1.1 The Two Central Concepts of Calculus

The two central concepts of calculus are

• the derivative of a function, and
• the definite integral of a function.

These two concepts are intimately related by means of The Fundamental Theorem of Calculus.
 

Newton (1642–1727) and Leibniz (1646–1716) are usually credited with the creation of calculus. This is due in great part to the fact that they were the first to fully understand the theorem and its significance, and then were able to apply it in the solution of many problems.

 
However, to reach this point in its development, calculus had already evolved over many centuries, going back, notably, for example, to the work of Archimedes (ca. 287–212 B.C.) who computed (among many other results) the area enclosed by a parabolic sector, and the surface area and volume of a sphere.

1.1.1 Additional Fundamental Concepts

A rigorous development of the two central concepts of calculus necessarily requires an understanding of two important additional concepts.

• the concept of the limit of a functions or the limit of a sequence, and
• the concept of continuity of a function at a point and on an interval.

Acquiring a good understanding of the two central concepts of calculus, plus these two additional concepts is precisely the goal of calculus I.

Prerequisites

 
For all practical purposes, we will study the above concepts in the context of elementary functions:

1. polynomial functions
2. rational functions
3. root functions
4. trigonometric functions
5. inverse trigonometric functions
6. exponential functions
7. logarithmic functions
8. their combinations

Their combinations include algebraic operations—addition, subtraction, multiplication, division— and their compositions. Thus, a good understanding of these functions is a necessary prerequisite to study calculus.

Naturally, you will need to be proficient on various types computations: factorization, simplifications, trigonometric identities, simplifying exponential and logarithmic expressions.

1.1.2 Two Geometric Problems Central to the Development of Calculus

 
Fortunately, the two central concepts of calculus can be readily formulated in the context of two geometric problems that are easy to state and understand. Namely,

• the problem of finding the slope of the tangent line to the graph of a given function \(f\) as shown in Figure 1.1 (a);
• the problem of finding the area under the graph of a given function \(f\) as shown in Figure 1.1 (b).

Figure 1.1: (a)

Figure 1.1: (b)

1.1.3 How to Solve the Tangent Line and the Area Problems: Approximation

Approximations to Tangent Lines by Means of Secant Lines

 
On one hand, the method to find the slope of the tangent line to the graph of a given function \(f\) consists on {approximating the slope} by means of slopes of secant lines to the graph. This can be accomplished by choosing a second point \((x,f(x))\) on the graph of the function, Figure 1.2(a) shows, and computing the slope of the secant line passing through

\((a,f(a))\) and \((x,f(x))\):
\[m_{\text{secant line}}=\frac{f(x)-f(a)}{x-a}.\]
Such an expression to compute the slope of a secant line is called a difference quotient.

 

Figure 1.2: (a)

Approximations to Area by Means of Rectangular Regions

 
On the other hand, given a function \(y=f(x)\), \(x \in [a,b]\), with \(f(x)\geq 0\) for all \(x \in [a,b]\), the approximation method to compute the area of the region enclosed by the graph of \(f\) and the \(x\)-axis consists in {approximating the area} by means of the area of rectangular regions, as Figure 1.2(b) shows. Such an approximation is called a rectangular approximation.
 

Figure 1.2: (b)

1.1.4 What is New in This Calculus Textbook

(1) Unlike standard textbooks, this textbook aims to introduce as early as possible the central concepts of calculus, thus allowing ample time to study them through a regular semester.

This means that the sequencing of the content is not the usual one from a standard calculus book.

The author strongly believes that this new sequencing leads to better integration of content.

(2) It is open source, and hence, the electronic copy is free.

1.1.5 Advise on How to Study

 
The structure of the book is such that each section is divided into several subsections. I have done so with the purpose of introducing only a small number of concepts per subsection and then provide the opportunity for students to solve problems related with such concepts Namely, this is focused practice on new material. Thus, you should use the textbook as follows:

– First, study each subsection.

Aim to understand the concepts and the examples.

Make sure to memorize definitions, statements of theorems, propositions, corollaries. Use flash cards. They will be your best friends in this context.

– Second, solve the subsequent exercises.

If you cannot solve a particular problem, go back to the subsection and look for a similar example. Return to try to solve the problem at hand. Strive to solve all exercises, even and odd, each subsection has a small number of problems.

If you still cannot solve a particular problem, ask the instructor (e-mail would be a good way to ask) or ask a tutor or a classmate for a hint.

1.1.7 Review of Difference Quotients

 
We first review the most common algebraic procedures used to find and simplify the difference quotients of polynomial, rational, and root functions. Then, we turn to compute the difference quotients of trigonometric functions.

