Lead City University,
Ibadan, Naigeria,
MTH 113
Elementary Mathematics II
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(CALCULUS)
Mr. M. T. AKOLADE
rs
t fo
No
2023/2024
COURSE DESCRIPTION
Functions of a real variable, graphs, limits and idea of continuity. The derivative, as limit of rate
of change. Techniques of differentiation, maxima and minima. Extreme curve sketching,
integration, Definite integrals, reduction formulae, application to areas, volumes (including
approximate integration: Trapezium and Simpson’s rule).
1 FUNCTIONS, LIMIT AND CONTINUITY
1 FUNCTIONS, LIMIT AND CONTINUITY
The branch of mathematics dealing with continuously varying quantities is call calculus. While
differential calculus deals with rates of change, integral calculus deals with addition of the
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effects of continuously varying quantities.
1.1 Functions and Limit
The concept of limit of a functions is very important in the subject of calculus. In fact, the
subject of calculus was evolved through the finding of limit. Limits are essential in the definition
of continuity, derivatives and integrals.
Goal:
1. define a function
2. define limit of function
3. evaluate limits of function
rs
4. evaluate limit of function at infinity.
t fo
Introduction
Function is an expression, rule or law that defines a relationship between a dependent variable
and an independent variable while limit is the value that a function approaches as the input
tends to a value.
Definition
No
Let 𝑓 : 𝐴 → 𝑅 and 𝑐 be a limit point of the domain A. We say that
lim 𝑓 (𝑥) = 𝐿
𝑥→𝑐
provided that for all 𝜖 > 0, there exists 𝛿 > 0 such that 0 ≤ |𝑥 −𝑐 | ≤ 𝛿 and x ∈ 𝐴 ⇒ |𝑓 (𝑥)−𝐿| ≤ 𝜖
Rules for limit of function
1. Limit of constant function
If 𝑓 (𝑥) = 𝑘, where k is a constant then lim𝑥→𝑎 𝑘 = 𝑘
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1 FUNCTIONS, LIMIT AND CONTINUITY
2. Let lim 𝑓 (𝑥) = 𝑙 and lim 𝑔(𝑥) = 𝑘 then Limits of the function 𝑓 (𝑥).𝑔(𝑥)
𝑥→𝑎 𝑥→𝑎
lim 𝑓 (𝑥) · 𝑔(𝑥) = lim 𝑓 (𝑥) · lim 𝑔(𝑥) = 𝑙𝑘
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎
3. Limit of rational function
𝑔(𝑥) lim𝑥→𝑎 𝑔(𝑥) 𝑙
If 𝑓 (𝑥) = , then = ,𝑘 ≠ 0
ℎ(𝑥) lim𝑥→𝑎 ℎ(𝑥) 𝑘
Activity 1
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Evaluate lim (𝑥 2 − 4𝑥 + 2)
𝑥→2
Solution:
lim𝑥→2 (22 − 4(2) + 2) = −2
Activity 2
Find lim (𝑥 + 3) (𝑥 2 − 5)
𝑥→−1
Solution:
lim (𝑥 + 3) lim (𝑥 2 − 5)
𝑥→−1 𝑥→−1
(−1 + 3)((−1) 2 − 5) = 2. − 4 = −8
rs
t fo
Activity 3
𝑥3 − 1
Evaluate lim
𝑥→1 𝑥 − 1
Solution:
0
If 𝑥 = 1 is inserted into the expression,it will give 0 = 0.Therefore ,we have to first reduce the
expression to its simplest form
(𝑥 − 1) (𝑥 2 + 𝑥 + 1)
lim
𝑥→1 𝑥 −1
⇒ lim (𝑥 2 + 𝑥 + 1)
No
𝑥→1
=(1) 2 + (1) + 1 = 3
Activity 4
9 − 𝑥2
Find the lim √
𝑥→3 4 − 𝑥 2 + 7
Solution:
Rationalize the denominator first
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1 FUNCTIONS, LIMIT AND CONTINUITY
√
© (9 − 𝑥 2 ) (4 + 𝑥 2 + 7) ª
lim √︃ √
®
𝑥→3 ®
(4 − 𝑥 2 + 7) (4 + 𝑥 2 + 7)
« √ ! ¬
(9 − 𝑥 2 ) (4 + 𝑥 2 + 7)
lim
𝑥→3 16 − (𝑥 2 + 7)
√ !
