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Vincent Selhorst-Jones

Vincent Selhorst-Jones

Area Under a Curve (Integrals)

Slide Duration:

Table of Contents

I. Introduction
Introduction to Precalculus

10m 3s

Intro
0:00
Title of the Course
0:06
Different Names for the Course
0:07
Precalculus
0:12
Math Analysis
0:14
Trigonometry
0:16
Algebra III
0:20
Geometry II
0:24
College Algebra
0:30
Same Concepts
0:36
How do the Lessons Work?
0:54
Introducing Concepts
0:56
Apply Concepts
1:04
Go through Examples
1:25
Who is this Course For?
1:38
Those Who Need eExtra Help with Class Work
1:52
Those Working on Material but not in Formal Class at School
1:54
Those Who Want a Refresher
2:00
Try to Watch the Whole Lesson
2:20
Understanding is So Important
3:56
What to Watch First
5:26
Lesson #2: Sets, Elements, and Numbers
5:30
Lesson #7: Idea of a Function
5:33
Lesson #6: Word Problems
6:04
What to Watch First, cont.
6:46
Lesson #2: Sets, Elements and Numbers
6:56
Lesson #3: Variables, Equations, and Algebra
6:58
Lesson #4: Coordinate Systems
7:00
Lesson #5: Midpoint, Distance, the Pythagorean Theorem and Slope
7:02
Lesson #6: Word Problems
7:10
Lesson #7: Idea of a Function
7:12
Lesson #8: Graphs
7:14
Graphing Calculator Appendix
7:40
What to Watch Last
8:46
Let's get Started!
9:48
Sets, Elements, & Numbers

45m 11s

Intro
0:00
Introduction
0:05
Sets and Elements
1:19
Set
1:20
Element
1:23
Name a Set
2:20
Order The Elements Appear In Has No Effect on the Set
2:55
Describing/ Defining Sets
3:28
Directly Say All the Elements
3:36
Clearly Describing All the Members of the Set
3:55
Describing the Quality (or Qualities) Each member Of the Set Has In Common
4:32
Symbols: 'Element of' and 'Subset of'
6:01
Symbol is ∈
6:03
Subset Symbol is ⊂
6:35
Empty Set
8:07
Symbol is ∅
8:20
Since It's Empty, It is a Subset of All Sets
8:44
Union and Intersection
9:54
Union Symbol is ∪
10:08
Intersection Symbol is ∩
10:18
Sets Can Be Weird Stuff
12:26
Can Have Elements in a Set
12:50
We Can Have Infinite Sets
13:09
Example
13:22
Consider a Set Where We Take a Word and Then Repeat It An Ever Increasing Number of Times
14:08
This Set Has Infinitely Many Distinct Elements
14:40
Numbers as Sets
16:03
Natural Numbers ℕ
16:16
Including 0 and the Negatives ℤ
18:13
Rational Numbers ℚ
19:27
Can Express Rational Numbers with Decimal Expansions
22:05
Irrational Numbers
23:37
Real Numbers ℝ: Put the Rational and Irrational Numbers Together
25:15
Interval Notation and the Real Numbers
26:45
Include the End Numbers
27:06
Exclude the End Numbers
27:33
Example
28:28
Interval Notation: Infinity
29:09
Use -∞ or ∞ to Show an Interval Going on Forever in One Direction or the Other
29:14
Always Use Parentheses
29:50
Examples
30:27
Example 1
31:23
Example 2
35:26
Example 3
38:02
Example 4
42:21
Variables, Equations, & Algebra

35m 31s

Intro
0:00
What is a Variable?
0:05
A Variable is a Placeholder for a Number
0:11
Affects the Output of a Function or a Dependent Variable
0:24
Naming Variables
1:51
Useful to Use Symbols
2:21
What is a Constant?
4:14
A Constant is a Fixed, Unchanging Number
4:28
We Might Refer to a Symbol Representing a Number as a Constant
4:51
What is a Coefficient?
5:33
A Coefficient is a Multiplicative Factor on a Variable
5:37
Not All Coefficients are Constants
5:51
Expressions and Equations
6:42
An Expression is a String of Mathematical Symbols That Make Sense Used Together
7:05
An Equation is a Statement That Two Expression Have the Same Value
8:20
The Idea of Algebra
8:51
Equality
8:59
If Two Things Are the Same *Equal), Then We Can Do the Exact Same Operation to Both and the Results Will Be the Same
9:41
Always Do The Exact Same Thing to Both Sides
12:22
Solving Equations
13:23
When You Are Asked to Solve an Equation, You Are Being Asked to Solve for Something
13:33
Look For What Values Makes the Equation True
13:38
Isolate the Variable by Doing Algebra
14:37
Order of Operations
16:02
Why Certain Operations are Grouped
17:01
When You Don't Have to Worry About Order
17:39
Distributive Property
18:15
It Allows Multiplication to Act Over Addition in Parentheses
18:23
We Can Use the Distributive Property in Reverse to Combine Like Terms
19:05
Substitution
20:03
Use Information From One Equation in Another Equation
20:07
Put Your Substitution in Parentheses
20:44
Example 1
23:17
Example 2
25:49
Example 3
28:11
Example 4
30:02
Coordinate Systems

35m 2s

Intro
0:00
Inherent Order in ℝ
0:05
Real Numbers Come with an Inherent Order
0:11
Positive Numbers
0:21
Negative Numbers
0:58
'Less Than' and 'Greater Than'
2:04
Tip To Help You Remember the Signs
2:56
Inequality
4:06
Less Than or Equal and Greater Than or Equal
4:51
One Dimension: The Number Line
5:36
Graphically Represent ℝ on a Number Line
5:43
Note on Infinities
5:57
With the Number Line, We Can Directly See the Order We Put on ℝ
6:35
Ordered Pairs
7:22
Example
7:34
Allows Us to Talk About Two Numbers at the Same Time
9:41
Ordered Pairs of Real Numbers Cannot be Put Into an Order Like we Did with ℝ
10:41
Two Dimensions: The Plane
13:13
We Can Represent Ordered Pairs with the Plane
13:24
Intersection is known as the Origin
14:31
Plotting the Point
14:32
Plane = Coordinate Plane = Cartesian Plane = ℝ²
17:46
The Plane and Quadrants
18:50
Quadrant I
19:04
Quadrant II
19:21
Quadrant III
20:04
Quadrant IV
20:20
Three Dimensions: Space
21:02
Create Ordered Triplets
21:09
Visually Represent This
21:19
Three-Dimension = Space = ℝ³
21:47
Higher Dimensions
22:24
If We Have n Dimensions, We Call It n-Dimensional Space or ℝ to the nth Power
22:31
We Can Represent Places In This n-Dimensional Space As Ordered Groupings of n Numbers
22:41
Hard to Visualize Higher Dimensional Spaces
23:18
Example 1
25:07
Example 2
26:10
Example 3
28:58
Example 4
31:05
Midpoints, Distance, the Pythagorean Theorem, & Slope

48m 43s

Intro
0:00
Introduction
0:07
Midpoint: One Dimension
2:09
Example of Something More Complex
2:31
Use the Idea of a Middle
3:28
Find the Midpoint of Arbitrary Values a and b
4:17
How They're Equivalent
5:05
Official Midpoint Formula
5:46
Midpoint: Two Dimensions
6:19
The Midpoint Must Occur at the Horizontal Middle and the Vertical Middle
6:38
Arbitrary Pair of Points Example
7:25
Distance: One Dimension
9:26
Absolute Value
10:54
Idea of Forcing Positive
11:06
Distance: One Dimension, Formula
11:47
Distance Between Arbitrary a and b
11:48
Absolute Value Helps When the Distance is Negative
12:41
Distance Formula
12:58
The Pythagorean Theorem
13:24
a²+b²=c²
13:50
Distance: Two Dimensions
14:59
Break Into Horizontal and Vertical Parts and then Use the Pythagorean Theorem
15:16
Distance Between Arbitrary Points (x₁,y₁) and (x₂,y₂)
16:21
Slope
19:30
Slope is the Rate of Change
19:41
m = rise over run
21:27
Slope Between Arbitrary Points (x₁,y₁) and (x₂,y₂)
22:31
Interpreting Slope
24:12
Positive Slope and Negative Slope
25:40
m=1, m=0, m=-1
26:48
Example 1
28:25
Example 2
31:42
Example 3
36:40
Example 4
42:48
Word Problems

56m 31s

Intro
0:00
Introduction
0:05
What is a Word Problem?
0:45
Describes Any Problem That Primarily Gets Its Ideas Across With Words Instead of Math Symbols
0:48
Requires Us to Think
1:32
Why Are They So Hard?
2:11
Reason 1: No Simple Formula to Solve Them
2:16
Reason 2: Harder to Teach Word Problems
2:47
You Can Learn How to Do Them!
3:51
Grades
7:57
'But I'm Never Going to Use This In Real Life'
9:46
Solving Word Problems
12:58
First: Understand the Problem
13:37
Second: What Are You Looking For?
14:33
Third: Set Up Relationships
16:21
Fourth: Solve It!
17:48
Summary of Method
19:04
Examples on Things Other Than Math
20:21
Math-Specific Method: What You Need Now
25:30
Understand What the Problem is Talking About
25:37
Set Up and Name Any Variables You Need to Know
25:56
Set Up Equations Connecting Those Variables to the Information in the Problem Statement
26:02
Use the Equations to Solve for an Answer
26:14
Tip
26:58
Draw Pictures
27:22
Breaking Into Pieces
28:28
Try Out Hypothetical Numbers
29:52
Student Logic
31:27
Jump In!
32:40
Example 1
34:03
Example 2
39:15
Example 3
44:22
Example 4
50:24
II. Functions
Idea of a Function

39m 54s

Intro
0:00
Introduction
0:04
What is a Function?
1:06
A Visual Example and Non-Example
1:30
Function Notation
3:47
f(x)
4:05
Express What Sets the Function Acts On
5:45
Metaphors for a Function
6:17
Transformation
6:28
Map
7:17
Machine
8:56
Same Input Always Gives Same Output
10:01
If We Put the Same Input Into a Function, It Will Always Produce the Same Output
10:11
Example of Something That is Not a Function
11:10
A Non-Numerical Example
12:10
The Functions We Will Use
15:05
Unless Told Otherwise, We Will Assume Every Function Takes in Real Numbers and Outputs Real Numbers
15:11
Usually Told the Rule of a Given Function
15:27
How To Use a Function
16:18
Apply the Rule to Whatever Our Input Value Is
16:28
Make Sure to Wrap Your Substitutions in Parentheses
17:09
Functions and Tables
17:36
Table of Values, Sometimes Called a T-Table
17:46
Example
17:56
Domain: What Goes In
18:55
The Domain is the Set of all Inputs That the Function Can Accept
18:56
Example
19:40
Range: What Comes Out
21:27
The Range is the Set of All Possible Outputs a Function Can Assign
21:34
Example
21:49
Another Example Would Be Our Initial Function From Earlier in This Lesson
22:29
Example 1
23:45
Example 2
25:22
Example 3
27:27
Example 4
29:23
Example 5
33:33
Graphs

58m 26s

Intro
0:00
Introduction
0:04
How to Interpret Graphs
1:17
Input / Independent Variable
1:47
Output / Dependent Variable
2:00
Graph as Input ⇒ Output
2:23
One Way to Think of a Graph: See What Happened to Various Inputs
2:25
Example
2:47
Graph as Location of Solution
4:20
A Way to See Solutions
4:36
Example
5:20
Which Way Should We Interpret?
7:13
Easiest to Think In Terms of How Inputs Are Mapped to Outputs
7:20
Sometimes It's Easier to Think In Terms of Solutions
8:39
Pay Attention to Axes
9:50
Axes Tell Where the Graph Is and What Scale It Has
10:09
Often, The Axes Will Be Square
10:14
Example
12:06
Arrows or No Arrows?
16:07
Will Not Use Arrows at the End of Our Graphs
17:13
Graph Stops Because It Hits the Edge of the Graphing Axes, Not Because the Function Stops
17:18
How to Graph
19:47
Plot Points
20:07
Connect with Curves
21:09
If You Connect with Straight Lines
21:44
Graphs of Functions are Smooth
22:21
More Points ⇒ More Accurate
23:38
Vertical Line Test
27:44
If a Vertical Line Could Intersect More Than One Point On a Graph, It Can Not Be the Graph of a Function
28:41
Every Point on a Graph Tells Us Where the x-Value Below is Mapped
30:07
Domain in Graphs
31:37
The Domain is the Set of All Inputs That a Function Can Accept
31:44
Be Aware That Our Function Probably Continues Past the Edge of Our 'Viewing Window'
33:19
Range in Graphs
33:53
Graphing Calculators: Check the Appendix!
36:55
Example 1
38:37
Example 2
45:19
Example 3
50:41
Example 4
53:28
Example 5
55:50
Properties of Functions

48m 49s

Intro
0:00
Introduction
0:05
Increasing Decreasing Constant
0:43
Looking at a Specific Graph
1:15
Increasing Interval
2:39
Constant Function
4:15
Decreasing Interval
5:10
Find Intervals by Looking at the Graph
5:32
Intervals Show x-values; Write in Parentheses
6:39
Maximum and Minimums
8:48
Relative (Local) Max/Min
10:20
Formal Definition of Relative Maximum
12:44
Formal Definition of Relative Minimum
13:05
Max/Min, More Terms
14:18
Definition of Extrema
15:01
Average Rate of Change
16:11
Drawing a Line for the Average Rate
16:48
Using the Slope of the Secant Line
17:36
Slope in Function Notation
18:45
Zeros/Roots/x-intercepts
19:45
What Zeros in a Function Mean
20:25
Even Functions
22:30
Odd Functions
24:36
Even/Odd Functions and Graphs
26:28
Example of an Even Function
27:12
Example of an Odd Function
28:03
Example 1
29:35
Example 2
33:07
Example 3
40:32
Example 4
42:34
Function Petting Zoo

29m 20s

Intro
0:00
Introduction
0:04
Don't Forget that Axes Matter!
1:44
The Constant Function
2:40
The Identity Function
3:44
The Square Function
4:40
The Cube Function
5:44
The Square Root Function
6:51
The Reciprocal Function
8:11
The Absolute Value Function
10:19
The Trigonometric Functions
11:56
f(x)=sin(x)
12:12
f(x)=cos(x)
12:24
Alternate Axes
12:40
The Exponential and Logarithmic Functions
13:35
Exponential Functions
13:44
Logarithmic Functions
14:24
Alternating Axes
15:17
Transformations and Compositions
16:08
Example 1
17:52
Example 2
18:33
Example 3
20:24
Example 4
26:07
Transformation of Functions

48m 35s

Intro
0:00
Introduction
0:04
Vertical Shift
1:12
Graphical Example
1:21
A Further Explanation
2:16
Vertical Stretch/Shrink
3:34
Graph Shrinks
3:46
Graph Stretches
3:51
A Further Explanation
5:07
Horizontal Shift
6:49
Moving the Graph to the Right
7:28
Moving the Graph to the Left
8:12
A Further Explanation
8:19
Understanding Movement on the x-axis
8:38
Horizontal Stretch/Shrink
12:59
Shrinking the Graph
13:40
Stretching the Graph
13:48
A Further Explanation
13:55
Understanding Stretches from the x-axis
14:12
Vertical Flip (aka Mirror)
16:55
Example Graph
17:07
Multiplying the Vertical Component by -1
17:18
Horizontal Flip (aka Mirror)
18:43
Example Graph
19:01
Multiplying the Horizontal Component by -1
19:54
Summary of Transformations
22:11
Stacking Transformations
24:46
Order Matters
25:20
Transformation Example
25:52
Example 1
29:21
Example 2
34:44
Example 3
38:10
Example 4
43:46
Composite Functions

33m 24s

Intro
0:00
Introduction
0:04
Arithmetic Combinations
0:40
Basic Operations
1:20
Definition of the Four Arithmetic Combinations
1:40
Composite Functions
2:53
The Function as a Machine
3:32
Function Compositions as Multiple Machines
3:59
Notation for Composite Functions
4:46
Two Formats
6:02
Another Visual Interpretation
7:17
How to Use Composite Functions
8:21
Example of on Function acting on Another
9:17
Example 1
11:03
Example 2
15:27
Example 3
21:11
Example 4
27:06
Piecewise Functions

51m 42s

Intro
0:00
Introduction
0:04
Analogies to a Piecewise Function
1:16
Different Potatoes
1:41
Factory Production
2:27
Notations for Piecewise Functions
3:39
Notation Examples from Analogies
6:11
Example of a Piecewise (with Table)
7:24
Example of a Non-Numerical Piecewise
11:35
Graphing Piecewise Functions
14:15
Graphing Piecewise Functions, Example
16:26
Continuous Functions
16:57
Statements of Continuity
19:30
Example of Continuous and Non-Continuous Graphs
20:05
Interesting Functions: the Step Function
22:00
Notation for the Step Function
22:40
How the Step Function Works
22:56
Graph of the Step Function
25:30
Example 1
26:22
Example 2
28:49
Example 3
36:50
Example 4
46:11
Inverse Functions

49m 37s

Intro
0:00
Introduction
0:04
Analogy by picture
1:10
How to Denote the inverse
1:40
What Comes out of the Inverse
1:52
Requirement for Reversing
2:02
The Basketball Factory
2:12
The Importance of Information
2:45
One-to-One
4:04
Requirement for Reversibility
4:21
When a Function has an Inverse
4:43
One-to-One
5:13
Not One-to-One
5:50
Not a Function
6:19
Horizontal Line Test
7:01
How to the test Works
7:12
One-to-One
8:12
Not One-to-One
8:45
Definition: Inverse Function
9:12
Formal Definition
9:21
Caution to Students
10:02
Domain and Range
11:12
Finding the Range of the Function Inverse
11:56
Finding the Domain of the Function Inverse
12:11
Inverse of an Inverse
13:09
Its just x!
13:26
Proof
14:03
Graphical Interpretation
17:07
Horizontal Line Test
17:20
Graph of the Inverse
18:04
Swapping Inputs and Outputs to Draw Inverses
19:02
How to Find the Inverse
21:03
What We Are Looking For
21:21
Reversing the Function
21:38
A Method to Find Inverses
22:33
Check Function is One-to-One
23:04
Swap f(x) for y
23:25
Interchange x and y
23:41
Solve for y
24:12
Replace y with the inverse
24:40
Some Comments
25:01
Keeping Step 2 and 3 Straight
25:44
Switching to Inverse
26:12
Checking Inverses
28:52
How to Check an Inverse
29:06
Quick Example of How to Check
29:56
Example 1
31:48
Example 2
34:56
Example 3
39:29
Example 4
46:19
Variation Direct and Inverse