As shown in 1.1.3, the slope of the secant line to the graph of a function \(f\) passing through the points \((a,f(a))\) and \((x,f(x))\) is given by
\begin{equation}\label{ch01-01-eq01}
m_{\text{secant line}}=\frac{f(x)-f(a)}{x-a}.\tag{1.1}
\end{equation}
Such an expression is called a difference quotient. An alternative form of the difference quotient is the following
\begin{equation}\label{ch01-01-eq02}
m_{\text{secant line}}=\frac{f(x+h)-f(x)}{h}.\tag{1.2}
\end{equation}
In this case, the secant line passes through the points \((x, f(x))\) and \((x+h, f(x+h))\).

 
Note. The goal of the algebraic process to find and simplify the difference quotient is to eliminate from the denominator \(x-a\) from (\ref{ch01-01-eq01}) or \(h\) from (\ref{ch01-01-eq02}).

In the first example, we use the factorization formulas:
\[a^2-b^2= (a-b)(a+b) \hspace{.75in} a^3-b^3= (a-b)(a^2+ab+b^2).\]

Example 1. Given \(y = f(x) = 3x^4-2x^3\), find and simplify both difference quotients (\ref{ch01-01-eq01}) and (\ref{ch01-01-eq02})

Solution.
\begin{align*}
\frac{f(x)-f(a)}{x-a} & = \frac{ 3x^4 -2x^3-\left( 3a^4 -2a^3\right)}{x-a}=\frac{3\left (x^4-a^4\right )}{x-a}-\frac{2(x^3-a^3) }{x-a}=
\vphantom{\frac{A}{A_A}}\\
&=3\frac{ \left (x -a \right )\left (x +a \right )\left (x^2+a^2\right )}{x-a}-2\frac{(x-a)\left ( x^2+ x a + a^2\right ) }{x-a}=
\vphantom{\frac{A}{A_A}}\\
&=3\left (x +a \right )\left (x^2+a^2\right ) -2 \left ( x^2+ x a + a^2\right ), \;\; \text{provided }\; x-a\neq 0
\end{align*}

\begin{align*}
\frac{f(x+h)-f(x)}{h} &= \frac{3(x+h)^4 -2(x+h)^3-\left (3x^4 -2x^3\right ) }{h}
=\vphantom{\frac{A}{\frac{A}{A}}}\\ &=
\frac{3(x+h)^4 – 3x^4 }{h} -\frac{ 2(x+h)^3-2x^3 }{h}
=\vphantom{\frac{A}{\frac{A}{A}}}\\ &=
\frac{3h(2x+h)\left ((x+h)^2 +x^2\right ) }{h} -\frac{ 2h\left ((x+h)^2+(x+h)x + x^2 \right )}{h}
=\vphantom{\frac{A}{\frac{A}{A}}}\\ &=
3(2x+h)\left ((x+h)^2 +x^2\right ) – 2 \left ((x+h)^2+(x+h)x + x^2 \right ), \;\; h\neq 0.
\end{align*}

Example 2. Given the function \(\displaystyle f(x) = \frac{x}{2x+3}\), find and simplify the difference quotient (\ref{ch01-01-eq02}).

Solution. Step (1) consists on finding the LCD of the numerator, \(\left (2(x+h)+3\right )\left (2x+3\right )\), and then multiplying both numerator and denominator by this LCD.

In step (2), rather than expanding all terms, we have grouped terms that involve \(\) and terms that do not involve \(h\). The terms that do not involve \(h\) cancel out.
\begin{align*}
\frac{f(x+h)-f(x)}{h} &\underset{\hphantom{(2)}}{=} \frac{\displaystyle \frac{x+h}{2(x+h)+3}-\frac{x}{2x+3}}{h}\underset{(1)}{=}\frac{(x+h)(2x+3)-x(2x+2h+3)}{h(2(x+h)+3)(2x+3)}=
\vphantom{\frac{A}{\frac{A}{A}}}\\
&\underset{(2)}{=}\frac{x(2x+3)+h(2x+3)-x(2x+3)-x(2h)}{h(2(x+h)+3)(2x+3)}=
\vphantom{\frac{A}{\frac{A}{A}}}\\
&\underset{\hphantom{(2)}}{=} \frac{3h}{h(2(x+h)+3)(2x+3)}=\frac{3}{(2(x+h)+3)(2x+3)},\;\; \text{provided }\; h\neq 0.
\end{align*}

Example 3. Given \(\displaystyle y = f(x) = \frac{2}{\sqrt{3x+1}} \), find and simplify the difference quotient (\ref{ch01-01-eq01}).

