Connecting...

This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of College Calculus: Level I

College Calculus: Level I Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus

Section 6: Integrals: Lecture 2 | 22:02 min

Lecture Description

Next Lecture

Previous Lecture

Discussion
Answer Engine
Study Guides
Practice Questions
Download Lecture Slides
Table of Contents
Related Books

Lecture Comments (10)

	0 answers Post by Israel Haile on October 23, 2015 I loved your methodology ,I wanna encourage you, I am 99.99% satisfied with your class,Stay Blessed !!!
	0 answers Post by Valeriya Pinkhasova on February 5, 2015 I am unable to watch additional example. Everytime I press on it ,lecture starts from the beginning .
	0 answers Post by Jason Kim on November 5, 2014 wait for the antideriviative of the lecture example 2 how do you exactly know that anti deriviative of 2x-5 is just x^2-5X. It could also be x^2-5x plus some number because the number just disappears when we derive it. So it could be x^2-5x+9 and the deriviative would be also 2x-5. How can you assume that the antideriviative is just x^2-5x?
	0 answers Post by Johnny Zamora on January 14, 2014 She is Amazing
	0 answers Post by Jose Gonzalez-Gigato on November 10, 2013 'Additional Example 5' was excellent!
	1 answer Last reply by: Martina Alvarez Fri Dec 9, 2011 4:28 AM Post by Martina Alvarez on December 9, 2011 Ex: 2 13 min. =(36-30) - (1-5) =6 - (-4) =10 not 8
	0 answers Post by David Bascom on September 4, 2011 In lecture example 3 where did 3xsquared come from?
	1 answer Last reply by: Jose Gonzalez-Gigato Sun Nov 10, 2013 11:30 AM Post by Okwudili Ezeh on July 21, 2011 The graph was supposed to be a sine graph not a cosine graph.

Answer EngineGet answers to any question!Ask any question related to College Calculus: Level I

Working on the solution...

Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus

Geometrically, Riemann sums represent sums of rectangle approximations.
The definite integral is a limit of Riemann sums.
For very simple functions, it is possible to directly compute Riemann sums and then take the limit.
Though this topic is very important theoretically, in practice we will compute integrals by using the First Fundamental Theorem of Calculus.

Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus

∫₁⁵ dx

∫₁⁵ dx = x |₁⁵
∫₁⁵ dx = 5 − 1

∫₁⁵ dx = 4

∫₀^π cosx (sinx)³ dx

u = sinx
du = cosx dx
∫₀^π cosx (sinx)³ dx = ∫₀^π u³ du
∫₀^π cosx (sinx)³ dx = [(u⁴)/4] |₀^π
Remember, the limits of the integral are dependent on x. We must plug back in for x before using the original limits
∫₀^π cosx (sinx)³ dx = [((sinx)⁴)/4] |₀^π
∫₀^π cosx (sinx)³ dx = [((sinπ)⁴)/4] − [((sin0)⁴)/4]
∫₀^π cosx (sinx)³ dx = [0/4] − [0/4]

∫₀^π cosx (sinx)³ dx = 0

∫_−[(π)/2]^[(π)/2] 5 cosx dx

∫_−[(π)/2]^[(π)/2] 5 cosx dx = 5 sinx |_−[(π)/2]^[(π)/2]
∫_−[(π)/2]^[(π)/2] 5 cosx dx = 5 sin[(π)/2] − 5 sin[(−π)/2]
∫_−[(π)/2]^[(π)/2] 5 cosx dx = 5 − (−5)

∫_−[(π)/2]^[(π)/2] 5 cosx dx = 10

Confirm that ∫₋₂² x² dx = 2 ∫₀² x² dx

∫₋₂² x² dx = [(x³)/3] |₋₂²
∫₋₂² x² dx = [(2³)/3] − [((−2)³)/3]
∫₋₂² x² dx = [8/3] + [8/3]
∫₋₂² x² dx = [16/3]
2 ∫₀² x² dx = 2 [(x³)/3] |₀²
2 ∫₀² x² dx = 2 [(2³)/3] − 2[(0³)/3]
2 ∫₀² x² dx = [16/3]
This is a property of integrals involving even functions
∫_−a^a f(x) dx = 2 ∫₀^a f(x) dx if f(x) is even

