Chabot Mathematics
§5.2 Integration
By Substitution
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Review §
5.1
 Any QUESTIONS About
• §5.1 → AntiDerivatives
 Any QUESTIONS About HomeWork
• §5.1 → HW-22
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
§5.2 Learning Goals
 Use the method of substitution to find
indefinite integrals
 Solve initial-value and boundary-value
problems using substitution
 Explore a price-adjustment model in
economics
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Recall: Fcn Integration Rules
1. Constant Rule:
for any constant, k
k
dx

k

x

C

n 1
x
 x dx  n  1  C
1
3. Logarithmic Rule:
dx

ln
x

C
x
for any x≠0
1 kx
4. Exponential Rule:
kx
e
dx

e

C

for any constant, k
k
2. Power Rule:
for any n≠−1
Chabot College Mathematics
4
n
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Recall: Integration Algebra Rules
1. Constant Multiple Rule: For any
constant, a




a

u
x
dx

u

u
x
dx


2. The Sum or Difference Rule:










u
x

v
x
dx

u
x
dx

v
x


 dx










p
x

q
x
dx

p
x
dx

q
x
dx



•
This often called the Term-by-Term Rule
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Integration by Substitution
 Sometimes it is MUCH EASIER to find
an AntiDerivative by allowing a new
variable, say u, to stand for an entire
expression in the original variable, x
 In the AntiDerivative expression ∫f(x)dx
substitutions must be made:
• Within the Integrand
• For dx
 Along Lines →
Chabot College Mathematics
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Let



u
d
u
dx
du
dx
du


G x 
d
Gx 
dx

g x 

g  x dx
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Investigate Substitution
 Compute the family of
AntiDerivatives given by


3
x

1

2
dx
a. by expanding (multiplying out) and
using rules of integration from Section
5.1
b. by writing the integrand in the form
u2and guessing at an antiderivative.
Chabot College Mathematics
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Investigate Substitution
 SOLUTION a:
 “Expand the BiNomial” by “FOIL”
Multiplication
 3x  1
2

 9 x
2
 6 x  1 dx
Power & Sum Rules

3x 3  3x 2  x  C
 SOLUTION b:
 Let: u  3x  1
2
 Sub u into Expression →  u  dx
Chabot College Mathematics
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
dx   3x  1  3x  1 dx   9 x 2  3x  3x  1 dx
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Investigate Substitution
 Examine the “substituted”
expression to find the
 u 
2
dx   u dx
2
• Integrand stated in terms of u
• Integrating factor (dx) stated in terms of x
 The Integrand↔IntegratingFactor
MisMatch does Not Permit the
AntiDerivation to move forward.
• Let’s persevere, with the understanding is
something missing by flagging
that with a (well-placed) question mark.
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Investigate Substitution
 Continuing
1 3
ò u dx = 3 u ×?+ C1
1
= ?× (3x +1)3 + C1
3
1
= ?× (27x 3 + 27x 2 + 9x +1) + C1
3
2
= ?×(9x 3 + 9x 2 + 3x +1/ 3)+C1
 ? (9 x  9 x  3x)  (?1 3  C1 )
3
2
 ? (9 x3  9 x 2  3x)  C & C  ?1 3  C1
Chabot College Mathematics
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Investigate Substitution
 The integral in part (b) (which is
speculative) agrees with the integral
calculated in part (a) (using established
techniques) when
3x 3 + 3x 2 + x + C = ?× (9x 3 + 9x 2 + 3x)+ C,
 By Correspondence observe that
?=⅓
• This Begs the Question: is there some
systematic, a-priori, method to determine
the value of the question-mark?
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
SubOut Integrating Factor, dx
 Let the single value, u,
represent an algebraic
u  G x 
expression in x, say:
d
 Then take the
u  Gx 
or
dx
derivative of both
d
d
sides
u 
G x 
dx
dx
 Then Isolate dx
 du
 dx
 dx
 g x 
 g x 
Chabot College Mathematics
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du
dx

g x 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
SubOut Integrating Factor, dx
 Then the Isolated dx: du  dx
g x 
 Thus the SubStitution Components
u  G x  and
du
du
dx 

g x  dG x  dx
 Consider the
2

3 x  1 dx

previous example
 Let: u  Gx  3x  1
2
u
 Then after subbing:  dx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
SubOut Integrating Factor, dx
d
d
 Now Use Derivation to
u  3 x  1
Find dx in terms of du → dx
dx
 Multiply both sides by dx/3 to isolate dx
du
 3 0 
dx
dx  du