Solution.
\begin{align*}
\frac{f(x)-f(a)}{x-a} &
= \frac{\displaystyle \frac{2}{\sqrt{3x+1}} – \frac{2}{\sqrt{3a+1}} }{x-a}
=\frac{2\sqrt{3a+1} – 2\sqrt{3x+1} }{(x-a)\sqrt{3x+1}\sqrt{3a+1}}
=\vphantom{\frac{A}{A_A}}\\&=
\frac{2\sqrt{3a+1} – 2\sqrt{3x+1} }{(x-a)\sqrt{3x+1}\sqrt{3a+1}} \cdot \frac{2\sqrt{3a+1} + 2\sqrt{3x+1} }{2\sqrt{3a+1} + 2\sqrt{3x+1}}
=\vphantom{\frac{A}{A_A}}\\&=
\frac{4(3a+1) – 4(3x+1) }{(x-a)\sqrt{3x+1}\sqrt{3a+1}\cdot 2\left (\sqrt{3a+1} + \sqrt{3x+1}\right )}
=\vphantom{\frac{A}{A_A}}\\&=
\frac{6(a-x) }{(x-a)\sqrt{3x+1}\sqrt{3a+1}\cdot\left (\sqrt{3a+1} + \sqrt{3x+1}\right )}
=\vphantom{\frac{A}{A_A}}\\&=
\frac{6}{\sqrt{3x+1}\sqrt{3a+1}\cdot\left (\sqrt{3a+1} + \sqrt{3x+1}\right )}, \;\; x-a\neq 0.
\end{align*}

Example 4. Find and simplify the difference quotients:   (a) \(\displaystyle \frac{\sqrt{x}-\sqrt{a}}{x-a}\);     (b) \(\displaystyle \frac{\sqrt[3]{x}-\sqrt[3]{a}}{x-a}\).

Solution. (a) We use the factorization formula \(a^2-b^2= (a-b)(a+b)\), and assume that \(x\neq a\).
\[\frac{\sqrt{x}-\sqrt{a}}{x-a}= \frac{\sqrt{x}-\sqrt{a}}{x-a}\cdot\frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}} = \frac{x-a}{(x-a)\left (\sqrt{x}+\sqrt{a}\right )}=\frac{1}{\sqrt{x}+\sqrt{a}}.\]
Note that we could have proceeded, alternatively, as follows:
\[\frac{\sqrt{x}-\sqrt{a}}{x-a}= \frac{\sqrt{x}-\sqrt{a}}{\left (\sqrt{x}\,\right )^2-\left (\sqrt{a}\,\right )^2}=
\frac{\sqrt{x}-\sqrt{a}}{\left (\sqrt{x}-\sqrt{a}\right )\left (\sqrt{x}+\sqrt{a}\right )}=\frac{1}{\sqrt{x}+\sqrt{a}}.\]

(b) We use the following factorization formula and rule of exponents
\[a^3-b^3= (a-b)\left (a^2+ab+b^2\right )\hspace{.75in} \sqrt[3]{x^2} = \left (\sqrt[3]{x}\,\right )^2.\]
Assume that \(x\neq a\). As in part (a), there are two alternative ways to proceed.

\begin{align*}
\frac{\sqrt[3]{x}-\sqrt[3]{a}}{x-a}
&=\frac{\sqrt[3]{x}-\sqrt[3]{a}}{x-a}\cdot\frac{\sqrt[3]{x^2}+ \sqrt[3]{x} \sqrt[3]{a} +\sqrt[3]{a^2}}{\sqrt[3]{x^2}+ \sqrt[3]{x} \sqrt[3]{a} +\sqrt[3]{a^2}}
=\\
&=\frac{x-a}{(x-a)\left (\sqrt[3]{x^2}+ \sqrt[3]{x} \sqrt[3]{a} +\sqrt[3]{a^2}\,\right )}
=\frac{1}{\sqrt[3]{x^2}+ \sqrt[3]{x} \sqrt[3]{a} +\sqrt[3]{a^2}}.
\end{align*}
Alternatively
\begin{align*}
\frac{\sqrt[3]{x}-\sqrt[3]{a}}{x-a} &= \frac{\sqrt[3]{x}-\sqrt[3]{a}}{\left (\sqrt[3]{x}\,\right )^3-\left (\sqrt[3]{a}\,\right )^3}
= \frac{\sqrt[3]{x}-\sqrt[3]{a}}{\left (\sqrt[3]{x}-\sqrt[3]{a}\,\right )\left (\sqrt[3]{x^2}+ \sqrt[3]{x} \sqrt[3]{a} +\sqrt[3]{a^2}\,\right )}=\\
&=\frac{1}{\sqrt[3]{x^2}+ \sqrt[3]{x} \sqrt[3]{a} +\sqrt[3]{a^2}}.
\end{align*}

