Math 138 - Week 1

Chapter 0 - Pre-Calc Review

0.1 Sigma Notation

Definition - Sigma Notation

Let $m$ and $n$ be integers with $m \leq n$ , and let $a_{m}, a_{m + 1}, a_{m + 2}, \dots, a_{n}$ be numbers. We define

\sum_{i = m}^{n} a_{i} = a_{m} + a_{m + 1} + a_{m + 2} + \dots + a_{n,}

which we refer to as a sum expressed in $sigma notation$ .

The symbol " $Σ$ " is the capital Greek letter $sigma$ , which indicates addition. $i$ is the $index of the summation$ and varies through all integer values from $i = m$ to $i = n$ , as specified by the $m$ and $n$ below and above the sigma, respectively. The expression $a_{i}$ represents the $i^{t h}$ term in the sum.

Remark!

Different letters can be used to represent the index of the summation. $i$ could be substituted for $j$ or $k$ , etc.

Example

(a) $\sum_{i = 1}^{4} 2^{i} = 2^{1} + 2^{2} + 2^{3} + 2^{4} = 30$

(b) $\sum_{j = 0}^{2} \frac{j}{j + 3} = \frac{0}{0 + 3} + \frac{1}{1 + 3} + \frac{2}{2 + 3} = 0 + \frac{1}{4} + \frac{2}{5} = \frac{13}{20}$

(d) $\sum_{k = 2}^{5} (\sqrt{k} - \sqrt{k + 1}) = (\sqrt{2} - \sqrt{3}) + (\sqrt{3} - \sqrt{4}) + (\sqrt{4} - \sqrt{5}) + (\sqrt{5} - \sqrt{6}) = \sqrt{2} - \sqrt{6}$

Finite sums written in sigma notation obey all of the usual rules we know for addition. These properties are summarized below.

Theorem 0.1.3 - Properties of Sums

Let $a_{m}, a_{m + 1}, a_{m + 2}, \dots, a_{n}$ and $b_{m}, b_{m + 1}, b_{m + 2}, \dots, b_{n}$ be real numbers.

(i) For any $c \in R$ , $\sum_{i = m}^{n} c a_{i} = c \sum_{i = m}^{n} a_{i}$
(ii) $\sum_{i = m}^{n} (a_{i} + b_{i}) = \sum_{i = m}^{n} a_{i} + \sum_{i = m}^{n} b_{i}$
(iii) $\sum_{i = m}^{n} (a_{i} - b_{i}) = \sum_{i = m}^{n} a_{i} - \sum_{i = m}^{n} b_{i}$

Example 0.1.4

If $\sum_{i = 1}^{n} a_{i} = 3$ and $\sum_{i = 1}^{n} a_{i} = 2$ , then what is the value of $\sum_{i = 1}^{n} (4 a_{i} - b_{i})$ ?

Solution: We have

\begin{aligned} \sum_{i = 1}^{n} (4 a_{i} - b_{i}) & = \sum_{i = 1}^{n} 4 a_{i} - \sum_{i = 1}^{n} b_{i} & (by Theorem 0.1.3 (iii)) \\ = 4 \sum_{i = 1}^{n} a_{i} - \sum_{i = 1}^{n} b_{i} & (by Theorem 0.1.3 (ii)) \\ = 4 (3) - 2 \\ = 10 \end{aligned}

We can reindex sums as there is not a unique way to represent a sum.

Method 1 - Add and subtract terms

Example 0.1.5

By adding and subtracting terms, reindex the sum $\sum_{i = 3}^{10} i^{2}$ to start at $i = 0$ .
Solution: We have

\sum_{i = 3}^{10} i^{2} = \sum_{i = 0}^{10} i^{2} - 0^{2} - 1^{2} - 2^{2} = - 5 + \sum_{i = 0}^{10} i^{2}

We added 3 terms and then subtracted them to fix the index. As intuition might tell you, with large numbers this becomes ridiculous and thus we turn to our second method.

Method 2 - Adjust the index within the sum itself

Example 0.1.5

Reindex the sum $\sum_{i = 100}^{200} \frac{1}{\sqrt{i}}$ to start at $i = 1$ .

Solution: We subtract 99 from the starting and ending index to get $i = 1$ and $101$ and then we add 99 to the i in the actual term. So,

\sum_{i = 100}^{200} \frac{1}{\sqrt{i}} = \sum_{i = 1}^{101} \frac{1}{\sqrt{i + 99}}

0.2 Completing the Square

How to complete the square:

Consider the polynomial $x^{2} + 6 x + 5$ .

