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Calculus BC Improper Integrals

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

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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????

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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]

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Answer

Improper Integrals

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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.

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John Zhu

Improper Integrals

Slide Duration:

Table of Contents

Section 1: Prerequisites

Parametric Curves

10m 10s

Intro