Section 4.1:  Describing Area and Summation Notation

Amy Graham Goodman

Section 4.1: Describing Area and Summation Notation

This content comes directly from Larry Musolino’s textbook Techniques of Calculus 1 Section 6.1 Antiderivatives.

Access this resource for free at https://psu.pb.unizin.org/math110/chapter/6-1-antiderivatives/

Learning Objectives

In this section, you will:

Expand summation notation.
Write sums in summation notation.
Use sigma (summation) notation to calculate sums using summation properties and power sums.

Let’s Get Started…

Archimedes was fascinated with calculating the areas of various shapes—in other words, the amount of space enclosed by the shape. He used a process that has come to be known as the method of exhaustion, which used smaller and smaller shapes, the areas of which could be calculated exactly, to fill an irregular region and thereby obtain closer and closer approximations to the total area. In this process, an area bounded by curves is filled with rectangles, triangles, and shapes with exact area formulas. These areas are then summed to approximate the area of the curved region.

Like Archimedes, mathematicians first approximated the area under the curve using shapes of known area (namely, rectangles). By using smaller and smaller rectangles, we get closer and closer approximations to the area. Taking a limit allows us to calculate the exact area under the curve.

Summation notation makes these calculations easier.

Sigma (Summation) Notation

As mentioned, we will use shapes of known area to approximate the area of an irregular region bounded by curves. This process often requires adding up long strings of numbers. To make it easier to write down these lengthy sums, we look at some new notation here, called sigma notation (also known as summation notation). The Greek capital letter $Σ,$ sigma, is used to express long sums of values in a compact form. For example, if we want to add all the integers from 1 to 20 without sigma notation, we have to write

$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20.$

We could probably skip writing a couple of terms and write

$1 + 2 + 3 + 4 + ⋯ + 19 + 20,$

which is better, but still cumbersome. With sigma notation, we write this sum as

$\underset{i = 1}{∑^{20}} i,$

which is much more compact.

Typically, sigma notation is presented in the form

$\underset{i = 1}{∑^{n}} a_{i}$

where $a_{i}$ describes the terms to be added, and the $i$ is called the index. Each term is evaluated, then we sum all the values, beginning with the value when $i = 1$ and ending with the value when $i = n .$ For example, an expression like

$\underset{i = 1}{∑^{7}} s_{i}$

is interpreted as

$s_{2} + s_{3} + s_{4} + s_{5} + s_{6} + s_{7} .$

Note that the index is used only to keep track of the terms to be added; it does not factor into the calculation of the sum itself. The index is therefore called a dummy variable. We can use any letter we like for the index. Typically, mathematicians use $i$ , $j$ , $k$ , $m$ , and $n$ for indices.

Let’s try an example.

EXAMPLE 1

Using Summation Notation

Write in sigma notation and evaluate the sum of terms $3^{i}$ for $i = 1, 2, 3, 4, 5.$

Write the sum in sigma notation: $1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} .$

Show/Hide Solution

Solution

$\begin{array}{lll} \underset{i = 1}{∑^{5}} 3^{i} & = & 3^{1} + 3^{2} + 3^{3} + 3^{4} + 3^{5} \\ = & 3 + 9 + 27 + 81 + 243 \\ = & 363 \end{array}$

The denominator of each term is a perfect square. Using sigma notation, this sum can be written as $\underset{i = 1}{∑^{5}} \frac{1}{i^{2}}$

There are rules which make the summation process even faster. In Example 1 (a), we calculated 3 raised to all 5 powers, then added them together. What if there had been 100 powers? 1000? The properties below help us find those sums quicker.

