Chabot Mathematics
§6.3 Improper
Integrals
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Review §
6.2
 Any QUESTIONS About
• §6.2 → Numerical Integration
 Any QUESTIONS
About
HomeWork
• §6.2 → HW-02
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Evaluating Improper Integrals
 Use LIMITS to Evaluate Improper
Integrals.
 Given AntiDerivative:  f x dx  g x 
 And assuming M, N are real Numbers


a

b




N  
b

f x dx  lim   f x dx  
N  N

N 
N


f  x dx  lim lim   f  x dx  
N   

 M  M
Chabot College Mathematics
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lim g  N   g a 
N

f  x dx  lim   f  x dx  
N   a

lim g b   g N 


lim lim g  N   g M 
N   M  
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  ImProper Population
 The Tasmanian Devil Population on The
Isolated Australian Island of Tasmania
CHANGES according to the Model
0.08t
P' t   e
 Where
• P ≡ the Tasmanian Devil
Population in k-Devils
• t ≡ the Number Years after Calendar year
2000; i.e; t = tcalendar − 2000
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  ImProper Population
 Then Calculate and Explain

t 1
t 0
P' (t ) dt and

t 
t 0
P' (t ) dt
 SOLUTION:
 First Convert to From La Grange Notation
to the MORE ILLUMINATING Lebniz form
1
P 1 dP dt

P   dP dt
0 P' t  dt P 0  dt 1 and 0 P' t  dt  P 0  dt 1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  ImProper Population
 The first Eqn is a Definite integral of a
type calculated many times in MTH15
 e
1
 0.08t
0
 dt

 P' t  dt
1
0
 By the
Transitive Property
 So 
P 1
P0
dP  P 
P 1
P0

P1  12.5 e
Chabot College Mathematics
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0.08

P 1
P0


P 1
P0
1

dP   e
0
dP
 0.08t
 dt

 e  12.51  0.9231  0.961
0
1
e

 P1  P0  P1  

  0.08  0
 0.08t
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  ImProper Population
 INTERPRETATION: In the Yr 2001 the
TD population will have Increased by
about 961 Devils compared to Yr 2000
 Now, assuming the Model Holds over
Loooong Time-Scales

P  
P0
dP 
 e

0
 0.08t
 dt
 AntiDerivating

P  
P0
N
lim
 e
 0.08t
N  0
 dt
e

0
 Nt
dP  P  lim 
 lim 12.5 e  e

N   0.08
N 

0
Chabot College Mathematics
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
N
 0.08t
 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx

Example  ImProper Population


 Or P  12.5 e0  e  12.51  0  12.5
 INTERPRETATION:
Tasmaniam Devil Population Change over Y2k
If the Model holds
for a Long Period
of Time then the
Tasmanian Devil
Population will
STABILIZE at
about 12.5 k-Devils
t = t - 2000
above theY2k Level
12
P (k-Devils)
10
8
6
4
2
0
Bruce May er, PE • BlueGreenBkGnd 140112.m
0
10
20
30
40
50
cal
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
60
70
80
Common ImProper Integ Limit
 Negative Exponentials Often Occur in
ImProper Integrals.
 A useful limit in these Circumstances:
For any Power, p, and Positive Number, k
p  kN
lim N e
N 
Chabot College Mathematics
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0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  UseFul Limit
3 x
lim
x
 Use x e

 Then the AUC for
𝒙𝟑 𝒆−𝒙

0


0
3 x
x e dx
3 x
x e dx
 Find By Repeated Use of
• Integration by Parts
p  kN
0
• The Limit: lim N e
N 
𝐀𝐔𝐂 = 𝟔


0


3 x
x
3
2
x
e
dx


e
x

3
x
 6x  6

x e
3 x


dx  e  x x 3  3x 2  6 x1  6 x 0

 
x 3e  x dx  e 0 03  3  0 2  6  0  6  lim e  N N 3  3N 2  6 N  6
N 
AUC   x3e x dx  16  0  0  0  0  6

0
Chabot College Mathematics
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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx

ImProper Integral Divergence
 ImProper Integration Often Times
FAILS to Return a Finite Value, that is:


a


f x dx  lim   f x dx   
N  a

 Example: Find the AUC for
 x2

AUC    xdx  lim  
0
N 
2

N
0


0
 xdx

 x2
   lim 
N  2



N
0

    0 


 Thus this ImProper Integral DIVERGES
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Double Infinity
3
 Find the value of
z   z
this ImProper Integral z  z 4 1dz
 SOLUTION:
 The integral can be divided into TWO
separate integrals, EACH containing
ONE infinite limit of integration.
 From the definition, we choose middleLimit c = 0 for convenience
• Note that c Can be ANY RealNumber
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Double Infinity
 Then the “Split” Integral
3
3
3
z   z
z 0
z


z
z
z  z 4  1dz  z  z 4  1dz  z 0 z 4  1dz
 Now Engage the SubStitution u  z 4  1
 Then
d
 du
 dz
4
3

u  z  1  
 4z  0 3
dz
 dz
 4z

du
 dz
3
4z
• And then z    u      1  
4
the Limits z  0 


u  0 1  1
4
z    u      1  
4
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Double Infinity
 Making the SubStitution
3
3
3
z   z
z 0 z
z


du
z du
z  z 4  1dz  z  u 4 z 3  z 0 u 4 z 3
1 u 1 1
1 u  1
 
du  
du
4 u  u
4 u 1 u
1
1
1
N
 lim ln u M  lim ln u 1
4 M 
4 N 
1
1
1
1
 ln 1  lim ln  M   lim ln  N   ln 1
4
4 M 
4 N 
4
Can Drop ABS bars as 1 & M & N are POS
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Double Infinity
 Continuing the Reduction
1
1
1
1
 ln 1  lim ln M   lim ln N   ln 1
4
4 M 
4 N 
4
1
1
1
1
 0  lim ln M   lim ln  N   0
4
4 M 
4 N 
4