(9 − 𝑥 2 ) (4 + 𝑥 2 + 7)
lim
𝑥→3 9 − 𝑥2
√ √
lim (4 + 𝑥 2 + 7) = 4 + 16 = 4 + 4 = 8
𝑥→3
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Limit at infinity
The limit of a function f(x) as x approaches infinity or becomes very large is denoted as
lim𝑥→∞ 𝑓 (𝑥)
Remark;
𝑔(𝑥)
If 𝑓 (𝑥) = , then lim𝑥→∞ 𝑓 (𝑥) can be found by firstly divide the numerator and denominator
ℎ(𝑥)
by the variable with the highest index
Activity 5
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5𝑥 2 − 1
Evaluate the lim
2𝑥 2 + 1
t fo
𝑥→∞
Solution2
5𝑥 1
!
𝑥2
− 𝑥2
lim 2𝑥 2 1
𝑥→∞
𝑥2
+ 𝑥!2
1
5− 𝑥2
lim 1
𝑥→∞2+ 𝑥2
5+0 5
=
2+0 2
No
Assessment Questions
1. What is a function?
2. What is a Limit?
3. Evaluate the following:
(i)lim (𝑥 2 + 4𝑥 + 2)
𝑥→1
(ii) lim (𝑥 2 + 𝑥 + 3)
𝑥→−2
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1 FUNCTIONS, LIMIT AND CONTINUITY
3𝑥 2 − 1
4. Evaluate lim
𝑥→∞ 2𝑥 2 + 1
Assignment 1
Evaluate the following:
3
• lim𝑥→∞ 2𝑥 𝑥−7𝑥+3
3 +2
3𝑥
• lim𝑥→∞ 𝑥+1
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2
• lim𝑥→9 𝑥√𝑥−81−3
2
• lim𝑥→2 𝑥 𝑥−3𝑥+2
2 −4
2
• lim𝑥→−1 𝑥 𝑥−7𝑥+10
2 −4
1.2 Continuous Function rs
A continuous function is a function in which a small change in the input results in an arbitrary
small change is the output and it affirm the existences of a given functions.
Goal
1. define continuity
t fo
2. define continuity of a function
3. establish continuity of functions.
A single valued function of 𝑥 is said to be continuous at 𝑥 = 𝑎 if the following conditions are
satisfied:
i lim 𝑓 (𝑥) exists.
𝑥→𝑎
No
ii the function is defined for the value 𝑥 = 𝑎 i.e.,
lim 𝑓 (𝑥) = 𝑓 (𝑎)
𝑥→𝑎
Rules for continuity at the point 𝑥 = 𝑎
1. 𝑓 (𝑎) should be well defined
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1 FUNCTIONS, LIMIT AND CONTINUITY
2. lim 𝑓 (𝑥) should exists
𝑥→𝑎
3. lim 𝑓 (𝑥) = 𝑓 (𝑎)
𝑥→𝑎
In other words,
A function 𝑓 (𝑥) is said to be continuous at 𝑥 0 if for every 𝜖 > 0 there exists a 𝛿 > 0 such that
|𝑓 (𝑥) − 𝑓 (𝑥 0 )| < 𝜖 whenever 0 < |𝑥 − 𝑥 0 | < 𝛿.