3


3  dx
du
 dx
3
 Now SubOut Integrating Factor, dx
 u dx
2
 du 
 u   
3
2
2
u
1
2
du

u
du
3

3
 Now can easily AntiDerivate (Integrate)
Chabot College Mathematics
14
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
SubOut Integrating Factor, dx
 Integrating
1
3
 u du

1 u3
u K


  K 
3 3
9 3

2
u
K

C & C 
9
3
 Recall: u  3x  1
 BackSub u=3x+1 into integration result


1 2
u
3 x  1
 3x  1 dx  3  u du  3  3  C  9  C
3
3
2
 Expanding the BiNomial find
3x  13  C  27 x3  27 x 2  9 x  1  C  3x3  3x 2  x   1  C 
9
Chabot College Mathematics
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9

9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx


SubOut Integrating Factor, dx
 Then
1

3
2


3
x

1
dx

3
x

3
x

x


C

3
x

3
x
 x  C0



9

2
3
2
 The Same Result as Expanding First
then Integrating Term-by-Term Using
the Sum Rule
Chabot College Mathematics
16
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
GamePlan: Integ by Substitution
1. Choose a (clever) substitution, u =
u(x), that “simplifies” the Integrand, f(x)
2. Find the Integrating Factor, dx, in
terms of x and du by:


 1 
du d
u  u x  
 u  x   dx  
du  hu du

d
dx dx
 u x 
 dx

Chabot College Mathematics
17
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
GamePlan: Integ by Substitution
3. After finding dx = r(h(u), du) Sub Out
the Integrand and Integrating Factor to
arrive at an equivalent
Integral of the form:
 f x dx   hu du
4. Evaluate the transformed integral by
finding the AntiDerivative H(u) for h(u)
5. BackSub u = u(x) into H(u) to eliminate
u in favor of x to obtain the x-Result:
 f x dx  H ux   C
Chabot College Mathematics
18
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example: Substitution with e
 Find Z x    e  x e  x  7 dx
 SOLUTION:
 First, note that none of the rules from the
Previous lecture on §5.1 will
immediately resolve this integral
 Need to choose a substitution that yields
a simpler integrand with which to work
• Perhaps if the radicand were simpler, the
§5.1 rules might apply
Chabot College Mathematics
19
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example: Substitution with e
 Try Letting: u = e- x + 7
 Take d/dx of Both Sides

d
du d  x
u

e 7
dx
dx dx
dx
Solving for dx:  x
e
  e   1  0  e
x
x
1
 du
x 
 dx  e   e  x du  dx

 Now from
u  e x  7  e x  u  7
u-Definition:
 Then
1
1
dx   x du 
du
dx →
e  u  7
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example: Substitution with e
 Now Sub Out in original AntiDerivative:
e
x
1
dx 
du
u 7
x
e 7 u
u 7
 This process yields
1
Z x    e  e  7  dx   u  7   u 
 du
u  7
x
x
 This works out VERY Well
Z x   Ru     u
12
u
du 
K
32
32
 Now can BackSub for u(x)
Chabot College Mathematics
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example: Substitution with e
 Using u(x) = e−x+7:
 2u 
Z  x   Ru  
3
32

 2 ex  7
K 
3

32
K
 Thus the Final Result:
Z x    e
x


32
2 x
e  7 dx   e  7  K
3
x
• This Result can be verified by taking the
derivative dZ/dx which should yield the
original integrand
Chabot College Mathematics
22
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example: Sub Rational Expression
 Find Z x   
3
4x  2
2x  2x 1
2
dx
 SOLUTION:
13
2
2
3
 Try: u  2 x  2 x  1  2 x  2 x  1
 Taking du/dx
3
3u 2
dx 
du 
du
2
find
4 x  2u
4 x  2
 This produces
Z x   
Chabot College Mathematics
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4 x  2
u
3u 2
du   3u du
4 x  2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example: Sub Rational Expression
 Solving



13 2
3u 2
3
2
Z x    3u du 
 C   2x  2x 1
C
2
2
 Thus the Answer
Z x   
4x  2
3
23
3
2
dx  2 x  2 x  1  C
2
2x2  2x 1
 An Alternative u:
u  2x  2x 1 
2
Chabot College Mathematics
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du
 4x  2
dx
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example: Sub Rational Expression
 SubOut x
du
2
u  2 x  2 x  1 & dx 
using:
4x  2
 Find
Z x   
4x  2
3
2x  2x 1
 Then
2
dx  
4 x  2 
3
u

