ = 2๐ฅ + 3 and ๐ ๐ฅ =1
๐ฅ, find
(i) the domain and range of ๐(๐ฅ)
(ii) the range of ๐(๐ฅ)
(iii) ๐ โ ๐ ๐ฅ (called the composition function)
Solution: (i) The domain and range of ๐(๐ฅ) is the set of all real numbers i.e โ.
(ii) Note that one way of determining the range is to express ๐ฅ as a function of
๐ฆ; ๐ฅ = ๐(๐ฆ), then the range is the set of all real values of ๐ฆ for which the function ๐
is defined.
Solve for ๐ฅ:
๐ฅ =1
๐ฆ define for all real numbers except when ๐ฆ = 0. Therefore the range of ๐(๐ฅ)
is the set of real numbers when zero is deleted i.e. Rng h = โ\ 0 .
(iii) ๐ โ ๐ ๐ฅ = ๐ ๐(๐ฅ) = 2๐ฅ + 3 2 โ 1 = 4๐ฅ2 + 12๐ฅ + 8 .
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Different between a Function and an Equation
Function Equation
Let ๐ฆ = ๐ ๐ฅ = ๐ฅ2 โ 3๐ฅ โ 10 = ๐ฅ โ 5 ๐ฅ + 2 .
This states the function. A function cannot be
solved but is used to find the images of ๐ฅ. The
values of ๐ฅ which make ๐ ๐ฅ = 0 are called the
zeros of ๐. The zeros of this function are
๐ฅ = 5 and ๐ฅ = โ2.
For example ๐ฅ2 โ 3๐ฅ โ 10 = 0. An
equation is solved to find its roots
which are the values of ๐ฅ which
satisfy the equation. The roots of
this equation are ๐ฅ = 5 or ๐ฅ = โ2.
Some important types of functions:
1. Linear function ๐ฆ = 2๐ฅ โ 4
2. Quadratic function ๐ฆ = ๐ฅ2 โ ๐ฅ + 12
3. Polynomial function ๐ฆ = ๐๐๐ฅ๐ + ๐๐โ1๐ฅ๐โ1 + โฏ + ๐1๐ฅ + ๐0
4. Trigonometric function ๐ฆ = sin ๐ฅ
5. Logarithmic function ๐ฆ = log ๐ฅ
6. Exponential function ๐ฆ = 3๐ฅ or ๐๐ฅ
Many other functions will be defined by formula. For example, area, A, of a circle is a
function of radius r, as ๐ด = ๐๐2, also Total value (T) = Price(P) ร Quantity (Q) and
so on.
Exercise 1
1. Find the values stated for these functions:
(a) ๐ ๐ฅ = 3๐ฅ โ 1 ; ๐(0), ๐(โ1), ๐(1)
(b) ๐ ๐ฅ = ๐ฅ2 + ๐ฅ + 1 ; ๐(โ1), ๐(0), ๐(๐ฅ + 1)
(c) ๐ ๐ฅ = ๐ฅ2 โ ๐ฅ โ 2 ; ๐ โ1 , ๐ โ1
2 , ๐ 0 , ๐ 1
2
(d) ๐ ๐ฅ =๐ฅ
๐ฅ+1; ๐ 0 , ๐ 1 , ๐(๐ฅ โ 1)
2. Find the zeros of the following functions:
(a) ๐ ๐ฅ = ๐ฅ โ 2 (b) ๐ ๐ฅ = ๐ฅ + 3 2
(c) ๐ ๐ฅ =๐ฅโ1
๐ฅ+6 (d) ๐ ๐ฅ =
๐ฅ2+2๐ฅโ15
๐ฅ2โ1
3. If ๐ ๐ฅ = 3๐ฅ + 2, what is the value of ๐ฅ whose image is 5? If also given that
๐ ๐ฅ = 2๐ฅ โ 3 and ๐ ๐ฅ + ๐ ๐ฅ = 4, find the value of ๐ฅ.
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4. Given the function ๐ ๐ฅ = 2๐ฅ2 โ ๐ฅ + 1, express ๐ ๐ฅ+๐ โ๐(๐ฅ)
๐ in its simplest form.
ยง 2.0 MATRICES AND DETERMINANTS
2.1 Basic Concepts
A matrix is an array (enclosed between brackets) of real numbers arranged in
๐ rows and ๐ columns. The matrix is then referred to as an ๐ by ๐ matrix or
(๐ ร ๐) matrix.
For example,
(a) ๐ด = 1 2 3 is a 1 by 3 matrix.
(b) ๐ต = 2 4 12 0 3
is a 2 by 3 matrix
(c) ๐ถ = 8 24 50 1
is a 3 by 2 matrix
(d) ๐ท = 1 2 34 5 67 8 9
is a 3 by 3 matrix.
Any ๐ by ๐ matrix is called a square matrix.
2.2 Types of Matrices
(I) Equal matrices: Two matrices ๐ด and ๐ต are said to be equal when
(i) The number of rows of ๐ด = the number of rows of ๐ต and
(ii) The number of columns of ๐ด = the number of columns of B and
(iii) The entries in corresponding position are the same in both ๐ด and ๐ต.
For example:
1. If ๐ด = 3 12 2
and ๐ต = 3 1 22 2 0
, then ๐ด โ ๐ต.
2. ๐ = 2 0 33 1 00 0 6
, ๐ =1
2 4 0 66 2 00 0 12
, then ๐ = ๐ since entries in
corresponding positions in both matrices are equal
(II) .Identity Matrix: A square matrix ๐ด is said to be an identity matrix when each
entry on the leading diagonal of ๐ด is unity and other entries off the leading
diagonal are zeros. For example, the following are identity matrices:
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I2 = 1 00 1
, ๐ผ3 = 1 0 00 1 00 0 1
, ๐ผ4 =
1000
0100
0010
0001
(III)Diagonal Matrix: A square matrix with zero entries off the leading diagonal is a
diagonal matrix provided the entries on the leading diagonal are non-zero. For
example,
๐ด = 3 0 00 6 00 0 9
, ๐ต =
4000
0500
0090
0002
are diagonal matrices.
(IV) Triangular Matrix
A matrix whose entries above or below the leading diagonal are zeros is
referred to as a triangular matrix. It is said to be upper triangular where all
entries below the leading diagonal are zeros. It is lower triangular when all
entries above the leading diagonal are zeros. Following are the two examples:
๐ = 1 4 30 7 50 0 8
, ๐ = 1 0 03 6 04 5 7
.
(V) The Transpose of a Matrix
A matrix ๐ต is the transpose of ๐ด if when the rows of ๐ต are turned into columns,
the resulting matrix is the same as ๐ด. For example, let ๐ด = 1 3 59 6 32 1 4
, if the
rows of ๐ด are now put into the columns of another matrix ๐ต, we have
๐ต = 1 3 59 6 32 1 4
. Then ๐ต is the transpose of ๐ด. This is written as ๐ต = ๐ด๐ .
(VI) Symmetric Matrix
A matrix ๐ is said to be symmetric if ๐ = ๐๐. In other words, a matrix is
symmetric if the matrix and its transpose are the same. For example,
๐ = 2 8 18 9 21 2 3
is such that ๐ = ๐๐. Hence ๐ is a symmetric matrix.
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(VII) Skew Symmetric Matrix
A matrix ๐ is said to be skew symmetric if ๐๐ = โ๐. We noted that this could
only be possible if all the leading diagonal entries of ๐ are zeros. For
example, ๐ = 0 1 โ4
โ1 0 64 โ6 0
, ๐๐ = 0 โ1 41 0 โ6
โ4 6 0 , โ๐ =
0 โ1 41 0 โ6
โ4 6 0 .
Observed that ๐๐ = โ๐. Therefore, ๐ is a skew symmetric matrix.
(VIII) Zero Matrix or Null Matrix
A null matrix is a matrix with all its entries zero. For example, ๐ = 0 0 00 0 00 0 0
.
2.3 Basic Operations on Matrices
(a) Multiplication of a Matrix by a Constant (Scalar)
Let ๐ be a constant and ๐ a matrix. The matrix ๐ = ๐๐, is obtained from ๐ by
multiplying each entry of ๐ by ๐.
Example: If ๐ด = 3 2 11 4 5
, then ๐๐ด = 3๐ 2๐ ๐๐ 4๐ 5๐
= 12 8 44 16 20
when
๐ = 4.
(b) Addition of Two Matrices
Let ๐ด and ๐ต be two matrices. The sum ๐ด + ๐ต exists only when the number
of rows of both matrices are the same, and the number of columns of both
matrices are also the same. That is, both matrices have the same order.
When ๐ด + ๐ต exists, then the sum is obtained by adding corresponding entries
of ๐ด and ๐ต and putting the sum in the corresponding positions of a new matrix
๐ด + ๐ต.
Example: Let ๐ด = 2 14 79 1
, ๐ต = 5 44 12 3
. Then
๐ด + ๐ต = 2 + 5 1 + 44 + 4 7 + 19 + 2 1 + 3
= 7 58 8
11 4 .
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(c) Difference of Two Matrices
The difference of two matrices ๐ด and ๐ต which are of the same order is defined
as the sum of the two matrices ๐ด and (โ๐ต) where โ๐ต = โ1 โ ๐ต. Thus if
๐ด =
๐11 โฏ ๐1๐
โฎ โฑ โฎ๐๐1 โฏ ๐๐๐
and ๐ต = ๐11 โฏ ๐1๐
โฎ โฑ โฎ๐๐1 โฏ ๐๐๐
then
๐ด โ ๐ต = ๐11 โ ๐11 โฏ ๐1๐ โ ๐1๐
โฎ โฑ โฎ๐๐1 โ ๐๐1 โฏ ๐๐๐ โ ๐๐๐
.
Example: If ๐ด = 5 2 13 8 7
and ๐ต = 6 1 โ39 7 5
, evaluate (i) ๐ด โ ๐ต (ii)
4๐ด โ 3๐ต.
Solution:
๐ด โ ๐ต = 5 โ 6 2 โ 1 1 โ (โ3)3 โ 9 8 โ 7 7 โ 5
= โ1 1 4โ6 1 2
.
4๐ด โ 3๐ต = 20 โ 18 8 โ 3 4 + 912 โ 27 32 โ 21 28 โ 15
= 2 5 13
โ15 11 13 .
Note: ๐ด + ๐ต = ๐ต + ๐ด, and ๐ต โ ๐ด = โ ๐ด โ ๐ต โ ๐ด โ ๐ต.
(d) Product of Two Matrices
Let ๐ด and ๐ต be two matrices. The condition that the product ๐ด๐ต (in that order)
exists is that the number of columns of ๐ด must be equal to the number of
rows of ๐ต. When the product ๐ด๐ต exists, then the entry in row ๐ and column ๐
of ๐ด๐ต is obtained as the scalar product of the entries in row ๐ of ๐ด and those
in the column ๐ of ๐ต. For instance, if ๐ด = ๐11 ๐12
๐21 ๐22 , ๐ต =
๐11 ๐12 ๐21 ๐22
๐13
๐23 .
Let ๐ด๐ต = ๐ถ = ๐11 ๐12 ๐13
๐21 ๐22 ๐23 then the matrices ๐ด and ๐ต are comfortable for
product where ๐11 = ๐11 ๐12 ๐11
๐21 = ๐11๐11 + ๐12๐21
๐12 = ๐11 ๐12 ๐12
๐22 = ๐11๐12 + ๐12๐22
๐13 = ๐11 ๐12 ๐13
๐23 = ๐11๐13 + ๐12๐23
๐21 = ๐21 ๐22 ๐11
๐21 = ๐21๐11 + ๐22๐21
๐22 = ๐21 ๐22 ๐12
๐22 = ๐21๐12 + ๐22๐22
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๐23 = ๐21 ๐22 ๐13
๐23 = ๐21๐13 + ๐22๐23.
Note: If ๐ด is an ๐ ร ๐ matrix and ๐ต is an ๐ ร ๐ matrix, then the product ๐ด๐ต is
an ๐ ร ๐ matrix.
Examples
1. If ๐ด = 6 1 45 3 27 0 5
and ๐ต = 2 48 61 7
, evaluate ๐ด๐ต.
Solution
๐ด๐ต =
6 2 + 1 8 + 4 (1) 6 4 + 1 6 + 4 (7) 5 2 + 3 8 + 2 (1) 5 4 + 3 6 + 2 (7) 7 2 + 0 8 + 5 (1) 7 4 + 0 6 + 5 (7)
= 24 5836 5219 63
.
2. Find the value of ๐ and ๐ so that 2 7
โ3 0
๐ ๐1 5
= 11 29โ6 9
.
Solution
2 7
โ3 0
๐ ๐1 5
= 2๐ + 7 2๐ + 35โ3๐ โ3๐
= 11 29โ6 9
โ 2๐ + 7 = 11, 2๐ + 35 = 29, โ 3๐ = โ6, โ 3๐ = 9
โ ๐ = 2 and ๐ = โ3.