Sum, Difference, and Constant Multiple Properties of Sigma Notation

Let $a_{1}, a_{2},…, a_{n}$ and $b_{1}, b_{2},…, b_{n}$ represent two sequences of terms and let $c$ be a constant. The following properties hold for all positive integers $n$ and for integers $m$ , with $1 \leq m \leq n .$

$\underset{i = 1}{∑^{n}} (a_{i} + b_{i}) = \underset{i = 1}{∑^{n}} a_{i} + \underset{i = 1}{∑^{n}} b_{i}$

$\underset{i = 1}{∑^{n}} c a_{i} = c \underset{i = 1}{∑^{n}} a_{i}$

$\underset{i = 1}{∑^{n}} c = n c$

$\underset{i = 1}{∑^{n}} (a_{i} - b_{i}) = \underset{i = 1}{∑^{n}} a_{i} - \underset{i = 1}{∑^{n}} b_{i}$

$\underset{i = 1}{∑^{n}} a_{i} = \underset{i = 1}{∑^{m}} a_{i} + \underset{i = m + 1}{∑^{n}} a_{i}$

These properties are helpful when our sums have many terms. In addition, there are some expressions that we find ourselves adding up very often when using summation notation to approximate areas. These frequently occurring sums are called Power Sums, and formulas for adding them quickly are provided below.

Power Sum Formulas

The sum of a constant $k$ $n$ times is given by
$\underset{i = 1}{∑^{n}} k = \underset{n times}{\underset{⏟}{k + k + k + k + ⋯ + k}} = k \cdot n .$

The sum of $n$ integers is given by
$\underset{i = 1}{∑^{n}} i = 1 + 2 + ⋯ + n = \frac{n (n + 1)}{2} .$

The sum of consecutive integers squared is given by
$\underset{i = 1}{∑^{n}} i^{2} = 1^{2} + 2^{2} + ⋯ + n^{2} = \frac{n (n + 1) (2 n + 1)}{6} .$

The sum of consecutive integers cubed is given by
$\underset{i = 1}{∑^{n}} i^{3} = 1^{3} + 2^{3} + ⋯ + n^{3} = \frac{n^{2} {(n + 1)}^{2}}{4} .$

EXAMPLE 2

Evaluating Sums Using Sigma Notation

Write using sigma notation and evaluate:

The sum of the values of $4 + 3 i$ for $i = 1, 2, ⋯, 100.$

The sum of the values of $f (x) = x^{3}$ for $i = 1, 2, ⋯, 10.$

Show/Hide Solution

Solution

a. First, let’s write $4 + 3 i$ for $i = 1, 2, ⋯, 100$ in summation notation: $\underset{i = 1}{∑^{100}} 4 + 3 i .$

Next, we can use Property #1 for the Sum of Summations to split up this sum.

$\underset{i = 1}{∑^{100}} 4 + 3 i = \underset{i = 1}{∑^{100}} 4 + \underset{i = 1}{∑^{100}} 3 i$

Then, we can use Power Sum Formula #1 to add up 4 100 times. That is,

$\underset{i = 1}{∑^{100}} 4 = 4 \cdot 100 = 400.$

For $\underset{i = 1}{∑^{100}} 3 i$ , we can use Property #3 for Constant Multiples to “kick out” the 3. That is,

$\underset{i = 1}{∑^{100}} 3 i = 3 \underset{i = 1}{∑^{100}} i .$

Now, we can use Power Sum Formula #2 to add up $i$ 100 times and multiply it by 3. That is,

$\begin{array}{lll} 3 \underset{i = 1}{∑^{100}} i & = & 3 \cdot \frac{100 (100 + 1)}{2} \\ = & 3 \cdot \frac{10, 000 + 100}{2} \\ = & \frac{3 \cdot 10, 100}{2} \\ = & \frac{30, 300}{2} \\ = & 15, 150 \end{array}$

Putting both pieces together, we have

$\begin{array}{lll} \underset{i = 1}{∑^{100}} 4 + 3 i & = & 400 + 15, 150 \\ = & 15, 550 \end{array}$

b. Adding the values of $f (x) = x^{3}$ for $i = 1, 2, ⋯, 10$ just means we need to add all the values of $x^{3}$ from $x = 1$ to $x = 10$ . In summation notation that is $\underset{x = 1}{∑^{10}} x^{3}$ .