Difference Quotients of Trigonometric Functions

 
Example 5. Use the addition formula \(\cos (x+y)= \cos x \cos y – \sin x\sin y\) to show that\[\frac{\cos a(x+h)-\cos ax}{h} = \cos ax \frac{\cos ah-1}{h} – \frac{a\sin ah}{ah}\sin ax , \;\; 0\neq a \in \mathbb{R}.\]

Solution.
\begin{align*}
\frac{\cos a(x+h)-\cos ax}{h}&=\frac{\cos ax \cos ah – \sin ax \sin ah -\cos ax}{h}=\\
&=\frac{\cos ax (\cos ah-1) + \sin ax \sin ah}{h} = \\
& = \cos ax \frac{\cos ah-1}{h} -\frac{\sin ah}{h}\sin ax.
\end{align*}

Example 6. Use the difference formula \(\sin (x-y)= \sin x \cos y – \cos x\sin y\) to show that
\[\frac{\tan x-\tan a}{x-a} =\frac{\sin (x-a)}{x-a}\sec x \sec a.\]

Solution.
\begin{align*}
\frac{\tan x-\tan a}{x-a} &= \frac{\displaystyle \frac{\sin x}{\cos x}-\frac{\sin a}{\cos a}}{x-a}
= \frac{\sin x \cos a -\sin a \cos x}{(x-a)\cos x \cos a}
=\frac{\sin (x-a)}{(x-a)\cos x \cos a} =\\
&=\frac{\sin (x-a)}{(x-a)}\sec x \sec a.
\end{align*}

Example 7. (a) Show that if \(\displaystyle \, a = -\frac{7\pi}{6}\), then \(\displaystyle \;\frac{\sin(a+h) -\sin a}{h} = \frac{1}{2}\cdot \frac{\cos h -1}{h}-\frac{\sqrt{3}}{2}\cdot\frac{\sin h}{h}\);
 

(b) show that if \(\displaystyle \, a = -\frac{5\pi}{4}\), then \(\displaystyle \;\frac{\cos(a+h) -\cos a}{h} = \frac{1}{\sqrt{2}}\cdot \frac{1-\cos h }{h}-\frac{1}{\sqrt{2}}\cdot\frac{\sin h}{h}\).

Solution.

(a) Using the difference formula \(\sin (x-y)= \sin x \cos y – \cos x\sin y\) and the fact that sine is odd, that is, \(\sin (-x)=-\sin x\), we have
\begin{align*}
\frac{\sin\left (a+h\right ) -\sin a}{h} &=
\frac{\displaystyle \sin\left (h -\frac{7\pi}{6}\right )- \sin\left ( -\frac{7\pi}{6}\right ) }{h} =
\frac{\displaystyle \sin h \cos \frac{7\pi}{6}-\cos h \sin \frac{7\pi}{6} + \sin\frac{7\pi}{6}}{h} =\\
&=
\frac{\displaystyle \sin h \cdot\left (- \frac{\sqrt{3}}{2}\right )-\cos h \cdot\left (-\frac{1}{2}\right )-\frac{1}{2}}{h} =
\frac{1}{2}\cdot \frac{\cos h -1}{h}-\frac{\sqrt{3}}{2}\cdot\frac{\sin h}{h}.
\end{align*}

(b) Using the difference formula \(\cos (x-y)= \cos x \cos y + \sin x\sin y\) and the fact that cosine is even, that is, \(\cos (-x)=\cos x\), we have
\begin{align*}
\frac{\cos(a+h) -\cos a}{h} &=
\frac{\displaystyle \cos\left (h -\frac{5\pi}{4}\right )- \cos\left ( -\frac{5\pi}{4}\right ) }{h} =
\frac{\displaystyle \cos h \cos \frac{5\pi}{4}+\sin h \sin \frac{5\pi}{4}- \cos \frac{5\pi}{4} }{h} =\\
&=
\frac{\displaystyle \cos h \left ( -\frac{1}{\sqrt{2}}\right )+\sin h \left (-\frac{1}{\sqrt{2}}\right )+ \frac{1}{\sqrt{2}} }{h}
= \frac{1}{\sqrt{2}}\cdot \frac{1-\cos h }{h}-\frac{1}{\sqrt{2}}\cdot\frac{\sin h}{h}.
\end{align*}