28m 49s

Intro
0:00
Introduction
0:06
Direct Variation
1:14
Same Direction
1:21
Common Example: Groceries
1:56
Different Ways to Say that Two Things Vary Directly
2:28
Basic Equation for Direct Variation
2:55
Inverse Variation
3:40
Opposite Direction
3:50
Common Example: Gravity
4:53
Different Ways to Say that Two Things Vary Indirectly
5:48
Basic Equation for Indirect Variation
6:33
Joint Variation
7:27
Equation for Joint Variation
7:53
Explanation of the Constant
8:48
Combined Variation
9:35
Gas Law as a Combination
9:44
Single Constant
10:33
Example 1
10:49
Example 2
13:34
Example 3
15:39
Example 4
19:48
III. Polynomials
Intro to Polynomials

38m 41s

Intro
0:00
Introduction
0:04
Definition of a Polynomial
1:04
Starting Integer
2:06
Structure of a Polynomial
2:49
The a Constants
3:34
Polynomial Function
5:13
Polynomial Equation
5:23
Polynomials with Different Variables
5:36
Degree
6:23
Informal Definition
6:31
Find the Largest Exponent Variable
6:44
Quick Examples
7:36
Special Names for Polynomials
8:59
Based on the Degree
9:23
Based on the Number of Terms
10:12
Distributive Property (aka 'FOIL')
11:37
Basic Distributive Property
12:21
Distributing Two Binomials
12:55
Longer Parentheses
15:12
Reverse: Factoring
17:26
Long-Term Behavior of Polynomials
17:48
Examples
18:13
Controlling Term--Term with the Largest Exponent
19:33
Positive and Negative Coefficients on the Controlling Term
20:21
Leading Coefficient Test
22:07
Even Degree, Positive Coefficient
22:13
Even Degree, Negative Coefficient
22:39
Odd Degree, Positive Coefficient
23:09
Odd Degree, Negative Coefficient
23:27
Example 1
25:11
Example 2
27:16
Example 3
31:16
Example 4
34:41
Roots (Zeros) of Polynomials

41m 7s

Intro
0:00
Introduction
0:05
Roots in Graphs
1:17
The x-intercepts
1:33
How to Remember What 'Roots' Are
1:50
Naïve Attempts
2:31
Isolating Variables
2:45
Failures of Isolating Variables
3:30
Missing Solutions
4:59
Factoring: How to Find Roots
6:28
How Factoring Works
6:36
Why Factoring Works
7:20
Steps to Finding Polynomial Roots
9:21
Factoring: How to Find Roots CAUTION
10:08
Factoring is Not Easy
11:32
Factoring Quadratics
13:08
Quadratic Trinomials
13:21
Form of Factored Binomials
13:38
Factoring Examples
14:40
Factoring Quadratics, Check Your Work
16:58
Factoring Higher Degree Polynomials
18:19
Factoring a Cubic
18:32
Factoring a Quadratic
19:04
Factoring: Roots Imply Factors
19:54
Where a Root is, A Factor Is
20:01
How to Use Known Roots to Make Factoring Easier
20:35
Not all Polynomials Can be Factored
22:30
Irreducible Polynomials
23:27
Complex Numbers Help
23:55
Max Number of Roots/Factors
24:57
Limit to Number of Roots Equal to the Degree
25:18
Why there is a Limit
25:25
Max Number of Peaks/Valleys
26:39
Shape Information from Degree
26:46
Example Graph
26:54
Max, But Not Required
28:00
Example 1
28:37
Example 2
31:21
Example 3
36:12
Example 4
38:40
Completing the Square and the Quadratic Formula

39m 43s

Intro
0:00
Introduction
0:05
Square Roots and Equations
0:51
Taking the Square Root to Find the Value of x
0:55
Getting the Positive and Negative Answers
1:05
Completing the Square: Motivation
2:04
Polynomials that are Easy to Solve
2:20
Making Complex Polynomials Easy to Solve
3:03
Steps to Completing the Square
4:30
Completing the Square: Method
7:22
Move C over
7:35
Divide by A
7:44
Find r
7:59
Add to Both Sides to Complete the Square
8:49
Solving Quadratics with Ease
9:56
The Quadratic Formula
11:38
Derivation
11:43
Final Form
12:23
Follow Format to Use Formula
13:38
How Many Roots?
14:53
The Discriminant
15:47
What the Discriminant Tells Us: How Many Roots
15:58
How the Discriminant Works
16:30
Example 1: Complete the Square
18:24
Example 2: Solve the Quadratic
22:00
Example 3: Solve for Zeroes
25:28
Example 4: Using the Quadratic Formula
30:52
Properties of Quadratic Functions

45m 34s

Intro
0:00
Introduction
0:05
Parabolas
0:35
Examples of Different Parabolas
1:06
Axis of Symmetry and Vertex
1:28
Drawing an Axis of Symmetry
1:51
Placing the Vertex
2:28
Looking at the Axis of Symmetry and Vertex for other Parabolas
3:09
Transformations
4:18
Reviewing Transformation Rules
6:28
Note the Different Horizontal Shift Form
7:45
An Alternate Form to Quadratics
8:54
The Constants: k, h, a
9:05
Transformations Formed
10:01
Analyzing Different Parabolas
10:10
Switching Forms by Completing the Square
11:43
Vertex of a Parabola
16:30
Vertex at (h, k)
16:47
Vertex in Terms of a, b, and c Coefficients
17:28
Minimum/Maximum at Vertex
18:19
When a is Positive
18:25
When a is Negative
18:52
Axis of Symmetry
19:54
Incredibly Minor Note on Grammar
20:52
Example 1
21:48
Example 2
26:35
Example 3
28:55
Example 4
31:40
Intermediate Value Theorem and Polynomial Division

46m 8s

Intro
0:00
Introduction
0:05
Reminder: Roots Imply Factors
1:32
The Intermediate Value Theorem
3:41
The Basis: U between a and b
4:11
U is on the Function
4:52
Intermediate Value Theorem, Proof Sketch
5:51
If Not True, the Graph Would Have to Jump
5:58
But Graph is Defined as Continuous
6:43
Finding Roots with the Intermediate Value Theorem
7:01
Picking a and b to be of Different Signs
7:10
Must Be at Least One Root
7:46
Dividing a Polynomial
8:16
Using Roots and Division to Factor
8:38
Long Division Refresher
9:08
The Division Algorithm
12:18
How It Works to Divide Polynomials
12:37
The Parts of the Equation
13:24
Rewriting the Equation
14:47
Polynomial Long Division
16:20
Polynomial Long Division In Action
16:29
One Step at a Time
20:51
Synthetic Division
22:46
Setup
23:11
Synthetic Division, Example
24:44
Which Method Should We Use
26:39
Advantages of Synthetic Method
26:49
Advantages of Long Division
27:13
Example 1
29:24
Example 2
31:27
Example 3
36:22
Example 4
40:55
Complex Numbers

45m 36s

Intro
0:00
Introduction
0:04
A Wacky Idea
1:02
The Definition of the Imaginary Number
1:22
How it Helps Solve Equations
2:20
Square Roots and Imaginary Numbers
3:15
Complex Numbers
5:00
Real Part and Imaginary Part
5:20
When Two Complex Numbers are Equal
6:10
Addition and Subtraction
6:40
Deal with Real and Imaginary Parts Separately
7:36
Two Quick Examples
7:54
Multiplication
9:07
FOIL Expansion
9:14
Note What Happens to the Square of the Imaginary Number
9:41
Two Quick Examples
10:22
Division
11:27
Complex Conjugates
13:37
Getting Rid of i
14:08
How to Denote the Conjugate
14:48
Division through Complex Conjugates
16:11
Multiply by the Conjugate of the Denominator
16:28
Example
17:46
Factoring So-Called 'Irreducible' Quadratics
19:24
Revisiting the Quadratic Formula
20:12
Conjugate Pairs
20:37
But Are the Complex Numbers 'Real'?
21:27
What Makes a Number Legitimate
25:38
Where Complex Numbers are Used
27:20
Still, We Won't See Much of C
29:05
Example 1
30:30
Example 2
33:15
Example 3
38:12
Example 4
42:07
Fundamental Theorem of Algebra

19m 9s

Intro
0:00
Introduction
0:05
Idea: Hidden Roots
1:16
Roots in Complex Form
1:42
All Polynomials Have Roots
2:08
Fundamental Theorem of Algebra
2:21
Where Are All the Imaginary Roots, Then?
3:17
All Roots are Complex
3:45
Real Numbers are a Subset of Complex Numbers
3:59
The n Roots Theorem
5:01
For Any Polynomial, Its Degree is Equal to the Number of Roots
5:11
Equivalent Statement
5:24
Comments: Multiplicity
6:29
Non-Distinct Roots
6:59
Denoting Multiplicity
7:20
Comments: Complex Numbers Necessary
7:41
Comments: Complex Coefficients Allowed
8:55
Comments: Existence Theorem
9:59
Proof Sketch of n Roots Theorem
10:45
First Root
11:36
Second Root
13:23
Continuation to Find all Roots
16:00
IV. Rational Functions
Rational Functions and Vertical Asymptotes

33m 22s

Intro
0:00
Introduction
0:05
Definition of a Rational Function
1:20
Examples of Rational Functions
2:30
Why They are Called 'Rational'
2:47
Domain of a Rational Function
3:15
Undefined at Denominator Zeros
3:25
Otherwise all Reals
4:16
Investigating a Fundamental Function
4:50
The Domain of the Function
5:04
What Occurs at the Zeroes of the Denominator
5:20
Idea of a Vertical Asymptote
6:23
What's Going On?
6:58
Approaching x=0 from the left
7:32
Approaching x=0 from the right
8:34
Dividing by Very Small Numbers Results in Very Large Numbers
9:31
Definition of a Vertical Asymptote
10:05
Vertical Asymptotes and Graphs
11:15
Drawing Asymptotes by Using a Dashed Line
11:27
The Graph Can Never Touch Its Undefined Point
12:00
Not All Zeros Give Asymptotes
13:02
Special Cases: When Numerator and Denominator Go to Zero at the Same Time
14:58
Cancel out Common Factors
15:49
How to Find Vertical Asymptotes
16:10
Figure out What Values Are Not in the Domain of x
16:24
Determine if the Numerator and Denominator Share Common Factors and Cancel
16:45
Find Denominator Roots
17:33
Note if Asymptote Approaches Negative or Positive Infinity
18:06
Example 1
18:57
Example 2
21:26
Example 3
23:04
Example 4
30:01
Horizontal Asymptotes

34m 16s

Intro
0:00
Introduction
0:05
Investigating a Fundamental Function
0:53
What Happens as x Grows Large
1:00
Different View
1:12
Idea of a Horizontal Asymptote
1:36
What's Going On?
2:24
What Happens as x Grows to a Large Negative Number
2:49
What Happens as x Grows to a Large Number
3:30
Dividing by Very Large Numbers Results in Very Small Numbers
3:52
Example Function
4:41
Definition of a Vertical Asymptote
8:09
Expanding the Idea
9:03
What's Going On?
9:48
What Happens to the Function in the Long Run?
9:51
Rewriting the Function
10:13
Definition of a Slant Asymptote
12:09
Symbolical Definition
12:30
Informal Definition
12:45
Beyond Slant Asymptotes
13:03
Not Going Beyond Slant Asymptotes
14:39
Horizontal/Slant Asymptotes and Graphs
15:43
How to Find Horizontal and Slant Asymptotes
16:52
How to Find Horizontal Asymptotes
17:12
Expand the Given Polynomials
17:18
Compare the Degrees of the Numerator and Denominator
17:40
How to Find Slant Asymptotes
20:05
Slant Asymptotes Exist When n+m=1
20:08
Use Polynomial Division
20:24
Example 1
24:32
Example 2
25:53
Example 3
26:55
Example 4
29:22
Graphing Asymptotes in a Nutshell

49m 7s

Intro
0:00
Introduction
0:05
A Process for Graphing
1:22
1. Factor Numerator and Denominator
1:50
2. Find Domain
2:53
3. Simplifying the Function
3:59
4. Find Vertical Asymptotes
4:59
5. Find Horizontal/Slant Asymptotes
5:24
6. Find Intercepts
7:35
7. Draw Graph (Find Points as Necessary)
9:21
Draw Graph Example
11:21
Vertical Asymptote
11:41
Horizontal Asymptote
11:50
Other Graphing
12:16
Test Intervals
15:08
Example 1
17:57
Example 2
23:01
Example 3
29:02
Example 4
33:37
Partial Fractions

44m 56s

Intro
0:00
Introduction: Idea
0:04
Introduction: Prerequisites and Uses
1:57
Proper vs. Improper Polynomial Fractions
3:11
Possible Things in the Denominator
4:38
Linear Factors
6:16
Example of Linear Factors
7:03
Multiple Linear Factors
7:48
Irreducible Quadratic Factors
8:25
Example of Quadratic Factors
9:26
Multiple Quadratic Factors
9:49
Mixing Factor Types
10:28
Figuring Out the Numerator
11:10
How to Solve for the Constants
11:30
Quick Example
11:40
Example 1
14:29
Example 2
18:35
Example 3
20:33
Example 4
28:51
V. Exponential & Logarithmic Functions
Understanding Exponents

35m 17s

Intro
0:00
Introduction
0:05
Fundamental Idea
1:46
Expanding the Idea
2:28
Multiplication of the Same Base
2:40
Exponents acting on Exponents
3:45
Different Bases with the Same Exponent
4:31
To the Zero
5:35
To the First
5:45
Fundamental Rule with the Zero Power
6:35
To the Negative
7:45
Any Number to a Negative Power
8:14
A Fraction to a Negative Power
9:58
Division with Exponential Terms
10:41
To the Fraction
11:33
Square Root
11:58
Any Root
12:59
Summary of Rules
14:38
To the Irrational
17:21
Example 1
20:34
Example 2
23:42
Example 3
27:44
Example 4
31:44
Example 5
33:15
Exponential Functions

47m 4s

Intro
0:00
Introduction
0:05
Definition of an Exponential Function
0:48
Definition of the Base
1:02
Restrictions on the Base
1:16
Computing Exponential Functions
2:29
Harder Computations
3:10
When to Use a Calculator
3:21
Graphing Exponential Functions: a>1
6:02
Three Examples
6:13
What to Notice on the Graph
7:44
A Story
8:27
Story Diagram
9:15
Increasing Exponentials
11:29
Story Morals
14:40
Application: Compound Interest
15:15
Compounding Year after Year
16:01
Function for Compounding Interest
16:51
A Special Number: e
20:55
Expression for e
21:28
Where e stabilizes
21:55
Application: Continuously Compounded Interest
24:07
Equation for Continuous Compounding
24:22
Exponential Decay 0<a<1
25:50
Three Examples
26:11
Why they 'lose' value
26:54
Example 1
27:47
Example 2
33:11
Example 3
36:34
Example 4
41:28
Introduction to Logarithms

40m 31s

Intro
0:00
Introduction
0:04
Definition of a Logarithm, Base 2
0:51
Log 2 Defined
0:55
Examples
2:28
Definition of a Logarithm, General
3:23
Examples of Logarithms
5:15
Problems with Unusual Bases
7:38
Shorthand Notation: ln and log
9:44
base e as ln
10:01
base 10 as log
10:34
Calculating Logarithms
11:01
using a calculator
11:34
issues with other bases
11:58
Graphs of Logarithms
13:21
Three Examples
13:29
Slow Growth
15:19
Logarithms as Inverse of Exponentiation
16:02
Using Base 2
16:05
General Case
17:10
Looking More Closely at Logarithm Graphs
19:16
The Domain of Logarithms
20:41
Thinking about Logs like Inverses
21:08
The Alternate
24:00
Example 1
25:59
Example 2
30:03
Example 3
32:49
Example 4
37:34
Properties of Logarithms

42m 33s

Intro
0:00
Introduction
0:04
Basic Properties
1:12
Inverse--log(exp)
1:43
A Key Idea
2:44
What We Get through Exponentiation
3:18
B Always Exists
4:50
Inverse--exp(log)
5:53
Logarithm of a Power
7:44
Logarithm of a Product
10:07
Logarithm of a Quotient
13:48
Caution! There Is No Rule for loga(M+N)
16:12
Summary of Properties
17:42
Change of Base--Motivation
20:17
No Calculator Button
20:59
A Specific Example
21:45
Simplifying
23:45
Change of Base--Formula
24:14
Example 1
25:47
Example 2
29:08
Example 3
31:14
Example 4
34:13
Solving Exponential and Logarithmic Equations

34m 10s

Intro
0:00
Introduction
0:05
One to One Property
1:09
Exponential
1:26
Logarithmic
1:44
Specific Considerations
2:02
One-to-One Property
3:30
Solving by One-to-One
4:11
Inverse Property
6:09
Solving by Inverses
7:25
Dealing with Equations
7:50
Example of Taking an Exponent or Logarithm of an Equation
9:07
A Useful Property
11:57
Bring Down Exponents
12:01
Try to Simplify
13:20
Extraneous Solutions
13:45
Example 1
16:37
Example 2
19:39
Example 3
21:37
Example 4
26:45
Example 5
29:37
Application of Exponential and Logarithmic Functions

48m 46s

Intro
0:00
Introduction
0:06
Applications of Exponential Functions
1:07
A Secret!
2:17
Natural Exponential Growth Model
3:07
Figure out r
3:34
A Secret!--Why Does It Work?
4:44
e to the r Morphs
4:57
Example
5:06
Applications of Logarithmic Functions
8:32
Examples
8:43
What Logarithms are Useful For
9:53
Example 1
11:29
Example 2
15:30
Example 3
26:22
Example 4
32:05
Example 5
39:19
VI. Trigonometric Functions
Angles

39m 5s

Intro
0:00
Degrees
0:22
Circle is 360 Degrees
0:48
Splitting a Circle
1:13
Radians
2:08
Circle is 2 Pi Radians
2:31
One Radian
2:52
Half-Circle and Right Angle
4:00
Converting Between Degrees and Radians
6:24
Formulas for Degrees and Radians
6:52
Coterminal, Complementary, Supplementary Angles
7:23
Coterminal Angles
7:30
Complementary Angles
9:40
Supplementary Angles
10:08
Example 1: Dividing a Circle
10:38
Example 2: Converting Between Degrees and Radians
11:56
Example 3: Quadrants and Coterminal Angles
14:18
Extra Example 1: Common Angle Conversions
-1
Extra Example 2: Quadrants and Coterminal Angles
-2
Sine and Cosine Functions

43m 16s

Intro
0:00
Sine and Cosine
0:15
Unit Circle
0:22
Coordinates on Unit Circle
1:03
Right Triangles
1:52
Adjacent, Opposite, Hypotenuse
2:25
Master Right Triangle Formula: SOHCAHTOA
2:48
Odd Functions, Even Functions
4:40
Example: Odd Function
4:56
Example: Even Function
7:30
Example 1: Sine and Cosine
10:27
Example 2: Graphing Sine and Cosine Functions
14:39
Example 3: Right Triangle
21:40
Example 4: Odd, Even, or Neither
26:01
Extra Example 1: Right Triangle
-1
Extra Example 2: Graphing Sine and Cosine Functions
-2
Sine and Cosine Values of Special Angles