1
 0  0  lim ln  N   lim ln M 
M 
4 N 
3
z   z
1
ln N  lim ln M
 So z  4 dz  Nlim
M 
z 1
4 

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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx

Example  Double Infinity
 But BOTH of the Limits DIVERGE
lim ln N   & lim ln M  
N 
M 
 Since the ∞ is NOT a Number, then the
subtraction, ∞ − ∞, is MEANINGLESS
 Thus This Expression has NO Number
Value
z  
z
3
dz
z  z 4 1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Double Disaster
 Be CareFul in this Case – It’s Easy to
Make a Disastrous Mistake
 From the Previous Reduction
1
 0  0  lim ln  N   lim ln M 
M 
4 N 
1
 lim ln  N   lim ln M 
M 
4 N 
 Now since M & N are DUMMY
Variables it can be written that


Chabot College Mathematics
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

Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Double Disaster
 Using M=P=N write the Limits
lim ln N   lim lnP   lim lnM 
N 
P 
M 
 Using above in the Reduction
1
 lim ln  N   lim ln M 
M 
4 N 
1
 lim ln P   lim ln P 
P 
4 P 
 ReCall (InCompletely) the Limit Property
lim f x   lim g x   lim  f x   g x 

x c
Chabot College Mathematics
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


x c
x c
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Double Disaster
 Apply the Difference-of-Limits Property
1
 lim ln P   lim ln P 
P 
4 P 
1
1
 lim ln P   ln P  
lim 0  0
4 P 
4 P 
 Thus One might be Tempted to Say
3
z   z
1
ln P   lim ln P   0!!!
z  z 4 1dz  4 Plim

P 





Chabot College Mathematics
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


Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Double Disaster
 Q) What’s WRONG with Assessment?
 A) ReCall From Limits Properties the
Qualifying Statement
• IF these
lim f  x  &
Limit EXIST x c
lim g  x 
x c
• (only) THEN
lim  f x   g x   lim f x   lim g x 
x c
x c
x c
 In the current case, BOTH individual
Limits did NOT exist
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Double Disaster
 Deceptive Plot Suggests Net AUC = 0
 It looks Like The
“Equal but Opposite”
areas “Cancel Each
Other Out”, adding to
ZERO → WRONG!
Negative
Area
Chabot College Mathematics
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Positive
Area
Infinite Areas
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Semi-Infinity
 Consider the same Fcn: z = 0→+∞
3
z  
z
AUC  
dz
4
z 0
z 1
 By the Same SubStitution u  z 4  1
z 
z3
1 u  1
N
AUC  
dz  
du  lim ln u 1
4
z 0 z  1
N 
4 u 1 u
1
1
 lim ln  N   ln 1    0  
4 N 
4
 Thus the “Semi-Infinite” ImProper
Integral DIVERGES
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Example  Semi-Infinity
Infinite Area
Area_Under_Curve_Hatch_Plot_BMayer_1401.mn
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
MuPAD Code
fOFx := fOFx := x^3/(x^4 + 1)
plot(fOFx, x=-50..50, GridVisible = TRUE,
LineWidth = 0.04*unit::inch,
Width = 320*unit::mm, Height = 180*unit::mm,
AxesTitleFont = ["sans-serif", 24],
TicksLabelFont=["sans-serif", 16],
BackgroundColor = RGB::colorName([0.8, 1, 1]))
Plot the AREA under the Integrand Curve (a very cool plot don't you think...) Need to CHECK
Graph Box: Scene2D→BackGroundColor
fArea := plot::Function2d(fOFx, x = 0..50, GridVisible = TRUE):
plot(plot::Hatch(fArea), fArea, Width = 320*unit::mm, Height =
180*unit::mm,
AxesTitleFont = ["sans-serif", 24], TicksLabelFont=["sans-serif", 12],
LineWidth = 0.04*unit::inch,BackgroundColor = RGB::colorName([0.8, 1,
1]) )
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
WhiteBoard Work
 Problems From §6.3
• P34 → Professorial Endowed Chair
• P42 → Waste Accumulation
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
All Done for Today
Break at
Asymptote
Location
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx
∞
1
ln 𝑥
𝑑𝑥
2
𝑥
Bruce Mayer, PE
MTH15 • 13Jan14
Area_Under_Curve_Hatch_Plot_BMayer_1401.mn
Plot the Integrand
fOFx := ln(x)/x^2
int(ln(x)/x^2, x)
AUC = int(ln(x)/x^2, x=1..infinity)
AUC = int(ln(x)/x^2, x=1..1E6)
Plot the AREA under the Integrand Curve • CHECK Graph Box: Scene2D→BackGroundColor
fArea := plot::Function2d(fOFx, x = 1..20, GridVisible = TRUE):
plot(plot::Hatch(fArea), fArea, Width = 320*unit::mm, Height =
180*unit::mm,
AxesTitleFont = ["sans-serif", 24], TicksLabelFont=["sans-serif", 16],
LineWidth = 0.04*unit::inch,BackgroundColor = RGB::colorName([0.8, 1, 1]) )
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH16_Lec-01_sec_6-1_Integration_by_Parts.pptx