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Activity 1
Test the function
𝑥 3 +1
if 𝑥 ≠ −1
𝑓 (𝑥) = 𝑥+1 , (1.1)
3 if 𝑥 = −1
for continuity
Solution:
1. lim
lim
𝑥→−1
𝑥3 + 1
𝑥→−1 𝑥 + 1
(𝑥 + 1) (𝑥 2 − 𝑥 + 1)
𝑥 +1
=3
rs
Therefore, lim 𝑓 (𝑥) exists
𝑥→−1
2.the function is defined at 𝑥 = −1 as 𝑓 (−1) = 3
t fo
3. lim 𝑓 (𝑥) = 𝑓 (−1) it is continuous i.e
𝑥→−1
lim 𝑓 (𝑥) = 𝑓 (−1) = 3
𝑥→−1
Activity 2
Test the function
2𝑥 2 +3𝑥+1
if𝑥 ≠ −1
𝑓 (𝑥) = 𝑥+1 , (1.2)
3 if𝑥 = −1
for continuity
No
Solution:
2𝑥 2 + 3𝑥 + 1
1. lim
𝑥→−1 𝑥 +1
(2𝑥 + 1) (𝑥 + 1)
lim
𝑥→−1 𝑥 +1
lim (2𝑥 + 1) = −1
𝑥→−1
2. the function is defined at 𝑥 = −1 as 𝑓 (−1) = 3
3. lim 𝑓 (𝑥) ≠ 𝑓 (−1), it is not continuous i.e
𝑥→−1
lim𝑛→−1 𝑓 (𝑥) = 𝑓 (−1) ≠ 3
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1 FUNCTIONS, LIMIT AND CONTINUITY
Activity 3
Test the continuity of a function 𝑓 (𝑥) = 2𝑥𝑥+2
−2
Solution:
The function 𝑓 (𝑥) is defined for all real numbers 𝑥 ≠ −2 and is continuous at x except at −2
Thus, 𝑓 (𝑥) is a continuous function.
Assessment Questions
1. Define the Continuity of a function
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2. Test the continuity of a function 𝑓 (𝑥) = 4𝑥 2 + 1 at a point 𝑥 = 1
3. Test the function
2𝑥 2 +3𝑥+1
if𝑥 ≠ −1
𝑓 (𝑥) = 𝑥+1 , (1.3)
3 if𝑥 = −1
for continuity
Assignments 2
rs
1. Test for continuity the function
𝑥 2, if𝑥 = −3
𝑓 (𝑥) = (1.4)
6𝑥 − 9 if𝑥 =≥ 3
t fo
2. Test for continuity the function
9−𝑥 2
( )
√ , if𝑥 = 3
𝑓 (𝑥) = 2
4− 𝑥 +7 (1.5)
5𝑥 − 7 if𝑥 = 3
3. Test the continuity of a function 𝑓 (𝑥) = 5𝑥 2 + 𝑥 − 2 at a point 𝑥 = −2
No
1.3 Derivatives as Limit of Change
Derivative is the instantaneous rate of change of a function at any point. Graphically, the
derivative is the slope of the tangent line through the point.
Goal:
1. define derivative;
2. determine the derivative of a function; and
3. differentiate from the First Principle.
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1 FUNCTIONS, LIMIT AND CONTINUITY
Introduction
Derivative is regarded as a limit of change for an assumed function 𝑦 = 𝑓 (𝑥). In order to know
how this function changes, if the independent variable x change. That is, if 𝑥 has a little change
say 𝛿𝑥 to become 𝑥 + 𝛿𝑥, then, 𝑦 will also change from 𝑓 (𝑥) to 𝑓 (𝑥 + 𝛿𝑥). For example, if the
initial value of Cassava is denoted 𝑥, then changes in 𝑥 amount to 𝑥 + 𝛿𝑥. Therefore, Garri
production will change from 𝑓 (𝑥) to 𝑓 (𝑥 + 𝛿𝑥). The change in y will then be
𝛿𝑦 = 𝑓 (𝑥 + 𝛿𝑥) − 𝑓 (𝑥) (1.6)
the average rate of change will be
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𝛿𝑦 𝑓 (𝑥 + 𝛿𝑥) − 𝑓 (𝑥)
= (1.7)
𝛿𝑥 𝛿𝑥
To have the rate of change of the function 𝑓 at 𝑥. 𝛿𝑥 is taking so very small that it is close to
zero knowing that average rate of change over little increase will get closer to some number. If
this happens, then, that limiting number is called the rate of change of 𝑓 at 𝑥 or, the derivative
of 𝑓 at 𝑥. It is denoted as 𝑓 ′ (𝑥) and mathematically written as
rs
𝑓 ′ (𝑥) = 𝑙𝑖𝑚
𝛿𝑦
𝛿𝑥→0 𝛿𝑥
= 𝑙𝑖𝑚
𝛿𝑥→0
𝑓 (𝑥 + 𝛿𝑥) − 𝑓 (𝑥)
𝛿𝑥
(1.8)
1.4 Differentiation from the First Principle
t fo
This is known as the first technique that can be used to obtain the derivative of a function. It
could be the instantaneous rate of change in volume, discharge of liquid in a container with
time, in value of Naira with time, distance 𝑦 with 𝑥 on a curve (known as Slope). It is also
known as Delta Method. It is worth mentioning that when you are asked to "Differentiate a
function", it simply means to find the derivative of the function.