33m 5s

Intro
0:00
45-45-90 Triangle and 30-60-90 Triangle
0:08
45-45-90 Triangle
0:21
30-60-90 Triangle
2:06
Mnemonic: All Students Take Calculus (ASTC)
5:21
Using the Unit Circle
5:59
New Angles
6:21
Other Quadrants
9:43
Mnemonic: All Students Take Calculus
10:13
Example 1: Convert, Quadrant, Sine/Cosine
13:11
Example 2: Convert, Quadrant, Sine/Cosine
16:48
Example 3: All Angles and Quadrants
20:21
Extra Example 1: Convert, Quadrant, Sine/Cosine
-1
Extra Example 2: All Angles and Quadrants
-2
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D

52m 3s

Intro
0:00
Amplitude and Period of a Sine Wave
0:38
Sine Wave Graph
0:58
Amplitude: Distance from Middle to Peak
1:18
Peak: Distance from Peak to Peak
2:41
Phase Shift and Vertical Shift
4:13
Phase Shift: Distance Shifted Horizontally
4:16
Vertical Shift: Distance Shifted Vertically
6:48
Example 1: Amplitude/Period/Phase and Vertical Shift
8:04
Example 2: Amplitude/Period/Phase and Vertical Shift
17:39
Example 3: Find Sine Wave Given Attributes
25:23
Extra Example 1: Amplitude/Period/Phase and Vertical Shift
-1
Extra Example 2: Find Cosine Wave Given Attributes
-2
Tangent and Cotangent Functions

36m 4s

Intro
0:00
Tangent and Cotangent Definitions
0:21
Tangent Definition
0:25
Cotangent Definition
0:47
Master Formula: SOHCAHTOA
1:01
Mnemonic
1:16
Tangent and Cotangent Values
2:29
Remember Common Values of Sine and Cosine
2:46
90 Degrees Undefined
4:36
Slope and Menmonic: ASTC
5:47
Uses of Tangent
5:54
Example: Tangent of Angle is Slope
6:09
Sign of Tangent in Quadrants
7:49
Example 1: Graph Tangent and Cotangent Functions
10:42
Example 2: Tangent and Cotangent of Angles
16:09
Example 3: Odd, Even, or Neither
18:56
Extra Example 1: Tangent and Cotangent of Angles
-1
Extra Example 2: Tangent and Cotangent of Angles
-2
Secant and Cosecant Functions

27m 18s

Intro
0:00
Secant and Cosecant Definitions
0:17
Secant Definition
0:18
Cosecant Definition
0:33
Example 1: Graph Secant Function
0:48
Example 2: Values of Secant and Cosecant
6:49
Example 3: Odd, Even, or Neither
12:49
Extra Example 1: Graph of Cosecant Function
-1
Extra Example 2: Values of Secant and Cosecant
-2
Inverse Trigonometric Functions

32m 58s

Intro
0:00
Arcsine Function
0:24
Restrictions between -1 and 1
0:43
Arcsine Notation
1:26
Arccosine Function
3:07
Restrictions between -1 and 1
3:36
Cosine Notation
3:53
Arctangent Function
4:30
Between -Pi/2 and Pi/2
4:44
Tangent Notation
5:02
Example 1: Domain/Range/Graph of Arcsine
5:45
Example 2: Arcsin/Arccos/Arctan Values
10:46
Example 3: Domain/Range/Graph of Arctangent
17:14
Extra Example 1: Domain/Range/Graph of Arccosine
-1
Extra Example 2: Arcsin/Arccos/Arctan Values
-2
Computations of Inverse Trigonometric Functions

31m 8s

Intro
0:00
Inverse Trigonometric Function Domains and Ranges
0:31
Arcsine
0:41
Arccosine
1:14
Arctangent
1:41
Example 1: Arcsines of Common Values
2:44
Example 2: Odd, Even, or Neither
5:57
Example 3: Arccosines of Common Values
12:24
Extra Example 1: Arctangents of Common Values
-1
Extra Example 2: Arcsin/Arccos/Arctan Values
-2
VII. Trigonometric Identities
Pythagorean Identity

19m 11s

Intro
0:00
Pythagorean Identity
0:17
Pythagorean Triangle
0:27
Pythagorean Identity
0:45
Example 1: Use Pythagorean Theorem to Prove Pythagorean Identity
1:14
Example 2: Find Angle Given Cosine and Quadrant
4:18
Example 3: Verify Trigonometric Identity
8:00
Extra Example 1: Use Pythagorean Identity to Prove Pythagorean Theorem
-1
Extra Example 2: Find Angle Given Cosine and Quadrant
-2
Identity Tan(squared)x+1=Sec(squared)x

23m 16s

Intro
0:00
Main Formulas
0:19
Companion to Pythagorean Identity
0:27
For Cotangents and Cosecants
0:52
How to Remember
0:58
Example 1: Prove the Identity
1:40
Example 2: Given Tan Find Sec
3:42
Example 3: Prove the Identity
7:45
Extra Example 1: Prove the Identity
-1
Extra Example 2: Given Sec Find Tan
-2
Addition and Subtraction Formulas

52m 52s

Intro
0:00
Addition and Subtraction Formulas
0:09
How to Remember
0:48
Cofunction Identities
1:31
How to Remember Graphically
1:44
Where to Use Cofunction Identities
2:52
Example 1: Derive the Formula for cos(A-B)
3:08
Example 2: Use Addition and Subtraction Formulas
16:03
Example 3: Use Addition and Subtraction Formulas to Prove Identity
25:11
Extra Example 1: Use cos(A-B) and Cofunction Identities
-1
Extra Example 2: Convert to Radians and use Formulas
-2
Double Angle Formulas

29m 5s

Intro
0:00
Main Formula
0:07
How to Remember from Addition Formula
0:18
Two Other Forms
1:35
Example 1: Find Sine and Cosine of Angle using Double Angle
3:16
Example 2: Prove Trigonometric Identity using Double Angle
9:37
Example 3: Use Addition and Subtraction Formulas
12:38
Extra Example 1: Find Sine and Cosine of Angle using Double Angle
-1
Extra Example 2: Prove Trigonometric Identity using Double Angle
-2
Half-Angle Formulas

43m 55s

Intro
0:00
Main Formulas
0:09
Confusing Part
0:34
Example 1: Find Sine and Cosine of Angle using Half-Angle
0:54
Example 2: Prove Trigonometric Identity using Half-Angle
11:51
Example 3: Prove the Half-Angle Formula for Tangents
18:39
Extra Example 1: Find Sine and Cosine of Angle using Half-Angle
-1
Extra Example 2: Prove Trigonometric Identity using Half-Angle
-2
VIII. Applications of Trigonometry
Trigonometry in Right Angles

25m 43s

Intro
0:00
Master Formula for Right Angles
0:11
SOHCAHTOA
0:15
Only for Right Triangles
1:26
Example 1: Find All Angles in a Triangle
2:19
Example 2: Find Lengths of All Sides of Triangle
7:39
Example 3: Find All Angles in a Triangle
11:00
Extra Example 1: Find All Angles in a Triangle
-1
Extra Example 2: Find Lengths of All Sides of Triangle
-2
Law of Sines

56m 40s

Intro
0:00
Law of Sines Formula
0:18
SOHCAHTOA
0:27
Any Triangle
0:59
Graphical Representation
1:25
Solving Triangle Completely
2:37
When to Use Law of Sines
2:55
ASA, SAA, SSA, AAA
2:59
SAS, SSS for Law of Cosines
7:11
Example 1: How Many Triangles Satisfy Conditions, Solve Completely
8:44
Example 2: How Many Triangles Satisfy Conditions, Solve Completely
15:30
Example 3: How Many Triangles Satisfy Conditions, Solve Completely
28:32
Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
-1
Extra Example 2: How Many Triangles Satisfy Conditions, Solve Completely
-2
Law of Cosines

49m 5s

Intro
0:00
Law of Cosines Formula
0:23
Graphical Representation
0:34
Relates Sides to Angles
1:00
Any Triangle
1:20
Generalization of Pythagorean Theorem
1:32
When to Use Law of Cosines
2:26
SAS, SSS
2:30
Heron's Formula
4:49
Semiperimeter S
5:11
Example 1: How Many Triangles Satisfy Conditions, Solve Completely
5:53
Example 2: How Many Triangles Satisfy Conditions, Solve Completely
15:19
Example 3: Find Area of a Triangle Given All Side Lengths
26:33
Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
-1
Extra Example 2: Length of Third Side and Area of Triangle
-2
Finding the Area of a Triangle

27m 37s

Intro
0:00
Master Right Triangle Formula and Law of Cosines
0:19
SOHCAHTOA
0:27
Law of Cosines
1:23
Heron's Formula
2:22
Semiperimeter S
2:37
Example 1: Area of Triangle with Two Sides and One Angle
3:12
Example 2: Area of Triangle with Three Sides
6:11
Example 3: Area of Triangle with Three Sides, No Heron's Formula
8:50
Extra Example 1: Area of Triangle with Two Sides and One Angle
-1
Extra Example 2: Area of Triangle with Two Sides and One Angle
-2
Word Problems and Applications of Trigonometry

34m 25s

Intro
0:00
Formulas to Remember
0:11
SOHCAHTOA
0:15
Law of Sines
0:55
Law of Cosines
1:48
Heron's Formula
2:46
Example 1: Telephone Pole Height
4:01
Example 2: Bridge Length
7:48
Example 3: Area of Triangular Field
14:20
Extra Example 1: Kite Height
-1
Extra Example 2: Roads to a Town
-2
IX. Systems of Equations and Inequalities
Systems of Linear Equations

55m 40s

Intro
0:00
Introduction
0:04
Graphs as Location of 'True'
1:49
All Locations that Make the Function True
2:25
Understand the Relationship Between Solutions and the Graph
3:43
Systems as Graphs
4:07
Equations as Lines
4:20
Intersection Point
5:19
Three Possibilities for Solutions
6:17
Independent
6:24
Inconsistent
6:36
Dependent
7:06
Solving by Substitution
8:37
Solve for One Variable
9:07
Substitute into the Second Equation
9:34
Solve for Both Variables
10:12
What If a System is Inconsistent or Dependent?
11:08
No Solutions
11:25
Infinite Solutions
12:30
Solving by Elimination
13:56
Example
14:22
Determining the Number of Solutions
16:30
Why Elimination Makes Sense
17:25
Solving by Graphing Calculator
19:59
Systems with More than Two Variables
23:22
Example 1
25:49
Example 2
30:22
Example 3
34:11
Example 4
38:55
Example 5
46:01
(Non-) Example 6
53:37
Systems of Linear Inequalities

1h 13s

Intro
0:00
Introduction
0:04
Inequality Refresher-Solutions
0:46
Equation Solutions vs. Inequality Solutions
1:02
Essentially a Wide Variety of Answers
1:35
Refresher--Negative Multiplication Flips
1:43
Refresher--Negative Flips: Why?
3:19
Multiplication by a Negative
3:43
The Relationship Flips
3:55
Refresher--Stick to Basic Operations
4:34
Linear Equations in Two Variables
6:50
Graphing Linear Inequalities
8:28
Why It Includes a Whole Section
8:43
How to Show The Difference Between Strict and Not Strict Inequalities
10:08
Dashed Line--Not Solutions
11:10
Solid Line--Are Solutions
11:24
Test Points for Shading
11:42
Example of Using a Point
12:41
Drawing Shading from the Point
13:14
Graphing a System
14:53
Set of Solutions is the Overlap
15:17
Example
15:22
Solutions are Best Found Through Graphing
18:05
Linear Programming-Idea
19:52
Use a Linear Objective Function
20:15
Variables in Objective Function have Constraints
21:24
Linear Programming-Method
22:09
Rearrange Equations
22:21
Graph
22:49
Critical Solution is at the Vertex of the Overlap
23:40
Try Each Vertice
24:35
Example 1
24:58
Example 2
28:57
Example 3
33:48
Example 4
43:10
Nonlinear Systems

41m 1s

Intro
0:00
Introduction
0:06
Substitution
1:12
Example
1:22
Elimination
3:46
Example
3:56
Elimination is Less Useful for Nonlinear Systems
4:56
Graphing
5:56
Using a Graphing Calculator
6:44
Number of Solutions
8:44
Systems of Nonlinear Inequalities
10:02
Graph Each Inequality
10:06
Dashed and/or Solid
10:18
Shade Appropriately
11:14
Example 1
13:24
Example 2
15:50
Example 3
22:02
Example 4
29:06
Example 4, cont.
33:40
X. Vectors and Matrices
Vectors

1h 9m 31s

Intro
0:00
Introduction
0:10
Magnitude of the Force
0:22
Direction of the Force
0:48
Vector
0:52
Idea of a Vector
1:30
How Vectors are Denoted
2:00
Component Form
3:20
Angle Brackets and Parentheses
3:50
Magnitude/Length
4:26
Denoting the Magnitude of a Vector
5:16
Direction/Angle
7:52
Always Draw a Picture
8:50
Component Form from Magnitude & Angle
10:10
Scaling by Scalars
14:06
Unit Vectors
16:26
Combining Vectors - Algebraically
18:10
Combining Vectors - Geometrically
19:54
Resultant Vector
20:46
Alternate Component Form: i, j
21:16
The Zero Vector
23:18
Properties of Vectors
24:20
No Multiplication (Between Vectors)
28:30
Dot Product
29:40
Motion in a Medium
30:10
Fish in an Aquarium Example
31:38
More Than Two Dimensions
33:12
More Than Two Dimensions - Magnitude
34:18
Example 1
35:26
Example 2
38:10
Example 3
45:48
Example 4
50:40
Example 4, cont.
56:07
Example 5
1:01:32
Dot Product & Cross Product

35m 20s

Intro
0:00
Introduction
0:08
Dot Product - Definition
0:42
Dot Product Results in a Scalar, Not a Vector
2:10
Example in Two Dimensions
2:34
Angle and the Dot Product
2:58
The Dot Product of Two Vectors is Deeply Related to the Angle Between the Two Vectors
2:59
Proof of Dot Product Formula
4:14
Won't Directly Help Us Better Understand Vectors
4:18
Dot Product - Geometric Interpretation
4:58
We Can Interpret the Dot Product as a Measure of How Long and How Parallel Two Vectors Are
7:26
Dot Product - Perpendicular Vectors
8:24
If the Dot Product of Two Vectors is 0, We Know They are Perpendicular to Each Other
8:54
Cross Product - Definition
11:08
Cross Product Only Works in Three Dimensions
11:09
Cross Product - A Mnemonic
12:16
The Determinant of a 3 x 3 Matrix and Standard Unit Vectors
12:17
Cross Product - Geometric Interpretations
14:30
The Right-Hand Rule
15:17
Cross Product - Geometric Interpretations Cont.
17:00
Example 1
18:40
Example 2
22:50
Example 3
24:04
Example 4
26:20
Bonus Round
29:18
Proof: Dot Product Formula
29:24
Proof: Dot Product Formula, cont.
30:38
Matrices

54m 7s

Intro
0:00
Introduction
0:08
Definition of a Matrix
3:02
Size or Dimension
3:58
Square Matrix
4:42
Denoted by Capital Letters
4:56
When are Two Matrices Equal?
5:04
Examples of Matrices
6:44
Rows x Columns
6:46
Talking About Specific Entries
7:48
We Use Capitals to Denote a Matrix and Lower Case to Denotes Its Entries
8:32
Using Entries to Talk About Matrices
10:08
Scalar Multiplication
11:26
Scalar = Real Number
11:34
Example
12:36
Matrix Addition
13:08
Example
14:22
Matrix Multiplication
15:00
Example
18:52
Matrix Multiplication, cont.
19:58
Matrix Multiplication and Order (Size)
25:26
Make Sure Their Orders are Compatible
25:27
Matrix Multiplication is NOT Commutative
28:20
Example
30:08
Special Matrices - Zero Matrix (0)
32:48
Zero Matrix Has 0 for All of its Entries
32:49
Special Matrices - Identity Matrix (I)
34:14
Identity Matrix is a Square Matrix That Has 1 for All Its Entries on the Main Diagonal and 0 for All Other Entries
34:15
Example 1
36:16
Example 2
40:00
Example 3
44:54
Example 4
50:08
Determinants & Inverses of Matrices

47m 12s

Intro
0:00
Introduction
0:06
Not All Matrices Are Invertible
1:30
What Must a Matrix Have to Be Invertible?
2:08
Determinant
2:32
The Determinant is a Real Number Associated With a Square Matrix
2:38
If the Determinant of a Matrix is Nonzero, the Matrix is Invertible
3:40
Determinant of a 2 x 2 Matrix
4:34
Think in Terms of Diagonals
5:12
Minors and Cofactors - Minors
6:24
Example
6:46
Minors and Cofactors - Cofactors
8:00
Cofactor is Closely Based on the Minor
8:01
Alternating Sign Pattern
9:04
Determinant of Larger Matrices
10:56
Example
13:00
Alternative Method for 3x3 Matrices
16:46
Not Recommended
16:48
Inverse of a 2 x 2 Matrix
19:02
Inverse of Larger Matrices
20:00
Using Inverse Matrices
21:06
When Multiplied Together, They Create the Identity Matrix
21:24
Example 1
23:45
Example 2
27:21
Example 3
32:49
Example 4
36:27
Finding the Inverse of Larger Matrices
41:59
General Inverse Method - Step 1
43:25
General Inverse Method - Step 2
43:27
General Inverse Method - Step 2, cont.
43:27
General Inverse Method - Step 3
45:15
Using Matrices to Solve Systems of Linear Equations

58m 34s

Intro
0:00
Introduction
0:12
Augmented Matrix
1:44
We Can Represent the Entire Linear System With an Augmented Matrix
1:50
Row Operations
3:22
Interchange the Locations of Two Rows
3:50
Multiply (or Divide) a Row by a Nonzero Number
3:58
Add (or Subtract) a Multiple of One Row to Another
4:12
Row Operations - Keep Notes!
5:50
Suggested Symbols
7:08
Gauss-Jordan Elimination - Idea
8:04
Gauss-Jordan Elimination - Idea, cont.
9:16
Reduced Row-Echelon Form
9:18
Gauss-Jordan Elimination - Method
11:36
Begin by Writing the System As An Augmented Matrix
11:38
Gauss-Jordan Elimination - Method, cont.
13:48
Cramer's Rule - 2 x 2 Matrices
17:08
Cramer's Rule - n x n Matrices
19:24
Solving with Inverse Matrices
21:10
Solving Inverse Matrices, cont.
25:28
The Mighty (Graphing) Calculator
26:38
Example 1
29:56
Example 2
33:56
Example 3
37:00
Example 3, cont.
45:04
Example 4
51:28
XI. Alternate Ways to Graph
Parametric Equations

