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Summary introduction to calculus 1

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Course
Calculus 1

Institution
Harvard University

Book
Essential Calculus Skills Practice Workbook with Full Solutions

this is a document that contains a clear and comprehensible well-structured notes and hypothesis for engineers as well as other students that need proper skills in the course. it will allow one to be able to understand the applications of calculus in concepts such as stationary points, calculating ...

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introduction
calculus one
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Book Title:Essential Calculus Skills Practice Workbook with Full Solutions

Author(s):Chris McMullen

Edition:2018
ISBN:9781941691243
Edition:Unknown

Institution
Harvard University
Course
Calculus 1

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SMA 2101: CALCULUS I
⃝Francis
c O. Ochieng

Department of Pure and Applied Mathematics
Jomo Kenyatta University of Agriculture and Technology

Course content
• Trigonometric functions: addition, multiple angle and factor formulae.
• Limits, continuity and diﬀerentiability.
• Diﬀerentiation by ﬁrst principles and by rule for xn (integral and fractional n), sums, products,
quotients, chain rule, trigonometric, logarithmic and exponential functions of a single variable.
Parametric diﬀerentiation.
• Applications: equations of tangent and normal, kinematics, rates of change and stationary points,
economics and ﬁnancial models.
• Integration: anti-derivatives and their applications to areas, volumes, and economics and
ﬁnancial models.

References
[1] Calculus and Analytical Geometry by Thomas and Funnery
[2] Advanced Engineering Mathematics (10th ed.) by Erwin Kreyszig
[3] Calculus by Larson Hostellem

Lecture 1

1 Introduction
Calculus is a Latin name meaning stone used for calculation. We therefore deal with calculus as
inﬁnitesimal calculus meaning calculation of numbers which are inﬁnitesimally small. For example,
consider the growth of a sidling - it grows gradually and continuously. If observed after few days,
the growth can be measured, but if observed after few minutes, although growth has taken place the
amount is too small to be distinguished. Therefore, it has an inﬁnitesimal growth. In calculus, the
amount of growth is not important but the rate of growth is important.

1.1 Functions
The growth of a sidling is an instance of a functional relation, in that the growth may be aﬀected
by variations in temperature, moisture, sunlight, etc. If all these factors remain constant, then the
growth is a function of time. To understand the word function, we ﬁrst look the following deﬁnitions.
Deﬁnition 1.1 (Variables and constants). Consider the formula used for calculating the volume of a
sphere of radius r.
4
V = πr3 (1)
3
Then,

1

, i) V and r vary with diﬀerent spheres. Hence, they are called variables.
4
ii) π and are constants, irrespective of the size of the sphere.
3
Deﬁnition 1.2 (Dependent and independent variables). From the formula (1), the volume V depends
on value of the radius, r, of the sphere. In this case, r is called the independent variable (input) since
it is not aﬀected while V is called the dependent variable (output) since it is aﬀected by the variation
of r. Similarly, in a second degree equation y = ax2 + bx + c. Here, a, b, c are constants, x is the
independent variable, and y is the dependent variable.
Deﬁnition 1.3 (Calculus). This is the study of the eﬀect of change in the input (independent variable)
on the output (dependent variable).
Deﬁnition 1.4 (Function). This is a mathematical relation between the dependent and independent
variable.
For example,
i) If the variable y dependents and the variable x e.g., y = 2x2 − 3x + 5, then we say that
dependent variable y is a function of the independent variable x and we denote y = f (x), where
f is a function and x is the independent variable. We may also denote this functional relation
as y = ϕ(x) or y = ψ(x) or y = g(x), etc.
ii) From formula (1), the volume V of a sphere is a function of the radius r - i.e., V = f (r).
1
iii) In the formula S = gt2 , the distance S moved by a falling body is a function of time t.
2
iv) The volume of a ﬁxed mass of a gas is a function of the absolute temperature while the pressure
remains constant, and vise versa.
v) The sine, cosine, tangent of an angle are functions of the angle.

