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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????

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
  • 1k [3/x]dx
  • Integrate
  • 1k [3/x]dx = 3[ln|x|]1k
  • = 3lnk − 3ln1
= 3lnk = limk → ∞ 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)] ]k0.
  • = − [1/(0 + 5)] + [1/(k + 5)]
= limx → − ∞ [ [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∫0k [7/((x − 3)2)]dx
  • Integrate
  • 2∫0k [7/((x − 3)2)]dx = 14∫0k [1/((x − 3)2)]dx
  • = 14[ − [1/(x − 3)] ]0k
  • = 14[ − [1/(k − 3)] + [1/(0 − 3)] ]
  • = 14[ − [1/(k − 3)] − [1/3] ]
  • = 14[ − limk → ∞ [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
− 2k [2/(x√{x2 − 1} )]dx
  • Replace the bound 1 with k
  • − 2k [2/(x√{x2 − 1} )]dx = ∫ − 2k [2/(x√{x2 − 1} )]dx
  • Integrate using inverse trigonmetric identity
  • − 2k [2/(x√{x2 − 1} )]dx = 2∫ − 2k [1/(x√{x2 − 1} )]dx
  • = 2[ sec − 1|x| ] − 2k
  • = 2[ sec − 1|k| − sec − 1| − 2| ]
= 2[ limk1 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 = ∫0k [1/((x + 1))]dx
  • Integrate
  • 0k [1/((x + 1))]dx = [ ln|x + 1| ]0k
  • = ln|k + 1| − ln|0 + 1|
  • = ln|k + 1| − ln|1|
  • = ln|k + 1| − 0
  • = ln|k + 1|
  • = limk → ∞ 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|
= limk → − ∞ [ − 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 = ∫0k [4/((x + 1)(x + 2))] dx
  • Integrate
  • 0k [4/((x + 1)(x + 2))] dx = ∫0k [4/(x + 1)] − [4/(x + 2)] dx
  • = 4∫0k [1/(x + 1)] − [1/(x + 2)] dx
  • = 4[ ln|x + 1| − ln|x + 2| ]0k
= 4[ limk → ∞ ln[(k + 1)/(k + 2)] − ln[1/2] ] = 4[ 0 − ln[1/2] ] = − 4ln[1/2]
− 22 [1/(√{4 − x2} )]dx
  • Note the function is symmetrical and simplify
  • − 22 [1/(√{4 − x2} )]dx = 2∫02 [1/(√{4 − x2} )]dx
  • Replace the bound 2 with k
  • 2∫02 [1/(√{4 − x2} )]dx = 2∫0k [1/(√{4 − x2} )]dx
  • Apply the inverse trignometric integration identity to integrate
  • 2∫0k [1/(√{4 − x2} )]dx = 2[ sin − 1[x/2] ]0k
  • = 2[ sin − 1[k/2] − sin − 1[0/2] ]
  • = 2[ sin − 1[k/2] ]
  • = 2[ limk → 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 = ∫0k [1/(( x − 2 )3)] dx
  • Integrate
  • 0k [1/(( x − 2 )3)] dx = [ − [1/(2(x − 2)2)] ]0k
  • = [ − limk → ∞ [1/(2(k − 2)2)] + [1/(2(0 − 2)2)] ]
  • = 0 + [1/2(4)]
= [1/8]

*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 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