This is already written in the form of Power Sum Formula #4. So, we can evaluate this as:

$\begin{array}{lll} \underset{x = 1}{∑^{10}} x^{3} & = & \frac{10^{2} {(10 + 1)}^{2}}{4} \\ = & \frac{100 \cdot 121}{4} \\ = & \frac{12, 100}{4} \\ = & 3, 025. \end{array}$

EXAMPLE 3

Evaluating Sums Using Sigma Notation

Write using sigma notation and evaluate:

The sum of the terms ${(i - 3)}^{2}$ for $i = 1, 2, ⋯, 200.$

The sum of the terms $i^{3} - i^{2}$ for $i = 1, 2, ⋯, 200.$

Show/Hide Solution

Solution

a. First, let’s write ${(i - 3)}^{2}$ for $i = 1, 2, ⋯, 200$ in summation notation: $\underset{i = 1}{∑^{200}} {(i - 3)}^{2} .$

This does not match and of our Properties or Power Sums. So, we will have to FOIL out ${(i - 3)}^{2}$ to get it into a form we know.

$\begin{array}{lll} {(i - 3)}^{2} & = & (i - 3) (i - 3) \\ = & i^{2} - 6 i + 9 \end{array}$

So, we can substitute that into our Summation Notation:

$\underset{i = 1}{∑^{200}} {(i - 3)}^{2} = \underset{i = 1}{∑^{200}} i^{2} - 6 i + 9.$

Next, we can use Properties #1, #2, #3 for the Sum of Summations/Differences and Constant Multiples to split up this sum of 3 terms into 3 sums.

$\underset{i = 1}{∑^{200}} i^{2} - 6 i + 9 = \underset{i = 1}{∑^{200}} i^{2} - 6 \underset{i = 1}{∑^{200}} i + \underset{i = 1}{∑^{200}} 9$

From here, we can use Power Sum Formula #3 to evaluate $\underset{i = 1}{∑^{200}} i^{2} .$

$\begin{array}{lll} \underset{i = 1}{∑^{200}} i^{2} & = & \frac{200 (200 + 1) (2 \cdot 200 + 1)}{6} \\ = & \frac{200 \cdot 201 \cdot 401}{6} \\ = & \frac{16, 120, 200}{6} \\ = & 2, 686, 700 \end{array}$

Then, we can use Power Sum Formula #2 to evaluate $- 6 \underset{i = 1}{∑^{200}} i .$

$\begin{array}{lll} - 6 \underset{i = 1}{∑^{200}} i & = & - 6 \cdot \frac{200 (200 + 1)}{2} \\ = & - 6 \cdot \frac{40, 000 + 200}{2} \\ = & \frac{- 6 \cdot 40, 200}{2} \\ = & \frac{241, 200}{2} \\ = & - 120, 600 \end{array}$

And, for the third term, we can use Power Sum Formula #1 to evaluate $\underset{i = 1}{∑^{200}} 9$ .

$\underset{i = 1}{∑^{200}} 9 = 9 \cdot 200 = 1800$ .

Putting all three pieces together, we have

$\begin{array}{lll} \underset{i = 1}{∑^{200}} {(i - 3)}^{2} & = & \underset{i = 1}{∑^{200}} i^{2} - 6 i + 9 \\ = & 2, 686, 700 - 120, 600 + 1800 \\ = & 2, 567, 900 \end{array}$

b. Writing the sum of the terms of the terms $i^{3} - i^{2}$ for $i = 1, 2, ⋯, 200$ in summation notation gives us:

$\underset{i = 1}{∑^{200}} i^{3} - i^{2} .$

This is a combination of two Power Sum formulas. We just need to use Property #2 for the Sum of a Difference to split up these 2 terms.

$\underset{i = 1}{∑^{200}} i^{3} - i^{2} = \underset{i = 1}{∑^{200}} i^{3} - \underset{i = 1}{∑^{200}} i^{2}$

Now we can use Power Sum Formulas #4 and #3 to evaluate these.

$\begin{array}{lll} \underset{i = 1}{∑^{200}} i^{3} - \underset{i = 1}{∑^{200}} i^{2} & = & \frac{200^{2} {(200 + 1)}^{2}}{4} - \frac{200 (200 + 1) (2 \cdot 200 + 1)}{6} \\ = & \frac{40, 00 \cdot 40, 401}{4} - \frac{200 \cdot 201 \cdot 401)}{6} \\ = & \frac{1, 616, 040, 000}{4} - \frac{16, 120, 200}{6} \\ = & 404, 010, 000 - 2, 686, 700 \\ = & 401, 323, 300 \end{array}$

Section 5.1 Exercises

These problems come directly from the OpenStax textbook Calculus Volume 1 Section 5.1 Approximating Areas.