53m 33s

Intro
0:00
Introduction
0:06
Definition
1:10
Plane Curve
1:24
The Key Idea
2:00
Graphing with Parametric Equations
2:52
Same Graph, Different Equations
5:04
How Is That Possible?
5:36
Same Graph, Different Equations, cont.
5:42
Here's Another to Consider
7:56
Same Plane Curve, But Still Different
8:10
A Metaphor for Parametric Equations
9:36
Think of Parametric Equations As a Way to Describe the Motion of An Object
9:38
Graph Shows Where It Went, But Not Speed
10:32
Eliminating Parameters
12:14
Rectangular Equation
12:16
Caution
13:52
Creating Parametric Equations
14:30
Interesting Graphs
16:38
Graphing Calculators, Yay!
19:18
Example 1
22:36
Example 2
28:26
Example 3
37:36
Example 4
41:00
Projectile Motion
44:26
Example 5
47:00
Polar Coordinates

48m 7s

Intro
0:00
Introduction
0:04
Polar Coordinates Give Us a Way To Describe the Location of a Point
0:26
Polar Equations and Functions
0:50
Plotting Points with Polar Coordinates
1:06
The Distance of the Point from the Origin
1:09
The Angle of the Point
1:33
Give Points as the Ordered Pair (r,θ)
2:03
Visualizing Plotting in Polar Coordinates
2:32
First Way We Can Plot
2:39
Second Way We Can Plot
2:50
First, We'll Look at Visualizing r, Then θ
3:09
Rotate the Length Counter-Clockwise by θ
3:38
Alternatively, We Can Visualize θ, Then r
4:06
'Polar Graph Paper'
6:17
Horizontal and Vertical Tick Marks Are Not Useful for Polar
6:42
Use Concentric Circles to Helps Up See Distance From the Pole
7:08
Can Use Arc Sectors to See Angles
7:57
Multiple Ways to Name a Point
9:17
Examples
9:30
For Any Angle θ, We Can Make an Equivalent Angle
10:44
Negative Values for r
11:58
If r Is Negative, We Go In The Direction Opposite the One That The Angle θ Points Out
12:22
Another Way to Name the Same Point: Add π to θ and Make r Negative
13:44
Converting Between Rectangular and Polar
14:37
Rectangular Way to Name
14:43
Polar Way to Name
14:52
The Rectangular System Must Have a Right Angle Because It's Based on a Rectangle
15:08
Connect Both Systems Through Basic Trigonometry
15:38
Equation to Convert From Polar to Rectangular Coordinate Systems
16:55
Equation to Convert From Rectangular to Polar Coordinate Systems
17:13
Converting to Rectangular is Easy
17:20
Converting to Polar is a Bit Trickier
17:21
Draw Pictures
18:55
Example 1
19:50
Example 2
25:17
Example 3
31:05
Example 4
35:56
Example 5
41:49
Polar Equations & Functions

38m 16s

Intro
0:00
Introduction
0:04
Equations and Functions
1:16
Independent Variable
1:21
Dependent Variable
1:30
Examples
1:46
Always Assume That θ Is In Radians
2:44
Graphing in Polar Coordinates
3:29
Graph is the Same Way We Graph 'Normal' Stuff
3:32
Example
3:52
Graphing in Polar - Example, Cont.
6:45
Tips for Graphing
9:23
Notice Patterns
10:19
Repetition
13:39
Graphing Equations of One Variable
14:39
Converting Coordinate Types
16:16
Use the Same Conversion Formulas From the Previous Lesson
16:23
Interesting Graphs
17:48
Example 1
18:03
Example 2
18:34
Graphing Calculators, Yay!
19:07
Plot Random Things, Alter Equations You Understand, Get a Sense for How Polar Stuff Works
19:11
Check Out the Appendix
19:26
Example 1
21:36
Example 2
28:13
Example 3
34:24
Example 4
35:52
XII. Complex Numbers and Polar Coordinates
Polar Form of Complex Numbers

40m 43s

Intro
0:00
Polar Coordinates
0:49
Rectangular Form
0:52
Polar Form
1:25
R and Theta
1:51
Polar Form Conversion
2:27
R and Theta
2:35
Optimal Values
4:05
Euler's Formula
4:25
Multiplying Two Complex Numbers in Polar Form
6:10
Multiply r's Together and Add Exponents
6:32
Example 1: Convert Rectangular to Polar Form
7:17
Example 2: Convert Polar to Rectangular Form
13:49
Example 3: Multiply Two Complex Numbers
17:28
Extra Example 1: Convert Between Rectangular and Polar Forms
-1
Extra Example 2: Simplify Expression to Polar Form
-2
DeMoivre's Theorem

57m 37s

Intro
0:00
Introduction to DeMoivre's Theorem
0:10
n nth Roots
3:06
DeMoivre's Theorem: Finding nth Roots
3:52
Relation to Unit Circle
6:29
One nth Root for Each Value of k
7:11
Example 1: Convert to Polar Form and Use DeMoivre's Theorem
8:24
Example 2: Find Complex Eighth Roots
15:27
Example 3: Find Complex Roots
27:49
Extra Example 1: Convert to Polar Form and Use DeMoivre's Theorem
-1
Extra Example 2: Find Complex Fourth Roots
-2
XIII. Counting & Probability
Counting

31m 36s

Intro
0:00
Introduction
0:08
Combinatorics
0:56
Definition: Event
1:24
Example
1:50
Visualizing an Event
3:02
Branching line diagram
3:06
Addition Principle
3:40
Example
4:18
Multiplication Principle
5:42
Example
6:24
Pigeonhole Principle
8:06
Example
10:26
Draw Pictures
11:06
Example 1
12:02
Example 2
14:16
Example 3
17:34
Example 4
21:26
Example 5
25:14
Permutations & Combinations

44m 3s

Intro
0:00
Introduction
0:08
Permutation
0:42
Combination
1:10
Towards a Permutation Formula
2:38
How Many Ways Can We Arrange the Letters A, B, C, D, and E?
3:02
Towards a Permutation Formula, cont.
3:34
Factorial Notation
6:56
Symbol Is '!'
6:58
Examples
7:32
Permutation of n Objects
8:44
Permutation of r Objects out of n
9:04
What If We Have More Objects Than We Have Slots to Fit Them Into?
9:46
Permutation of r Objects Out of n, cont.
10:28
Distinguishable Permutations
14:46
What If Not All Of the Objects We're Permuting Are Distinguishable From Each Other?
14:48
Distinguishable Permutations, cont.
17:04
Combinations
19:04
Combinations, cont.
20:56
Example 1
23:10
Example 2
26:16
Example 3
28:28
Example 4
31:52
Example 5
33:58
Example 6
36:34
Probability

36m 58s

Intro
0:00
Introduction
0:06
Definition: Sample Space
1:18
Event = Something Happening
1:20
Sample Space
1:36
Probability of an Event
2:12
Let E Be An Event and S Be The Corresponding Sample Space
2:14
'Equally Likely' Is Important
3:52
Fair and Random
5:26
Interpreting Probability
6:34
How Can We Interpret This Value?
7:24
We Can Represent Probability As a Fraction, a Decimal, Or a Percentage
8:04
One of Multiple Events Occurring
9:52
Mutually Exclusive Events
10:38
What If The Events Are Not Mutually Exclusive?
12:20
Taking the Possibility of Overlap Into Account
13:24
An Event Not Occurring
17:14
Complement of E
17:22
Independent Events
19:36
Independent
19:48
Conditional Events
21:28
What Is The Events Are Not Independent Though?
21:30
Conditional Probability
22:16
Conditional Events, cont.
23:51
Example 1
25:27
Example 2
27:09
Example 3
28:57
Example 4
30:51
Example 5
34:15
XIV. Conic Sections
Parabolas

41m 27s

Intro
0:00
What is a Parabola?
0:20
Definition of a Parabola
0:29
Focus
0:59
Directrix
1:15
Axis of Symmetry
3:08
Vertex
3:33
Minimum or Maximum
3:44
Standard Form
4:59
Horizontal Parabolas
5:08
Vertex Form
5:19
Upward or Downward
5:41
Example: Standard Form
6:06
Graphing Parabolas
8:31
Shifting
8:51
Example: Completing the Square
9:22
Symmetry and Translation
12:18
Example: Graph Parabola
12:40
Latus Rectum
17:13
Length
18:15
Example: Latus Rectum
18:35
Horizontal Parabolas
18:57
Not Functions
20:08
Example: Horizontal Parabola
21:21
Focus and Directrix
24:11
Horizontal
24:48
Example 1: Parabola Standard Form
25:12
Example 2: Graph Parabola
30:00
Example 3: Graph Parabola
33:13
Example 4: Parabola Equation
37:28
Circles

21m 3s

Intro
0:00
What are Circles?
0:08
Example: Equidistant
0:17
Radius
0:32
Equation of a Circle
0:44
Example: Standard Form
1:11
Graphing Circles
1:47
Example: Circle
1:56
Center Not at Origin
3:07
Example: Completing the Square
3:51
Example 1: Equation of Circle
6:44
Example 2: Center and Radius
11:51
Example 3: Radius
15:08
Example 4: Equation of Circle
16:57
Ellipses

46m 51s

Intro
0:00
What Are Ellipses?
0:11
Foci
0:23
Properties of Ellipses
1:43
Major Axis, Minor Axis
1:47
Center
1:54
Length of Major Axis and Minor Axis
3:21
Standard Form
5:33
Example: Standard Form of Ellipse
6:09
Vertical Major Axis
9:14
Example: Vertical Major Axis
9:46
Graphing Ellipses
12:51
Complete the Square and Symmetry
13:00
Example: Graphing Ellipse
13:16
Equation with Center at (h, k)
19:57
Horizontal and Vertical
20:14
Difference
20:27
Example: Center at (h, k)
20:55
Example 1: Equation of Ellipse
24:05
Example 2: Equation of Ellipse
27:57
Example 3: Equation of Ellipse
32:32
Example 4: Graph Ellipse
38:27
Hyperbolas

38m 15s

Intro
0:00
What are Hyperbolas?
0:12
Two Branches
0:18
Foci
0:38
Properties
2:00
Transverse Axis and Conjugate Axis
2:06
Vertices
2:46
Length of Transverse Axis
3:14
Distance Between Foci
3:31
Length of Conjugate Axis
3:38
Standard Form
5:45
Vertex Location
6:36
Known Points
6:52
Vertical Transverse Axis
7:26
Vertex Location
7:50
Asymptotes
8:36
Vertex Location
8:56
Rectangle
9:28
Diagonals
10:29
Graphing Hyperbolas
12:58
Example: Hyperbola
13:16
Equation with Center at (h, k)
16:32
Example: Center at (h, k)
17:21
Example 1: Equation of Hyperbola
19:20
Example 2: Equation of Hyperbola
22:48
Example 3: Graph Hyperbola
26:05
Example 4: Equation of Hyperbola
36:29
Conic Sections

18m 43s

Intro
0:00
Conic Sections
0:16
Double Cone Sections
0:24
Standard Form
1:27
General Form
1:37
Identify Conic Sections
2:16
B = 0
2:50
X and Y
3:22
Identify Conic Sections, Cont.
4:46
Parabola
5:17
Circle
5:51
Ellipse
6:31
Hyperbola
7:10
Example 1: Identify Conic Section
8:01
Example 2: Identify Conic Section
11:03
Example 3: Identify Conic Section
11:38
Example 4: Identify Conic Section
14:50
XV. Sequences, Series, & Induction
Introduction to Sequences

57m 45s

Intro
0:00
Introduction
0:06
Definition: Sequence
0:28
Infinite Sequence
2:08
Finite Sequence
2:22
Length
2:58
Formula for the nth Term
3:22
Defining a Sequence Recursively
5:54
Initial Term
7:58
Sequences and Patterns
10:40
First, Identify a Pattern
12:52
How to Get From One Term to the Next
17:38
Tips for Finding Patterns
19:52
More Tips for Finding Patterns
24:14
Even More Tips
26:50
Example 1
30:32
Example 2
34:54
Fibonacci Sequence
34:55
Example 3
38:40
Example 4
45:02
Example 5
49:26
Example 6
51:54
Introduction to Series

40m 27s

Intro
0:00
Introduction
0:06
Definition: Series
1:20
Why We Need Notation
2:48
Simga Notation (AKA Summation Notation)
4:44
Thing Being Summed
5:42
Index of Summation
6:21
Lower Limit of Summation
7:09
Upper Limit of Summation
7:23
Sigma Notation, Example
7:36
Sigma Notation for Infinite Series
9:08
How to Reindex
10:58
How to Reindex, Expanding
12:56
How to Reindex, Substitution
16:46
Properties of Sums
19:42
Example 1
23:46
Example 2
25:34
Example 3
27:12
Example 4
29:54
Example 5
32:06
Example 6
37:16
Arithmetic Sequences & Series

31m 36s

Intro
0:00
Introduction
0:05
Definition: Arithmetic Sequence
0:47
Common Difference
1:13
Two Examples
1:19
Form for the nth Term
2:14
Recursive Relation
2:33
Towards an Arithmetic Series Formula
5:12
Creating a General Formula
10:09
General Formula for Arithmetic Series
14:23
Example 1
15:46
Example 2
17:37
Example 3
22:21
Example 4
24:09
Example 5
27:14
Geometric Sequences & Series

39m 27s

Intro
0:00
Introduction
0:06
Definition
0:48
Form for the nth Term
2:42
Formula for Geometric Series
5:16
Infinite Geometric Series
11:48
Diverges
13:04
Converges
14:48
Formula for Infinite Geometric Series
16:32
Example 1
20:32
Example 2
22:02
Example 3
26:00
Example 4
30:48
Example 5
34:28
Mathematical Induction

49m 53s

Intro
0:00
Introduction
0:06
Belief Vs. Proof
1:22
A Metaphor for Induction
6:14
The Principle of Mathematical Induction
11:38
Base Case
13:24
Inductive Step
13:30
Inductive Hypothesis
13:52
A Remark on Statements
14:18
Using Mathematical Induction
16:58
Working Example
19:58
Finding Patterns
28:46
Example 1
30:17
Example 2
37:50
Example 3
42:38
The Binomial Theorem

1h 13m 13s

Intro
0:00
Introduction
0:06
We've Learned That a Binomial Is An Expression That Has Two Terms
0:07
Understanding Binomial Coefficients
1:20
Things We Notice
2:24
What Goes In the Blanks?
5:52
Each Blank is Called a Binomial Coefficient
6:18
The Binomial Theorem
6:38
Example
8:10
The Binomial Theorem, cont.
10:46
We Can Also Write This Expression Compactly Using Sigma Notation
12:06
Proof of the Binomial Theorem
13:22
Proving the Binomial Theorem Is Within Our Reach
13:24
Pascal's Triangle
15:12
Pascal's Triangle, cont.
16:12
Diagonal Addition of Terms
16:24
Zeroth Row
18:04
First Row
18:12
Why Do We Care About Pascal's Triangle?
18:50
Pascal's Triangle, Example
19:26
Example 1
21:26
Example 2
24:34
Example 3
28:34
Example 4
32:28
Example 5
37:12
Time for the Fireworks!
43:38
Proof of the Binomial Theorem
43:44
We'll Prove This By Induction
44:04
Proof (By Induction)
46:36
Proof, Base Case
47:00
Proof, Inductive Step - Notation Discussion
49:22
Induction Step
49:24
Proof, Inductive Step - Setting Up
52:26
Induction Hypothesis
52:34
What We What To Show
52:44
Proof, Inductive Step - Start
54:18
Proof, Inductive Step - Middle
55:38
Expand Sigma Notations
55:48
Proof, Inductive Step - Middle, cont.
58:40
Proof, Inductive Step - Checking In
1:01:08
Let's Check In With Our Original Goal
1:01:12
Want to Show
1:01:18
Lemma - A Mini Theorem
1:02:18
Proof, Inductive Step - Lemma
1:02:52
Proof of Lemma: Let's Investigate the Left Side
1:03:08
Proof, Inductive Step - Nearly There
1:07:54
Proof, Inductive Step - End!
1:09:18
Proof, Inductive Step - End!, cont.
1:11:01
XVI. Preview of Calculus
Idea of a Limit

40m 22s

Intro
0:00
Introduction
0:05
Motivating Example
1:26
Fuzzy Notion of a Limit
3:38
Limit is the Vertical Location a Function is Headed Towards
3:44
Limit is What the Function Output is Going to Be
4:15
Limit Notation
4:33
Exploring Limits - 'Ordinary' Function
5:26
Test Out
5:27
Graphing, We See The Answer Is What We Would Expect
5:44
Exploring Limits - Piecewise Function
6:45
If We Modify the Function a Bit
6:49
Exploring Limits - A Visual Conception
10:08
Definition of a Limit
12:07
If f(x) Becomes Arbitrarily Close to Some Number L as x Approaches Some Number c, Then the Limit of f(x) As a Approaches c is L.
12:09
We Are Not Concerned with f(x) at x=c
12:49
We Are Considering x Approaching From All Directions, Not Just One Side
13:10
Limits Do Not Always Exist
15:47
Finding Limits
19:49
Graphs
19:52
Tables
21:48
Precise Methods
24:53
Example 1
26:06
Example 2
27:39
Example 3
30:51
Example 4
33:11
Example 5
37:07
Formal Definition of a Limit

57m 11s

Intro
0:00
Introduction
0:06
New Greek Letters
2:42
Delta
3:14
Epsilon
3:46
Sometimes Called the Epsilon-Delta Definition of a Limit
3:56
Formal Definition of a Limit
4:22
What does it MEAN!?!?
5:00
The Groundwork
5:38
Set Up the Limit
5:39
The Function is Defined Over Some Portion of the Reals
5:58
The Horizontal Location is the Value the Limit Will Approach
6:28
The Vertical Location L is Where the Limit Goes To
7:00
The Epsilon-Delta Part
7:26
The Hard Part is the Second Part of the Definition
7:30
Second Half of Definition
10:04
Restrictions on the Allowed x Values
10:28
The Epsilon-Delta Part, cont.
13:34
Sherlock Holmes and Dr. Watson
15:08
The Adventure of the Delta-Epsilon Limit
15:16
Setting
15:18
We Begin By Setting Up the Game As Follows
15:52
The Adventure of the Delta-Epsilon, cont.
17:24
This Game is About Limits
17:46
What If I Try Larger?
19:39
Technically, You Haven't Proven the Limit
20:53
Here is the Method
21:18
What We Should Concern Ourselves With
22:20
Investigate the Left Sides of the Expressions
25:24
We Can Create the Following Inequalities
28:08
Finally…
28:50
Nothing Like a Good Proof to Develop the Appetite
30:42
Example 1
31:02
Example 1, cont.
36:26
Example 2
41:46
Example 2, cont.
47:50
Finding Limits