2.4 Determinants
The determinant of a square matrix ๐ ร ๐ is a specific number associated
with the matrix, and is usually denoted by ๐ด or ๐๐๐ก ๐ด . The determinant may be
positive, negative or zero. A square matrix ๐ด is called singular matrix if its
determinant ๐ด = 0.
Given matrix ๐ด = ๐11 ๐12
๐21 ๐22 , ๐ด can be obtained as
๐ด = ๐11 ๐12
๐21 ๐22 = ๐11๐22 โ ๐12๐21. Also given a 3 ร 3 matrix ๐ด =
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
.
The determinant of ๐ด is given by
๐๐๐ก ๐ด =
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
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= ๐11 ๐22 ๐23
๐32 ๐33 โ ๐12
๐21 ๐23
๐31 ๐33 + ๐13
๐21 ๐22
๐31 ๐32
= ๐11๐22๐33 + ๐12๐23๐31 + ๐13๐21๐32 โ ๐11๐23๐32 โ ๐12๐21๐33 โ ๐13๐22๐31.
Examples
1. If ๐ด = 2 โ35 4
then ๐ด = 2 4 โ โ3 5 = 23.
2. If ๐ = โ2 28 4
then ๐ = โ2 4 โ 2 8 = โ24.
3. Given that ๐ = 1 0 โ24 6 12 โ3 1
, find ๐ .
Solution ๐ = +1 6 1
โ3 1 โ 0
4 12 1
+ โ2 4 62 โ3
= 6 โ โ3 โ 0 โ 2 โ12 โ 12
= 9 + 48 = 57
4. Given that ๐ฅ + 1 1โ1 ๐ฅ โ 1
= 4, find the value of ๐ฅ.
Solution
๐ฅ + 1 ๐ฅ โ 1 โ โ1 1 = 4
โ ๐ฅ2 โ ๐ฅ + ๐ฅ โ 1 + 1 = 4
โ ๐ฅ2 = 4
โ ๐ฅ = ยฑ2
2.5 Minor, Cofactor and Adjoint of a Matrix
I. A minor of a matrix ๐ด is any square submatrix of ๐ด. For example, if
๐ด = 1 2 34 5 67 8 9
, then 1 24 5
, 1 34 6
, 5 68 9
, 4 57 8
are submatrices of ๐ด.
Each of them is therefore a minor of ๐ด. Some particular types of minors of a
matrix ๐ด are obtained by deleting from ๐ด its ๐th row and the ๐ th column. The
determinant of such a minor is denoted by ๐ด๐๐ .
Example 1: Given the matrix ๐ด = 1 2 34 5 67 8 9
, find the determinant of all 2 ร 2
minors of ๐ด.
Solution: The (1, 1) minor is obtained by deleting from ๐ด the first row and the
first column. The determinants are as follows:
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๐ด11 = 5 68 9
= 45 โ 48 = โ3
๐ด12 = 4 67 9
= 36 โ 42 = โ6
๐ด13 = 4 57 8
= 32 โ 35 = โ3
๐ด21 = 2 38 9
= 18 โ 24 = โ6
๐ด22 = 1 37 9
= 9 โ 21 = โ12
๐ด23 = 1 27 8
= 8 โ 14 = โ6
๐ด31 = 2 35 6
= 12 โ 15 = โ3
๐ด32 = 1 34 6
= 6 โ 12 = โ6
๐ด33 = 1 24 5
= 5 โ 8 = โ3.
II. Cofactor of a Matrix
The (๐, ๐ ) cofactor of a matrix is the determinant of the matrix obtained by
deleting the ๐th and the ๐ th column of ๐ด and then multiplying the result by
โ1 ๐+๐ . Let ๐ถ๐๐ denote the (๐, ๐ ) cofactor. Then ๐ถ๐๐ = โ1 ๐+๐ ๐ด๐๐ .
Example 2: Find the cofactor of ๐ด in the example 1 above.
Solution: Using ๐ถ๐๐ = โ1 ๐+๐ ๐ด๐๐ , we have the following.
๐ถ11 = โ1 1+1๐ด11 = โ3
๐ถ12 = โ1 1+2๐ด12 = 6
๐ถ13 = โ1 1+3๐ด13 = โ3
๐ถ21 = โ1 2+1๐ด21 = 6
๐ถ22 = โ1 2+2๐ด22 = โ12
๐ถ23 = โ1 2+3๐ด22 = 6
๐ถ31 = โ1 3+1๐ด31 = โ3
๐ถ32 = โ1 3+2๐ด32 = 6
๐ถ33 = โ1 3+3๐ด33 = โ3.
Therefore, the cofactor of ๐ด is given by
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๐ถ = โ3 6 โ36 โ12 6
โ3 6 โ3 .
III. The Adjoint of a Matrix
The Adjoint of a matrix ๐ด is the transpose of its matrix of cofactors. Let ๐ด be a
matrix whose matrix of cofactor is ๐ถ. Then ๐ด๐๐ ๐ด = ๐ถ๐. Thus in the example
two above,
๐ด๐๐ ๐ด = โ3 6 โ36 โ12 6
โ3 6 โ3
๐
= โ3 6 โ36 โ12 6
โ3 6 โ3 .
2.6 The Inverse of a Matrix ๐จ ๐จโ๐
Let ๐ด be an ๐ ร ๐. Let ๐ผ๐ be the ๐ ร ๐ identity matrix. Then ๐ด๐ผ = ๐ผ๐ด = ๐ด.
Furthermore, ๐ดโ1๐ด = ๐ด๐ดโ1 = ๐ผ. Note that we do not write ๐ดโ1 as 1
๐ด since
1
๐ด has no
meaning in the theory of matrices.
Let ๐ด be an ๐ ร ๐ matrix such that ๐ดโ1 exist. Then, the expression for the
inverse of matrix ๐ด is given as:
๐ดโ1 =๐ด๐๐ ๐ด
๐ด provided ๐ด โ 0.
Matrix ๐ด is said to be singular when ๐ด = 0 and so inverse not exist. While matrix ๐ด
is non singular when ๐ด โ 0.
Examples
1. Find ๐โ1 if ๐ = 8 43 1
.
Solution:
๐ = 8 43 1
= 8 โ 12 = โ4.
Cofactors are:
๐ถ11 = โ1 1+1๐ด11 = 1 ๐ถ12 = โ1 1+2๐ด12 = โ3
๐ถ21 = โ1 2+1๐ด21 = โ4 ๐ถ22 = โ1 2+2๐ด22 = 8.
Then ๐ถ = 1 โ3
โ4 8
๐ด๐๐ ๐ = 1 โ3
โ4 8
๐
= 1 โ4
โ3 8
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โด ๐โ1 =๐ด๐๐ ๐
๐ =
1 โ4
โ3 8
โ4=
โ1
41
3
4โ2
.
Check: ๐ฟ๐ฟโ๐ = ๐ฐ๐. As exercise.
2. If ๐ด = 2 1 01 3 23 1 โ1
, find ๐ดโ1.
Solution:
๐ด = 2 3 21 โ1
โ 1 1 23 โ1
+ 0 1 33 1
= 2 โ3 โ 2 โ โ1 โ 6 + 0 = โ3
Cofactor of matrix ๐ด:
๐ถ =
+
3 21 โ1
โ 1 23 โ1
+ 1 33 1
โ 1 01 โ1
+ 2 03 โ1
โ 2 13 1
+ 1 03 2
โ 2 01 โ1
+ 2 11 3
= โ5 7 โ81 โ2 12 โ4 5
๐ด๐๐ ๐ด = ๐ถ๐ = โ5 1 27 โ2 โ4
โ8 1 5
โด ๐ดโ1 =๐ด๐๐ ๐ด
๐ด =
โ5 1 27 โ2 โ4
โ8 1 5
โ3=
1
3
5 โ1 โ2โ7 2 48 โ1 โ5
.
Check that ๐ด๐ดโ1 = ๐ผ3
2.7 SYSTEMS OF LINEAR EQUATIONS
A system of linear equations is of the form:
๐11๐ฅ1 + ๐12๐ฅ2 + โฏ + ๐1๐๐ฅ๐ = ๐1
๐21๐ฅ1 + ๐22๐ฅ2 + โฏ + ๐2๐๐ฅ๐ = ๐2
โฎ
๐๐1๐ฅ1 + ๐๐2๐ฅ2 + โฏ + ๐๐๐๐ฅ๐ = ๐๐
The method of solution could be
(a) Solution by direct algebraic method (substitution method or elimination
method)
(b) Solution using the matrix inverse
(c) Solution using Cramerโs rule.
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For solution by direct algebraic method, it has been discussed in MTS 107 and so
we omit it here.
(b) Solution using Inverse Method
In matrix form:
๐11 โฏ ๐1๐
โฎ โฑ โฎ๐๐1 โฏ ๐๐๐
๐ฅ1
โฎ๐ฅ๐
= ๐1
โฎ๐๐
๐ด๐ = ๐ต
๐ดโ1๐ด๐ = ๐ดโ1๐ต
๐ผ๐ = ๐ดโ1๐ต
โ ๐ = ๐ดโ1๐ต provided ๐ด โ 0
Example: Solve the following equations by inverse method:
1. ๐ฅ1 + 4๐ฅ2 = 22
๐ฅ1 + ๐ฅ2 = 7
2. 3๐ฅ1 โ 2๐ฅ2 + ๐ฅ3 = 2
๐ฅ1 + 3๐ฅ2 + 4๐ฅ3 = 19
5๐ฅ1 + 4๐ฅ2 โ 3๐ฅ3 = 4.
Solution:
1. In matrix form:
1 41 1
๐ฅ1
๐ฅ2 =
227
Where ๐ด = 1 41 1
, ๐ = ๐ฅ1
๐ฅ2 and ๐ต =
227
.
The solution of which is
๐ฅ1
๐ฅ2 =
1 41 1
โ1
227
= โ1
3
1 โ4โ1 1
227
= โ1
3
โ6โ15
= 25
That is, ๐ฅ1 = 2 and ๐ฅ2 = 5.
2. This can be written in the form
3 โ2 11 3 45 4 โ3
๐ฅ1
๐ฅ2
๐ฅ3
= 2
194
Or ๐ด๐ = ๐ต
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Hence, ๐ = ๐ดโ1๐ต or
๐ฅ1
๐ฅ2
๐ฅ3
= 3 โ2 11 3 45 4 โ3
โ1
2
194
thus we need ๐ดโ1.
๐ด = 3๐ด11 โ โ2 ๐ด12 + (1)๐ด13
= 3 โ25 + 2 โ23 + (โ11)
= โ132.
Cof ๐ด =
+
3 44 โ3
โ 1 45 โ3
+ 1 35 4
โ โ2 14 โ3
+ 3 15 โ3
โ 3 โ25 4
+ โ2 13 4
โ 3 11 4
+ 3 โ21 3
= โ25 23 โ11โ2 โ14 โ22โ11 โ11 11
๐ด๐๐ ๐ด = ๐ถ๐ = โ25 โ2 โ1123 โ14 โ11
โ11 โ22 11
๐ดโ1 =๐ด๐๐ ๐ด
๐ด =
โ25 โ2 โ1123 โ14 โ11
โ11 โ22 11
โ132=
1
132
25 2 11โ23 14 1111 22 โ11
Therefore,
๐ฅ1
๐ฅ2
๐ฅ3
= ๐ดโ1 2
194
=1
132
25 2 11โ23 14 1111 22 โ11
2
194
=1
132
50 + 38 + 44โ46 + 266 + 4422 + 418 โ 44
=1
132 132264392
= 123 .
(c) Solution by Cramerโs rule
1. The 2 ร 2 matrix case:
๐ฅ1 + 4๐ฅ2 = 22
๐ฅ1 + ๐ฅ2 = 7 Or 1 41 1
๐ฅ1
๐ฅ2 =
227
i.e ๐ด๐ = ๐ต.
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By Cramerโs rule,
๐ฅ1 = 22 47 1
1 41 1
, ๐ฅ2 =
1 221 7
1 41 1
๐ฅ1 =โ1
โ , ๐ฅ2 =
โ2
โ
Where โโ ๐ด , โ1 is the determinant obtained by replacing the first column of ๐ด by
the matrix 227
and โ2 is the determinant obtained by replacing the second column
of ๐ด by the matrix 227
. Thus,
โ= 1 41 1
= 1 โ 4 = โ3
โ1= 22 47 1
= 22 โ 28 = โ6
โ2= 1 221 7
= 7 โ 22 = โ15
Hence,
๐ฅ1 =โ1
โ=
โ6
โ3= 2 and ๐ฅ2 =
โ2
โ=
โ15
โ3= 5 .
Therefore, ๐ฅ1
๐ฅ2 =
25 .
