Starting up...

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 Calculus BC

Calculus BC Improper Integrals

Section 3: Integration: Lecture 3 | 19:01 min

Lecture Description

Next Lecture

Previous Lecture

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

Lecture Comments (8)

	0 answers Post by Kevin Hong on April 27, 2014 at about 5:45, you said inverse tangent of infinity is pi over 2 but how did you come up with pi over 2?
	1 answer Last reply by: Huijie Shen Sun Apr 19, 2015 8:55 PM Post by Mohamad Oda on April 6, 2014 2*sqrt(2) is the right answer
	2 answers Last reply by: Thomas Zhang Mon Mar 17, 2014 1:10 PM Post by Raudel Ulloa on August 6, 2012 I checked in the calcuator for number 5 and the interagal is 2 radical 2, what i got, not 4
	1 answer Last reply by: Raudel Ulloa Mon Aug 6, 2012 9:48 AM Post by Chung Teak Joon on April 29, 2012 isnt progress on number four kind of wrong? or am I just misunderstanding the steps? If we integrate root u, doesnt it have to become 0.5u^0.5??? How did it suddenly become 2u^0.5????

Answer EngineGet answers to any question!Ask any question related to Calculus BC

Working on the solution...

Improper Integrals

Use method of improper integrals when:

There is an infinity in either bound of the integral
Function evaluated at any bound is undefined

Improper Integrals

∫₁^∞ [3/x]dx

Replace the improper bound ∞ with k
∫₁^k [3/x]dx
Integrate
∫₁^k [3/x]dx = 3[ln|x|]₁^k
= 3lnk − 3ln1

= 3lnk = lim_{k → ∞} 3lnk = ∞

∫_{− ∞}⁰ [1/(( x + 5 )²)]dx

Replace the improperbound ∞ with k
∫_k⁰ [1/(( x + 5 )²)]dx
Integrate
∫_k⁰ [1/(( x + 5 )²)]dx = [ − [1/(x + 5)] ]_k^0.
= − [1/(0 + 5)] + [1/(k + 5)]

= lim_{x → − ∞} [ [1/(k + 5)] ] − [1/5] = 0 − [1/5] = − [1/5]

∫_{− ∞}^∞ [7/((x − 3)²)]dx

Note that [7/((x − 3)²)] is an even function,and the limits are the same magnitude
∫_{− ∞}^∞ [7/((x − 3)²)]dx = 2∫₀^∞ [7/((x − 3)²)]dx
Replace the improper limit ∞ with k
2∫₀^∞ [7/((x − 3)²)]dx = 2∫₀^k [7/((x − 3)²)]dx
Integrate
2∫₀^k [7/((x − 3)²)]dx = 14∫₀^k [1/((x − 3)²)]dx
= 14[ − [1/(x − 3)] ]₀^k
= 14[ − [1/(k − 3)] + [1/(0 − 3)] ]
= 14[ − [1/(k − 3)] − [1/3] ]
= 14[ − lim_{k → ∞} [1/(k − 3)] − [1/3] ]
= 14[ 0 − [1/3] ]

= [( − 14)/3]

∫₃⁷− [4/(√{x − 3} )]dx

Replace the bound 3 with k
∫₃⁷− [4/(√{x − 3} )]dx = ∫_k⁷− [4/(√{x − 3} )]dx
Integrate
∫_k⁷− [4/(√{x − 3} )]dx = − 4∫_k⁷ [1/(√{x − 3} )]dx
= − 4[ 2√{x − 3} ]_k⁷
= − 4[2√{7 − 3} − 2√{k − 3} ]
= − 4[2√4 − 2√{k − 3} ]

= − 4[4 − 0] = − 4[4] = − 16

∫_{− 2}^k [2/(x√{x² − 1} )]dx

Replace the bound 1 with k
∫_{− 2}^k [2/(x√{x² − 1} )]dx = ∫_{− 2}^k [2/(x√{x² − 1} )]dx
Integrate using inverse trigonmetric identity
∫_{− 2}^k [2/(x√{x² − 1} )]dx = 2∫_{− 2}^k [1/(x√{x² − 1} )]dx
= 2[ sec^{− 1}|x| ]_{− 2}^k
= 2[ sec^{− 1}|k| − sec^{− 1}| − 2| ]

= 2[ lim_k1 sec^{− 1}|k| − [(π)/3] ] = 2[ 0 − [(π)/3] ] = − [(2π)/3]

∫₀^∞ [(x − 4)/(x² − 3x − 4)]dx

Factor x² − 3x − 4 to simplify
∫₀^∞ [(x − 4)/(x² − 3x − 4)]dx = ∫₀^∞ [(x − 4)/((x − 4)(x + 1))]dx
= ∫₀^∞ [1/((x + 1))]dx
Replace the improper limit ∞ with k
∫₀^∞ [1/((x + 1))]dx = ∫₀^k [1/((x + 1))]dx
Integrate
∫₀^k [1/((x + 1))]dx = [ ln|x + 1| ]₀^k
= ln|k + 1| − ln|0 + 1|
= ln|k + 1| − ln|1|
= ln|k + 1| − 0
= ln|k + 1|
= lim_{k → ∞} ln|k + 1|