Example(s):
(a) If a function y = f (x) = x2 − 4x + 3. Find (i) f (1), (ii) f (2), (iii) f (a), and (iv) f (a + h).

Solution
i) f (x) = x2 − 4x + 3 ⇒ f (1) = 12 − 4(1) + 3 = 0
ii) f (x) = x2 − 4x + 3 ⇒ f (2) = 22 − 4(2) + 3 = −1
iii) f (x) = x2 − 4x + 3 ⇒ f (a) = a2 − 4a + 3
iv) f (x) = x2 − 4x + 3 ⇒ f (a + h) = (a + h)2 − 4(a + h) + 3
(b) If a function ϕ(θ) = 2 sin θ, ﬁnd (i) ϕ( π2 ), (ii) ϕ(0), and (iii) ϕ( π3 ).

Solution
( )
i) ϕ(θ) = 2 sin θ ⇒ ϕ( π2 ) = 2 sin π2 = 2
ii) ϕ(θ) = 2 sin θ ⇒ ϕ(0) = 2 sin (0) = 0
√
(π) 3 √
iii) ϕ(θ) = 2 sin θ ⇒ ϕ( π3 ) = 2 sin 3 =2× = 3
2

Exercise:
(a) If f (x) = 2x2 − 4x + 1, ﬁnd (i) f (1), (ii) f (0), (iii) f (2), (iv) f (a), and f (x + h).
(b) If f (x) = (x − 1)(x + 5), ﬁnd (i) f (1), (ii) f (0), (iii) f (2), (iv) f (a + 1), and f ( a1 ).
(c) If f (θ) = cos θ, ﬁnd (i) f ( π2 ), (ii) f (0), (iii) f ( π3 ), (iv) f ( π6 ), and (v) f (π).
f (3.001) − f (3)
(d) If f (x) = x2 , ﬁnd (i) f (3), (ii) f (3.1), (iii) f (3.01), (iv) f (3.001), and .
0.001
(e) If ϕ(x) = 2x , ﬁnd (i) ϕ(0), (ii) ϕ(1), and (iii) ϕ(0.5).

2

, 1.2 Limits of functions
Deﬁnition 1.5 (Basic limit deﬁnition). Let f (x) be a function and let a and L be real numbers. If
f (x) approaches L as x approaches a (but is not equal to a), then we say that f (x) has limit L as x
approaches a and we denote
. = L.
lim f (x)
x→a

→ Note: lim f (x) is the value that f (x) approaches as x approaches a, and a does not have to be in
x→a
the domain of f .

1.3 Properties of limits
Theorem 1.1. Suppose lim f (x) = L1 and lim g(x) = L2 , then
x→a x→a

1. [Addition rule] lim [f (x) + g(x)] = L1 + L2
x→a

2. [Scalar multiple] lim [λf (x)] = λL1 , where λ is a constant.
x→a

3. [Product rule] lim [f (x) · g(x)] = L1 · L2
x→a
[ ]
f (x) L1
4. [Quotient rule] lim =
x→a g(x) L2

1.4 Methods of evaluating limits of functions
Direct substitution
The required limit is obtained by just plugging in the value of input, say x, into the given
function, say f (x).

Example(s):

(a) Evaluate lim 3x3 − x2 + 2x + 5.
x→2
Solution

lim (3x3 − x2 + 2x + 5) = 3 lim x3 − lim x2 + 2 lim x + lim 5)
x→2 x→2 x→2 x→2 x→2
= 3(2 ) − (2 ) + 2(2) + 5
3 2

= 29

x2 − 1
(b) Evaluate lim .
x→1 x + 1
Solution
x2 − 1 D.S 12 − 1 0
lim = = =0
x→1 x + 1 1+1 2

Factorization
If on direct substitution we get the indeterminate form 0/0, then it means that there is a
common factor in both the numerator and denominator. In this case, we perform factorization
ﬁrst so as to simplify the given function.

→ Note: if the polynomial in the numerator is of degree greater than the degree of the polynomial
in the denominator, we ﬁrst need to perform long division.

Example(s):

3

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