2. The (3 ร 3) matrix case:
3 โ2 11 3 45 4 โ3
๐ฅ1
๐ฅ2
๐ฅ3
= 2
194
โ= 3 โ2 11 3 45 4 โ3
= 132
โ1= 2 โ2 1
19 3 44 4 โ3
= โ132
โ2= 3 2 11 19 45 4 โ3
= โ264
โ3= 3 โ2 21 3 195 4 4
= โ396
Hence,
๐ฅ1 =โ1
โ=
โ132
โ132= 1
๐ฅ2 =โ2
โ=
โ264
โ132= 2
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๐ฅ3 =โ3
โ=
โ396
โ132= 3.
2.8 BUSINESS APPLICATION
Example: There are 30 secondary schools and 60 primary schools in a local
government area in a certain state. Each of the secondary schools and the primary
schools has 1 Messenger (ME), 5 Clerks (CL) and 1 Cashier (CA). Each secondary
school in addition has 1 Accountant (AC) and 1 Head Clerk (HC). The monthly salary
(Nโ000) of each of them is as follows:
Messenger โ N20; Clerk โ N40, Cashier โ N35, Accountant โ N50, and Head clerk โ
N60. Use matrices to find
(i) The total number of posts of each kind of primary and secondary schools
taken together.
(ii) The total monthly salary of each primary school and each secondary
school separately.
(iii) The total monthly salary bill of all secondary and primary schools.
Solution: Consider the row matrix of the two types of schools:
๐ = 30 60
Representing the number of secondary schools and primary schools respectively,
we have
Let ME CL CA AC HC
๐ = 1 5 11 5 1
10
10
This is the matrix of the number of person in each position for each type of
school. Consider the product matrix:
๐๐ = 30 60 1 5 11 5 1
10
10
= 90 450 90 30 30 .
Then, there are in the two types of schools, 90 Messengers, 450 Clerks, 90
Cashiers, 30 Accountants and 30 Head Clerks.
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(ii) The monthly salaries (Nโ000) can be put as a column matrix ๐ =
2040355060
.
Total monthly bill of each kind of school is
๐๐ = 1 5 11 5 1
10
10
2040355060
= 20 + 200 + 35 + 50 + 60
20 + 200 + 35 + 0 + 0
= 365255
Total monthly salary bill for secondary schools is N365,000 while total monthly for
primary schools is N255,000.
(iii) ๐ ๐๐ gives the total monthly salary bill for all secondary and primary schools.
๐ ๐๐ = 30 60 365255
= ๐26,250 (โ000)
= ๐26,250,000.
Exercise 2
1. Use the following information to answer question (i) โ (iv):
๐ด = 2 4 13 โ5 0
and ๐ต = 6 7 10 0 2
.
(i) Find ๐ด + ๐ต
(ii) Find ๐ด๐ต
(iii) ๐ด๐๐ต
(iv) ๐ด๐๐ต .
2. Evaluate 3 5 18 0 24 0 6
, โ10 15
2 2 ,
1 โ2 3โ4 5 67 8 โ9
.
3. Use the following information to answer question (a) โ (c):
2 34 5
๐ฅ๐ฆ =
2543
is written as ๐ด๐ = ๐ต.
(a) Write down ๐ด and ๐ต.
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(b) Find ๐ดโ1
(c) Find the matrix ๐.
(d) Find ๐ด โ ๐๐ผ
4. A woman invested different amounts at 8%, 83
4% and 9%, all at simple
interest. Altogether she invested N40,000 and earns N3,455 per year. How
much does she have invested at each rate if she has N4,000 more invested at
9% than 8%? Solve by using matrices.
5. A salesman has the following record of sales during three months for three
items ๐ด, ๐ต and ๐ถ which have different rates of commission.
Months Sales units Total commission drawn
(in ยฃ) A B C
May 90 100 20 800
June 130 50 40 900
July 60 100 30 850
Find out the rates of commission on the items A, B and C. Solve by Cramerโs
rule (determination method)
ยง 3.0 STATISTICAL DISTRIBUTION
3.1 BINOMIAL DISTRIBUTION
An experiment consisting of ๐ repeated trials such that
a) the trials are independent and identical
b) each trial result in only one or two possible outcomes
c) the probability of success ๐ remains constant
d) the random variable of interest is the total number of success.
The binomial distribution is one of the widely used in statistics and it used to
find the probability that an outcome would occur ๐ฅ times in ๐ performances of an
experiment and its probability density function is given by
๐ ๐ฅ; ๐, ๐ = ๐๐ฅ ๐๐ฅ 1 โ ๐ ๐โ๐ฅ , ๐ฅ = 0,1, โฆ , ๐
Where ๐ = Total number of trials
๐ = Probability of success
1 โ ๐ = Probability of failure
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๐ โ ๐ฅ = Number of failure in ๐ โ trials
To find the probability of ๐ฅ success in ๐ trials, the only values we need are that of ๐
and ๐.
Properties of Binomial Distribution
Mean ๐ = ๐๐
Variance ๐2 = ๐๐๐ ; ๐ = 1 โ ๐
Standard deviation ๐ = ๐๐๐
Note: ๐~๐ต(๐, ๐) reads as ๐ follows binomial distribution.
Examples:
1. Observation over a long period of time has shown that a particular sales man
can make a sale on a single contact with the probability of 20%. Suppose the
same person contact four prospects,
(a) What is the probability that exactly 2 prospects purchase the product?
(b) What is the probability that at least 2 prospects purchase the product?
(c) What is the expected value (mean) of the prospects that would purchase the
product?
Solution: Let ๐ denote the number of prospect.
Let ๐ denote the probability of (success) purchase = 0.2
Then ๐~๐ต(4,0.2).
๐ ๐ฅ = ๐๐ฅ ๐๐ฅ๐๐โ๐ฅ ๐ฅ = 0,1,2,3,4
= 0, ๐๐๐ ๐๐ค๐๐๐๐
๐ ๐ฅ = 4
๐ฅ 0.2 ๐ฅ 0.8 4โ๐ฅ
(a) ๐ ๐ฅ = 2 = 42 0.2 2 0.8 4โ2 = 0.1536
(b) ๐ ๐ฅ โฅ 2 = ๐ 2 + ๐ 3 + ๐(4) ( OR 1 โ ๐ ๐ฅ โค 1 = 1 โ {๐ 0 + ๐(1)})
= 0.1536 + 43 0.2 3 0.8 4โ3 + 4
4 0.2 4 0.8 4โ4
= 0.1536 + 0.0256 + 0.0016
= 0.1808
(c) Expected value = ๐๐
= 4 ร 0.2
= 0.8
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2. The probability that a student is accepted to a prestigious college is 0.3. If 5
students from the same school apply, what is the probability that at most 2 are
accepted?
Solution:
๐ ๐ฅ โค 2 = ๐ 0 + ๐ 1 + ๐(2)
= 5C0 0.3 0 0.7 5โ0 + 5C1 0.3 1 0.7 5โ1 + 5C2 0.3 2 0.7 5โ2
= 0.1681 + 0.3602 + 0.3087
= 0.837
3.2 POISSON DISTRIBUTION
When the size of the sample (๐) is very large and the probability of obtaining
success in any one trial very small, then Poisson distribution is adopted. Given an
interval of real numbers, assumed counts of occur at random throughout interval, if
the interval can be partition into sub interval of small enough length such that
a) The probability of more than one count sub interval is 0
b) The probability of one count in a sub interval is the same for all sub intervals
and proportional to the length of the sub interval
c) The count in each of the sub interval is independent of all other sub intervals.
A random experiment of this type is called a Poisson Process. If the mean
number of count in an interval is ๐ > 0, the random variable ๐ฅ that equals the number
of count in an interval has a Poisson distribution with parameter ๐ and the probability
density function is given by
๐ ๐ฅ; ๐ =๐๐ฅ๐โ๐
๐ฅ !; ๐ฅ = 0,1, โฆ , ๐.
For a Poisson distribution, the mean and the variance is given by ๐ธ ๐ฅ = ๐, ๐ ๐ฅ = ๐.
Poisson distribution probability density function can also be used to approximate
binomial distribution when ๐ is large (๐ โฅ 30) and ๐๐ < 5.
Examples
1. Attendance in a factory shows 7 absences. What is the probability that on a
given day there will be more than 8 people absent?
Solution
๐ ๐ > 8 = 1 โ ๐ ๐ โค 8
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= 1 โ ๐ 1 + ๐ 2 + ๐ 3 + ๐ 4 + ๐ 5 + ๐ 6 + ๐ 7 + ๐(8)
= 1 โ 71๐โ7
1!+
72๐โ7
2!+
73๐โ7
3!+
74๐โ7
4!+
75๐โ7
5!+
76๐โ7
6!+
77๐โ7
7!+
78๐โ7
8!
= 1 โ 0.0064 + 0.0223 + 0.0521 + 0.0912 + 0.1277 + 0.149
+0.149 + 0.1304
= 0.2709
2. An automatic production line breaks down every 2 hours. Special production
requires uninterrupted operation for 8 hours. What is the probability that this
can be achieved?
Solution ๐ =8
2= 4, ๐ฅ = 0 (no breaks down)
๐ ๐ฅ = 0 =40๐โ4
0! = 0.0183 = 1.83%
3.3 NORMAL DISTRIBUTION
The normal distribution is the most important and the most widely used
among all continuous distribution in the statistics. The graph of a normal distribution
is a bell โ shaped curved that extends indefinitely in both direction.
Normal curve
1 or 100%
๐
3.3.1 Features (Properties) of Normal Curve
1. The curve is symmetrical about the vertical axis through the mean ๐.
2. The mode is the highest point on the horizontal axis where the curve is
maximum and occurs where ๐ฅ = ๐.
3. The normal curve approaches the horizontal axis asymptotically.
4. The total area under the curve is one (1) or 100%.
It is clear from these properties that a knowledge of the population means and
standard deviation gives a complete picture of the distribution of all the values.
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Notation: Instead of saying that the values of a variable ๐ฅ are normally distributed
with mean ๐ and standard deviation ๐, we simply say that ๐ฅ has an ๐ ๐, ๐2 or ๐ฅ is
๐ ๐, ๐2 or ๐~๐ ๐, ๐2 .
A random variable ๐ฅ is said to have a normal distribution if its probability
density function is given by
๐ ๐ฅ; ๐, ๐ =1
2๐๐ ๐โ1
2 ๐ฅโ๐
๐
2
, โโ < ๐ฅ < โ
Where ๐ and ๐ are the parameters of the distribution (standard deviation and mean
respectively).
3.3.2 The Standard Normal Curve
The standard normal distribution is a special case. If ๐~๐ 0,1 , then ๐ฅ is
called the standard normal variable with probability density function ๐ ๐ง =
1
2๐๐โ๐ง2
2 , โโ < ๐ง < โ.
The table usually used to determine the probability that a random variable ๐ฅ
drawn from a normal population with no mean and standard deviation 1 is the
standard normal distribution table (or ๐ง- scores table).
We shall note the following point when using the table:
1. ๐ ๐ง โค 0 = 0.5
2. ๐ ๐ง โค โ๐ง = 1 โ ๐ ๐ง โค ๐ง
= ๐(๐ง โฅ ๐ง)
3. ๐ โ๐ง1 < ๐ง < ๐ง2 = ๐ ๐ง < ๐ง2 โ ๐ ๐ง > โ๐ง1
= ๐ ๐ง < ๐ง2 โ [1 โ ๐ ๐ง < ๐ง1 ]
= ๐ ๐ง < ๐ง2 + ๐ ๐ง < ๐ง1 โ 1
Example 1: Find the probability that a random variable having the standard normal
distribution will take a value
(a) Less than 1.53 (b) Less than โ0.82
(c) Between 1.25 and 2.34 (d) between โ0.35 and 1.26
Solution:
(a) ๐ ๐ง < 1.53 = 0.9370 (using the table)
(b) ๐ ๐ง < โ0.82 = 0.2061
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(c) ๐ 1.25 โค ๐ง โค 2.34 = ๐ ๐ง โค 2.34 โ ๐ ๐ง โฅ 1.25
= ๐ ๐ง โค 2.34 โ 1 โ ๐(๐ง < 1.25)
= 0.9904 โ 1 + 0.8944
= 0.8848
(d) ๐ โ0.35 โค ๐ง โค 1.26 = ๐ ๐ง โค 1.26 โ ๐ ๐ง โฅ โ0.35
= ๐ ๐ง โค 1.26 โ 1 โ ๐(๐ง < 0.35)
= ๐ ๐ง โค 1.26 + ๐ ๐ง < 0.35 โ 1
= 0.8962 + 0.6368 โ 1
= 0.533
3.3.3 STANDARDIZED NORMAL VARIABLE
In real world application, the given continuous random variable may have a
normal distribution in value of a mean and standard deviation different from 0 and 1.
To overcome this difficulty, we obtain a new variable denoted by ๐ง and this is given
by
๐ง =Value ๐ฅโMean
Standard deviation =
๐โ๐
๐ (for ๐ โ 0 and ๐ โ 1)
Example 2: If ๐~๐(510๐, 6.25๐), calculate ๐(๐ = 507.5๐)
Solution: ๐ง =๐โ๐
๐
=507.5โ510
2.5
= โ1
โด ๐ ๐ = 507.5๐ = ๐ ๐ง โค โ1
= 0.1587
Example 3: It is known from the previous examination results that the marks of
candidates have a normal distribution with mean 55 and standard deviation 10. If the
pass mark in a new examination is set at 40, what percentages of the candidates will
be expected to fail?