1
du   u 1 3 du
4 x  2

23
u
3 23 3
1 3
Z x    u du 
 u  2x2  2x 1  C
23 2
2
• The Same Result as before
Chabot College Mathematics
25
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example  DE Model for Annuities
 Li Mei is a Government Worker who
has an annuity referred to as a 403b.
She deposits money continuously into
the 403b at a rate of $40,000 per year,
and it earns 2.6% annual interest.
 Write a differential equation modeling
the growth rate of the net worth of the
annuity, solve it, and determine how
much the annuity is worth at the end of
10 years.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example  DE Model for Annuities
 SOLUTION:
 TRANSLATE: The 403b has two ways
in which it grows yearly:
• The annual Deposit by Li Mei = $40k
• The annual interest accrued = 0.026·A
– Where A is the current Amount in the 403b
 Then the yearly Rate of Change for the
Amount in the 403b account
dA
 $40k  0.026 A
dt
Chabot College Mathematics
27
in $k/yr
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example  DE Model for Annuities
dA
 This DE is
= 40 + 0.026A
Variable Separable dt
 Affecting the Separation and Integrating
dA
= dt
40 + 0.026A
1
 40  0.026 A dA   dt
 Find the AntiDerivative by Substitution
 Let: u  40  0.026 A
 Then: du  0  0.026  dA  1 du
dA
Chabot College Mathematics
28
0.026
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example  DE Model for Annuities
1
dA = ò dt
 SubOut A in ò
40 + 0.026A
favor of u:
1
1
ò u × 0.026 du = t + C1
 Integrating:
1
1
du  t  C1

0.026 u
1
ln u + C2 = t + C1
0.026
1
ln u + C2 = t + C1
0.026
ln u  0.026t  C & C  0.026C1  C2 
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example  DE Model for Annuities
 Note that u = $40k + 0.026A is always
positive, so the ABS-bars can be
dispensed with
 Now BackSub ln u   ln 40  0.026 A  0.026t  C
 Solve for A(t) by
e ln400.026 A   e 0.026t C
raising e to the
40 + 0.026A = e0.026t+C
power of both sides
0.026t+C
e
- 40
Find the General
A=
0.026
(Includes C) solution:
Chabot College Mathematics
30
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example  DE Model for Annuities
 Use the KNOWN data that at year-Zero
there is NO money in the 403b; i.e.;
(t0,A0) = (0,A(0)) = (0,0)
 Sub (0,0) into the General Soln to find C
e0.0260 C  40
0
 eC  40  ln eC  40
0.026


C
ln
e
 C  ln 40  3.689
 Or
 Thus the
e0.026t ln 40  40 e 0.026t 3.689  40

particular soln A 
0.026
0.026
Chabot College Mathematics
31
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Example  DE Model for Annuities
 Using the Log property a u v  a u  a v
 Find
e0.026t ln 40  40 eln 40  e0.026t  40 40  e 0.026t  40
A


0.026
0.026
0.026
e0.026t  1
 Factoring
A  40
0.026
Out the 40
 Then at 10 years the 403b Amount
e0.02610  1
0.2969
A  40
 $40k
 $456.816k
0.026
0.026
Chabot College Mathematics
32
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
WhiteBoard Work
 Problems From §5.2
• P61 → Retirement Income vs. Outcome
• P66 → Price Sensitivity to Supply &
Demand
Chabot College Mathematics
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
All Done for Today
Substitution
City
Chabot College Mathematics
34
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.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
35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
ConCavity Sign Chart
ConCavity
Form
d2f/dx2 Sign
++++++
Critical (Break)
Points
Chabot College Mathematics
36
−−−−−−
a
Inflection
−−−−−−
b
NO
Inflection
++++++
c
Inflection
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
x
Chabot College Mathematics
37
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Chabot College Mathematics
38
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Chabot College Mathematics
39
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Chabot College Mathematics
40
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Chabot College Mathematics
41
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Chabot College Mathematics
42
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Chabot College Mathematics
43
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Chabot College Mathematics
44
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Chabot College Mathematics
45
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Chabot College Mathematics
46
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx
Chabot College Mathematics
47
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-23_sec_5-2_Integration_Substitution.pptx