Suppose 𝑦 is a function of 𝑥, say
𝑦 = 𝑓 (𝑥) (1.9)
No
Whenever there is change in 𝑥 by 𝛿𝑥, there is also a change in 𝑦 by 𝛿𝑦. Then you write
𝑦 + 𝛿𝑦 = 𝑓 (𝑥 + 𝛿𝑥) (1.10)
Then, make 𝛿𝑦 as the subject of formula. This is done by subtracting 𝑦 from both sides of the
equation then, substitute the value of y as 𝑓 (𝑥) on the right hand side so you have
𝛿𝑦 = 𝑓 (𝑥 + 𝛿𝑥) − 𝑓 (𝑥) (1.11)
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1 FUNCTIONS, LIMIT AND CONTINUITY
Divide both sides by change in 𝑥, 𝛿𝑥. Since change in 𝑦 with respect to 𝑥 is taking by introducing
the limit of 𝛿𝑥 as it close to zero. This results to
𝑑𝑦 𝛿𝑦 𝑓 (𝑥 + 𝛿𝑥) − 𝑓 (𝑥)
= 𝑙𝑖𝑚 = 𝑙𝑖𝑚 (1.12)
𝑑𝑥 𝛿𝑥→0 𝛿𝑥 𝛿𝑥→0 𝛿𝑥
This is the derivative of 𝑦 with respect to 𝑥 by using first principle technique.
Activity 1
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Use first principle to obtain the derivative of a function 𝑦 = 𝑥 + 2
Solution
Given equation 𝑦 = 𝑥 + 2 as a function. Recall that, for every little change in independent
variable there is also simultaneous change in 𝑦 so that
𝑦 + 𝛿𝑦 = 𝑥 + 𝛿𝑥. Note: There is no need to find change in 2 because it is a constant.
Therefore,
𝛿𝑦 = 𝑥 + 𝛿𝑥 − (𝑥 + 2) rs
By expansion and further simplification we have
𝛿𝑦 = 𝛿𝑥. Then take the average rate of change by dividing through by 𝛿𝑥 to have
𝛿𝑦 𝛿𝑥
=
𝛿𝑥 𝛿𝑥
Then, take the limit as 𝛿𝑥 goes to zero to have
𝑑𝑦 𝛿𝑥
= 𝑙𝑖𝑚 =1
t fo
𝑑𝑥 𝛿𝑥→0 𝛿𝑥
Therefore, the derivative of 𝑦 = 𝑥 + 2 is 1
Activity 2
Find the derivative of a function 𝑦 = 𝑥 2 by first principle
Solution
The given equation is 𝑦 = 𝑥 2
No
For every little changes in 𝑥, there occur change in 𝑦 as well. This implies
𝑦 + 𝛿𝑦 = (𝑥 + 𝛿𝑥) 2
Then 𝛿𝑦 becomes
𝛿𝑦 = (𝑥 + 𝛿𝑥) 2 − 𝑥 2
By expanding the factors, you have
𝛿𝑦 = 𝑥 2 + 2𝑥𝛿𝑥 + (𝛿𝑥) 2 − 𝑥 2
if you simplify further it becomes
𝛿𝑦 = 2𝑥𝛿𝑥 + (𝛿𝑥) 2
divide through by 𝛿𝑥, average rate of change thus become
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1 FUNCTIONS, LIMIT AND CONTINUITY
𝛿𝑦
= 2𝑥 + 𝛿𝑥
𝛿𝑥
If 𝛿𝑥 is considered so small that it closes to zero, then
𝑑𝑦 𝛿𝑦
= 𝑙𝑖𝑚 = 𝑙𝑖𝑚 (2𝑥 + 𝛿𝑥)
𝑑𝑥 𝛿𝑥→0 𝛿𝑥 𝛿𝑥→0
𝑑𝑦
= 2𝑥
𝑑𝑥
Therefore, the derivative of 𝑦 = 𝑥 2 using first principle is 2𝑥.