Solution: We have ๐ฅ~๐(55,100) and the required proportion of ๐ฅ โvalues that are
below 40 is ๐(๐ฅ < 40).
๐ ๐ฅ < 40 = ๐(๐ง <๐โ๐
๐)
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= ๐ ๐ง <40โ55
10
= ๐(๐ง โค โ1.5)
= 0.0668
โด 6.68% of the candidates are expected to fail.
Exercise 3
1. A surgery is successful for 75% patients. What is the probability of its success
in at least 7 cases out of randomly selected 9 patients?
2. If the 5% of the electric bulbs manufactured by a company are defective. Find
the probability that in the sample of 130 bulbs, at most 3 doors are defective.
a) Use binomial to solve the problem
b) Use Poisson distribution and compare your results.
3. It is known that the marks in a University direct entry examination are normally
distributed with mean 70 and standard deviation 8. Given that your score is
69, what percentage of all the candidates will be expected to score more than
you?
ยง 4.0 LIMIT AND CONTINUITYOF FUNCTIONS
4.1 Definition
A given function ๐(๐ฅ) is said to have a limit ๐ฟ as ๐ฅ approaches ๐ (in symbol
lim๐ฅโ๐ ๐(๐ฅ) = ๐ฟ) if and only if, ๐(๐ฅ) is as near to ๐ฟ as we please for all values of ๐ฅ โ ๐
but sufficiently near to ๐.
We say that ๐(๐ฅ) approaches the limit in real number as ๐ฅ approaches a fixed
number ๐, if and only if, lim๐ฅโ๐โ ๐(๐ฅ) = ๐ฟ = lim๐ฅโ๐+ ๐(๐ฅ). When this is so, we
write lim๐ฅโ๐ ๐(๐ฅ) = ๐ฟ. Alternatively, this simply means that ๐(๐ฅ) is arbitrary close to ๐ฟ
whenever the variable ๐ฅ is sufficiently near to the number ๐.
Remark: A function ๐(๐ฅ) cannot have more than one limit.
4.2 Rules for evaluating limits
If lim๐ฅโ๐ ๐(๐ฅ) and lim๐ฅโ๐ ๐(๐ฅ) exist, then the following rules hold.
(i) lim๐ฅโ๐ ๐ = ๐ where ๐ is a constant.
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(ii) lim๐ฅโ๐ ๐๐ ๐ฅ = ๐ lim๐ฅโ๐ ๐(๐ฅ) for any ๐ real number
(iii) lim๐ฅโ๐ ๐(๐ฅ) ยฑ ๐(๐ฅ) = lim๐ฅโ๐ ๐(๐ฅ) ยฑ lim๐ฅโ๐ ๐(๐ฅ)
(iv) lim๐ฅโ๐ ๐(๐ฅ)
๐(๐ฅ) =
lim ๐ฅโ๐ ๐(๐ฅ)
lim ๐ฅโ๐ ๐(๐ฅ) , provided lim๐ฅโ๐ ๐(๐ฅ) โ 0.
Example: Evaluate the following limits:
(1) lim๐ฅโ0(๐ฅ3 โ 5๐ฅ2 + 2) (2) lim๐ฅโ0 9+๐ฅโ3
๐ฅ (3) lim๐ฅโ2
๐ฅ2โ๐ฅโ2
๐ฅ2โ4 (4) lim๐ฅโโ
4๐ฅ3
๐ฅ3+3 .
Solution
(1) lim๐ฅโ0(๐ฅ3 โ 5๐ฅ2 + 2) = lim๐ฅโ0
๐ฅ3 โ lim๐ฅโ0
5๐ฅ2 + lim๐ฅโ0
2
= 0 2 โ 5 02 + 2
= 2
(2) lim๐ฅโ0 9+๐ฅโ3
๐ฅ= lim
๐ฅโ0
9+๐ฅโ3
๐ฅร
9+๐ฅ+3
9+๐ฅ+3
= lim๐ฅโ0
๐ฅ
๐ฅ 9+๐ฅ+3
= lim๐ฅโ0
1
9+๐ฅ+3
=1
9+3
=1
6
(3) lim๐ฅโ2๐ฅ2โ๐ฅโ2
๐ฅ2โ4= lim๐ฅโ2
๐ฅ+1 ๐ฅโ2
๐ฅ+2 ๐ฅโ2
= lim๐ฅโ2
๐ฅ+1
๐ฅ+2
=lim๐ฅโ2
(๐ฅ+1)
lim๐ฅโ2
(๐ฅ+2)
=3
4
(4) lim๐ฅโโ4๐ฅ3
๐ฅ3+3= lim
๐ฅโโ
4๐ฅ3
๐ฅ3
๐ฅ3
๐ฅ3+3
๐ฅ3
= lim๐ฅโโ
4
1+3
๐ฅ3
=lim๐ฅโโ
4
lim๐ฅโโ
1+lim3
๐ฅ3๐ฅโโ
= 4
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4.3 CONTINUITY OF FUNCTION
We say that the function ๐ is continuous at ๐ฅ = ๐ if lim๐ฅโ๐ ๐(๐ฅ) = ๐(๐). We say
that ๐ is continuous if it is continuous at each point of its domain.
Remark:
(1) The continuity definition requires that the following conditions be met if ๐ is to
be continuous at ๐ (a point): (a) ๐(๐) is defined as a finite real number, (b)
lim๐ฅโ๐โ ๐(๐ฅ) exists and equals ๐(๐), (c) lim๐ฅโ๐โ ๐(๐ฅ) = f(c) = lim๐ฅโ๐+ ๐(๐ฅ) .
When a function ๐ is not continuous at ๐, one or more, of these conditions are
not met.
(2) All polynomials, sin ๐ฅ, cos ๐ฅ , ๐๐ฅ , sinh ๐ฅ, cosh ๐ฅ, ๐๐ฅ , ๐ โ 1 are continuous for all
real values of ๐ฅ. All logarithmic functions, log๐ ๐ฅ , ๐ > 0, ๐ โ 1 are continuous
for all ๐ฅ > 0. Each rational function, ๐(๐ฅ)/๐(๐ฅ), is continuous where ๐(๐ฅ) โ 0.
Examples
1. Determine whether the function ๐ ๐ฅ = 3๐ฅ2 + 4๐ฅ โ 10 is continuous at ๐ฅ = 3
or not.
Solution:
๐ 3 = 3 32 + 4 3 โ 10 = 29.
lim๐ฅโ3 ๐ ๐ฅ = lim๐ฅโ3 3๐ฅ2 + lim๐ฅโ3 4๐ฅ โ lim๐ฅโ3 10
= 27 + 12 โ 10
= 29
โด lim๐ฅโ3
๐ ๐ฅ = ๐ 3 = 29
๐ ๐ฅ is continuous at ๐ฅ = 3.
2. Verify the continuity of the function ๐ ๐ฅ =๐ฅโ2
๐ฅ2โ5๐ฅ+6 at ๐ฅ = 2.
Solution:
๐ ๐ฅ =๐ฅโ2
๐ฅ2โ5๐ฅ+6=
๐ฅโ2
๐ฅโ2 ๐ฅโ3 =
1
๐ฅโ3
๐ 2 =1
2โ3= โ1
lim๐ฅโ2 ๐ ๐ฅ = lim๐ฅโ31
๐ฅโ3
=lim ๐ฅโ2 1
lim ๐ฅโ2 ๐ฅโ3
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=1
2โ3= โ1
โด lim๐ฅโ2
๐ ๐ฅ = ๐ 2 = โ1
๐ ๐ฅ is continuous at ๐ฅ = 2.
3. The function ๐ ๐ฅ =๐ฅ2โ๐ฅโ6
๐ฅโ3 is not continuous at ๐ฅ = 3 since ๐ 3 is not
defined.
Exercise 4
1. Evaluate each of the following limits.
(a) lim๐ฅโ1๐ฅ2โ1
๐ฅ3โ1
(b) lim๐ฅโโ๐ฅ2โ๐ฅ
๐ฅ3+2๐ฅ
(c) lim๐ฅโ4 ๐ฅโ2
๐ฅโ4
2. Is function ๐ ๐ฅ =2๐ฅ2+3
2๐ฅ+1 continuous at the point ๐ฅ = โ
1
2?
ยง 5.0 BASIC CONCEPT OF DIFFERENTIATION
Differentiation is a mathematical concept dealing with the rate of change. This
rate of change is generally termed the derivative, slope, gradient or marginal
measure in various areas of usage. That is, the process of finding the derivative of a
function is known as differentiation. Formally, differentiation (the differential calculus)
is the process of finding the derivative of a function. And it has a lot of applications in
many areas of human endeavours such as the field of engineering, science,
economics, business, and so on.
5.1 Determination of Derivative from First Principle
There is need to apply the limits theory to the incremental principle of the
variables involved in the function. For instance, let us consider the function ๐ฆ = ๐(๐ฅ).
A small increment in ๐ฅ (denoted by โ๐ฅ) will cause a small increment in ๐ฆ (also
denoted by โ๐ฆ). Then for
๐ฆ = ๐(๐ฅ) โฆโฆโฆโฆโฆโฆโฆโฆ. (i)
We have ๐ฆ + โ๐ฆ = ๐ ๐ฅ + โ๐ฅ โฆโฆโฆโฆ (ii)
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Subtract equation (i) from equation (ii), gives
๐ฆ + โ๐ฆ โ ๐ฆ = ๐ ๐ฅ + โ๐ฅ โ ๐(๐ฅ)
โ๐ฆ = ๐ ๐ฅ + โ๐ฅ โ ๐(๐ฅ)
Divide all through by โ๐ฅ, we have
ฮ๐ฆ
ฮ๐ฅ=
๐ ๐ฅ+โ๐ฅ โ๐(๐ฅ)
โ๐ฅ โฆโฆโฆโฆโฆ. (iii)
As โ๐ฅ โ 0, we have
limโ๐ฅโ0ฮ๐ฆ
ฮ๐ฅ= limโ๐ฅโ0
๐ ๐ฅ+โ๐ฅ โ๐(๐ฅ)
โ๐ฅ โฆ. (iv)
By the limitโs theory, we have
๐๐ฆ
๐๐ฅ= limโ๐ฅโ0
๐ ๐ฅ+โ๐ฅ โ๐(๐ฅ)
โ๐ฅ โฆโฆโฆ.. (v) (i.e.
๐๐ฆ
๐๐ฅโก limโ๐ฅโ0
ฮ๐ฆ
ฮ๐ฅ)
Note that ๐๐ฆ
๐๐ฅ is also represented by ๐ โฒ (๐ฅ) which is called the derivative of ๐(๐ฅ).
Others are ๐
๐๐ฅ ๐ ,
๐๐
๐๐ฅ, ๐ฆโฒ .
If the left hand side of the equation (v) exists, then ๐๐ฆ
๐๐ฅ is called the differential
coefficient of ๐ฆ with respect to ๐ฅ.
Example: Find the derivative of (i) ๐ฆ = ๐ฅ3 + 2 (ii) ๐ฆ = ๐๐ฅ๐ from the first principle.
Solution:
(i) ๐ฆ = ๐ฅ3 + 2 ------------- (*)
๐ฆ + โ๐ฆ = ๐ฅ + โ๐ฅ 3 + 2 -------------- (**)
Then subtracting equation (*) from equation (**), we have
๐ฆ + โ๐ฆ โ ๐ฆ = ๐ฅ + โ๐ฅ 3 + 2 โ ๐ฅ3 + 2
โ๐ฆ = ๐ฅ3 + 3๐ฅ2 โ๐ฅ + 3๐ฅ โ๐ฅ 2 + โ๐ฅ 3 + 2 โ ๐ฅ3 โ 2
โ๐ฆ = 3๐ฅ2 โ๐ฅ + 3๐ฅ โ๐ฅ 2 + โ๐ฅ 3
Divide through by โ๐ฅ, we have
ฮ๐ฆ
ฮ๐ฅ= 3๐ฅ2 + 3๐ฅ โ๐ฅ + โ๐ฅ 2
limฮ๐ฅโ0
ฮ๐ฆ
ฮ๐ฅ= lim
ฮ๐ฅโ0 3๐ฅ2 + 3๐ฅ โ๐ฅ + โ๐ฅ 2
โด๐๐ฆ
๐๐ฅ= 3๐ฅ2 + 0 + 0
= 3๐ฅ2
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(ii) Given ๐ฆ = ๐๐ฅ๐ .