32m 40s

Intro
0:00
Introduction
0:08
Method - 'Normal' Functions
2:04
The Easiest Limits to Find
2:06
It Does Not 'Break'
2:18
It Is Not Piecewise
2:26
Method - 'Normal' Functions, Example
3:38
Method - 'Normal' Functions, cont.
4:54
The Functions We're Used to Working With Go Where We Expect Them To Go
5:22
A Limit is About Figuring Out Where a Function is 'Headed'
5:42
Method - Canceling Factors
7:18
One Weird Thing That Often Happens is Dividing By 0
7:26
Method - Canceling Factors, cont.
8:16
Notice That The Two Functions Are Identical With the Exception of x=0
8:20
Method - Canceling Factors, cont.
10:00
Example
10:52
Method - Rationalization
12:04
Rationalizing a Portion of Some Fraction
12:05
Conjugate
12:26
Method - Rationalization, cont.
13:14
Example
13:50
Method - Piecewise
16:28
The Limits of Piecewise Functions
16:30
Example 1
17:42
Example 2
18:44
Example 3
20:20
Example 4
22:24
Example 5
24:24
Example 6
27:12
Continuity & One-Sided Limits

32m 43s

Intro
0:00
Introduction
0:06
Motivating Example
0:56
Continuity - Idea
2:14
Continuous Function
2:18
All Parts of Function Are Connected
2:28
Function's Graph Can Be Drawn Without Lifting Pencil
2:36
There Are No Breaks or Holes in Graph
2:56
Continuity - Idea, cont.
3:38
We Can Interpret the Break in the Continuity of f(x) as an Issue With the Function 'Jumping'
3:52
Continuity - Definition
5:16
A Break in Continuity is Caused By the Limit Not Matching Up With What the Function Does
5:18
Discontinuous
6:02
Discontinuity
6:10
Continuity and 'Normal' Functions
6:48
Return of the Motivating Example
8:14
One-Sided Limit
8:48
One-Sided Limit - Definition
9:16
Only Considers One Side
9:20
Be Careful to Keep Track of Which Symbol Goes With Which Side
10:06
One-Sided Limit - Example
10:50
There Does Not Necessarily Need to Be a Connection Between Left or Right Side Limits
11:16
Normal Limits and One-Sided Limits
12:08
Limits of Piecewise Functions
14:12
'Breakover' Points
14:22
We Find the Limit of a Piecewise Function By Checking If the Left and Right Side Limits Agree With Each Other
15:34
Example 1
16:40
Example 2
18:54
Example 3
22:00
Example 4
26:36
Limits at Infinity & Limits of Sequences

32m 49s

Intro
0:00
Introduction
0:06
Definition: Limit of a Function at Infinity
1:44
A Limit at Infinity Works Very Similarly to How a Normal Limit Works
2:38
Evaluating Limits at Infinity
4:08
Rational Functions
4:17
Examples
4:30
For a Rational Function, the Question Boils Down to Comparing the Long Term Growth Rates of the Numerator and Denominator
5:22
There are Three Possibilities
6:36
Evaluating Limits at Infinity, cont.
8:08
Does the Function Grow Without Bound? Will It 'Settle Down' Over Time?
10:06
Two Good Ways to Think About This
10:26
Limit of a Sequence
12:20
What Value Does the Sequence Tend to Do in the Long-Run?
12:41
The Limit of a Sequence is Very Similar to the Limit of a Function at Infinity
12:52
Numerical Evaluation
14:16
Numerically: Plug in Numbers and See What Comes Out
14:24
Example 1
16:42
Example 2
21:00
Example 3
22:08
Example 4
26:14
Example 5
28:10
Example 6
31:06
Instantaneous Slope & Tangents (Derivatives)

51m 13s

Intro
0:00
Introduction
0:08
The Derivative of a Function Gives Us a Way to Talk About 'How Fast' the Function If Changing
0:16
Instantaneous Slop
0:22
Instantaneous Rate of Change
0:28
Slope
1:24
The Vertical Change Divided by the Horizontal
1:40
Idea of Instantaneous Slope
2:10
What If We Wanted to Apply the Idea of Slope to a Non-Line?
2:14
Tangent to a Circle
3:52
What is the Tangent Line for a Circle?
4:42
Tangent to a Curve
5:20
Towards a Derivative - Average Slope
6:36
Towards a Derivative - Average Slope, cont.
8:20
An Approximation
11:24
Towards a Derivative - General Form
13:18
Towards a Derivative - General Form, cont.
16:46
An h Grows Smaller, Our Slope Approximation Becomes Better
18:44
Towards a Derivative - Limits!
20:04
Towards a Derivative - Limits!, cont.
22:08
We Want to Show the Slope at x=1
22:34
Towards a Derivative - Checking Our Slope
23:12
Definition of the Derivative
23:54
Derivative: A Way to Find the Instantaneous Slope of a Function at Any Point
23:58
Differentiation
24:54
Notation for the Derivative
25:58
The Derivative is a Very Important Idea In Calculus
26:04
The Important Idea
27:34
Why Did We Learn the Formal Definition to Find a Derivative?
28:18
Example 1
30:50
Example 2
36:06
Example 3
40:24
The Power Rule
44:16
Makes It Easier to Find the Derivative of a Function
44:24
Examples
45:04
n Is Any Constant Number
45:46
Example 4
46:26
Area Under a Curve (Integrals)

45m 26s

Intro
0:00
Introduction
0:06
Integral
0:12
Idea of Area Under a Curve
1:18
Approximation by Rectangles
2:12
The Easiest Way to Find Area is With a Rectangle
2:18
Various Methods for Choosing Rectangles
4:30
Rectangle Method - Left-Most Point
5:12
The Left-Most Point
5:16
Rectangle Method - Right-Most Point
5:58
The Right-Most Point
6:00
Rectangle Method - Mid-Point
6:42
Horizontal Mid-Point
6:48
Rectangle Method - Maximum (Upper Sum)
7:34
Maximum Height
7:40
Rectangle Method - Minimum
8:54
Minimum Height
9:02
Evaluating the Area Approximation
10:08
Split the Interval Into n Sub-Intervals
10:30
More Rectangles, Better Approximation
12:14
The More We Us , the Better Our Approximation Becomes
12:16
Our Approximation Becomes More Accurate as the Number of Rectangles n Goes Off to Infinity
12:44
Finding Area with a Limit
13:08
If This Limit Exists, It Is Called the Integral From a to b
14:08
The Process of Finding Integrals is Called Integration
14:22
The Big Reveal
14:40
The Integral is Based on the Antiderivative
14:46
The Big Reveal - Wait, Why?
16:28
The Rate of Change for the Area is Based on the Height of the Function
16:50
Height is the Derivative of Area, So Area is Based on the Antiderivative of Height
17:50
Example 1
19:06
Example 2
22:48
Example 3
29:06
Example 3, cont.
35:14
Example 4
40:14
XVII. Appendix: Graphing Calculators
Buying a Graphing Calculator

10m 41s

Intro
0:00
Should You Buy?
0:06
Should I Get a Graphing Utility?
0:20
Free Graphing Utilities - Web Based
0:38
Personal Favorite: Desmos
0:58
Free Graphing Utilities - Offline Programs
1:18
GeoGebra
1:31
Microsoft Mathematics
1:50
Grapher
2:18
Other Graphing Utilities - Tablet/Phone
2:48
Should You Buy a Graphing Calculator?
3:22
The Only Real Downside
4:10
Deciding on Buying
4:20
If You Plan on Continuing in Math and/or Science
4:26
If Money is Not Particularly Tight for You
4:32
If You Don't Plan to Continue in Math and Science
5:02
If You Do Plan to Continue and Money Is Tight
5:28
Which to Buy
5:44
Which Graphing Calculator is Best?
5:46
Too Many Factors
5:54
Do Your Research
6:12
The Old Standby
7:10
TI-83 (Plus)
7:16
TI-84 (Plus)
7:18
Tips for Purchasing
9:17
Buy Online
9:19
Buy Used
9:35
Ask Around
10:09
Graphing Calculator Basics

10m 51s

Intro
0:00
Read the Manual
0:06
Skim It
0:20
Play Around and Experiment
0:34
Syntax
0:40
Definition of Syntax in English and Math
0:46
Pay Careful Attention to Your Syntax When Working With a Calculator
2:08
Make Sure You Use Parentheses to Indicate the Proper Order of Operations
2:16
Think About the Results
3:54
Settings
4:58
You'll Almost Never Need to Change the Settings on Your Calculator
5:00
Tell Calculator In Settings Whether the Angles Are In Radians or Degrees
5:26
Graphing Mode
6:32
Error Messages
7:10
Don't Panic
7:11
Internet Search
7:32
So Many Things
8:14
More Powerful Than You Realize
8:18
Other Things Your Graphing Calculator Can Do
8:24
Playing Around
9:16
Graphing Functions, Window Settings, & Table of Values

10m 38s

Intro
0:00
Graphing Functions
0:18
Graphing Calculator Expects the Variable to Be x
0:28
Syntax
0:58
The Syntax We Choose Will Affect How the Function Graphs
1:00
Use Parentheses
1:26
The Viewing Window
2:00
One of the Most Important Ideas When Graphing Is To Think About The Viewing Window
2:01
For Example
2:30
The Viewing Window, cont.
2:36
Window Settings
3:24
Manually Choose Window Settings
4:20
x Min
4:40
x Max
4:42
y Min
4:44
y Max
4:46
Changing the x Scale or y Scale
5:08
Window Settings, cont.
5:44
Table of Values
7:38
Allows You to Quickly Churn Out Values for Various Inputs
7:42
For example
7:44
Changing the Independent Variable From 'Automatic' to 'Ask'
8:50
Finding Points of Interest

9m 45s

Intro
0:00
Points of Interest
0:06
Interesting Points on the Graph
0:11
Roots/Zeros (Zero)
0:18
Relative Minimums (Min)
0:26
Relative Maximums (Max)
0:32
Intersections (Intersection)
0:38
Finding Points of Interest - Process
1:48
Graph the Function
1:49
Adjust Viewing Window
2:12
Choose Point of Interest Type
2:54
Identify Where Search Should Occur
3:04
Give a Guess
3:36
Get Result
4:06
Advanced Technique: Arbitrary Solving
5:10
Find Out What Input Value Causes a Certain Output
5:12
For Example
5:24
Advanced Technique: Calculus
7:18
Derivative
7:22
Integral
7:30
But How Do You Show Work?
8:20
Parametric & Polar Graphs

7m 8s

Intro
0:00
Change Graph Type
0:08
Located in General 'Settings'
0:16
Graphing in Parametric
1:06
Set Up Both Horizontal Function and Vertical Function
1:08
For Example
2:04
Graphing in Polar
4:00
For Example
4:28
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Lecture Comments (4)

1 answer

Last reply by: Professor Selhorst-Jones
Sun Mar 8, 2015 9:10 PM

Post by Jamal Tischler on March 8, 2015

If we try to evaluate the integral of 1/x from a=1 to b=2 with the summation we get the limit of n*(1/(n+1)+1/(n+2)+...+1/(n+n)). How do we evaluate this ?

1 answer

Last reply by: Professor Selhorst-Jones
Mon Aug 26, 2013 11:09 PM

Post by Richard Gregory on August 22, 2013

Thank you very much Vincent. I really enjoyed this course and will be moving on to further maths classes. Your teaching style is excellent and I admire your enthusiasm for the topic. I always struggled to understand the meaning of calculus in school and I believe the meaning is more important than the technical application. Your lectures have filled in all the holes school created.

Area Under a Curve (Integrals)

  • The integral is a way to find the area underneath some portion of a curve.
  • We can approximate an integral by using rectangles. We can break an interval up into sub-intervals, and then put a rectangle the width of each sub-interval in place. By taking the area of each rectangle and summing them all up, we have an approximation of the area under the curve in that interval.
  • The width of each rectangle is the same, because we cut the sub-intervals evenly. However, the height of each rectangle can vary depending on the height of the function in that sub-interval. Furthermore, the function has different heights throughout the sub-interval, so we have to come up with some method to choose an xi in each sub-interval to find the height of its associated rectangle: f(xi).
  • Here are some of the most common methods for choosing the xi in each sub-interval:
    • Left-Most Point: We choose xi such that it is the left-most point in each sub-interval. The height of the rectangle is based on its left side.
    • Right-Most Point: We choose xi such that it is the right-most point in each sub-interval. The height of the rectangle is based on its right side.
    • Mid-Point: We choose xi such that it is the mid-point in each sub-interval. The height of the rectangle is based on its middle.
    • Maximum (Upper Sum): We choose xi such that the rectangle is the highest possible for the sub-interval. The height of the rectangle is the highest place the function achieves in that sub-interval.
    • Minimum (Lower Sum): We choose xi such that the rectangle is the lowest possible for the sub-interval. The height of the rectangle is the lowest place the function achieves in that sub-interval.
  • We can evaluate the area approximation by summing up all of the rectangles. This comes out to be


     
     
    b−a

    n


     



     
    ·f(xi).
    [However, if you are told to approximate area, you normally won't need the above formula. Just figure out the area for each rectangle, then add them all together.]
  • Since the above approximation becomes more accurate as n→ ∞, we can take the limit at infinity to find area.
    Area under f(x)from a tob:      
    lim
     
     

     
     
    b−a

    n


     



     
    ·f(xi)
  • If the above limit exists, it is called the integral from a to b. It is denoted by



     
    f(x)   dx
    The process of finding integrals is called integration.
  • The really amazing part is that the integral of f(x) is based on the antiderivative of f(x): that is, the derivative process done in reverse on f(x). We can symbolize the antiderivative of f(x) with F(x). With this notation, we have



     
    f(x)   dx    =   F(b) − F(a).
  • We can see the above is true, because we can think of the area underneath the curve as a function A(x). Notice that the rate of change for the area is based on the height of the function: f(x). Thus, height is the derivative of area, so area is based on the antiderivative of height.

Area Under a Curve (Integrals)

Find an approximation to the area under the curve of f(x) = 10−x2 from a=0 to b=3 by using three equal width rectangles (n=3), where the height of each rectangle is determined by the left-hand side of each interval.
  • If you didn't watch the video lesson, make sure to watch it first. The idea of finding area under a curve is difficult to explain just using words, but the video has a lot of helpful diagrams to clarify the idea.
  • Start off by figuring out the width of the intervals you'll be using. It's the total length being covered, divided by the number of rectangles to be used:
    Width of each rectangle:     b−a

    n
        ⇒     3−0

    3
        =     1
    Thus the width of each interval/rectangle is 1 for this problem. Next, figure out where you'll be determining the height for each rectangle. The problem told us to determine the heights using the "left-hand side of each interval". Since we have a starting location of a=0 and an interval width of 1, our intervals are:
    [0,  1]               [1,  2]               [2,  3]
    The left-hand side of each rectangle are the locations of x=0,   1,  2. We will use these locations to determine the heights.
  • Compute the height of each rectangle by applying the function to each of the locations we're using:
        f(0)    
        f(1)    
        f(2)    
    10
    9
    6
    Now that we know the height of each rectangle, we can find the area for each one by multiplying by the width (1), then sum them up to find the approximate area:
    Aapprox     =     10 ·1   +   9 ·1   +   6 ·1

        =     25
25
Find an approximation to the area under the curve of f(x) = 10−x2 from a=0 to b=3 by using three equal width rectangles (n=3), where the height of each rectangle is determined by the right-hand side of each interval.
  • If you didn't watch the video lesson, make sure to watch it first. The idea of finding area under a curve is difficult to explain just using words, but the video has a lot of helpful diagrams to clarify the idea.
  • Start off by figuring out the width of the intervals you'll be using. It's the total length being covered, divided by the number of rectangles to be used:
    Width of each rectangle:     b−a

    n
        ⇒     3−0

    3
        =     1
    Thus the width of each interval/rectangle is 1 for this problem. Next, figure out where you'll be determining the height for each rectangle. The problem told us to determine the heights using the "right-hand side of each interval". Since we have a starting location of a=0 and an interval width of 1, our intervals are:
    [0,  1]               [1,  2]               [2,  3]
    The right-hand side of each rectangle are the locations of x=1,   2,  3. We will use these locations to determine the heights.
  • Compute the height of each rectangle by applying the function to each of the locations we're using:
        f(1)    
        f(2)    
        f(3)    
    9
    6
    1
    Now that we know the height of each rectangle, we can find the area for each one by multiplying by the width (1), then sum them up to find the approximate area:
    Aapprox     =     9 ·1   +   6 ·1   +   1 ·1

        =     16
16
Find an approximation to the area under the curve of f(x) = 10−x2 from a=0 to b=3 by using six equal width rectangles (n=6), where the height of each rectangle is determined by the left-hand side of each interval.
  • If you didn't watch the video lesson, make sure to watch it first. The idea of finding area under a curve is difficult to explain just using words, but the video has a lot of helpful diagrams to clarify the idea.
  • Start off by figuring out the width of the intervals you'll be using. It's the total length being covered, divided by the number of rectangles to be used:
    Width of each rectangle:     b−a

    n
        ⇒     3−0

    6
        =     1

    2
    Thus the width of each interval/rectangle is [1/2] for this problem. Next, figure out where you'll be determining the height for each rectangle. The problem told us to determine the heights using the "left-hand side of each interval". Since we have a starting location of a=0 and an interval width of [1/2], our intervals are:
    [0,   1

    2
    ]            [ 1

    2
    ,  1]            [1,   3

    2
    ]            [ 3

    2
    ,  2]            [2,   5

    2
    ]            [ 5

    2
    ,  3]
    The left-hand side of each rectangle are the locations of x=0,   [1/2],  1,   [3/2],   2,   [5/2]. We will use these locations to determine the heights.
  • Compute the height of each rectangle by applying the function to each of the locations we're using:
        f(0)    
        f( 1

    2
    )    
        f(1)    
        f( 3

    2
    )    
        f(2)    
        f( 5

    2
    )    
    10
    9.75
    9
    7.75
    6
    3.75
    Now that we know the height of each rectangle, we can find the area for each one by multiplying by the width ([1/2]), then sum them up to find the approximate area:
    Aapprox     =     10 · 1

    2
       +   9.75 · 1

    2
       +   9 · 1

    2
       +   7.75 · 1

    2
       +   6 · 1

    2
       +   3.75 · 1

    2

        =    
    10  +  9.75  +  9  +  7.75  +  6  +  3.75
    · 1

    2

        =     46.25 · 1

    2

        =     23.125
23.125
Find an approximation to the area under the curve of g(x) = cos(x) + 1 from a=0 to b=2π by using four equal width rectangles (n=4), where the height of each rectangle is determined by the maximum height in each interval (also called the upper sum).
  • To help us understand what we're doing, begin by graphing the function. Seeing it graphed will assist us later on by making it easier to find where we should determine the heights of the rectangles. Also, to help with that, let's mark the intervals that will be used by each rectangle. There should be four rectangles, so we mark out four intervals (the dashed blue lines, below).
  • Notice that the problem told us to determine the rectangle heights by using the "maximum height in each interval." This means that the place we use to determine rectangle height can vary from one interval to the next. Look at the graph from the previous step to figure out where the locations should be. In the first two intervals, the maximum height occurs at the left of each interval, while the last two intervals have the maximum height at the right of each interval. This means we'll use the following x-values to determine the height of each rectangle:
    x=   0,   π

    2
    ,    

    2
    ,     2 π
  • Compute the height of each rectangle by applying the function to each of the locations we're using:
        g(0)    
        g(π

    2
    )    
        g(

    2
    )    
        g(2π)    
    2
    1
    1
    2
    Now that we know the height of each rectangle, we can find the area for each one by multiplying by the width ([(π)/2]), then sum them up to find the approximate area:
    Aapprox     =     2 ·π

    2
       +   1 ·π

    2
       +   1 ·π

    2
       +   2 ·π

    2

        =    
    2  +  1  +  1  +  2
    ·π

    2

        =     6 ·π

    2

        =     3π
Find an approximation to the area under the curve of g(x) = cos(x) + 1 from a=0 to b=2π by using four equal width rectangles (n=4), where the height of each rectangle is determined by the minimum height in each interval (also called the lower sum).
  • To help us understand what we're doing, begin by graphing the function. Seeing it graphed will assist us later on by making it easier to find where we should determine the heights of the rectangles. Also, to help with that, let's mark the intervals that will be used by each rectangle. There should be four rectangles, so we mark out four intervals (the dashed blue lines, below).
  • Notice that the problem told us to determine the rectangle heights by using the "minimum height in each interval." This means that the place we use to determine rectangle height can vary from one interval to the next. Look at the graph from the previous step to figure out where the locations should be. In the first two intervals, the minimum height occurs at the right of each interval, while the last two intervals have the minimum height at the left of each interval. This means we'll use the following x-values to determine the height of each rectangle:
    x=   π

    2
    ,    π,     π,    

    2
  • Compute the height of each rectangle by applying the function to each of the locations we're using:
        g(π

    2
    )    
        g(π)    
        g(π)    
        g(

    2
    )    
    1
    0
    0
    1
    Now that we know the height of each rectangle, we can find the area for each one by multiplying by the width ([(π)/2]), then sum them up to find the approximate area:
    Aapprox     =     1 ·π

    2
       +   0 ·π

    2
       +   0 ·π

    2
       +   1 ·π

    2

        =    
    1  +  0  +  0  +  1
    ·π

    2

        =     2 ·π

    2

        =     π
π
Consider approximating the area under the curve f(x) = 10−x2 from a=0 to b=3 by using equal-width rectangles. Find a formula that gives the area approximation for using n rectangles, where the height of each rectangle is determined by the right-hand side of each interval.