= ∞

∫_{− ∞}¹ [2/(x² + 2x)]dx

Replace the bound − ∞ with k
∫_{− ∞}¹ [2/(x² + 2x)]dx = ∫_k¹ [2/(x² + 2x)]dx
Integrate
∫_k¹ [2/(x² + 2x)]dx = ∫_k¹ [1/x] − [1/(x + 2)]dx
= ∫_k¹ [1/x] − ∫_k¹ [1/(x + 2)]dx
= [ ln|x| ]_k¹ − [ ln|x + 2| ]_k¹
= ln1 − lnk − ln3 + ln|k + 2|

= lim_{k → − ∞} [ − ln3 + ln|k + 2| − lnk ] = − ln3 − ∞− ∞ = − ln3

∫₀^∞ [4/((x + 1)(x + 2))] dx

Replace the bound ∞ with k
∫₀^∞ [4/((x + 1)(x + 2))] dx = ∫₀^k [4/((x + 1)(x + 2))] dx
Integrate
∫₀^k [4/((x + 1)(x + 2))] dx = ∫₀^k [4/(x + 1)] − [4/(x + 2)] dx
= 4∫₀^k [1/(x + 1)] − [1/(x + 2)] dx
= 4[ ln|x + 1| − ln|x + 2| ]₀^k

= 4[ lim_{k → ∞} ln[(k + 1)/(k + 2)] − ln[1/2] ] = 4[ 0 − ln[1/2] ] = − 4ln[1/2]

∫_{− 2}² [1/(√{4 − x²} )]dx

Note the function is symmetrical and simplify
∫_{− 2}² [1/(√{4 − x²} )]dx = 2∫₀² [1/(√{4 − x²} )]dx
Replace the bound 2 with k
2∫₀² [1/(√{4 − x²} )]dx = 2∫₀^k [1/(√{4 − x²} )]dx
Apply the inverse trignometric integration identity to integrate
2∫₀^k [1/(√{4 − x²} )]dx = 2[ sin^{− 1}[x/2] ]₀^k
= 2[ sin^{− 1}[k/2] − sin^{− 1}[0/2] ]
= 2[ sin^{− 1}[k/2] ]
= 2[ lim_{k → 2} sin^{− 1}[k/2] ]
= 2[ [(π)/2] ]

= π

10.∫₀^∞ [1/(( x − 2 )³)] dx

Replace the bound ∞ with k
∫₀^∞ [1/(( x − 2 )³)] dx = ∫₀^k [1/(( x − 2 )³)] dx
Integrate
∫₀^k [1/(( x − 2 )³)] dx = [ − [1/(2(x − 2)²)] ]₀^k
= [ − lim_{k → ∞} [1/(2(k − 2)²)] + [1/(2(0 − 2)²)] ]
= 0 + [1/2(4)]

= [1/8]

« 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

Improper 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

Improper Integrals

3 Steps

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Intro 0:00
Improper Integrals 0:06

3 Steps

Example 1 1:46
Example 2 3:20
Example 3 6:05
Example 4 9:02
Example 5 11:21
Example 6 15:54

Mathematics: AP Calculus BC

Section 1: Prerequisites
	Parametric Curves	10:10
	Polar Coordinates	14:54
	Vector Functions	12:05
Section 2: Differentiation
	Parametric Differentiation	13:15
	Polar Differentiation	10:14
	L'Hopital's Rule	6:43
Section 3: Integration
	Integration By Parts	19:04
	Integration By Partial Fractions	22:04
	Improper Integrals	19:01
Section 4: Applications of Integration
	Logistic Growth	19:16
	Arc Length for Parametric & Polar Curves	17:40
	Area for Parametric & Polar Curves	4:33
Section 5: Sequences, Series, & Approximations
	Definition & Convergence	7:10
	Geometric Series	10:59
	Harmonic & P Series	3:58
	Integral Test	11:31
	Comparison Test	7:47
	Limit Comparison Test	9:40
	Ratio Test	11:20
	Alternating Series	9:11
	Absolute Convergence	7:13
	Power Series Convergence	12:52
	Taylor Series	15:11
	Power Series Operations	16:40
	Lagrange Error	7:09
Section 6: Practice Test
	AP Calc BC Practice Test Section 1: Multi Choice Part 1	15:45
	AP Calc BC Practice Test Section 1: Multi Choice Part 2	21:38
	AP Calc BC Practice Test Section 1: Multi Choice Part 3	17:31
	AP Calc BC Practice Test Section 1: Multi Choice Part 4	14:17
	AP Calc BC Practice Test Section 1: Multi Choice Part 5	22:45
	AP Calc BC Practice Test Section 1: Free Response Part 1	9:55
	AP Calc BC Practice Test Section 1: Free Response Part 2	10:08
	AP Calc BC Practice Test Section 1: Free Response Part 3	13:58

Related Books

	Cracking the AP Calculus BC Exam, 2015 Edition Author: Princeton Review ISBN: 804124825 Publisher: Princeton Review Year: 2014 The book includes 3 full length practice tests with detailed explanations, a review of all the key concepts, and targeted strateeies to ace th exam for your highest score.
	Cliffs AP: Calculus AB & BC, 3rd Edition Authors: Dale W Johnson, Kerry J King ISBN: 764586831 Publisher: Cliffs Notes Year: 2001 The book has a topic-by topic breakdown and lots of problem approach suggestions for both free response and multiple choice calculus questions. The book also includes two full length tests.

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

John Zhu

Improper Integrals

Slide Duration:

Table of Contents

Section 1: Prerequisites

Parametric Curves

10m 10s

Intro