4 5 Substitution
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Transcript 4 5 Substitution
4.5: Integration by Substitution
Objectives:
Assignment:
1. To use a change of
variables to evaluate
indefinite and definite
integrals
β’ P. 304-305: 7-33 odd, 3747 odd, 61-67 odd, 70,
71-81 eoo, 83, 87
2. To use symmetry as an
aid in integration
β’ P. 306: 101-107 odd
Warm Up 1
Find the derivative of β π₯ = π₯ 2 + π₯.
β π₯ = π₯2 + π₯
ββ² π₯ =
ββ² π₯ =
1/2
1 2
π₯ +π₯
2
2π₯ + 1
β1/2
2π₯ + 1
πβ² π₯
2 π₯2 + π₯
π π₯
π π₯ = π₯
The Chain Rule
If π¦ = π π’ is a differentiable function of π’ and
π’ = π π₯ is a differentiable function of π₯, then
π¦ = π π π₯ is a differentiable function of π₯
and
ππ¦ ππ¦ ππ’
=
β
ππ₯ ππ’ ππ₯
π
π π π₯
ππ₯
= πβ² π π₯
β πβ² π₯
The derivative of the exterior function times
the derivative of the interior function
Warm Up 2
2π₯+1
β²
Find the antiderivative of β π₯ =
ββ² π₯ =
2π₯ + 1
2
π₯ 2 +π₯
πβ² π₯
2 π₯2 + π₯
π π₯ = π₯
π π₯
ββ² π₯ = πβ² π π₯ πβ² π₯
β π₯ =π π π₯
+πΆ
=
π₯2 + π₯ + πΆ
.
Objective 1
You will be able to
use a change of
variables to
evaluate indefinite
and definite
integrals
Composite Functions
When integrating a composite function π,
realize that function π was the product of the
Chain Rule.
πΉβ² π π₯
πβ² π₯ ππ₯ = πΉ π π₯
+πΆ
Antidifferentiation of Composite Functions
Let π be a function whose range is an
interval πΌ, and let π be a function that is
continuous on πΌ. If π is differentiable on its
domain and πΉ is an antiderivative of π on πΌ,
then
π π π₯
πβ² π₯ ππ₯ =
πΉ π π₯
+πΆ
Substitution Rule
If π’ = π(π₯), then
ππ’
ππ₯
= πβ² π₯ .
ππ’ = πβ² π₯ ππ₯
π π π₯
πβ² π₯ ππ₯ =
π π’ ππ’ = πΉ π’ + πΆ
This substitution is accomplished with a change of
variable, where π’ is a function of π₯.
Exercise 1
Find
1 + π₯ 2 2π₯ ππ₯
2π₯ 1 + π₯ 2 ππ₯ =
Let π’ = 1 + π₯ 2
ππ’
= 2π₯
ππ₯
=
Always remember
that your answer
should be in terms of
the original variable.
ππ’ = 2π₯ ππ₯
π’ ππ’ =
π’1/2 ππ’
2 3/2
2
= π’ + πΆ = 1 + π₯2
3
3
3/2
+πΆ
Check your answer
by taking its
derivative.
Protip
When changing variables, try letting π’ =
Some complicated part of the integrand
The inner function of a composite function
Some function whose derivative is part of the integrand
Use Constant Multiple Rule
Except for a constant factor
If your first
guess doesnβt
work, try
something
different.
Exercise 2
Find
2π₯ π₯ 2 + 1
2
ππ₯.
Exercise 3
Find
π₯ π₯2 + 1
2
ππ₯.
Exercise 4
Find
5cos 5π₯ ππ₯.
Exercise 5
Find
2π₯ β 1 ππ₯.
Exercise 6
Find
π₯ 2π₯ β 1 ππ₯.
Protip:
When changing
variables,
sometimes you
have to use the
equation for π’ to
solve for π₯ to do
another
substitution.
Exercise 7
Find
sin2 3π₯ cos 3π₯ ππ₯.
General Power Rule for Integration
If π is a differentiable function of π₯, then
π+1
π
π₯
π π₯ π πβ² π₯ ππ₯ =
+ πΆ,
π+1
If π’ = π π₯ , then
π+1
π’
π’π ππ’ =
+ πΆ,
π+1
π β β1
π β β1
Exercise 8
Find the following antiderivatives.
1.
3 3π₯ β 4
4 ππ₯
2.
2π₯ + 1 π₯ 2 + π₯
4 ππ₯
Exercise 8
Find the following antiderivatives.
3.
3π₯ 2
π₯3
β 2 ππ₯
4.
β4π₯
ππ₯
1β2π₯ 2 2
Exercise 9
Find
cos2 π₯ sin π₯ ππ₯.
Exercise 10
Evaluate
1
π₯
0
π₯2 + 1
3
ππ₯.
Method 1:
π’ = π₯2 + 1
ππ’ = 2π₯ ππ₯
ππ’
= π₯ ππ₯
2
π
π
ππ’
1
3
π’
=
2
2
π
π
1 4
3
π’ ππ’ = π’
8
π
π
1 2
= π₯ +1
8
=
15
8
1
4
0
Exercise 10
Evaluate
1
π₯
0
π₯2 + 1
3
ππ₯.
Method 2:
π’ = π₯2 + 1
ππ’ = 2π₯ ππ₯
ππ’
= π₯ ππ₯
2
2
1
ππ’
1
3
π’
=
2
2
New Limits of Integration:
π’ = 12 + 1 = 2
π’ = 02 + 1 = 1
2
1
1 4
3
π’ ππ’ = π’
8
15
=
8
2
1
Definite Integrals
If the function
π’ = π π₯ has a
continuous
derivative on the
π, π and π is
continuous on
the range of π,
then
π
π π
π π π₯ πβ² π₯ ππ₯ =
π
π π’ ππ’
π π
Exercise 11
Evaluate π΄ =
4
0
2π₯ + 1 ππ₯.
Objective 2
You will be able to use
symmetry as an aid in
integration
Exercise 12
Evaluate each definite integral.
1.
2 6
π₯
β2
+ 1 ππ₯
2.
2 6
π₯
0
+ 1 ππ₯
Exercise 13
Evaluate
π/3
sin
3π₯
ππ₯.
βπ/3
Symmetry
Recall that the graphs of even and odd functions
have a useful kind of symmetry:
Even = π¦-axis Symmetry
π βπ₯ = π π₯
Odd = Origin Symmetry
π βπ₯ = βπ π₯
Integration of Even/Odd Functions
Let π be integrable on βπ, π .
If π is an even function, then
π
π
π
π₯
ππ₯
=
2
π π₯ ππ₯.
βπ
0
If π is an odd function, then
π
π π₯ ππ₯ = 0.
βπ
Exercise 14
Evaluate
π/2
3
sin
π₯
cos
π₯
βπ/2
+ sin π₯ cos π₯ ππ₯.
4.5: Integration by Substitution
Objectives:
Assignment:
1. To use a change of
variables to evaluate
indefinite and definite
integrals
β’ P. 304-305: 7-33 odd, 3747 odd, 61-67 odd, 70,
71-81 eoo, 83, 87
2. To use symmetry as an
aid in integration
β’ P. 306: 101-107 odd