Hint: In doing this problem, you will come up with a summation of all the rectangles added together. To convert this into an easily usable formula, you will need to use the below identities:
n

i=1 
kai   =  kn

i=1 
ai,    where k is a constant

n

i=1 

ai ±bi
 = n

i=1 
ai  ± n

i=1 
bi

n

i=1 
c  = cn,    where c is a constant

n

i=1 
i2  =  n(n+1)(2n+1)

6
  • Notice this problem has not given us a specific number of rectangles to work with. Instead, we need to come up with a formula that gives the approximate area for an arbitrary number n of rectangles. We are going to make a formula that will tell us the approximate area for any number of rectangles we choose to plug in for n. With this in mind, we need to figure out a way to talk about how wide and tall the rectangles we will be adding up are. Since the number of rectangles is a variable (n), these things will be based off the variable as well.
  • Begin by figuring out how wide each rectangle must be. The total length being covered is from a to b, and the number of rectangles is n. Since the rectangles are all the same width, we have
    Width of each rectangle:    b−a

    n
        ⇒     3−0

    n
        =     3

    n
  • Next, start working out the heights of the rectangles. This is a little bit more difficult since the height of the rectangle varies depending on which one we consider. Still, we can at least figure out where we will determine their heights from. The problem said we should base the heights off of the "right-hand side of each interval," so we start from the right side of our first interval. The first interval begins at a=0 and has the width we figured out above, so it will be [0, [3/n]]. The right side of that gives [3/n] for our first location. The second interval will be based off the right side of the next interval, so one width forward: 2·[3/n]. Similarly, the third interval will be based off the right side again, so another width forward to 3·[3/n]. This pattern continues until we've got all n locations:
    x =        3

    n
    ,        2· 3

    n
    ,        3· 3

    n
    ,        4· 3

    n
    ,     …    n· 3

    n
  • Now that we know the locations to find the rectangle heights from, we can find the area of each rectangle by multiplying the height (plug location in to f(x)) by the width (what we figured out before, [3/n]):
    Aapprox =     f
    3

    n

    · 3

    n
       +   f
    3

    n

    · 3

    n
       +   f
    3

    n

    · 3

    n
      +  …  +  f
    n · 3

    n

    · 3

    n
    This is an awful lot to write out, so let's instead use sigma summation notation. [If you're unfamiliar with summation notation, watch the lesson Introduction to Series.]
    Aapprox =    n

    i=1 
      f
    i · 3

    n

    · 3

    n
    Make sure that the above sigma notation matches the original, expanded version. It does, so we can continue.
  • We can now apply the function and simplify to get
    Aapprox =    n

    i=1 
      f
    i · 3

    n

    · 3

    n
        =    n

    i=1 
      
    10−
    i · 3

    n

    2

     

    · 3

    n

        =    n

    i=1 
      
    10−
    3i

    n

    2

     

    · 3

    n

        =    n

    i=1 
      
    10− 9i2

    n2

    · 3

    n
  • Great, we have a formula in summation notation now:
    Aapprox =    n

    i=1 
      
    10− 9i2

    n2

    · 3

    n
    However, it would be very difficult to actually compute a value using the summation notation. This is where all the identities given in the problem hint come into play. We apply these identities to break the summation into something that we could calculate more easily. First, notice that while n is a variable, it doesn't actually change. It's assumed to be fixed while we work with it, so we can treat it as a constant. This means that the [3/n] multiplied on the right can be "pulled out" since it's just multiplying by a constant. Identity: ∑i=1n kai   =  k ∑i=1n ai,    where k is a constant
    n

    i=1 
      
    10− 9i2

    n2

    · 3

    n
        =     3

    n
    · 
    n

    i=1 
      
    10− 9i2

    n2


    Next, we have two pieces of the summation separated by subtraction, so we can split them into two different summations: Identity: ∑i=1n ( ai ±bi )  = ∑i=1n ai  ± ∑i=1n bi
    3

    n
    · 
    n

    i=1 
      
    10− 9i2

    n2


        =     3

    n
    · 
    n

    i=1 
     10    −    n

    i=1 
    9i2

    n2
     
  • Continue to break apart the expression by using the identities. Notice that [9/(n2)] is also a constant (remember that n is constant for the sum, even if it is a variable), so we can pull it out of its sum, too:
    3

    n
    · 
    n

    i=1 
     10    −    n

    i=1 
    9i2

    n2
     
        =     3

    n
    · 
    n

    i=1 
     10    −     9

    n2
    ·n

    i=1 
    i2  
    Finally, we can use the last two identities from the problem to convert the summations entirely: Identity: ∑i=1n c  = cn,    where c is a constant
    3

    n
    · 
    n

    i=1 
     10    −     9

    n2
    ·n

    i=1 
    i2  
        =     3

    n
    · 
    10 ·n    −     9

    n2
    ·n

    i=1 
    i2  
    Identity: ∑i=1n i2  = [(n(n+1)(2n+1))/6]
    3

    n
    · 
    10 ·n    −     9

    n2
    ·n

    i=1 
    i2  
        =     3

    n
    · 
    10 ·n    −     9

    n2
    · n(n+1)(2n+1)

    6
     
  • Finally, now that our expression for area has no sum, we can simplify to finish finding our formula:
    Aapprox = 3

    n
    · 
    10 ·n    −     9

    n2
    · n(n+1)(2n+1)

    6
     

    = 30   −    27(n)(n+1)(2n+1)

    6n3

    =30   −    9(n)(2n2+3n+1)

    2n3

    =30   −    9(2n3+3n2+n)

    2n3


    [Remark: In a previous problem, we figured out the approximate area under f(x) = 10−x2 from a=0 to b=3 using three right-side rectangles. The approximation we got was 16. We can use this to check the formula we just made. We should be able to plug in n=3 and get the same value:
    30  −   9(2(3)3+3(3)2+(3))

    2(3)3
        =     30  −  9(84)

    54
        =    30−14     =     16
    Great! Using our newly created formula gives us the same value for n=3 compared to when we found it before, so our formula checks out against our previous work!]
30  −  [(9(2n3+3n2+n))/(2n3)]
Consider the area under the curve of f(x) = 10 −x2 from a=0 to b=3. In the previous problem, you found a formula for the right-side approximation using n equal-width rectangles. This formula is
30  −   9(2n3+3n2+n)

2n3
.
Using this formula, find the exact value of the area under the curve by taking the limit as the number of rectangles goes to infinity.
  • The area approximation of a curve becomes more and more accurate as the number of rectangles (n) being used to approximate the area becomes larger and larger. Thus, if we consider the limit as n→ ∞, we can find the exact area under the curve.
  • Since we found a formula for the approximate area using n rectangles in the previous problem, we can take the limit of that as n→ ∞ to find the precise area:

    lim
    n→ ∞ 
      30  −   9(2n3+3n2+n)

    2n3
  • If you're unfamiliar with taking the limit as a variable approaches infinity, check out the lesson Limits at Infinity and Limits of Sequences. Distribute the 9 to make it easier to see what happens:

    lim
    n→ ∞ 
      30  −   9(2n3+3n2+n)

    2n3
        =    
    lim
    n→ ∞ 
      30  −   18n3+27n2+9n

    2n3
    Remember, as n→ ∞, the only "important" parts of the fraction are the n3 parts, since they are the fastest growing parts:

    lim
    n→ ∞ 
      30  −   18n3+27n2+9n

    2n3
        =    
    lim
    n→ ∞ 
      30  −   18n3

    2n3
    We can now simplify, canceling out the n3 on top and bottom:

    lim
    n→ ∞ 
      30  −   18n3

    2n3
        =    
    lim
    n→ ∞ 
      30  −   18

    2
    Finally, since these are now just constants, they are unaffected by the limit:

    lim
    n→ ∞ 
      30  −   18

    2
        =     30  −   18

    2
        =     30  −  9     =     21
21
From the lesson, we learned that the area under the curve of f(x) = 10−x2 from a=0 to b=3 can be represented by the integral below:

3

0 
 10−x2   dx
Furthermore, we also learned about the fundamental theorem of calculus. Given some function f(x) where F(x) is its antiderivative, then

b

a 
f(x)  dx    =   F(b) − F(a).
Use the above theorem to find the area under f(x) = 10−x2 from a=0 to b=3 and confirm the value you found for the area in the previous problem.
  • To use the fundamental theorem of calculus we first need to understand what an antiderivative is. The antiderivative to a function is the opposite of its derivative: in effect, it is using the derivative process in "reverse" on the function f(x). The function F(x) is set up in such a way that when we take its derivative, it produces f(x). Thus, we need to think about what expressions would produce each part of f(x) when we take their derivatives. Notice that both 10 and x2 could have the power rule [which we learned about at the very end of the previous lesson] applied to them. This realization allows us to see that we can create them with the power rule in "reverse", so to speak. Instead of multiplying by the exponent and then subtracting 1 from the exponent, we will add 1 to the exponent and divide by the new exponent:
    10
    x2
           Antiderivative       
    10x
    1

    3
    x3
  • Combining these two pieces, we have found our antiderivative:
    f(x) = 10−x2        ⇒        F(x) = 10x − x3

    3
    Do a quick check on F(x) to make sure that if we take its derivative, we do indeed get f(x). This is the case, so we are ready to move on.

    [Remark: Don't worry if the above process was a little confusing-if you take a calculus class, lots of time will be spent on this idea to make it clear. The important part to realize is that finding the integral ( ∫ ) of a function is based around figuring out its antiderivative. Also, it should be pointed out that F(x) above is technically not the only antiderivative to f(x). Notice that we could add any constant (+C) to F(x) and, since constants "disappear" when we take the derivative, it would still produce the same f(x). Don't worry: this will not affect using the integral to find the area under f(x). Still, the idea of adding an unknown constant (+C) is an important point that will be addressed if/when you take a calculus class.]
  • Once we know the antiderivative F(x), we're ready to apply the fundamental theorem of calculus. Just plug in based on how the theorem is set up:

    b

    a 
    f(x)  dx    =   F(b) − F(a)


    3

    0 
     10−x2   dx    =   F(3) − F(0)

       =   
    10(3)− (3)3

    3

      −  
    10(0) − (0)3

    3


       =   
    30−32
      −  
    0

       =   30 − 9

       =   21
    Therefore, by using the fundamental theorem of calculus, we have found that the area underneath the curve of f(x) = 10−x2 from a=0 to b=3 is 21. In the previous problem, we considered approximating the area under the curve by using n rectangles, then, since the approximation grows more accurate as n becomes larger, we looked at what happened as n→ ∞. Doing so we found that the area under the curve was 21. Thus the two different ways of approaching the problem give the same answer, so they check out against each other.
21
Evaluate the integral ∫−31 −2x+3   dx  by using a graph of f(x) = −2x+3.
  • Remember what the integral (∫) means: it's a way of talking about the area underneath a curve. In general,

    b

    a 
    f(x)   dx        ⇔        Area under f(x) from a to b
    That means, for the problem we're working on, we want to find the area under f(x) = −2x+3 from the starting location of a=−3 to the ending location of b=1.
  • The problem told us to find the value of the integral by using a graph of the function, so graph f(x)=−2x+3. We also mark where the integral starts (a=−3) and stops (b=1) on the graph, then shade in the area we're looking to find.
  • Notice that we can make it easier to find the area by splitting it into two chunks: an upper triangle and a lower rectangle, as in the below picture:
  • With the area now split into two pieces, we can easily find the area of each piece, then add them together to find the total area. For the upper triangle, we have a height of 8 (bottom at 1, top at 9) and a width of 4 (left at −3, right at 1). This gives us
    Atriangle     =     h·w

    2
        =     8·4

    2
        =     16
    For the lower rectangle, we have a height of 1 (bottom at 0, top at 1) and a width of 4 (left at −3, right at 1). This gives us
    Arectangle     =     h·w     =     1 ·4     =     4
    Since the total area is just made up of these two pieces, we add them together to find the area:
    Atotal     =     16 + 4     =     20
    Since   ∫−31 −2x+3   dx  just represents this area, we've found its value.
  • If we want to, we can also check this answer by using the fundamental theorem of calculus (which appeared in the previous problem), which says

    b

    a 
    f(x)  dx    =   F(b) − F(a),
    where F(x) is the antiderivative of f(x). Using it, we start by finding the antiderivative of f(x):

    1

    −3 
    −2x+3   dx     =     −x2 +3x   
    1
    −3 
    Then we plug in the endpoints of b=1 and a=−3 as done in the theorem (the bar with the numbers on the right of it indicates that this plugging in still needs to be done):

    1

    −3 
    −2x+3   dx    =   −x2 +3x   
    1
    −3 

       =   
    −(1)2 +3(1)
      −   
    −(−3)2 +3(−3)

       =   
    −1 +3
      −   
    −9 −9

       =   
    2
      −   
    −18

       =   20    
    Thus the fundamental theorem of calculus gives the same value for the area, which confirms our previous answer.
20
A large water tank currently stands empty. At time t=0 (where t is measured in hours), water begins to flow into the tank from a pipe. However, the rate of flow from the pipe varies moment by moment. The below table tells us the flow rate at the beginning of every hour:
t (in hours)
   0   
   1   
   2   
   3   
   4   
   5   
   6   
   7   
V′ (liters/hour)
1250
1134
1018
902
786
670
554
438
Using the table, approximate the amount of water in the tank eight hours after the water begins to flow from the pipe.
  • While we can't figure out the precise amount of water that enters the tank from the pipe for every hour, we can come up with an approximate value for each hour by using the rate at the beginning of the hour. For example, for the first hour of t=0↔1, we know that the hour begins at t=0 with the water flowing at a rate of 1250 liters per hour. Since we're interested in knowing the total volume during that hour, we multiply the rate by the amount of time (1 hour):
    V0, approx     =     1250 ·1     =     1250
  • We can carry this process out for each hour, finding the approximate volume to flow through for every hour. Since we're multiplying 1 hour by each of the flow rates (which are in liters/hour), the approximate volume is the same number as the flow rate at the beginning of the hour. Once we know the approximate volumes for each hour, we sum them all up to find the total approximate volume:
    Vtotal, approx   =  1250 + 1134 + 1018 + 902 + 786 + 670 + 554 + 438

      =  6752
    Thus the approximate volume in the tank at the end of eight hours is 6752 liters.
  • Remark: You might wonder at this point, "How is this problem connected to calculus and integration?" To answer this, notice that the problem was very similar to approximating the area under a curve by using equal-width rectangles on the left side. While we never mentioned finding the area under the curve, that is effectively what we did. We found an approximation to the area under the curve of the flow rate. (Although we weren't explicitly given the flow rate as a function, we were given all the points we needed to make an approximation with rectangles.) We found (an approximation of) the integral for the flow rate from a=0 to b=8. Furthermore, notice that this integral gave us the total volume. In other words, by taking the integral of a function that describes how some quantity changes, we find the total quantity at the end of that period. This is a crucial idea in calculus and one you should hold on to. You can visualize the integral as the area under a curve, but you can also think of it as a way of "totaling up" the change over time.
Vapprox =  6752 liters [Note: If you're not sure how this problem is connected to calculus or the idea of an integral, check out the last step to the problem. It's a remark that explains how this word problem is based on the same ideas.]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Area Under a Curve (Integrals)

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:06
    • Integral
  • Idea of Area Under a Curve 1:18
  • Approximation by Rectangles 2:12
    • The Easiest Way to Find Area is With a Rectangle
  • Various Methods for Choosing Rectangles 4:30
  • Rectangle Method - Left-Most Point 5:12
    • The Left-Most Point
  • Rectangle Method - Right-Most Point 5:58
    • The Right-Most Point
  • Rectangle Method - Mid-Point 6:42
    • Horizontal Mid-Point
  • Rectangle Method - Maximum (Upper Sum) 7:34
    • Maximum Height
  • Rectangle Method - Minimum 8:54
    • Minimum Height
  • Evaluating the Area Approximation 10:08
    • Split the Interval Into n Sub-Intervals
  • More Rectangles, Better Approximation 12:14
    • The More We Us , the Better Our Approximation Becomes
    • Our Approximation Becomes More Accurate as the Number of Rectangles n Goes Off to Infinity
  • Finding Area with a Limit 13:08
    • If This Limit Exists, It Is Called the Integral From a to b
    • The Process of Finding Integrals is Called Integration
  • The Big Reveal 14:40
    • The Integral is Based on the Antiderivative
  • The Big Reveal - Wait, Why? 16:28
    • The Rate of Change for the Area is Based on the Height of the Function
    • Height is the Derivative of Area, So Area is Based on the Antiderivative of Height
  • Example 1 19:06
  • Example 2 22:48
  • Example 3 29:06
    • Example 3, cont.
  • Example 4 40:14