Yes, ∫₋₂² x² dx = 2 ∫₀² x² dx

∫₀^ln5 e^2x dx

u = 2x
du = 2 dx
∫₀^ln5 e^5x dx = [1/2] ∫₀^ln5 e^u du
∫₀^ln5 e^5x dx = [1/2] e^u |₀^ln5
∫₀^ln5 e^5x dx = [1/2] e^2x |₀^ln5
∫₀^ln5 e^5x dx = [1/2] e^{2 ln5} − [1/2] e²⁽⁰⁾
∫₀^ln5 e^5x dx = [1/2] e^ln5² − [1/2] (1)
∫₀^ln5 e^5x dx = [25/2] − [1/2]

∫₀^ln5 e^5x dx = 12

∫₋₁¹ x² + 3x + 1 dx

∫₋₁¹ x² + 3x + 1 dx = ∫₋₁¹ x² dx + ∫₋₁¹ 3x dx + ∫₋₁¹ dx
∫₋₁¹ x² + 3x + 1 dx = [(x³)/3] |₋₁¹ + [(3x²)/2] |₋₁¹ + x |₋₁¹
∫₋₁¹ x² + 3x + 1 dx = [1/3] + [1/3] + [3/2] − [3/2] + 1 + 1

∫₋₁¹ x² + 3x + 1 dx = [8/3]

∫₋₁¹ x³ dx

∫₋₁¹ x³ dx = [(x⁴)/4] |₋₁¹
∫₋₁¹ x³ dx = [1/4] − [1/4]
Property of odd functions
∫_−a^a f(x) = 0 if f(x) is odd

∫₋₁¹ x³ dx = 0

∫₀² [1/(√{4 − x²})] dx

∫₀² [1/(√{4 − x²})] dx = ∫₀² [1/(√{2² − x²})] dx
∫₀² [1/(√{4 − x²})] dx = sin⁻¹ [x/2] |₀²
∫₀² [1/(√{4 − x²})] dx = sin⁻¹ 1 − sin⁻¹ 0
∫₀² [1/(√{4 − x²})] dx = [(π)/2] − 0

∫₀² [1/(√{4 − x²})] dx = [(π)/2]

∫₀¹ [1/(√{4 − x²})] dx + ∫₁² [1/(√{4 − x²})] dx

∫₀¹ [1/(√{4 − x²})] dx + ∫₁² [1/(√{4 − x²})] dx = sin⁻¹ [x/2] |₀¹ + sin⁻¹ [x/2] |₁²
∫₀¹ [1/(√{4 − x²})] dx + ∫₁² [1/(√{4 − x²})] dx = sin⁻¹ [1/2] − sin⁻¹ 0 + sin⁻¹ 1 − sin⁻¹ [1/2]
∫₀¹ [1/(√{4 − x²})] dx + ∫₁² [1/(√{4 − x²})] dx = sin⁻¹ 1 − sin⁻¹ 0
Another property of integrals.
∫_a^b f(x) dx + ∫_b^c f(x) dx = ∫_a^c f(x) dx

∫₀¹ [1/(√{4 − x²})] dx + ∫₁² [1/(√{4 − x²})] dx = [(π)/2]

Show that ∫₁² e^x dx = − ∫₂¹ e^x dx

∫₁² e^x dx = e^x |₁²
∫₁² e^x dx = e² − e¹
− ∫₂¹ e^x dx = − e^x |₂¹
− ∫₂¹ e^x dx = −e¹ − (−e²)
− ∫₂¹ e^x dx = e² − e¹

∫₁² e^x dx = − ∫₂¹ e^x dx

Given the values below, use right-hand Riemann Sum with 4 intervals to approximate ∫₀¹⁰ f(x) dx
f(0) = 2 f(3) = 5 f(4) = 5.5 f(7) = 3 f(10) = 7

∫₀¹⁰ f(x) dx ≈ 5(3 − 0) + 5.5 (4 − 3) + 3(7 − 4) + 7(10 − 7)
∫₀¹⁰ f(x) dx ≈ 15 + 5.5 + 9 + 21