Let ๐ฆ + โ๐ฆ = ๐ ๐ฅ + โ๐ฅ ๐
Then
๐ฆ + โ๐ฆ โ ๐ฆ = ๐ ๐ฅ + โ๐ฅ ๐ โ ๐๐ฅ๐
โ๐ฆ = ๐๐ฅ๐ 1 +โ๐ฅ
๐ฅ
๐
โ 1 Using 1 + ๐ฅ ๐ = 1 + ๐ ๐โ1 ๐โ2 โฆ(๐โ๐+1)
๐ !โ๐=1 ๐ฅ๐
โ๐ฆ = ๐๐ฅ๐ 1 + ๐ โ๐ฅ
๐ฅ +
๐(๐ โ 1)
2! โ๐ฅ
๐ฅ
2
+ โฏ โ 1
= ๐ ๐๐ฅ๐โ1 โ๐ฅ +๐(๐โ1)
2!๐ฅ๐โ2 โ๐ฅ 2 + โฏ
โดฮ๐ฆ
ฮ๐ฅ= ๐ ๐๐ฅ๐โ1 + terms containing higher power of โ๐ฅ
limฮ๐ฅโ0
ฮ๐ฆ
ฮ๐ฅ= lim
โ๐ฅโ0 ๐๐๐ฅ๐โ1 + terms containing higher power of โ๐ฅ
โด๐๐ฆ
๐๐ฅ= ๐ ๐๐ฅ๐โ1 โฆโฆโฆโฆโฆโฆโฆโฆ. (vi)
We note that equation (vi) is the derivative of ๐ = ๐๐๐ and it shall be applied as
a rule when the derivative from first principle is not required.
Example: (a) Obtain the derivative for the following. (i) ๐ฆ = ๐ฅ4 (ii) ๐ฆ = 4๐ฅ5.
(b) Obtain the value of the derivative for the following functions at ๐ฅ = 2: (i) ๐ฆ = 2๐ฅ3
(ii) ๐ฆ = 3๐ฅ6.
Solution:
(a) (i) ๐๐ฆ
๐๐ฅ= 4 ๐ฅ4โ1 = 4๐ฅ3
(ii) ๐๐ฆ
๐๐ฅ= 4 5 ๐ฅ5โ1 = 20๐ฅ4
(b) (i) ๐๐ฆ
๐๐ฅ= 6๐ฅ2
At ๐ฅ = 2, ๐๐ฆ
๐๐ฅ= 6 22 = 24
(ii) ๐๐ฆ
๐๐ฅ= 18๐ฅ5
At ๐ฅ = 2,๐๐ฆ
๐๐ฅ= 18 25 = 576
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5.2 Rules of Differentiation
The following rules shall be applied to various functions:
Rule 1: If a polynomial ๐ฆ = ๐ข ยฑ ๐ฃ ยฑ ๐ค ยฑ โฏ where ๐ข, ๐ฃ, ๐ค are functions of ๐ฅ, then
๐๐ฆ
๐๐ฅ=
๐
๐๐ฅ ๐ข ยฑ ๐ฃ ยฑ ๐ค ยฑ โฏ
=๐๐ข
๐๐ฅยฑ
๐๐ฃ
๐๐ฅยฑ
๐๐ค
๐๐ฅยฑ โฏ
Hence, the derivative of a sum is the of the derivatives and the derivative of a
difference is the difference of the derivatives.
Example: Find the derivative of each of the following functions.
(i) ๐ฆ = 2๐ฅ5 โ 3๐ฅ6 + 6๐ฅ2; (ii) ๐ ๐ฅ = 2๐ฅ3 + 3๐ฅ4 โ2
๐ฅ .
Solution:
(i) ๐๐ฆ
๐๐ฅ= 2 5 ๐ฅ5โ1 โ 3 6 ๐ฅ6โ1 + 6(2) ๐ฅ2โ1
= 10๐ฅ4 โ 18๐ฅ5 + 12๐ฅ
(ii) ๐๐
๐๐ฅ= 2 3 ๐ฅ3โ1 + 3 4 ๐ฅ4โ1 โ 2 โ1 ๐ฅโ1โ1
= 6๐ฅ2 + 12๐ฅ3 +2
๐ฅ2 .
Rule 2: If ๐ฆ = ๐ถ, where ๐ถ is a constant, then ๐๐ฆ
๐๐ฅ= 0. E.g. (a) ๐ฆ = 2,
๐๐ฆ
๐๐ฅ= 0.
(b) ๐ฆ = ๐2 , ๐๐ฆ
๐๐ฅ= 0 since ๐ is a constant and has no degree of ๐ฅ from 1 and above.
Rule 3: If a product ๐ฆ = ๐ข๐ฃ, where ๐ข and ๐ฃ are functions of ๐ฅ, then
๐๐ฆ
๐๐ฅ= ๐ฃ
๐๐ข
๐๐ฅ+ ๐ข
๐๐ฃ
๐๐ฅ .
By extension, if ๐ฆ = ๐ข๐ฃ๐ค, where ๐ข, ๐ฃ and ๐ค are functions of ๐ฅ, then
๐๐ฆ
๐๐ฅ= ๐ฃ๐ค
๐๐ข
๐๐ฅ+ ๐ข๐ค
๐๐ฃ
๐๐ฅ+ ๐ฃ๐ข
๐๐ค
๐๐ฅ .
Example: Find the derivative of the following functions:
(a) ๐ฆ = 2๐ฅ + 2 3๐ฅ2 + 2๐ฅ (b) ๐ฆ = 2๐ฅ + 4 ๐ฅ2 + 3๐ฅ 2๐ฅ3 + 5๐ฅ .
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Solution:
(a) Let ๐ข = 2๐ฅ + 2, ๐๐ข
๐๐ฅ= 2
๐ฃ = 3๐ฅ2 + 2๐ฅ, ๐๐ฃ
๐๐ฅ= 6๐ฅ + 2 .
By product rule,
๐๐ฆ
๐๐ฅ= ๐ฃ
๐๐ข
๐๐ฅ+ ๐ข
๐๐ฃ
๐๐ฅ
= 3๐ฅ2 + 2๐ฅ 2 + 2๐ฅ + 2 6๐ฅ + 2
= 6๐ฅ2 + 4๐ฅ + 12๐ฅ2 + 4๐ฅ + 12๐ฅ + 4
= 18๐ฅ2 + 20๐ฅ + 4 .
(b) Let ๐ข = 2๐ฅ + 4, ๐๐ข
๐๐ฅ= 2
๐ฃ = ๐ฅ2 + 3๐ฅ, ๐๐ฃ
๐๐ฅ= 2๐ฅ + 3
๐ค = 2๐ฅ3 + 5๐ฅ, ๐๐ค
๐๐ฅ= 6๐ฅ2 + 5
But ๐๐ฆ
๐๐ฅ= ๐ฃ๐ค
๐๐ข
๐๐ฅ+ ๐ข๐ค
๐๐ฃ
๐๐ฅ+ ๐ฃ๐ข
๐๐ค
๐๐ฅ
= ๐ฅ2 + 3๐ฅ 2๐ฅ3 + 5๐ฅ 2 + 2๐ฅ + 4 2๐ฅ3 + 5๐ฅ 2๐ฅ + 3
+ ๐ฅ2 + 3๐ฅ 2๐ฅ + 4 6๐ฅ2 + 5
Rule 4: If quotient ๐ฆ =๐ข
๐ฃ, where ๐ข and ๐ฃ are functions of ๐ฅ, then
๐๐ฆ
๐๐ฅ=
๐ฃ ๐๐ข
๐๐ฅ โ ๐ข
๐๐ฃ
๐๐ฅ
๐ฃ2 .
Example: Differentiate the following with respect to ๐ฅ.
(i) ๐ฆ =1+๐ฅ2
3๐ฅ (ii) ๐ฆ =
2+3๐ฅ
๐ฅ2+3๐ฅ+5
Solution:
(i) Let ๐ข = 1 + ๐ฅ2 , ๐๐ข
๐๐ฅ= 2๐ฅ
๐ฃ = 3๐ฅ, ๐๐ฃ
๐๐ฅ= 3
By Quotient Rule:
๐๐ฆ
๐๐ฅ=
๐ฃ ๐๐ข
๐๐ฅ โ ๐ข
๐๐ฃ
๐๐ฅ
๐ฃ2
=3๐ฅ 2๐ฅ โ 1+๐ฅ2 (3)
3๐ฅ 2
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=6๐ฅ2โ3โ3๐ฅ2
9๐ฅ2
=๐ฅ2โ1
3๐ฅ2 .
(ii) Let ๐ข = 2 + 3๐ฅ, ๐๐ข
๐๐ฅ= 3
๐ฃ = ๐ฅ2 + 3๐ฅ + 5, ๐๐ฃ
๐๐ฅ= 2๐ฅ + 3
But
๐๐ฆ
๐๐ฅ=
๐ฃ ๐๐ข
๐๐ฅ โ ๐ข
๐๐ฃ
๐๐ฅ
๐ฃ2
= ๐ฅ2+3๐ฅ+5 (3)โ 2+3๐ฅ (2๐ฅ+3)
๐ฅ2+3๐ฅ+5 2
=3๐ฅ2+9๐ฅ+15โ4๐ฅโ6โ6๐ฅ2โ9๐ฅ
๐ฅ2+3๐ฅ+5 2
=โ3๐ฅ2โ4๐ฅ+9
๐ฅ2+3๐ฅ+5 2 .
Rule 5: If a function of function ๐ฆ = ๐๐ฅ + ๐ ๐ where ๐ and ๐ are constants and ๐ is
the index, then ๐๐ฆ
๐๐ฅ=
๐
๐๐ฅ ๐๐ฅ + ๐ ๐ = ๐ ๐๐ฅ + ๐ ๐โ1
๐
๐๐ฅ(๐๐ฅ + ๐) .
This rule is also called the composite or chain rule. In another way, this can
be written as: ๐๐ฆ
๐๐ฅ=
๐๐ฆ
๐๐ข
๐๐ข
๐๐ฅ
Where ๐ฆ is a function of ๐ข and ๐ข is a function of ๐ฅ. It is called function of a
function.
Example: Differentiate the following: (i) ๐ฆ = ๐ฅ + 3 4 (ii) ๐ฆ = 3๐ฅ2 โ 5 6 .
Solution:
(i) Let ๐ข = ๐ฅ + 3; ๐๐ข
๐๐ฅ= 1 then ๐ฆ = ๐ข4 ,
๐๐ฆ
๐๐ข= 4๐ข3 .
โด๐๐ฆ
๐๐ฅ=
๐๐ฆ
๐๐ขร
๐๐ข
๐๐ฅ
= 4๐ข3(1)
= 4๐ข3
= 4 ๐ฅ + 3 3 .
(ii) Let ๐ข = 3๐ฅ2 โ 5; ๐๐ข
๐๐ฅ= 6๐ฅ then ๐ฆ = ๐ข6 ,
๐๐ฆ
๐๐ข= 6๐ข5 .
โด๐๐ฆ
๐๐ฅ=
๐๐ฆ
๐๐ขร
๐๐ข
๐๐ฅ
= 6๐ข5 (6๐ฅ)
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= 36๐ฅ๐ข5
= 36๐ฅ 3๐ฅ2 โ 5 5 .
Rule 6: For the implicit function, the relationship between ๐ฆ and ๐ฅ variables may not
be expressed explicitly (e.g. ๐ฅ๐ฆ2, ๐ฅ๐ฆ + ๐ฅ2 = 3๐ฅ2๐ฆ5). In this case, treat ๐ฆ as if
it is a function of ๐ฅ and the rules of differentiation discussed above are
applied as appropriate.
Example: Differentiate the following implicitly:
(i) ๐ฅ3 + ๐ฆ4 = 7 (ii) 6๐ฆ2๐ฅ โ 4๐ฅ2๐ฆ3 + 2๐ฆ = 0
Solution:
(i) Differentiate the function term by term with respect to ๐ฅ
3๐ฅ2 ๐๐ฅ
๐๐ฅ + 4๐ฆ3
๐๐ฆ
๐๐ฅ= 0
3๐ฅ2 + 4๐ฆ3 ๐๐ฆ
๐๐ฅ= 0
โ 4๐ฆ3 ๐๐ฆ
๐๐ฅ= โ3๐ฅ2
โด๐๐ฆ
๐๐ฅ=
โ3๐ฅ2
4๐ฆ3 .
(ii) 6 ๐ฆ2 ๐๐ฅ
๐๐ฅ+ 2๐ฆ๐ฅ
๐๐ฆ
๐๐ฅ โ 4 2๐ฅ
๐๐ฅ
๐๐ฅ๐ฆ3 + 3๐ฅ2๐ฆ2 ๐๐ฆ
๐๐ฅ + 2
๐๐ฆ
๐๐ฅ= 0
6๐ฆ2 + 12๐ฅ๐ฆ๐๐ฆ
๐๐ฅโ 8๐ฅ๐ฆ3 โ 12๐ฅ2๐ฆ2
๐๐ฆ
๐๐ฅ+ 2
๐๐ฆ
๐๐ฅ= 0
โ 12๐ฅ๐ฆ โ 12๐ฅ2๐ฆ2 + 2 ๐๐ฆ
๐๐ฅ= 8๐ฅ๐ฆ3 โ 6๐ฆ2
โด๐๐ฆ
๐๐ฅ=
8๐ฅ๐ฆ3 โ 6๐ฆ2
12๐ฅ๐ฆ โ 12๐ฅ2๐ฆ2 + 2
=4๐ฅ๐ฆ3โ3๐ฆ2
6๐ฅ๐ฆโ6๐ฅ2๐ฆ2+1 .