Transcription: Area Under a Curve (Integrals)

Hi--welcome back to Educator.com.0000

Today, we are going to talk about area under a curve--also known as integrals.0002

The other major idea in calculus is the notion of the integral, a way to find the area underneath some portion of a curve.0006

Like the derivative, at first glance, this might not seem terribly useful.0013

Of course, it is somewhat interesting to be able to know the area under a curve.0017

But what can we do with it that has any real use?--as it turns out, a huge amount.0020

It allows us to consider what a function has done, in total, over the span of some interval.0025

For example, if we have a function that gives the velocity of an object, how fast an object is moving around, the integral will tell us the object's location.0031

The velocity of an object, as a function--we can take the integral of that, and we can get location out of it.0039

Why is that? Well, velocity tells us motion--how fast we are moving around.0045

Well, if we look at how fast something has moved around in total, that ends up telling us where the thing lands at the end.0049

Motion, looked at over the long term--if we put all of the motion together--if we look at all it has done in total,0055

all of the motion in total, we end up seeing where we end up; we end up seeing location out of it.0062

So, that is just one example of how useful integrals can be.0067

Trust me on this: this is really useful stuff--let's check it out.0070

Consider if we had some function f(x) that has the graph right here.0075

We could ask, "What is the area between the curve and the x-axis on the interval going from a to b?"0079

So, we can fill that in; and the shaded-in portion that we have right here represents that area.0086

But how could we go about finding how much area that actually is?0090

Let's look for some way to approximate it; and notice that that is what we did with derivatives, as well.0095

What we did is thought, "Is there some way to approximate close to what the slope is?" and then,0101

"Is there a way to improve on that approximation?" and then,0105

"Is there a way to basically see what that improvement will eventually go to in the infinite version?"0107

As we got closer and closer to being right up against it with the derivative, as h went to 0, we were able to see0115

what it became perfectly--what the slope was in that instant.0120

And we will see sort of a similar idea as we work with integrals.0123

All right, we are getting a little ahead of ourselves; so let's start looking at our approximation.0126

The easiest way to find area is with a rectangle; rectangles are a great way to find area, because it is so easy to figure out what the area of a rectangle is.0130

It is simply width times height: area is just width times height.0138

Let's approximate the area under our curve with rectangles.0143

We begin by breaking our interval into n equal sub-intervals.0146

In this case, we have broken it into n = 4; here are 1, 2, 3, 4 equal-width sub-intervals.0150

So, we go from a to b, and we break it into n equal-width chunks.0158

In each one of these sub-intervals, we will base the height of each rectangle off of some xi in each sub-interval.0164

That is to say, we will choose some horizontal location in the sub-interval; and then we will see what height that horizontal location goes to.0170

That horizontal location, as a point--what height is that point at?0178

And then, that is going to set the height of that rectangle for that sub-interval.0183

So, for the ith sub-interval, we will use xi to determine that height.0189

The ith rectangle has a height of f evaluated at xi;0195

so xi is the horizontal reference location that we use to determine the height for the ith rectangle.0200

For example, in our third sub-interval, we might look at x3, the horizontal reference location for our third sub-interval.0207

We check, and we see what height that ends up going to; and so, that would be f evaluated at x3.0215

And then, that determines the height for our third rectangle.0222

We use xi to determine the horizontal reference location for our ith sub-interval,0226

which then determines the height for our ith rectangle.0232

We use this xi to determine height by getting f(xi) to tell us what that rectangle's height is.0238

Now, there is a variety of ways to choose our xi.0244

In any given sub-interval, there is a continuum: there are infinitely many different locations we could pick to be our x that we are going to choose.0247

We could choose this location, but we could also choose this location.0255

Or we could choose the location between them, or the location in between them, or over here,0258

or at the very far edge, or the other very far edge, or somewhere else.0261

There are all sorts of different ways that we can choose this.0265

And notice: the way we decide to choose our xi, for each one of these sub-intervals:0268

because the xi, that horizontal reference location, determines the height for that rectangle,0272

it is going to affect the height of the ith rectangle,0278

because height is just f evaluated at whatever horizontal reference location we chose, our xi.0281

By using a different method to choose our xi, we will get a different approximation for the area,0287

because we are going to change the height of that rectangle; and if we change the heights of all of our rectangles,0294

we will end up having a totally new version for our area approximation.0298

So, we are going to now look at some of the most common methods to choose xi for creating the heights of our rectangles.0302

Our first common method that we see is the leftmost point method,0309

where we end up evaluating the height of each rectangle by just evaluating where the leftmost point gets mapped to.0313

How high is the left side of each sub-interval?0320

We can see an example of this right here: what we do for our first sub-interval--0323

we look: where would the leftmost point of that sub-interval get mapped to?0327

It gets mapped to a height of here; and so, that determines the height of our entire rectangle.0330

For our second sub-interval, we look at the leftmost location here; it gets mapped to this height, and that determines the height of that rectangle.0335

For our third sub-interval, we look at this leftmost location; that determines that height.0342

Our fourth and final: we look at the leftmost location, and that determines that height.0347

We are determining the height of each rectangle based on the leftmost location in each of the rectangles.0351

Basically the same idea, but flipped: we can look at the rightmost point.0357

We can say, "For each sub-interval, what is the height that the rightmost horizontal location gets mapped to?"0361

For our first sub-interval, we look at the rightmost location for that one;0367

that ends up getting mapped to here, so that determines the height of our first rectangle.0370

For our second sub-interval, we look at the rightmost location in that one; that is here, so that determines the height of our second rectangle.0374

For our third sub-interval, we look at the rightmost location of that sub-interval; that determines the height of our third rectangle.0382

And for our fourth and final one, we look at the rightmost location, and that determines the height of that fourth and final rectangle.0389

We are just looking at the rightmost horizontal location to determine the height for each one of our rectangles.0396

We could also do this based on the horizontal midpoint.0402

So far, we have looked at the two extremes: the far left side of the sub-interval and the far right side of the sub-interval.0404

But we could also ask what happens to the one in the middle.0409

For our first sub-interval, we look at the one that is in the middle of that; we go up, and the midpoint gets mapped to this height.0412

So, that determines the height of that first rectangle.0419

For our second sub-interval, we look at the midpoint of that sub-interval.0422

We go up, and that tells us the height for our rectangle here.0426

For our third sub-interval, we look at the midpoint; we see what height that horizontal midpoint gets mapped to.0430

And that determines the height of that entire third rectangle.0437

And the same thing for our fourth one: we look at the midpoint.0440

What value does that midpoint get mapped to? That determines the height of that entire rectangle.0442

So, so far, we have looked at the far left side, the far right side, and the middle.0448

But there is also another way of looking at this; we can also look at the maximum height in each one of these.0452

We look at the sub-interval, and we see which one of these points gets mapped to the highest possible location.0457

We look over the entire sub-interval, and we see which one is highest in this portion.0463

So, in this case, the highest location that we get for this sub-interval is here; so that determines the height of the entire rectangle.0468

For our second sub-interval, the highest possible location we reach in this sub-interval is this location right here.0474

So, that determines the height of that entire second rectangle.0480

For our third sub-interval, we see that the highest possible height we reach in that sub-interval is here.0483

So, that determines the height of the entire third rectangle.0488

And for our fourth and final one, the highest possible point we get is here; so that determines the height of the entire rectangle.0491

The maximum height method is also sometimes known as the upper sum, because, notice:0497

by choosing the maximum height in each of our sub-intervals, we will always end up getting an approximation0502

that is going to be at least the value of the area under the curve, and more than that area, usually.0508

We can see all of the places that we ended up going above it.0514

And so, we end up having something that is above what we are actually going to end up having as the area underneath the curve.0518

And so, we call it an upper sum, because we are getting something that is above the value of the actual area underneath the curve.0525

We can also talk about the minimum height for each one of our sub-intervals.0533

We can look and see, for our first sub-interval, what is the lowest possible height that we have in here.0537

We see that it is here, and so we end up getting this as the height for it.0542

We can also do this for each one of our sub-intervals.0547

We can have our second sub-interval; the lowest possible height in that second sub-interval is right here,0551

so we end up evaluating that as the height of our second rectangle.0558

For our third sub-interval, the lowest possible height is right here; so that gives us the height of our third rectangle.0561

And for our fourth and final sub-interval, the lowest possible height over that sub-interval is right there.0566

And so, that gives us the height of our fourth rectangle.0571

This one is sometimes called the lower sum, because it is always going to end up giving us a value0574

that is below the actual area underneath the curve, because, since we are choosing based on minimum heights,0580

each of our rectangles is actually under-cutting the area.0586

We can see area that we are missing each time.0589

For each of our rectangles, we are not fully filling out that area, because we are always below it.0592

So, we end up having less than the area that is actually underneath the curve.0596

And so, it is sometimes also called the lower sum.0601

How can we find what this approximation actually ends up giving us?0605

How can we find this area? Well, first we want to figure out what the width of each one of these rectangles is.0610

That is the easier thing to figure out: we split the interval into n sub-intervals.0615

That is how we did this, right from the beginning; we split it into n equal-width sub-intervals.0620

So, what is the total length of the interval we have?0624

Well, that is going to be b - a, where we end minus where we started.0627

If we have split it into n equal-width sub-intervals, the width of each one of them must be the total length, divided by how many intervals we have.0632

So, our width is equal to b - a, divided by n; that is the width of one of our rectangles, b - a over n.0641

To figure out the height for each rectangle...well, the ith rectangle's height0651

depends on the specific xi we chose for it, with the different methods that we were just talking about.0654

There are various different ways that we can choose that xi.0659

But whatever xi we end up choosing will tell us the height; so just by definition,0661

our height is equal to f evaluated at the various xi that we chose.0665

So, with these two things in mind, we now have our width, and we have our height.0670

The area for the ith rectangle, any given rectangle, is going to be that rectangle's width,0675

which is b - a divided by n, times that rectangle's height, which was determined by the xi we chose, so our height right here.0680

So, the area of our ith rectangle is equal to the width, b - a over n, times the height of that ith rectangle,0689

which is going to be f evaluated at xi--whatever horizontal reference location we chose.0695

If we want to figure out what the total approximation is, we need to add up each one of our rectangles.0700

We aren't concerned with just one of our rectangles; we want to add up all of the rectangles in our approximation.0705

So, we sum them all up; and we can use sigma notation for that.0710

So, we have i = 1, our first rectangle, to n, the nth rectangle, which is our last rectangle,0713

since we are using a total of n sub-intervals; and then we just end up adding up the area from each one of these.0720

So, it is going to be b - a over n, the width of each one of these, times the height of each one of these, f(xi).0726

Notice that whatever method we choose to determine the height of the rectangles, the more rectangles we use, the better our approximation becomes.0734

Over here, we have n = 4; but over here, we have n = 8.0742

And notice: in this second one, we end up having a closer approximation.0747

We have less missing chunks; the more rectangles we use for our approximation,0752

the closer our approximation becomes to actually giving us the area underneath the curve.0757

Thus, our approximation becomes ever more accurate as the number of rectangles, n, goes off to infinity.0763

So, as we increase our n-value higher and higher and higher and higher, our approximation becomes better and better and better and better.0769

As our n slides off to infinity, we will be able to get, effectively, what the perfect value is underneath that curve.0776

With this realization in mind, remember: the approximation we just figured out for the area with n rectangles0784

was the sum of i = 1 up until n of our width for each rectangle, (b - a)/n,0789

times the height of each rectangle, f evaluated at its specific xi.0795

And this approximation becomes more accurate as our n goes off to infinity,0799

as the number of rectangles we have becomes more and more and more and more.0803

We get a finer and finer sense of the area underneath the curve.0806

With that idea in mind, we take the limit at infinity to find the area underneath the curve.0809

The area under our function f(x) from a to b, in that interval from a to b, is the limit as n goes to infinity,0815

as our number of rectangles slides off to infinity, of our approximation formula, the sum from i = 1 to n of the width, (b - a)/n, times the height, f(xi).0822

It is the limit as our number of rectangles slides off to infinity of our normal approximation formula.0834

If this limit exists (and it might not for some weird functions, but for most of the functions we are used to,0841

it will end up existing), what that limit is: it is called the integral from a to b.0846

It is denoted with the integral from a to b...that is the integral sign there,0853

that new sign that you probably haven't seen before: integral from a to b of f(x)dx.0856

The process of finding integrals is called integration.0861

The integral from a to b tells us the area underneath the curve, f(x), from a to b (that interval a to b)--really cool stuff here.0865

Here is the really amazing part: the integral from a to b of f(x)dx is based on the anti-derivative of f(x)--0875

that is, the derivative process done in reverse on f(x).0885

So, we talked about the idea of taking the derivative of some function, of being able to see some function,0889

and then turn it into another function that talks about the rate of change of that.0894

For that, we had f(x) become, through the derivative, f'(x).0898

But we can also talk about if we did the reverse of this process.0905

Instead of taking a derivative, we did the anti-derivative, where we worked our way up the chain.0910

And we symbolized that with the capital F(x); F(x) is the one that you can take the derivative of, and it becomes f(x).0914

So, F(x), the anti-derivative, is the thing where, if you take its derivative, you just get f(x),0925

the function that we are starting with, what is inside of our integral.0932

We can take the derivative process and reverse it, and we are able to start talking about the area underneath the curve.0935

This is really cool stuff; armed with this notation of F(x), it turns out that the integral of a to b of f(x)dx--0941

that is, the area underneath the curve from a to b of some f(x) function--0950

is equal to the anti-derivative of f, evaluated at b, minus the anti-derivative of f, evaluated at a.0955

Wow, that is amazing; it is just so incredibly elegant that there is this way to talk about the area underneath the curve0963

with this thing that we just talked about that had rate of change.0970

And it seems really shocking that these things are connected at all.0972

But it is absolutely amazing that it allows us to really easily find area for something that would be otherwise very difficult to work through,0975

that infinite limit that we were just talking about.0982

So, you might be wondering why this happens; why is there this connection between the area underneath the curve,0985

and this anti-derivative, where we are talking about what the height is...the anti-derivative of the height of it0991

gives us the area underneath the curve...it seems really surprising at first.0996

In short, what we can do is think of the area underneath the curve as being a special function, a(x).1000

Now, notice: we have this area underneath the curve, the shaded portion, as being a(x).1007

Notice that the rate of change for the area is based on the height of the function at any given location.1013

For example, if we consider this x location right here, how fast our area function is changing is based on how tall our function is in that moment.1020

And how tall is a function? Well, that is just what you get when you evaluate the function, f(x).1029

Notice: if we had gone to some other horizontal location, where we had a different height,1034

we would end up having a very different speed that our area was growing at.1039

If the height of the function is small, our area grows at a slow rate.1044

If the height of the function is high, our area grows at a fast rate.1048

So, the height of the function changes how fast our area grows.1052

How fast does something grow? Well, that is what we are talking about: rate of change.1056

That means that since height is connected to area through rate of change, since height gives us the rate of change of area,1060

well, rate of change was a derivative that we were just talking about.1069

So, height is the derivative of area, because height is the rate of change of our area.1073

That means that the reverse works as well; we can look at the symmetric version of this.1078

Area is based on the anti-derivative of height; since height is the derivative of area,1083

that means that area must be the anti-derivative of height.1088

Since we go in one direction, if we go in the opposite direction, doing the opposite thing, we get the other version.1091

Since a'(x) = f(x)--that is to say, the derivative of x equals little f(x), if we do the anti-derivative of f,1095

we end up getting the anti-derivative of a', which is just our area function that we started with.1104

If this a little bit confusing to you now, don't worry; it very well might be, and it would be perfectly reasonable.1109

Just think about it when you end up getting exposed to this new idea when you actually take a calculus class.1114

This ends up being a fair bit of a way into calculus class, but I think it is a really cool idea.1119

And now that you understand what is going on intuitively, or at least have the seed ready to blossom at a future date,1123

when you see it later, you will think, "Oh, now I understand what that guy was talking about--1130

it is now starting to make sense!" and there is really cool stuff here.1135

I really, really love this stuff; and I hope that you are getting some sense of just how amazingly beautiful all of this stuff in math is.1138

All right, let's start looking at some examples.1144

Find an approximation to the area under f(x) = x2 from a = 0 to b = 3 by using three equal-width rectangles--1147

that is, n = 3--on the left-hand point of each interval.1155

The first thing: let's just get a quick sketch here, so that we can see what is going on.1159

f(x) = x2: we will just, really quickly, sketch what we are looking at here.1163

And so, let's say here is a = 0; here is b = 3.1167

So, if we are going to use three equal-width rectangles, n = 3, we are breaking it up into three chunks.1173

And we are going to evaluate each one of these rectangles on the left-hand point of each interval.1179

Three equal chunks here...if it is the left-hand point, then that first one there, this one here, and this one here give us the area for each one of these.1183

Notice that that first rectangle won't have any area at all, because we are evaluating the left-hand point of each interval.1196

First, let's check and see what the width is: the width is going to be...1203

since it is equal width for each one of these...how long is our interval?1208

That is b - a, divided by...how many sub-intervals do we break it into? n.1212

So, we have 3 for our ending location, minus 0 for our starting location, divided by 3 (is the total number of sub-intervals).1216

3/3 gets us 1, so we have a width of 1 for each one of these.1223

So, at this point, we can go in and see where we have our first location; the sub-interval will go from 0 to 1;1228

the next one will go from 1 to 2; the next one, the last one, goes from 2 to 3.1235

All right, with all of this in mind, we can now see about evaluating each one of these rectangles.1240

Our first rectangle will be i = 1; our second rectangle, i = 2; and our last, final rectangle, i = 3.1244

Where will we evaluate our first rectangle? Well, that is the left-hand point.1253

If our first rectangle...remember: it is going from 0 to 1; 0 to 1 is the sub-interval it is evaluating.1257

The left-hand point is 0; so it is going to have a height of f at 0.1266

What is the width? The width is 1, so 1 times f(0).1271

In general, notice that that is also just the same thing as saying the width, (b - a)/n, times our function evaluated at...however we determined our xi.1275

In this case, we are determining based on left-hand points.1287

So, 1 times f at 0...our f(x) equals x2, so we have 1(0)2,1290

which comes out to be just 0 for the area of our first rectangle, this right here.1296

And we can see that it is going to have to be completely flat, not really a rectangle,1301

just a chunk of line, because it doesn't have any height, because we are evaluating at the left-hand side.1305

For our second rectangle, once again, there is a width of 1, times the height (will be evaluated at the left-hand side here); it will be 1.1309