∫₀¹⁰ f(x) dx ≈ 50.5

Given the values below, use left-hand Riemann Sum with 4 intervals to approximate ∫₀¹⁰ f(x) dx
f(0) = 2 f(3) = 5 f(4) = 5.5 f(7) = 3 f(10) = 7

∫₀¹⁰ f(x) dx ≈ 2(3 − 0) + 5(4 − 3) + 5.5(7 − 4) + 3(10 − 7)
∫₀¹⁰ f(x) dx ≈ 6 + 5 + 16.5 + 9

∫₀¹⁰ f(x) dx ≈ 36.5

Given the values below, use right-hand Riemann Sum with 2 intervals to approximate ∫₀⁴ f(x) dx
f(0) = 2 f(3) = 5 f(4) = 5.5 f(7) = 3 f(10) = 7

∫₀⁴ f(x) dx ≈ 5(3 − 0) + 5.5(4 − 3)
∫₀⁴ f(x) dx ≈ 15 + 5.5

∫₀⁴ f(x) dx ≈ 20.5

Given the values below, use right-hand Riemann Sum with 5 intervals to approximate ∫₋₅⁵ f(x) dx
f(−5) = 25 f(−1) = 1 f(0) = 0 f(2) = 4 f(4) = 16 f(5) = 25

∫₋₅⁵ f(x) dx ≈ 1(−1 − (−5)) + 0(0 − (−1)) + 4(2 − 0) + 16(4 − 2) + 25(5 − 4)
∫₋₅⁵ f(x) dx ≈ 4 + 0 + 8 + 32 + 25

∫₋₅⁵ f(x) dx ≈ 69

Given the values below, use right-hand Riemann Sum with 3 intervals to approximate ∫₀⁵ f(x) dx
f(−5) = 25 f(−1) = 1 f(0) = 0 f(2) = 4 f(4) = 16 f(5) = 25

∫₀⁵ f(x) dx ≈ 4(2 − 0) + 16(4 − 2) + 25(5 − 4)
∫₀⁵ f(x) dx ≈ 8 + 32 + 25

∫₀⁵ f(x) dx ≈ 65

Given the values below, use left-hand Riemann Sum with 2 intervals to approximate ∫₋₅⁰ f(x) dx
f(−5) = 25 f(−1) = 1 f(0) = 0 f(2) = 4 f(4) = 16 f(5) = 25

∫₋₅⁰ f(x) dx ≈ 25(−1 − (−5)) + 1(0 − (−1))
∫₋₅⁰ f(x) dx ≈ 100 + 1

∫₋₅⁰ f(x) dx ≈ 101

Given the values below, use right-hand Riemann Sum with 4 intervals to approximate ∫₋₂² f(x) dx
f(−2) = −8 f(−1) = −1 f(0) = 0 f(1) = 1 f(2) = 8

∫₋₂² f(x) dx ≈ −1(−1 −(−2)) + 0(0 − (−1)) + 1(1 − 0) 8(2 − 1)
∫₋₂² f(x) dx ≈ −1 + 0 + 1 + 8

∫₋₂² f(x) dx ≈ 8

Given the values below, use left-hand Riemann Sum with 4 intervals to approximate ∫₋₂² f(x) dx
f(−2) = −8 f(−1) = −1 f(0) = 0 f(1) = 1 f(2) = 8

∫₋₂² f(x) dx ≈ −8(−1 − (−2)) + −1(0 − (−1)) + 0(1 − 0) + 1(2 − 1)
∫₋₂² f(x) dx ≈ −8 − 1 + 0 + 1

∫₋₂² f(x) dx ≈ −8

Given the values below, use right-hand Riemann Sum with 6 intervals to approximate ∫₀¹² f(x) dx
f(−2) = −8 f(−1) = −1 f(0) = 0 f(1) = 1 f(2) = 8 f(4) = 64 f(8) = 512 f(10) = 1000 f(12) = 1728

∫₀¹² f(x) dx ≈ 1(1 − 0) + 8(2 − 1) + 64(4 − 2) + 512(8 − 4) + 1000(10 − 8) + 1728(12 − 10)
∫₀¹² f(x) dx ≈ 1 + 8 + 128 + 2048 + 2000 + 3456