The goal is to add and subtract certain constant that will make part of the quadratic recognizable as a perfect square; hence the name "completing the square":

(x^{2} + 6 x + ◻) + 5 - ◻

We divide the coefficient of the $x$ -term by $2$ and square the result. In this case, dividing $6$ by $2$ , we get $3$ , squaring we get $9$ . Replacing the blank box above with 9, we get:

(x^{2} + 6 x + 9) - 4

So, we have:

\begin{aligned} x^{2} + 6 x + 5 & = (x^{2} + 6 x + 9) - 4 \\ = (x + 3)^{2} - 4 \end{aligned}

Example 0.2.1

Rewrite $x^{2} - 2 x + 9$ by completing the square.

Solution: We take the coefficient of the $b$ -term, in this case $2$ . We divide by $2$ and square the result to get $1$ .
So,

\begin{aligned} x^{2} - 2 x + 9 & = (x^{2} - 2 x + 1) + 9 - 1 \\ = (x - 1)^{2} + 8 \end{aligned}

There is one more thing we must keep in mind. You always want to factor out the $a$ -term from all terms that involve $x$ .

For example, consider the polynomial $2 x^{2} - 16 x + 33$ . Factoring the coefficient of $a = 2$ from all terms involving $x$ , we get

2 (x^{2} - 8 x) + 33.

We divide the coefficient ( $8$ ) by $2$ and square the result to get $16$ . Since we factored out $2$ , we must multiply $16$ by $2$ when we subtract on the outside. This turns to be

2 (x^{2} - 8 x + 16) + 33 - 32

Finalizing the steps, we get

2 (x - 4)^{2} + 1.

Chapter 1 - Integration

The motivation is to compute the area between the graph of a function and the $x$ -axis over an interval [ $a$ , $b$ ]. This leads us to the definition of the definite integral. Integrals are used to calculate volumes of 3D solids, the wok done in applying force to an object over some distance, probabilities and expected values, forces and stresses on structures, fluid flow and heat transfer, and much more.

1.1 The Definite Integral

1.1.1 Area Under a Curve

If we are trying to find the area under a line, it is simple. We can just break the graph into separate polygons. In this example, the area of $A_{1}$ is $\frac{b h}{2} = \frac{(6) (3)}{2} = 9$ and the area of $A_{2}$ is $(6) (2) = 8,$ and so the total area is $A_{T o t a l} = 9 + 8 = 17$

However, if we try to find the area under a curve like $y = x^{2}$ over [ $- 2, 5$ ], it becomes much harder because we cannot use geometry.

Riemann Sum Motivation

Let $Δ x = \frac{b - a}{n}$ where $Δ x$ is the width of each rectangle.
And, $n$ is the number of intervals in the domain of [ $a, b$ ].
Therefore, $A r e a_{e a c h r e c t a n g l e} = Δ x \cdot f (x_{i})$
$T o t a l a r e a = \sum_{i = 1}^{n} Δ x \cdot f (x_{i})$

Overestimate

When the function is decreasing and we use left-point Riemann sums
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When the function is increasing and we use right-point Riemann sums
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Underestimate

When the function is decreasing and we use right-point Riemann sums
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When the function is increasing and we use left-point Riemann sums
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Unable to tell if over or under estimate

When we use the midpoint
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The big Idea
We want to increase the amount of subintervals ( $n$ ) to the point that the width of each rectangle gets closer and closer to $0$ . At this rate, the left and right point Riemann sums almost don't even matter because they're basically the same point.
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Definition - Regular Partition

Consider an interval [ $a, b$ ] with $a < b$ and let $n$ be a positive integer.
If we define $Δ x = \frac{b - a}{n}$ , then points $a = x_{0} < x_{1} < x_{2} < \dots < x_{n - 1} < x_{n} = b$ given by

x_{i} = a + i Δ x, i = 0, 1, 2, \dots, n

separate [ $a, b$ ] into $n$ subintervals of equal width, $Δ x$ :

[a, b] = [x_{0}, x_{1}] \cup

Consider the function $y = x^{2}$ on the interval [ $0$ , $1$ ]. We want to find the area between the curve and the $x$ -axis. If we divide the region into 4 rectangles of equal length