Access this resource for free at https://openstax.org/books/calculus-volume-1/pages/5-1-approximating-areas

[Answers to odd problem numbers are provided at the end of the problem set. Just scroll down!]

Verbal

1. State whether the given sums are equal or unequal.

a. $\underset{i = 1}{∑^{10}} i$ and $\underset{k = 1}{∑^{10}} k$

b. $\underset{i = 1}{∑^{10}} i$ and $\underset{i = 6}{∑^{15}} (i - 5)$

c. $\underset{i = 1}{∑^{10}} i (i - 1)$ and $\underset{j = 0}{∑^{9}} (j + 1) j$

d. $\underset{i = 1}{∑^{10}} i (i - 1)$ and $\underset{k = 1}{∑^{10}} (k^{2} - k) j$

In the following exercises, compute the sums.

2. $\underset{i = 5}{∑^{10}} i$

3. $\underset{i = 5}{∑^{10}} i^{2}$

Suppose that $\underset{i = 1}{∑^{100}} a_{i} = 15$ and $\underset{i = 1}{∑^{100}} b_{i} = 12$ . In the following exercises, compute the sums.

4. $\underset{i = 1}{∑^{100}} (a_{i} + b_{i})$

5. $\underset{i = 1}{∑^{100}} (a_{i} - b_{i})$

6. $\underset{i = 1}{∑^{100}} (3 a_{i} - 4 b_{i})$

7. $\underset{i = 1}{∑^{100}} (5 a_{i} + 4 b_{i})$

In the following exercises, use summation properties and formulas to rewrite and evaluate the sums.

8. $\underset{k = 1}{∑^{200}} 100 (k^{2} - 5 k + 1)$

9. $\underset{j = 1}{∑^{50}} (j^{2} - 2 j)$

10. $\underset{j = 11}{∑^{20}} (j^{2} - 10 j)$

11. $\underset{k = 1}{∑^{25}} ({(2 k)}^{2} - 100 k)$

Answers to Section 5.1 Odd Problems

1. a. They are equal; both represent the sum of the first 10 whole numbers.

b. They are equal; both represent the sum of the first 10 whole numbers.

c. They are equal by substituting $j = i - 1$ .

d. They are equal; the first sum factors the terms of the second.

3. 385−30=355

5. 15−(−12)=27

7. 5(15)+4(−12)=27

9. $\underset{j = 1}{∑^{50}} j^{2} - 2 \underset{j = 1}{∑^{50}} j = \frac{(50) (51) (101)}{6} - \frac{2 (50) (51)}{2} = 40, 375$

11. $4 \underset{k = 1}{∑^{25}} k^{2} - 100 \underset{k = 1}{∑^{25}} k = \frac{4 (25) (26) (51)}{6} - \frac{100 (25) (26)}{2} = - 10, 400$

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Baylor University's Co-requisite Supplement for Calculus I Copyright © 2023 by Amy Graham Goodman is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

This content comes directly from Larry Musolino’s textbook Techniques of Calculus 1 Section 6.1 Antiderivatives.

Access this resource for free at https://psu.pb.unizin.org/math110/chapter/6-1-antiderivatives/

Learning Objectives

Let’s Get Started…

Sigma (Summation) Notation

EXAMPLE 1

Using Summation Notation

Solution

Sum, Difference, and Constant Multiple Properties of Sigma Notation

Power Sum Formulas

EXAMPLE 2

Evaluating Sums Using Sigma Notation

Solution

EXAMPLE 3

Evaluating Sums Using Sigma Notation

Solution

Section 5.1 Exercises

These problems come directly from the OpenStax textbook Calculus Volume 1 Section 5.1 Approximating Areas.

Access this resource for free at https://openstax.org/books/calculus-volume-1/pages/5-1-approximating-areas

Verbal

Answers to Section 5.1 Odd Problems

License

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