So, f evaluated at 1...1 times 12 gets us just 1; so there is an area of 1 for our second rectangle.1316

And our third and final rectangle: 1 times f evaluated at 2, because the left-hand from 2 to 3 is going to be 2;1324

1 times f(2) is 1 times 22, when we evaluate that function; and that comes out to 4.1332

So, the total amount of area that we get for our approximation, the total approximation we get,1338

is going to be equal to each of these added up together--the first rectangle, 0,1343

plus the second rectangle, 1, plus the third, final rectangle (4 there)--each of our areas.1349

0 + 1 + 4 gets us a total area of 5 for our entire approximation of a = 0 to b = 3 with three sub-intervals and the left-hand point for each one.1356

Our second example is very similar to our previous example.1368

We are finding an approximation to the area under f(x) = x2 from a = 0 to b = 31371

by using four equal-width rectangles, n = 4, on the maximum point of each interval.1378

The only difference here is that we are now using four rectangles, and we are doing it based on the maximum point,1383

the highest location for each one of these intervals.1390

We draw in our curve...x2...we have something like this; we are going, once again, from a = 0 to b = 3.1392

So, there is our interval; and we are going to be looking for the maximum point of each interval.1400

If it is four equal-width rectangles (let's draw that in really quickly: 1, 2, 3, 4),1404

notice: the maximum point of each interval--where is that going to be here?--well, that is going to be here.1409

Where is that going to be for this one?--that will be here.1414

Where will that be for this one?--that will be here; and where for this one?--that will be here.1416

The maximum point for this one is basically the same thing as saying the right-hand side.1420

Now, I want to point out that this is not always true.1428

As we saw when we were working through this lesson, right-hand point and maximum can give us totally different rectangle pictures.1430

However, for something like f(x) = x2, where it is just constantly growing, constantly growing, constantly growing,1436

since it is always growing as it goes off to the right, that means that for any sub-interval,1442

the right-most point will be at the highest height; so the right-most point1446

is the same thing as maximum for the specific case of the function x2.1449

With a different function, you might end up having different things; so it is something that you have to think about.1454

But in this specific case, it will be the same thing as just evaluating at the right-hand side.1458

What is the width of each one of these rectangles?1462

Well, the width is (b - a)/n; in that case, we have a width of 3 total, divided by 4 for each one.1464

And let's put that in decimal for ease, because we are going to end up using a calculator to crunch these numbers,1470

because we will have a lot of decimals showing up otherwise.1474

Our very first horizontal location is 0, because we start at 0.1477

The next one will be 0.75; we have a width of 0.75.1482

The next one will be at 1.5, because we are another 0.75 step ahead of that.1486

The next one will be at 2.25, another 0.75 step ahead of that; and finally, we finish at 3, which makes sense.1491

We have to start at 0 and end at 3; and we work by step after step of our width, 0.75, to make it up each time.1498

Our first rectangle...figure out the area for that one; our second rectangle; our third rectangle; and our fourth rectangle.1506

OK, so our first rectangle: remember, the right-hand point is the same thing as the maximum point in the specific case of the function x2.1514

So, what is the width here? The width is 3/4; remember, in general, the area of any rectangle is the width times the height.1523

So, in this case, it will be (b - a)/n, because that is always going to be the width if we have n equal sub-intervals going from a to b.1530

b - a is the total length; divide by n for the width.1537

Multiply that times the height of each one of them, our f evaluated at xi.1541

And xi will depend on how we are choosing the point to look at.1545

In this case, we are looking at the maximum point, which ends up being the right-hand point;1548

so we will always end up looking at the right side of each one of these sub-intervals.1551

OK, back to our first one: 3/4, our width, times the height--where does the height end up getting evaluated?1555

Well, if we are going from 0 to 0.75 (that is our first sub-interval), we are going to end up looking at the most right-hand part, which is 0.75.1561

So, f...plug in 0.75; 3/4 times f(0.75) is the same thing as 0.75 times 0.752.1570

Since f is just x2, just a squaring function, it is 0.75 times 0.752.1583

We work that out with a calculator, and we get 0.422; great.1589

The second rectangle: once again, we evaluate 3/4, the width, times f evaluated now at 1.5,1594

because the right side of this one, the right side of our second rectangle, will be 1.5.1601

Here was our first rectangle; now we are on our second rectangle; its right side is 1.5.1607

So, 3/4 times 1.52: work that out with a calculator; we get 1.688.1612

The third rectangle is this one right here: the right side of that rectangle, the maximum height, is 2.25.1623

It is still the same width; the width will never end up changing, because we set the width to be equal for all of them.1629

So, f...plug in 2.25 from the right side (in this specific case);3/4...our function is squared, so 2.252;1634

that comes out to be 2.797; and our fourth and final rectangle...width times the height that it evaluates at...1641

that is going to be 3, because the far right-hand side of our final interval is 3.1650

3/4 times f(3)...remember, the right-hand side, in this specific case, happens to be the maximum.1654

If you worked with a different function, you would have to think about what was going to be the maximum location there.1661

3/4 times 32 comes out to be 6.75.1666

If we want to know what the total area approximation is, we add up all of these numbers.1671

So, it is going to be area equals 0.422 + 1.688 + 3.797 + 6.75.1676

We add up all four of the rectangles, and that tells us the total approximation we got by using this specific method.1686

And that comes out to be 12.657; so that is the total approximation we end up getting here.1693

Notice: at this point, we now have the lower sum and the upper sum.1701

In our first example, we chose minimum, because we chose the left-hand side; and we ended up getting 5 out of that.1705

At this point, we now have just chosen the maximum, so we got an upper sum,1711

something that is the most area it could possibly be; and we ended up getting 12.657.1715

So, whatever the actual area underneath that curve is, we know that it has to be between 51721

(because that was the lower sum in our first example) and 12.657 (because we just got an upper sum in this, our second example).1725

So, whatever the actual value of the area underneath that curve between 0 and 3 is, it has to be somewhere between 5 and 12.657.1731

Now, in our third example, we are going to actually figure out what it is precisely by taking an infinite limit.1739

For our third example, find the precise area under f(x) = x2 from a = 0 to b = 3,1745

using the limit as the number of rectangles, n, goes off to infinity.1751

To do this, we will also end up needing this specific identity, this sum of i = 1 to n of i2 = n(n + 1)(2n + 1), all divided by 6.1755

But we won't actually end up using that until we get about halfway through this thing.1766

So, first, how do we set this thing up?1769

Once again, here is just a quick picture to help us clarify things.1772

We have x2; we are working our way from a = 0 to b = 3.1777

All right, and we are going to end up cutting it into some number of sub-intervals.1785

We are going to end up cutting it into n sub-intervals.1789

And then, from there, we will let n go off to infinity.1790

But we have to start by figuring out what the area would be if we had just some actual value for n, and then we let n run off to infinity.1793

First, what is our width going to end up being?1801

Our width will end up being (b - a)/n, because that is always the case: (b - a)/n.1803

In this case, a = 0; b = 3; so our width is going to be 3, divided by the number of rectangles we choose to use, 3 divided by n.1809

That is the width of each one of these.1816

Now, we need to decide how we are going to determine where we choose our xi.1818

It is going to end up actually making our notation just a little bit easier if we choose right-most; so we are just going to arbitrarily choose right-most.1823

I want to point out to you, though, that it doesn't actually matter which one we choose: right-most, left-most, midpoint, upper sum, lower sum...1831

If it does eventually converge to a single value, if our limit does exist as n goes off to infinity, then we will have found the area underneath it.1838

And no matter how we choose to set up our cuts and our heights, as the number of rectangles goes farther and farther off to infinity,1846

our area has to get closer and closer to the actual thing,1854

no matter how we set up the heights, if it can get to an actual value under the area.1856

So, it doesn't matter which one we choose, specifically.1861

Right-most is nice, because it is easy to do it notation-wise; so if you end up having to do a similar problem,1863

you can basically just copy the method I am doing here to set up.1867

And it will end up working for you in notation, as well.1870

All right, let's get back to this: we will set up our right-most xi for each one.1873

So, what will xi end up being? Well, our first xi will be...1878

a is the far left side, so what would be the next one?1886

Well, that would be a plus...what is our width? Our width is 3/n...3/n for the first xi.1888

So, in general, to get out to the ith xi, we start at a.1896

And then, if we are at the ith xi, we will be...3/n is our width each time we go forward a step.1900

How many steps did we take forward? i steps.1905

If we start at a, and then we want to get to the xi, which is the far right side of any one,1908

well, the first one would show up at +1(3/n); the second one would show up at +2(3/n); and the third one would show up at +3(3/n).1912

It is the number of steps, times our width each time, 3/n.1924

So, the number of steps that we have taken, if we are at the right-most of each side, is just going to be i, whatever sub-interval we are at.1928

If we are in our first sub-interval, we have taken one width-step to get to the right side.1935

If we are in our tenth sub-interval, we have taken ten width-steps to get to the right side of that tenth sub-interval.1938

In general, xi is equal to a, our starting location, plus 3/n, times i.1944

In the specific case of this problem, we can plug in what our a equals, 0.1950

We have xi = 0 + 3/n(i); so 3 over n times i gives us our xi for this specific problem.1955

But if you were doing just any problem in general, you would want to use that first part, a plus...1964

and also, you wouldn't use 3/n; you would end up using whatever your specific width ended up being, which would be (b - a)/n.1969

It is not necessarily 3/n; that is this specific problem that we are working right here.1975

All right, if we have xi for each one, what will end up being the height for each one of our rectangles?1980

The height of the ith rectangle is going to be f evaluated at xi, which to say f evaluated at 3/n times i.1985

So, in general, what we do next is set up our limit: the limit as n goes off to infinity of our sum.2001

The approximation for n rectangles will be to start at our first rectangle, i = 1, and go out to our last rectangle, n.2009

And it is going to be the width of each one of these rectangles, (b - a)/n, times the height of each one of these rectangles, f(xi).2015

Now, this formula here will always end up working for any problem that you have set up.2023

So, that is a useful thing to work with.2028

Now, we are going to start using what we have in our specific problem--we will start plugging things in.2030

Our limit is still the same; our summation is still the same; we are going from the first to the last.2034

Our (b - a)/n...we figured out that that was a width of 3, divided by the number of rectangles we chose, 3 divided by n,2040

times f evaluated at xi...well, what is f evaluated at xi?2046

Well, f(x) = x2; so if our xi is 3/n times i (that was what we figured out that our xi has to be),2050

then f evaluated at 3/n, times i, is 3/n times i squared.2061

And we have that right here; we now plug that in: 3/n times i, the whole thing squared.2070

At this point, we can now simplify that just a little bit, and we have the limit as n goes off to infinity of the sum, i = 1 to n,2077

our first rectangle to our last rectangle; the square distributes--we have 32/n2 times i2.2086

We also have that 3/n there; so we simplify that to 27, over n3, times i2.2093

All right, at this point, we are now ready to move on to the second half of this.2103

We can now start working to figure out what this infinite series ends up coming out to be.2107

All right, continuing on with our example: we have that the area is equal to the limit as n goes off to infinity from i = 1 to n2113

of 27/n3 times i2, whatever that ends up happening to be.2121

And at a later point, we will end up putting this identity into the problem so that we can actually solve this thing out.2126

All right, the first thing to notice is that the 27/n3 part doesn't actually do anything inside of the sum.2131

The n, as far as the sum is concerned, is actually a fixed value.2140

The n here is just a constant; remember when we were working with sigma notation--the number on top was just some number.2144

It didn't change around during the course of doing things.2149

So, since the n isn't changing inside of the sigma (it will change over here, because n will go off to infinity,2152

but as far as the sigma is concerned, it doesn't change--the limit has to do something--2158

so since the sigma doesn't have anything happening), we can actually pull it outside of the sigma.2162

The first step here is to see that what we really have is: we can write this as limit as n goes off to infinity.2168

We pull out the 27/n3, because it is not affected if it does it on the inside or the outside.2173

It is just a scalar constant, as far as our series here is concerned.2179

So, i = 1 up to n of i2...at this point, look: we have the sum as i = 1 goes off to n of i2.2184

So, we can now swap it out for this portion right here.2195

So, we swap it out for that portion right here; we have the limit as n goes off to infinity of 27/n3;2199

and we swap out n times n + 1 times 2n + 1, all over 6.2207

Great; let's work to simplify things out a bit: the limit as n goes off to infinity...2217

we will keep our 27/n3 off to the side for just a moment.2222

Well, let's actually put it into the thing, but we will deal with this part first.2228

n times n + 1 times 2n + 1...let's expand this a little bit.2231

We do a little bit of expanding; we have n2 + n, times 2n + 1, all over n3 times 6; great.2234

Limit as n goes off to infinity...we can finish expanding our factors there.2249

We have 27, times n2 times 2n (becomes 2n3); n2 + 1...n2 times +1; n times 2n gets 2n2...2253

So, we have a total of +3n2; and n times 1 gets us + n, all over 6n3.2264

Limit as n goes off to infinity...we have 27 times 2n3; that comes out to be 54n3;2274

plus 3 times n2...that comes out to be 81n2; plus 27n, all divided by 6n3.2285

All right, at this point, let's move this all up here, so that we can keep working on it.2300

Notice that, if we want, we can break this into two separate fractions.2306

We have the limit as n goes to infinity of 54n3, over 6n3, plus 81n2, plus 27n, all over 6n3.2310

Now, this portion right here, the n3 and the n3, will end up canceling out.2331

We have 54 divided by 6; but the limit as n goes off to infinity for this portion...2336

well, we have n3 on the bottom; that is a 3 degree on the bottom; but on the top, we only have n2 and n.2343

So, that is a 2 degree and a 1 degree; those can't compete with an n3 on the bottom.2350

A degree of 3 on the bottom is going to end up crushing that in the long term.2355

As n rides off to infinity, our denominator is going to get so much bigger than our numerator that this whole part here just crushes down to 0.2359

That means that we are left, as our n goes off to infinity, with simply having 54 divided by 6.2366

And 54 divided by 6 gets us 9; so the total area underneath that curve is actually precisely equal to 9--pretty cool.2372

Notice that it does take a little bit of challenging effort to work through this limit as n goes to infinity.2381

We can work through it slowly, but it is not easy.2387

We had to pull up this kind of arcane formula.2389

It is not a very difficult summation formula, but it is not one that we probably know immediately.2392

So, this stuff isn't super easy to work with (limit as n goes off to infinity), if we are trying to do this infinite cut method.2396

And that is why that fundamental theorem of calculus, that integral with anti-derivative stuff, is so useful.2403

Let's see just how powerful that is now in our final example of the course.2409

Using the amazing fact that the integral of a to b of f(x)dx is equal to the anti-derivative of f evaluated at b,2413

minus the anti-derivative of f evaluated at a (and remember: the integral is just the area underneath the curve, from a to b,2421

for some curve defined by f(x)), find the area under f(x) = x2 from a = 0 to b = 3.2428

This part right here is the area that we are looking for.2435

So, what we want to do is figure out what F(b) - F(a) is; F(x) is the anti-derivative of f(x).2441

So, the very first step that we need to figure out here is what the anti-derivative is to f(x).2452

f(x) = x2; when we worked through the derivative examples, we talked about the power rule,2457

where you take the exponent, and you move it down.2465

So, for example, if we have f(x) = x5, the derivative of x is equal to...move that exponent, the 5, down...2467

we have 5 times x, and then we subtract 1 from the exponent; so - 1 from it at that point...it will be 5x4.2481

There is this nice, easy rule for taking derivatives with the power rule.2489

If we have x2, and we want to reverse the process, well, since it drops down by 12493

each time we end up taking a derivative, that means that the anti-derivative must push our exponent up by 1.2498

Since it will drop down one when we take the derivative, when we go to the derivative of F(x), remember:2506

since we can take the derivative of F(x) to just get f(x), we have to have this relationship2513

of our exponent dropping down when we take the derivative of F(x).2520

The derivative of F(x), that 3, must drop down to a square, since 3 - 1 will go down to 2.2525

So, we know that the exponent that must start there must be 3.2531

However, if we were to work this out, we would end up getting x3; if we took the derivative of this,2534

we would bring the 3 down, and we would get 3 times x3 - 1, or 3x2.2539

But there is this problem here: it is not 3x2; it is just plain x2.2545

So, how can we get rid of this 3 coefficient at the front? We just divide the whole thing by 3.2549

So, we divide the whole thing by 3, and now we have the anti-derivative to it.2555

Let's check and make sure that that works: let's check: if F(x) is equal to x3/3, then what is the derivative of F(x)?2560

Well, it should turn out to be f(x); we have this cubed exponent; we bring that down and out to the front.2570

That is going to be equal to 3 times what we started with; 3 - 1 becomes a 2; it is still divided by 3;2576

3 and 3 on the bottom cancel out, and we have x2, which is what we started with; that checks out--great.2584

So now, we have figured out what our anti-derivative is.2591

At this point, we can now actually use this portion of the formula:2594

F evaluated at...what is our b?...3, minus F evaluated at...what is our a?...0:2598

so, F(3) - F(0)...the anti-derivative of our function evaluated at the far end, minus the anti-derivative at our starting location.2608

The ending location, minus the starting location--what is our thing?2616

It is x3/3; that is what our F(x) is; so we have 33/3 - 03/3.2619

03/3 just disappears; so we have 33/3; 3 cancels this into a 2;2630

the 3 on the bottom cancels our top exponent down by 1; we have just 32; 32 is 9--nice.2637

And that is the exact same thing that we got by working through that long infinite series method,2645

where we had the limit as n goes to infinity of this approximation.2650

And notice how much faster this was than that previous method.2653

And this was with me carefully explaining a bunch of ideas that we have just seen for the very first time.2656

If you were used to doing this...when you get used to this method, you can fly through it.2661

You can do it so much faster than trying to work through the limits.2664

That precise, formal limit method is something that you learn at the very beginning, just so we can introduce this really, really amazing fact.2667

This is the fundamental theorem of calculus; this idea is so important that it gets the name "fundamental theorem of calculus."2674

It is really cool stuff here.2679

All right, that finishes this course; it has been a pleasure teaching you.2681

I hope that you have not just learned how these things work and how to use them,2684

but that you have also started to gain some idea of how math works on a deeper, more intuitive level.2688

Things are starting to make more sense to you than they were at the beginning of the course.2693

Beyond just the specific material that you have worked with, you are starting to get a better sense of just how math works in a personal way.2696

And even if you don't decide to continue on with math, I hope you have started to develop2702

a little bit more of an appreciation for just how cool it is--how many uses it ends up having.2705

This stuff is really great, and it forms the basis for so much of the society that we have.2709

The basis of technology, in many ways, is mathematics; plus, I think that this stuff is just beautiful and cool for itself, just to study it.2714

All right, we will see you at Educator.com later.2721

And I wish you luck in whatever you end up doing--goodbye!2723