∫₀¹² f(x) dx ≈ 7641

Given the values below, use right-hand Riemann Sum with 6 intervals to approximate ∫₀⁸ f(x) dx
f(−2) = −8 f(−1) = −1 f(0) = 0 f(1) = 1 f(2) = 8 f(4) = 64 f(8) = 512 f(10) = 1000 f(12) = 1728

∫₀⁸ f(x) dx ≈ 1(1 − 0) + 8(2 − 1) + 64(4 − 2) + 512(8 − 4)
∫₀⁸ f(x) dx ≈ 1 + 8 + 128 + 2048

∫₀⁸ f(x) dx ≈ 2185

« Previous Question | Next Question »

*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

Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus

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

Important Equations

Lecture Example 1

Lecture Example 2

Lecture Example 3

Additional Example 4

Additional Example 5

Intro 0:00
Important Equations 0:22

Riemann Sum
Integral
Integrand
Limits of Integration (Upper Limit, Lower Limit)
Other Equations
Fundamental Theorem of Calculus

Lecture Example 1 5:04
Lecture Example 2 10:43
Lecture Example 3 13:52
Additional Example 4
Additional Example 5

College Calculus 1 Online Course

Section 1: Overview of Functions
	Review of Functions	26:29
	Compositions of Functions	12:29
Section 2: Limits
	Average and Instantaneous Rates of Change	20:59
	Limit Investigations	22:37
	Algebraic Evaluation of Limits	28:19
	Formal Definition of a Limit	23:39
	Continuity and the Intermediate Value Theorem	19:09
Section 3: Derivatives, part 1
	Limit Definition of the Derivative	22:52
	The Power Rule	26:01
	The Product Rule	14:54
	The Quotient Rule	19:17
	Applications of Rates of Change	17:43
	Trigonometric Derivatives	26:58
	The Chain Rule	23:47
	Inverse Trigonometric Functions	27:05
	Equation of a Tangent Line	15:52
Section 4: Derivatives, part 2
	Implicit Differentiation	30:05
	Higher Derivatives	13:16
	Logarithmic and Exponential Function Derivatives	17:42
	Hyperbolic Trigonometric Function Derivatives	14:30
	Related Rates	29:05
	Linear Approximation	23:52
Section 5: Application of Derivatives
	Absolute Minima and Maxima	18:57
	Mean Value Theorem and Rolle's Theorem	20:00
	First Derivative Test, Second Derivative Test	27:11
	L'Hopital's Rule	23:09
	Curve Sketching	40:16
	Applied Optimization	25:37
	Newton's Method	25:13
Section 6: Integrals
	Approximating Areas and Distances	36:50
	Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus	22:02
	Substitution Method for Integration	23:19
Section 7: Application of Integrals, part 1
	Area Between Curves	19:59
	Volume by Method of Disks and Washers	24:22
	Volume by Method of Cylindrical Shells	30:29
	Average Value of a Function	16:31
Section 8: Extra
	Graphs of f, f', f''	23:58
	Slope Fields for Differential Equations	18:32
	Separable Differential Equations	17:04

Related Books

	Calculus by Larson, 9th Edition Authors: Ron Larson, Bruce H. Edwards ISBN: 0547167024 Publisher: Brooks Cole Year: 2009 The book is filled with examples and detailed explanations. Capstone exercises are also included for each section and synthesize the main concepts into a single example.
	Calculus by Larson, 9th Edition, Solutions Manual, Volume 1 Author: Ron Larson ISBN: 547213093 Publisher: Brooks Cole Year: 2008 This solutions manual includes worked out solutions to every odd-numbered exercise in Calculus of a Single Variable, 9th edition.
	Calculus by Larson, 9th Edition, Solutions Manual, Volume 2 Authors: Ron Larson, Bruce H. Edwards ISBN: 547213107 Publisher: Brooks Cole Year: 2009 This solutions manual includes worked out solutions to every odd-numbered exercise in Multivariable Calculus, 9th edition.

Related Books

Name	Description	Link
BookRenter.com	BookRenter.com is simply the most reliable online textbook rental service.	Visit BookRenter.com
PhysicsForums.com Homework Help	Physics Forums is a scientific community for students looking for math & science help.	Visit PhysicsForums.com Homework Help

Professor Switkes

Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus

Slide Duration:

Table of Contents

Section 1: Overview of Functions

Review of Functions

26m 29s

Intro