Rule 7: If the exponential function ๐ฆ = ๐๐๐ฅ +๐ , where ๐ and ๐ are constants, then
๐๐ฆ
๐๐ฅ=
๐
๐๐ฅ ๐๐๐ฅ+๐
=๐
๐๐ฅ ๐๐ฅ + ๐ ๐๐๐ฅ +๐
= ๐๐๐๐ฅ +๐
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Example: Differentiate the following with respect to ๐ฅ. (i) ๐ฆ = 6๐2๐ฅ2+๐ฅ (ii) ๐ฆ = ๐ฅ3๐๐ฅ3.
Solution:
(i) ๐๐ฆ
๐๐ฅ=
๐
๐๐ฅ 2๐ฅ2 + ๐ฅ โ 6๐2๐ฅ2+๐ฅ
= 4๐ฅ + 1 โ 6๐2๐ฅ2+๐ฅ
= 6 4๐ฅ + 1 ๐2๐ฅ2+๐ฅ .
(ii) Applying product rule:
Let ๐ข = ๐ฅ3; ๐๐ข
๐๐ฅ= 3๐ฅ2
๐ฃ = ๐๐ฅ3;
๐๐ฃ
๐๐ฅ= 3๐ฅ2๐๐ฅ3
But ๐๐ฆ
๐๐ฅ= ๐ฃ
๐๐ข
๐๐ฅ+ ๐ข
๐๐ฃ
๐๐ฅ
= ๐๐ฅ3 3๐ฅ2 + ๐ฅ3 3๐ฅ2๐๐ฅ3
= 3๐ฅ2๐๐ฅ3+ 3๐ฅ5๐๐ฅ3
= ๐๐ฅ3 3๐ฅ2 + 3๐ฅ5 .
Rule 8: If the logarithmic function ๐ฆ = log๐ ๐๐ฅ + ๐ = ln ๐๐ฅ + ๐ where ๐ and ๐ are
constants, then ๐๐ฆ
๐๐ฅ=
1
๐๐ฅ +๐โ
๐
๐๐ฅ ๐๐ฅ + ๐ =
๐
๐๐ฅ +๐ .
Example: Differentiate the following with respect to ๐ฅ.
(i) ๐ฆ = ln ๐ฅ (ii) ๐ฆ = log๐ 5๐ฅ + 2 (iii) ๐ฆ = ln 6๐ฅ โ 1 2 .
Solution:
(i) ๐๐ฆ
๐๐ฅ=
1
๐ฅ
(ii) ๐๐ฆ
๐๐ฅ=
1
5๐ฅ+2โ
๐
๐๐ฅ 5๐ฅ + 2
=5
5๐ฅ+2
(iii) ๐๐ฆ
๐๐ฅ=
1
6๐ฅโ1 2 โ๐
๐๐ฅ 6๐ฅ โ 1 2
=1
6๐ฅโ1 2 6 2 (6๐ฅ โ 1)
=12
6๐ฅโ1 .
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Rule 9: Differentiation of Trigonometric functions:
(a) If ๐ฆ = sin ๐ฅ then ๐๐ฆ
๐๐ฅ= cos ๐ฅ
(b) If ๐ฆ = cos ๐ฅ then ๐๐ฆ
๐๐ฅ= โ sin ๐ฅ
(c) If ๐ฆ = tan ๐ฅ then ๐๐ฆ
๐๐ฅ= sec2 ๐ฅ
Examples:
1. ๐ฆ = cos 7๐ฅ , ๐๐ฆ
๐๐ฅ= โ7 sin 7๐ฅ
2. ๐ฆ = sin 3๐ฅ โ 4 , ๐๐ฆ
๐๐ฅ= 3 cos 3๐ฅ โ 4
3. ๐ฆ = cos ๐
4โ 2๐ฅ ,
๐๐ฆ
๐๐ฅ= 2 sin
๐
4โ 2๐ฅ
5.3 SECOND ORDER DERIVATIVE
Second order derivative is one of higher derivatives where successive
differentiations are carried out. The process of differentiating a function more that
one is called successive differentiation.
If ๐ฆ = ๐ ๐ฅ ,๐๐ฆ
๐๐ฅ is also a function of ๐ฅ. The derivative of
๐๐ฆ
๐๐ฅ with respect to ๐ฅ
๐
๐๐ฅ
๐๐ฆ
๐๐ฅ . The expression
๐
๐๐ฅ
๐๐ฆ
๐๐ฅ is called the second order derivative of ๐ฆ with respect
to ๐ฅ and it is denoted by ๐2๐ฆ
๐๐ฅ2 or ๐ฆโฒโฒ or ๐ โฒโฒ (๐ฅ). It is useful when determining the turning
point of a function.
Example: Find the first, second and third derivatives of the following.
(i) ๐ฆ = 4๐ฅ5 (ii) ๐ฆ = 3๐ฅ6 โ 2๐ฅ5 + ๐๐ฅ + 3๐ฅ2 โ 8 .
Solution:
(i) ๐๐ฆ
๐๐ฅ= 20๐ฅ4
๐2๐ฆ
๐๐ฅ2= 80๐ฅ3
๐3๐ฆ
๐๐ฅ3 = 240๐ฅ2 .
(ii) ๐๐ฆ
๐๐ฅ= 18๐ฅ5 โ 10๐ฅ4 + ๐๐ฅ + 6๐ฅ
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๐2๐ฆ
๐๐ฅ2= 90๐ฅ4 โ 40๐ฅ3 + ๐๐ฅ + 6
๐3๐ฆ
๐๐ฅ3 = 360๐ฅ3 โ 120๐ฅ2 + ๐๐ฅ .
5.4 APPLICATION OF DIFFERENTIATION
5.4.1 Application to maximum and minimum values
The concept of maximum and minimum values is termed the zero slope
analysis. The focus here is to find a point where the slope of the function to be
optimized is zero. It should be noted that the concept is not restricted to plotting of
curve alone, but more importantly to determine the maximum and minimum values of
a function such as maximizing profit or minimizing cost, etc.
We have the following procedures for the determination of maximum and
minimum values of a function ๐ฆ = ๐(๐ฅ) if it exists:
Step 1: Obtain the first derivative and set it equal to zero i.e. ๐๐ฆ
๐๐ฅ= 0. This occur at
turning point.
Step 2: From the ๐๐ฆ
๐๐ฅ= 0, determine the stationary point.
Step 3: Compute ๐2๐ฆ
๐๐ฅ2 at these stationary points.
If ๐2๐ฆ
๐๐ฅ2 < 0, the stationary point is maximum.
If ๐2๐ฆ
๐๐ฅ2 > 0, the stationary point is minimum.
If ๐2๐ฆ
๐๐ฅ2 = 0, (that is point of inflection) higher derivative (than second) can be
used to decide.
Example 1: ADAS company planning to have a new product in the market came up
with a total sales function ๐ = โ1000๐2 + 10000๐ and the total cost function
๐ถ = โ2000๐ + 2500, where ๐ and ๐ถ are respectively the price and cost (in US dollar)
of the new product. Find the optimal price for the new product and the maximum
profit expected for the company.
Solution: The profit (๐) is obtained by the difference in total sales and total cost.
โด ๐ = ๐ โ ๐ถ
= โ1000๐2 + 10000๐ โ โ2000๐ + 2500
= โ1000๐2 + 10000๐ + 2000๐ โ 2500
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= โ1000๐2 + 12000๐ โ 2500
๐๐
๐๐= โ2000๐ + 12000
But at turning point, ๐๐
๐๐= 0, then
0 = โ2000๐ + 12000
โ ๐ =12000
2000= 6.
But ๐2๐
๐๐2= โ2000 (which is less than zero, indicating maximum)
Hence, ๐ = 6, gives a maximum point. Therefore, the optimal price is $6. And the
maximum profit ๐ = โ1000 6 2 + 12000(6) โ 2500
= $33500 .
Example 2: ๐ articles are produced at a total cost of N 2๐2 + 30๐ + 20 , and each
one is sold for N ๐
3+ 100 . Determine the value of ๐ which gives the greatest profit,
and find this profit.
Solution: Total cost of articles = N 2๐2 + 30๐ + 20
Total money received for ๐ articles sold = N๐ ๐
3+ 100 .
Then the profit N๐ = N ๐ ๐
3+ 100 โ 2๐2 + 30๐ + 20 .
So, ๐ = 70๐ โ5
3๐2 โ 20.
The minimum value of ๐ will be given by ๐๐
๐๐= 0
๐๐
๐๐= 70 โ
10
3๐ = 0 โ ๐ = 21.
๐2๐
๐๐2= โ
10
3< 0.
Therefore, the maximum profit = N 70 ร 21 โ5
3 21 2
= N715 .
5.4.2 Application to Marginal Cost and Revenue
For the cost analysis (that is, relationship between average and marginal cost)
Let ๐ = Total cost
๐ = Quantity demand
Then, the Average Cost ๐ด๐ถ =๐
๐
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And the Marginal Cost ๐๐ถ =๐๐
๐๐, i.e.
๐
๐๐ ๐
Also the slope of average cost is obtained by:
๐
๐๐
๐
๐ =
๐ ๐๐
๐๐ โ๐
๐๐
๐๐
๐2 (Quotient Rule)
=
๐๐
๐๐โ
๐
๐
๐
=1
๐
๐๐
๐๐โ
๐
๐
=1
๐๐๐ถ โ
1
๐๐ด๐ถ
It is established that if the cost curve is U โ shape and the ๐ด๐ถ has a sloping
downward curve, then ๐
๐๐
๐
๐ < 0 which means that ๐๐ถ < ๐ด๐ถ. When ๐ด๐ถ decreases
๐๐ถ < ๐ด๐ถ and on the lowest point of ๐ด๐ถ curve, the tangent will be horizontal.
Therefore ๐
๐๐
๐
๐ = 0 i.e. ๐๐ถ = ๐ด๐ถ. Also when ๐ด๐ถ increasing, ๐๐ถ > ๐ด๐ถ and
๐
๐๐
๐
๐ > 0.
Relationship between Average and Marginal revenue
Let Total revenue = ๐ and we know that ๐ = ๐๐๐๐๐ ร ๐๐ข๐๐๐ก๐๐ก๐ฆ i.e. ๐ = ๐๐.
Then, Average Revenue ๐ด๐ =๐
๐
And, the Marginal Revenue ๐๐ =๐
๐๐ ๐ = ๐ + ๐
๐๐
๐๐. (Product Rule)
Example 1: Determine the minimum average cost if the cost function is given by:
๐ = 36๐ โ 20๐2 + 4๐3. Also obtain the marginal cost at the point of minimum
average cost.
Solution:
Given ๐ = 36๐ โ 10๐2 + 2๐3
โด ๐ด๐ถ =๐
๐= 36 โ 10๐ + 2๐2
For maximum or minimum, ๐
๐๐
๐
๐ = 0
i.e. โ10 + 4๐ = 0
โ ๐ =5
2.
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More so, ๐2
๐๐2 ๐
๐ at ๐ = 5
2 = 4, which is greater than zero.
Hence, ๐ =5
2 is at minimum point.
โด The Minimum Average Cost = 36 โ 10 5
2 + 2
5
2
2
= 23.5
For the Marginal Cost (๐๐ถ), we have
๐๐ถ =๐๐
๐๐= 36 โ 20๐ + 6๐2
๐๐ถ at ๐ =5
2 is equal to 36 โ 20
5
2 + 6
5
2
2
= 23.5 .
Example 2: Find the maximum profit of a company with revenue function
๐ = 200๐ โ 2๐2 and the cost function ๐ = 2๐3 โ 57๐2.
Solution:
Profit ๐ = ๐ โ ๐
= 200๐ + 55๐2 โ 2๐3
At turning point, ๐๐
๐๐= 0
๐๐
๐๐= 200 + 110๐ โ 6๐2 = 0
โ 100 + 55๐ โ 3๐2 = 0
3๐2 โ 55๐ โ 100 = 0
3๐2 โ 60๐ + 5๐ โ 100 = 0
3๐ ๐ โ 20 + 5 ๐ โ 20 = 0
3๐ + 5 ๐ โ 20 = 0
โ ๐ = โ5
3 or 20
But ๐2๐
๐๐2= 110 โ 12๐
At ๐ = โ5
3,
๐2๐
๐๐2 = 110 โ 12 5
3 = 90 > 0 indicating minimum point.
At ๐ = 20, ๐2๐
๐๐2 = 110 โ 12 20 = โ130 < 0 indicating maximum point.
Hence, at ๐ = 20, we have the maximum profit.