Activity 3
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Find the derivative of a function 𝑦 = 5𝑥 2 by first principle
Solution
You are given 𝑦 = 5𝑥 2 as the function. For every little changes in 𝑥, there occur change in 𝑦 as
well. This implies
𝑦 + 𝛿𝑦 = 5(𝑥 + 𝛿𝑥) 2
Then 𝛿𝑦 becomes
𝛿𝑦 = 5(𝑥 + 𝛿𝑥) 2 − 5𝑥 2
By expanding the factors, you have
𝛿𝑦 = 5𝑥 2 + 10𝑥𝛿𝑥 + 5(𝛿𝑥) 2 − 5𝑥 2
if you simplify further it becomes
rs
𝛿𝑦 = 𝛿𝑥 (10𝑥 + 5𝛿𝑥)
average rate of change thus become
𝛿𝑦
= 10𝑥 + 5𝛿𝑥
t fo
𝛿𝑥
If 𝛿𝑥 is considered so small that it closes to zero, then
𝑑𝑦 𝛿𝑦
= 𝑙𝑖𝑚 = 𝑙𝑖𝑚 (10𝑥 + 5𝛿𝑥)
𝑑𝑥 𝛿𝑥→0 𝛿𝑥 𝛿𝑥→0
𝑑𝑦
= 10𝑥
𝑑𝑥
This is the derivative of 𝑦 = 5𝑥 2 using first principle.
Assessment Questions
No
Differentiate from first principle:
(𝑖) 𝑦 = 𝑡 2 + 2𝑡 + 2. (𝑖𝑖) 𝑦 = 4𝑥 2 + 𝑥
Assignments
Differentiate, using first principle
(𝑖) 𝑉 = 𝑆𝑖𝑛𝑢.
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1 FUNCTIONS, LIMIT AND CONTINUITY
1.5 Derivative of some frequent functions
Goal:
1. identify the following forms of functions
i constant function
ii identity function 𝑉 (𝑢) = 𝑢
iii function of the form 𝑎𝑥 𝑛
2. derive the derivatives of the above listed functions.
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Introduction
A function is said to be a constant function if its value does not change no matter the value
of independent variable inputted. So the graph of a constant function is always parallel to the
independent variable axis.
Mathematically, it is 𝑦 (𝑥) = 𝑐 ∀ 𝑐 ∈ 𝑅, that is, where 𝑐 is some real number.
rs
The derivative of a constant function: It is worth mentioning again that the slope of a
function is also regarded as its derivative. Therefore, since for every inputed values from
independent variable yields a constant value for the function. This implies that, the function
has a straight line parallel to 𝑥 axis, and by implication, the slope or derivative becomes zero.
Activity 1
t fo
Suppose 𝑦 (𝑥) = 𝑐 where 𝑐 is a constant. Show that the derivative of 𝑦 (𝑥) is zero.
Solution
Recall that the derivative of a function is given as
𝑦 (𝑥 + 𝛿𝑥) − 𝑦 (𝑥)
𝑦 ′ (𝑥) = 𝑙𝑖𝑚
𝛿𝑥→0 𝛿𝑥
Now, because 𝛿𝑥 → 0
𝑦 (𝑥) − 𝑦 (𝑥) 𝑐 − 𝑐
𝑦 ′ (𝑥) = = =0
0 0
No
Therefore,
𝑦 ′ (𝑥) = 0
Activity 2
Identify the constant function(s) in the following
(𝑎) 𝑓 (𝑥) = 10 (𝑏) 𝑦 (𝑥) = 10 (𝑐) 𝑃 (𝑥) = 𝑥 + 5 (𝑑) 𝑓 (𝑡) = 2(5)
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1 FUNCTIONS, LIMIT AND CONTINUITY
Solution
The constant functions are
(𝑎) 𝑓 (𝑥) = 10 (𝑏) 𝑦 (𝑥) = 10 (𝑑) (𝑐) 𝑓 (𝑡) = 2(5)
Activity 3
The derivative of 𝑓 (𝑥) = 5, say (5) ′ = 0.