โด The maximum profit = 200 20 + 55 20 2 โ 2 20 3
= 10,000
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5.4.3 Application of differentiation to the elasticity of demand
The price elasticity of demand is the rate of change in response to quantity
demanded to the change in price. Let ๐ = ๐(๐) be the demand function, where ๐ is
the demand and ๐ is the price. Then the elasticity of demand is
๐๐ = โ๐
๐โ
๐๐
๐๐ When ๐๐ > 1, the demand is elastic and, when ๐๐ < 1, the demand is
inelastic.
For elasticity of supply, Let ๐ฅ = ๐ ๐ be the supply function, where ๐ฅ is the
supply and ๐ is the price. The elasticity of supply is defined as ๐๐ =๐
๐ฅโ
๐๐ฅ
๐๐ .
Example: Given a demand function ๐ท = 60 โ ๐ โ ๐2 , determine its elasticity of
demand when ๐ = 4
Solution: ๐๐ท
๐๐= โ1 โ 2๐
But ๐๐ = โ๐
๐ท
๐๐ท
๐๐
= โ๐
60โ๐โ๐2 โ1 โ 2๐
=2๐2+๐
60โ๐โ๐2 .
When ๐ = 4 we get
๐๐ = 0.9
Exercise 5
1. A firm produces ๐ฅ tonnes of output at a total cost ๐ถ =1
10๐ฅ3 โ 5๐ฅ2 + 10๐ฅ โ 32.
At what level of output will the marginal cost and the average variable cost
attain their respective minimum?
2. A certain manufacturing company has total cost function
๐ถ = 15 + 9๐ฅ โ 6๐ฅ2 + ๐ฅ3 . Find ๐ฅ, when the total cost is minimum.
3. The relationship between profit ๐ and advertising cost ๐ฅ is given by
๐ =4000๐ฅ
500+๐ฅโ ๐ฅ. Find ๐ฅ which maximizes ๐.
4. The total cost and total revenue of a firm are given by
๐ถ = ๐ฅ3 โ 12๐ฅ2 + 48๐ฅ + 11 and ๐ = 83๐ฅ โ 4๐ฅ2 โ 21.
Find the output (i) when the revenue is maximum (ii) when profit is maximum.
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5. A GSM company has a profit of N2 per SIM when the number of SIM in the
exchanges is not over 10,000. The profit per SIM decreases by 0.01 kobo for
each SIM over 10,000. What is the maximum profit?
6. The total cost function of a firm is ๐ถ =1
3๐ฅ3 โ 5๐ฅ2 + 28๐ฅ + 10 where ๐ฅ, is the
output. A tax at N2 per unit of output is imposed and the producer adds it to
his cost. If the market demand function is given by ๐ = 2530 โ 5๐ฅ, where N๐
is the price per unit of output. Determine the profit maximizing output and
price.
7. Investigate the maxima and minima of the function 2๐ฅ3 โ 3๐ฅ2 โ 36๐ฅ + 10 .
8. For the cost function ๐ถ = 2000 + 1800๐ฅ โ 75๐ฅ2 + ๐ฅ3 find when the total cost
(๐ถ) is increasing and when it is decreasing.
9. Find the equation of the tangent and normal to the demand curve
๐ฆ = 10 โ 3๐ฅ2 at (1, 7).
10. ADAS produces ๐ฅ tonnes of output at a total cost
๐ถ ๐ฅ = 1
10๐ฅ3 โ 4๐ฅ2 + 20๐ฅ + 5 . Find (i) Average Cost (ii) Average Variable
Cost (iii) Average Fixed Cost (iv) Marginal Cost and (v) Marginal Average
Cost.
11. The total cost ๐ of making ๐ units of product is
๐ = 0.00005๐3 โ 0.06๐2 + 10๐ + 79020. Find the marginal cost at 1000 units
of output.
12. Find the elasticity of the following functions.
(a) ๐ฅ = 2๐2 + 8๐ + 10
(b) ๐ฆ = 4๐ฅ โ 8 when ๐ฅ = 5.
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ยง 6.0 INTEGRAL CALCULUS
Integration is the reverse of differentiation. If ๐น(๐ฅ) is a function whose
derivative ๐นโฒ ๐ฅ = ๐(๐ฅ), then ๐น(๐ฅ) is called an integral of ๐(๐ฅ). For example,
๐น ๐ฅ = ๐ฅ3 is an integral of ๐ ๐ฅ = 3๐ฅ2 since ๐นโฒ ๐ฅ = 3๐ฅ2 = ๐(๐ฅ).
6.1 Indefinite Integral
Let ๐ฆ = ๐(๐ฅ) be a function of ๐ฅ. Suppose that ๐๐ฆ
๐๐ฅ= ๐(๐ฅ) where ๐(๐ฅ) is a
known function of ๐ฅ. Then ๐ฆ = ๐(๐ฅ) ๐๐ฅ is called an indefinite integral of ๐(๐ฅ) with
respect to ๐ฅ.
Remark
1. Let ๐ be an arbitrary constant. Since ๐
๐๐ฅ ๐ = 0, it follows that ๐ = 0 ๐๐ฅ .
2. If ๐(๐ฅ) is an indefinite integral of a function ๐(๐ฅ) and ๐ is an arbitrary constant,
then all positive indefinite integrals of ๐(๐ฅ) are of the form
๐(๐ฅ) ๐๐ฅ = ๐ ๐ฅ + ๐. In this case, ๐ is called a constant of integration.
For all rational values of ๐ except when ๐ = โ1, ๐
๐๐ฅ ๐ฅ๐+1
๐+1 =
1
๐+1โ ๐ + 1 ๐ฅ๐ .
Therefore, if ๐ โ โ1, the indefinite integral of ๐ฅ๐ is ๐ฅ๐ ๐๐ฅ =๐ฅ๐+1
๐+1+ ๐ . Note also that
constant different from zero, the integral is obtained by simply multiply the constant
by the independent variable and add the constant of integration. E.g. 2 ๐๐ฅ = 2๐ฅ + ๐.
Example
1. If ๐๐ฆ
๐๐ฅ= 4๐ฅ2, find ๐ฆ.
Solution
๐๐ฆ = 4๐ฅ2๐๐ฅ
๐๐ฆ = 4๐ฅ2 ๐๐ฅ
โ ๐ฆ =4๐ฅ2+1
2+1+ ๐ =
4
3๐ฅ3 + ๐
2. Integrate 6๐ก3 + 6 with respect to ๐ก.
Solution
6๐ก3 + 6 ๐๐ก =6
4๐ก4 + 6๐ก + ๐
=3
2๐ก4 + 6๐ก + ๐ .
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6.2 Some Standard form of Integration
(i) ๐๐ฅ ๐๐ฅ = ๐๐ฅ + ๐
(ii) ๐๐ฅ
๐ฅ= log๐ ๐ฅ + ๐
(iii) sin ๐ฅ ๐๐ฅ = โ cos ๐ฅ + ๐
(iv) cos ๐ฅ ๐๐ฅ = sin ๐ฅ + ๐
(v) ๐๐ฅ
๐ฅ2+๐2 =1
๐tanโ1 ๐ฅ
๐+ ๐
(vi) ๐๐ฅ
๐2โ๐ฅ2= sinโ1 ๐ฅ
๐+ ๐
Example: Integrate the following with respect to ๐ฅ.
(a) ๐๐ฅ
16โ๐ฅ2=
๐๐ฅ
42โ๐ฅ2= sinโ1 ๐ฅ
4+ ๐
(b) ๐๐ฅ
๐ฅ2+25=
1
5tanโ1 ๐ฅ
5+ ๐
6.3 Rules of Integration
Rule 1: The integration of a sum (difference) of a finite number of functions is the
sum (difference) of their separate integrals. E.g.
๐ฅ2 โ ๐ฅ + 5 ๐๐ฅ = ๐ฅ2 ๐๐ฅ โ ๐ฅ ๐๐ฅ + 5 ๐๐ฅ
=๐ฅ3
3โ
๐ฅ2
2+ 5๐ฅ + ๐ .
Rule 2: A constant factor may be brought outside the integral sign. E.g.
18๐ฅ3 ๐๐ฅ = 18 ๐ฅ3 ๐๐ฅ = 18 ๐ฅ4
4 + ๐ =
9
2๐ฅ4 + ๐ .
Rule 3: The addition of a constant to the variable makes no difference to the form of
the result. E.g.
(i) ๐๐ฅ
๐ฅโ3 2+32=
1
3tanโ1 ๐ฅโ3
3+ ๐
(ii) ๐๐ฅ
๐ฅ+7= log๐ ๐ฅ + 6 + ๐
Rule 4: Multiplying the variables by the constant also makes no difference to the
form of the result, but we have to divide by the constant. (Note integration by
substitution can be used to verify this claim)
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Examples
1. cos 3๐ฅ ๐๐ฅ =1
3sin 3๐ฅ + ๐
2. ๐4๐ฅ ๐๐ฅ =1
4๐4๐ฅ + ๐
3. ๐2๐ฅโ5 ๐๐ฅ =1
2๐2๐ฅโ5 + ๐
4. 2๐ฅ + 3 12 ๐๐ฅ =
1
2 2๐ฅ+3
32
32
+ ๐ =1
3 2๐ฅ + 3
32 + ๐
Rule 5: The integration of a fraction whose numerator is derivative of the
denominator is obtained by the logarithm of the denominator. i.e.
๐ โฒ ๐ฅ
๐ ๐ฅ ๐๐ฅ = log๐ ๐ ๐ฅ + ๐
Examples
(a) 2๐ฅ
๐ฅ2+12๐๐ฅ = ln ๐ฅ2 + 12 + ๐
(b) ๐ฅ2
๐ฅ3+2๐๐ฅ =
1
3
3๐ฅ2
๐ฅ3+2๐๐ฅ =
1
3ln ๐ฅ3 + 2 + ๐
(c) ๐ฅ๐๐ฅ2
๐๐ฅ2+3
๐๐ฅ =1
2
2๐ฅ๐๐ฅ2
๐๐ฅ2+3
๐๐ฅ =1
2ln ๐๐ฅ2
+ 3 + ๐
(d) ๐๐ฅ
๐ฅ log ๐ ๐ฅ=
1
๐ฅ๐๐ฅ
log ๐ ๐ฅ= log๐ log๐ ๐ฅ + ๐
Rule 6: (Integration by Substitution)
Consider the indefinite integral ๐(๐ฅ) ๐๐ฅ of an arbitrary function ๐ฆ = ๐(๐ฅ). Suppose
that ๐(๐ฅ) can be written as ๐ ๐ฅ = ๐(๐ข)๐๐ข
๐๐ฅ for some function ๐(๐ข) of suitably chosen
variable ๐ข = ๐ข(๐ฅ), then ๐(๐ฅ) ๐๐ฅ = ๐(๐ข) ๐๐ข. That is, we get ๐(๐ฅ) ๐๐ฅ by
evaluating ๐(๐ข) ๐๐ข as a function of ๐ข and then substituting ๐ข = ๐ข(๐ฅ) to get a
function of the original variable ๐ฅ.
Example 1: Evaluate ๐ฅ
1โ2๐ฅ2๐๐ฅ.
Solution: Set ๐ข = 1 โ 2๐ฅ2; ๐๐ข
๐๐ฅ= โ4๐ฅ โ ๐๐ฅ =
1
โ4๐ฅ๐๐ข.
Therefore, ๐ฅ
1โ2๐ฅ2๐๐ฅ =
๐ฅ
๐ข
1
โ4๐ฅ๐๐ข
= โ1
4
1
๐ข๐๐ข
= โ1
4 ๐ขโ1
2 ๐๐ข
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= โ1
4 2 ๐ข
12 + ๐
= โ1
2 1 โ 2๐ฅ2 + ๐
Example 2: Evaluate ๐2โ5๐ฅ ๐๐ฅ
Solution: Let ๐ข = 2 โ 5๐ฅ; ๐๐ข
๐๐ฅ= โ5 โ ๐๐ฅ =
๐๐ข
โ5.
Hence, ๐2โ5๐ฅ ๐๐ฅ = ๐๐ข ๐๐ข
โ5
= โ1
5 ๐๐ข ๐๐ข
= โ1
5๐๐ข + ๐
= โ1
5๐2โ5๐ฅ + ๐
Rule 7: (Integration by Parts)
If ๐ and ๐ are two differentiable functions of ๐ฅ. Then ๐
๐๐ฅ ๐๐ = ๐ ๐ฅ
๐๐
๐๐ฅ+ ๐(๐ฅ)
๐๐
๐๐ฅ
which implies that ๐ ๐ฅ ๐๐
๐๐ฅ=
๐
๐๐ฅ ๐๐ โ ๐(๐ฅ)
๐๐
๐๐ฅ . Using the formula for the integral of a
sum which is sum of the integrals, we obtain
๐ ๐ฅ ๐๐
๐๐ฅ๐๐ฅ = ๐ ๐ฅ ๐ ๐ฅ โ ๐(๐ฅ)
๐๐
๐๐ฅ๐๐ฅ
This is called the formula for integration by parts. If ๐ข = ๐(๐ฅ) and ๐ฃ = ๐(๐ฅ), the
formula can be written as ๐ข ๐๐ฃ = ๐ข๐ฃ โ ๐ฃ ๐๐ข .
Example: Evaluate the following integrals:
1. 2๐ฅ sin ๐ฅ ๐๐ฅ
Solution: Let ๐ข = 2๐ฅ, ๐๐ข
๐๐ฅ= 2 โ ๐๐ข = 2๐๐ฅ
๐๐ฃ = sin ๐ฅ ๐๐ฅ โ ๐ฃ = cos ๐ฅ
Hence, by integration by parts, we have
2๐ฅ sin ๐ฅ ๐๐ฅ = 2๐ฅ cos ๐ฅ โ cos ๐ฅ 2๐๐ฅ
= 2๐ฅ cos ๐ฅ โ 2 cos ๐ฅ ๐๐ฅ
= 2๐ฅ cos ๐ฅ โ 2 sin ๐ฅ + ๐
2. ๐ฅ๐๐ฅ ๐๐ฅ = ๐ฅ๐๐ฅ โ ๐๐ฅ ๐๐ฅ = ๐ฅ๐๐ฅ โ ๐๐ฅ + ๐
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3. ๐ฅ log๐ ๐ฅ ๐๐ฅ
Solution: Let ๐ข = log๐ ๐ฅ , ๐๐ข
๐๐ฅ=
1
๐ฅ โ ๐๐ข =
1
๐ฅ๐๐ฅ
๐๐ฃ = ๐ฅ ๐๐ฅ โ ๐ฃ =๐ฅ2
2
Hence, by integration by parts, we have
๐ฅ log๐ ๐ฅ ๐๐ฅ =๐ฅ2
2log๐ ๐ฅ โ
๐ฅ2
2
1
๐ฅ๐๐ฅ
=๐ฅ2
2log๐ ๐ฅ โ
1
2 ๐ฅ ๐๐ฅ
=๐ฅ2
2log๐ ๐ฅ โ
1
4๐ฅ2 + ๐
PP: (a) ๐ฅ2๐๐ฅ ๐๐ฅ (b) ๐๐ฅ sin ๐ฅ ๐๐ฅ .
6.4 Definite Integral
Let ๐ be a function of ๐ฅ and ๐, ๐ be an interval, ๐(๐ฅ)๐
๐๐๐ฅ is called definite
integral with respect to ๐ฅ within the limit ๐ and ๐.
THEOREM (Fundamental Theorem of Calculus): Suppose that ๐ is a strong
continuous function over a closed interval ๐, ๐ and that ๐น is anti โ derivative of ๐.
Then ๐(๐ฅ)๐
๐๐๐ฅ = ๐น(๐ฅ)
๐๐
= ๐น ๐ โ ๐น(๐).
Examples
1. ๐ฅ23
1๐๐ฅ =
๐ฅ3
3
31
= 33
3 โ
13
3 = 8
2
3
2. 4๐ฅ5
0๐๐ฅ = 2๐ฅ2 5
0= 50
PP: (a) ๐2โ5๐ฅ1
0๐๐ฅ (b) ๐ฅ ln ๐ฅ
1
0๐๐ฅ
6.5 APPLICATIONS OF INTEGRATION TO ECONOMICS AND COMMERCE
We learnt already that the marginal function is obtained by differentiating the
total function. We were given the total cost, total revenue or demand function and
we obtained the marginal cost, marginal revenue or elasticity of demand. Now we
shall obtain the total function when marginal function is given.
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6.5.1 The cost function and average cost function from marginal cost
function:
If ๐ถ is the cost of producing an output ๐ฅ, then marginal cost function,
๐๐ถ =๐๐ถ
๐๐ฅ. Using integration as reverse process of differentiation, we obtain,
Cost function, ๐ถ = ๐๐ถ ๐๐ฅ + ๐
where ๐ is the constant of integration which can be evaluated if the fixed cost is
known. If the fixed cost is not known, then ๐ = 0.
Average cost function, ๐ด๐ถ =๐ถ
๐ฅ, ๐ฅ โ 0
Example 1: The marginal cost function of manufacturing ๐ฅ units of a commodity is
8 + 16๐ฅ โ 9๐ฅ2. Find the total cost and average cost, given that the total cost of
producing 1 unit is 20.
Solution: Given that,
๐๐ถ = 8 + 16๐ฅ โ 9๐ฅ2
๐ถ = ๐๐ถ ๐๐ฅ + ๐
= 8 + 16๐ฅ โ 9๐ฅ2 ๐๐ฅ + ๐
= 8๐ฅ + 8๐ฅ2 โ 3๐ฅ3 + ๐ ------------------ (*)
Given, when ๐ฅ = 1, ๐ถ = 20
โด โ โ 20 = 8 1 + 8 12 โ 3 13 + ๐ โ ๐ = 13
โด Total Cost function, ๐ถ = 8๐ฅ + 8๐ฅ2 โ 3๐ฅ3 + 13
Average Cost function, ๐ด๐ถ =๐ถ
๐ฅ, ๐ฅ โ 0
= 8 + 8๐ฅ โ 23 +13
๐ฅ .
Example 2: The marginal cost function of manufacturing ๐ฅ units of a commodity is
3๐ฅ2 โ 2๐ฅ + 32. If there is no fixed cost, find the total cost and average cost
functions. Solution: Given that,
๐๐ถ = 3๐ฅ2 โ 2๐ฅ + 32
๐ถ = ๐๐ถ ๐๐ฅ + ๐
= 3๐ฅ2 โ 2๐ฅ + 32 ๐๐ฅ + ๐
= ๐ฅ3 โ ๐ฅ2 + 32๐ฅ + ๐
No fixed cost โ ๐ = 0
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โด Total cost, ๐ถ = ๐ฅ3 โ ๐ฅ2 + 32๐ฅ
Average Cost, ๐ด๐ถ =๐ถ
๐ฅ, ๐ฅ โ 0
= ๐ฅ2 โ ๐ฅ + 32 .
6.5.2 The revenue function and demand function from marginal revenue
function:
If ๐ is the total revenue function when the output is ๐ฅ, then marginal revenue
๐๐ =๐๐
๐๐ฅ.
Integrating this with respect to ๐ฅ we have
Revenue function, ๐ = ๐๐ ๐๐ฅ + ๐
where ๐ is the constant of integration which can be evaluated under given
conditions.
If the total revenue ๐ = 0, when ๐ฅ = 0,
Demand function, ๐ =๐
๐ฅ, ๐ฅ โ 0
Example 1: If the marginal revenue for a commodity is ๐๐ = 10 โ 6๐ฅ2 + 2๐ฅ, find
the total revenue and demand function.
Solution: Given that,
๐๐ = 10 โ 6๐ฅ2 + 2๐ฅ
๐ = ๐๐ ๐๐ฅ + ๐
= 10 โ 6๐ฅ2 + 2๐ฅ ๐๐ฅ + ๐
= 10๐ฅ โ 2๐ฅ3 + ๐ฅ2 + ๐
Since ๐ = 0 when ๐ฅ = 0, then ๐ = 0.
โด ๐ = 10๐ฅ โ 2๐ฅ3 + ๐ฅ2
๐ =๐
๐ฅ, ๐ฅ โ 0
โ ๐ = 10 โ 2๐ฅ2 + ๐ฅ.
Example 2: For the marginal revenue function ๐๐ = 5 + sin 4๐ฅ โ 2๐ฅ โ ๐ฅ2, find the
revenue function and demand function.
Solution: Given that,
๐๐ = 5 + sin 4๐ฅ โ 2๐ฅ โ ๐ฅ2
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๐ = ๐๐ ๐๐ฅ + ๐
= 5 + sin 4๐ฅ โ 2๐ฅ โ ๐ฅ2 ๐๐ฅ + ๐
= 5๐ฅ โ1
4coss 4๐ฅ + ๐ฅ2 +
๐ฅ3
3+ ๐
Since ๐ = 0 when ๐ฅ = 0, then ๐ = โ1
4.
โด ๐ = 5๐ฅ โ1
4coss 4๐ฅ + ๐ฅ2 +
๐ฅ3
3โ
1
4
๐ =๐
๐ฅ, ๐ฅ โ 0
โ ๐ = 5 โ1
4๐ฅcoss 4๐ฅ + ๐ฅ2 +
๐ฅ3
3โ
1
4๐ฅ.
6.5.3 The demand function when the elasticity of demand is given
We know that,
Elasticity of demand ๐๐ = โ๐
๐ฅโ
๐๐ฅ
๐๐
โ โ๐๐
๐=
๐๐ฅ
๐ฅ
1
๐๐
Integrating both sides
โ ๐๐
๐=
1
๐๐
๐๐ฅ
๐ฅ
This equation yields the demand function ๐ as a function of ๐ฅ.
The revenue function can be found out by using the relation, ๐ = ๐๐ฅ.
Example 1: The elasticity of demand with respect to price ๐ for a commodity is
๐ฅโ6
๐ฅ, ๐ฅ > 6 when the demand is ๐ฅ. Find the demand function if the price is 2 when
demand is 8. Also find the revenue function.
Solution: Given that,
Elasticity of demand, ๐๐ = โ๐
๐ฅโ
๐๐ฅ
๐๐=
๐ฅโ6
๐ฅ
โ ๐๐ฅ
๐ฅโ6= โ
๐๐
๐
Integrating both sides,
๐๐ฅ
๐ฅโ6= โ
๐๐
๐+ ๐
โ log๐ ๐ฅ โ 6 = โ log๐ ๐ + log๐ ๐ (since c is a constant, ๐ โก log๐ ๐)
โ log๐ ๐ฅ โ 6 + log๐ ๐ = log๐ ๐
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โ log๐๐ ๐ฅ โ 6 = log๐ ๐
โ ๐ ๐ฅ โ 6 = ๐ -------------- (1)
When ๐ = 2, ๐ฅ = 8, from (1) we get
๐ = 4.
โด The demand function is,
๐ =4
๐ฅโ6, ๐ฅ > 6.
Revenue, ๐ = ๐๐ฅ or 4๐ฅ
๐ฅโ6, ๐ฅ > 6.
Example 2: The elasticity of demand with respect to price for a commodity is a
constant and is equal to 2. Find the demand function and hence the total revenue
function, given that when the price is 1, the demand is 4.
Solution: Given that,
Elasticity of demand, ๐๐ = 2
โ โ๐
๐ฅโ
๐๐ฅ
๐๐= 2
โ ๐๐ฅ
๐ฅ= โ2
๐๐
๐
Integrating both sides,
โ ๐๐ฅ
๐ฅ= โ2
๐๐
๐
ln ๐ฅ = โ2 ln ๐ + ln ๐
ln ๐ฅ + ln ๐2 = ln ๐
ln ๐ฅ๐2 = ln ๐
๐ฅ๐2 = ๐ --------------- (1)
Given, when ๐ฅ = 4, ๐ = 1
From (1) we get ๐ = 4
โด (1) โ ๐ฅ๐2 = 4 or ๐ =2
๐ฅ, ๐ฅ > 0
Demand function ๐ =2
๐ฅ, ๐ฅ > 0 ;
Revenue ๐ = ๐๐ฅ =2๐ฅ
๐ฅ= 2 ๐ฅ, ๐ฅ > 0.
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Exercise 6
1. The marginal cost function of manufacturing ๐ฅ units of a commodity is
3 โ 2๐ฅ โ ๐ฅ2. If the fixed cost is 200, find the total cost and average cost
function.
2. If the marginal revenue for a commodity is ๐๐ =๐2๐ฅ
100+ ๐ฅ + ๐ฅ2, find the
revenue function.
3. The marginal cost and marginal revenue with respect to a commodity of a firm
are given by ๐๐ถ = 4 + 0.8๐ฅ and ๐๐ = 12. Find the total profit, given that the
total cost at zero output is zero. Hence, determine the profit when the output
is 50.
4. The marginal revenue function (in thousands of naira) of a commodity is
7 + ๐โ0.05๐ฅ where ๐ฅ is the number of units sold. Find the total revenue from
the sale of 100 units ๐โ5 = 0.0067
5. The marginal cost and marginal revenue are given by ๐๐ถ = 20 +๐ฅ
20 and
๐๐ = 30. The fixed cost is N200. Find the maximum profit.
6. ADAS company determines that the marginal cost of producing ๐ฅ units is
๐๐ถ = 10.6๐ฅ. the fixed cost is N50. The selling price per unit is N5. Find
(i) Total Cost Function
(ii) Total Revenue Function
(iii) Profit Function
7. Find the cost of producing 3000 units of commodity if the marginal cost in
naira per unit is ๐ถโฒ ๐ฅ =๐ฅ
3000+ 2.50.
8. The marginal cost of a production level of ๐ฅ units is given by ๐ถโฒ ๐ฅ = 85 +375
๐ฅ2 .
Find the cost of producing 10 incremental units after 15 units have been
produced.