ale
Derivative of the identity function 𝑉 (𝑢) = 𝑢
If 𝑉 (𝑢) = 𝑢 is the identity function of 𝑢. Then the derivative of 𝑉 (𝑢), say 𝑑𝑉
𝑑𝑢 𝑜𝑟 𝑉 ′ (𝑢) will be
𝑉 ′ (𝑢) = 1
By first principle
𝑉 (𝑢) + 𝛿𝑉 = 𝑢 + 𝛿𝑢
𝛿𝑉 = 𝑢 + 𝛿𝑢 − 𝑉 (𝑢)
= 𝑢 + 𝛿𝑢 − 𝑢
𝛿𝑉 = 𝛿𝑢
divide through by 𝛿𝑢, it becomes
𝛿𝑉 𝛿𝑢
𝛿𝑢 𝛿𝑢
=
Therefore,
𝑑𝑉
= 𝑙𝑖𝑚
=1
𝛿𝑉
=1
rs
𝑑𝑢 𝛿𝑢→0 𝛿𝑢
t fo
Activity 4
Find the derivative of 𝑓 (𝑥) = 𝑥
Solution
Given 𝑓 (𝑥) = 𝑥
From the first principle
𝑓 (𝑥) + 𝛿 𝑓 (𝑥) = 𝑥 + 𝛿𝑥
𝛿 𝑓 (𝑥) = 𝑥 + 𝛿𝑥 − 𝑓 (𝑥)
No
= 𝑥 + 𝛿𝑥 − 𝑥
𝛿 𝑓 (𝑥) = 𝛿𝑥
divide through by 𝛿𝑥, it becomes
𝛿 𝑓 (𝑥) 𝛿𝑥
= =1
𝛿𝑥 𝛿𝑥
Therefore,
𝑑 𝑓 (𝑥) 𝛿 𝑓 (𝑥)
= 𝑙𝑖𝑚 =1
𝑑𝑥 𝛿𝑥→0 𝛿𝑥
𝑓 (𝑥) = 1
′
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1 FUNCTIONS, LIMIT AND CONTINUITY
Derivative of the function of the form 𝑎𝑥 𝑛
Suppose we have a function define as 𝑦 = 𝑎𝑥 𝑛 . if we choose to use first principle approach
𝑦 = 𝑎𝑥 𝑛 , when there is changes in the variable 𝑥, then
𝑦 + 𝛿𝑦 = 𝑎(𝑥 + 𝛿𝑥)𝑛
expand (𝑥 + 𝛿𝑥)𝑛 by Binomial theorem, you have
𝑦 + 𝛿𝑦 = 𝑎[𝑥 𝑛 + 𝑛𝑥 𝑛−1𝛿𝑥 + 𝑛(𝑛 − 1)𝑥 𝑛−2𝛿𝑥 2 + · · · + 𝛿𝑥 𝑛 ]
subtract 𝑦 as 𝑎𝑥 𝑛 from both sides
𝛿𝑦 = 𝑎[𝑛𝑥 𝑛−1𝛿𝑥 + 𝑛(𝑛 − 1)𝑥 𝑛−2𝛿𝑥 2 + · · · + 𝛿𝑥 𝑛 ]
ale
then, divide through by 𝛿𝑥, it becomes
𝛿𝑦
= 𝑎[𝑛𝑥 𝑛−1 + 𝑛(𝑛 − 1)𝑥 𝑛−2𝛿𝑥 + · · · + 𝛿𝑥 𝑛−1 ].
𝛿𝑥
Thus, the derivative of the function becomes
𝑑𝑦 𝛿𝑦
= 𝑙𝑖𝑚 = 𝑎𝑛𝑥 𝑛−1
𝑑𝑥 𝛿𝑥→0 𝛿𝑥
Activity 5 rs
Find the derivative of 𝑦 (𝑥) = 5𝑥 3 + 𝑥 2
Solution
Given 𝑦 (𝑥) = 5𝑥 3 + 𝑥 2 then
t fo
from the above approach, we have
𝑑𝑦
= 5(3)𝑥 3−1 + (2)𝑥 2−1
𝑑𝑥
therefore,
𝑦 ′ (𝑥) = 15𝑥 2 + 2𝑥
Assessment Questions
1. Find the derivative of 𝑦 (𝑥) = 5𝑥 5 + 𝑥 3
No
2. Find the derivative of 𝑦 (𝑥) = 4𝑥 3 − 𝑥 2
3. Derivative the function of form 𝑎𝑥 𝑛 form the first principle.
Assignments
Find the derivative of the functions:
1
(𝐼 ) 𝑦 = (2𝑥) − 2
12
