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„Ž π‘₯ = π‘₯ 2 + π‘₯.
β„Ž π‘₯ = π‘₯2 + π‘₯
β„Žβ€² π‘₯ =
β„Žβ€² π‘₯ =
1/2
1 2
π‘₯ +π‘₯
2
2π‘₯ + 1
βˆ’1/2
2π‘₯ + 1
𝑔′ π‘₯
2 π‘₯2 + π‘₯
𝑔 π‘₯
𝑓 π‘₯ = π‘₯
The Chain Rule
If 𝑦 = 𝑓 𝑒 is a differentiable function of 𝑒 and
𝑒 = 𝑔 π‘₯ is a differentiable function of π‘₯, then
𝑦 = 𝑓 𝑔 π‘₯ is a differentiable function of π‘₯
and
𝑑𝑦 𝑑𝑦 𝑑𝑒
=
βˆ™
𝑑π‘₯ 𝑑𝑒 𝑑π‘₯
𝑑
𝑓 𝑔 π‘₯
𝑑π‘₯
= 𝑓′ 𝑔 π‘₯
βˆ™ 𝑔′ π‘₯
The derivative of the exterior function times
the derivative of the interior function
Warm Up 2
2π‘₯+1
β€²
Find the antiderivative of β„Ž π‘₯ =
β„Žβ€² π‘₯ =
2π‘₯ + 1
2
π‘₯ 2 +π‘₯
𝑔′ π‘₯
2 π‘₯2 + π‘₯
𝑓 π‘₯ = π‘₯
𝑔 π‘₯
β„Žβ€² π‘₯ = 𝑓′ 𝑔 π‘₯ 𝑔′ π‘₯
β„Ž π‘₯ =𝑓 𝑔 π‘₯
+𝐢
=
π‘₯2 + π‘₯ + 𝐢
.
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 𝑓,
realize that function 𝑓 was the product of the
Chain Rule.
𝐹′ 𝑔 π‘₯
𝑔′ π‘₯ 𝑑π‘₯ = 𝐹 𝑔 π‘₯
+𝐢
Antidifferentiation of Composite Functions
Let 𝑔 be a function whose range is an
interval 𝐼, and let 𝑓 be a function that is
continuous on 𝐼. If 𝑔 is differentiable on its
domain and 𝐹 is an antiderivative of 𝑓 on 𝐼,
then
𝑓 𝑔 π‘₯
𝑔′ π‘₯ 𝑑π‘₯ =
𝐹 𝑔 π‘₯
+𝐢
Substitution Rule
If 𝑒 = 𝑔(π‘₯), then
𝑑𝑒
𝑑π‘₯
= 𝑔′ π‘₯ .
𝑑𝑒 = 𝑔′ π‘₯ 𝑑π‘₯
𝑓 𝑔 π‘₯
𝑔′ π‘₯ 𝑑π‘₯ =
𝑓 𝑒 𝑑𝑒 = 𝐹 𝑒 + 𝐢
This substitution is accomplished with a change of
variable, where 𝑒 is a function of π‘₯.
Exercise 1
Find
1 + π‘₯ 2 2π‘₯ 𝑑π‘₯
2π‘₯ 1 + π‘₯ 2 𝑑π‘₯ =
Let 𝑒 = 1 + π‘₯ 2
𝑑𝑒
= 2π‘₯
𝑑π‘₯
=
Always remember
that your answer
should be in terms of
the original variable.
𝑑𝑒 = 2π‘₯ 𝑑π‘₯
𝑒 𝑑𝑒 =
𝑒1/2 𝑑𝑒
2 3/2
2
= 𝑒 + 𝐢 = 1 + π‘₯2
3
3
3/2
+𝐢
Check your answer
by taking its
derivative.
Protip
When changing variables, try letting 𝑒 =
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 doesn’t
work, try
something
different.
Exercise 2
Find
2π‘₯ π‘₯ 2 + 1
2
𝑑π‘₯.
Exercise 3
Find
π‘₯ π‘₯2 + 1
2
𝑑π‘₯.
Exercise 4
Find
5cos 5π‘₯ 𝑑π‘₯.
Exercise 5
Find
2π‘₯ βˆ’ 1 𝑑π‘₯.
Exercise 6
Find
π‘₯ 2π‘₯ βˆ’ 1 𝑑π‘₯.
Protip:
When changing
variables,
sometimes you
have to use the
equation for 𝑒 to
solve for π‘₯ to do
another
substitution.
Exercise 7
Find
sin2 3π‘₯ cos 3π‘₯ 𝑑π‘₯.
General Power Rule for Integration
If 𝑔 is a differentiable function of π‘₯, then
𝑛+1
𝑔
π‘₯
𝑔 π‘₯ 𝑛 𝑔′ π‘₯ 𝑑π‘₯ =
+ 𝐢,
𝑛+1
If 𝑒 = 𝑔 π‘₯ , then
𝑛+1
𝑒
𝑒𝑛 𝑑𝑒 =
+ 𝐢,
𝑛+1
𝑛 β‰  βˆ’1
𝑛 β‰  βˆ’1
Exercise 8
Find the following antiderivatives.
1.
3 3π‘₯ βˆ’ 4
4 𝑑π‘₯
2.
2π‘₯ + 1 π‘₯ 2 + π‘₯
4 𝑑π‘₯
Exercise 8
Find the following antiderivatives.
3.
3π‘₯ 2
π‘₯3
βˆ’ 2 𝑑π‘₯
4.
βˆ’4π‘₯
𝑑π‘₯
1βˆ’2π‘₯ 2 2
Exercise 9
Find
cos2 π‘₯ sin π‘₯ 𝑑π‘₯.
Exercise 10
Evaluate
1
π‘₯
0
π‘₯2 + 1
3
𝑑π‘₯.
Method 1:
𝑒 = π‘₯2 + 1
𝑑𝑒 = 2π‘₯ 𝑑π‘₯
𝑑𝑒
= π‘₯ 𝑑π‘₯
2
𝑏
π‘Ž
𝑑𝑒
1
3
𝑒
=
2
2
𝑏
π‘Ž
1 4
3
𝑒 𝑑𝑒 = 𝑒
8
𝑏
π‘Ž
1 2
= π‘₯ +1
8
=
15
8
1
4
0
Exercise 10
Evaluate
1
π‘₯
0
π‘₯2 + 1
3
𝑑π‘₯.
Method 2:
𝑒 = π‘₯2 + 1
𝑑𝑒 = 2π‘₯ 𝑑π‘₯
𝑑𝑒
= π‘₯ 𝑑π‘₯
2
2
1
𝑑𝑒
1
3
𝑒
=
2
2
New Limits of Integration:
𝑒 = 12 + 1 = 2
𝑒 = 02 + 1 = 1
2
1
1 4
3
𝑒 𝑑𝑒 = 𝑒
8
15
=
8
2
1
Definite Integrals
If the function
𝑒 = 𝑔 π‘₯ has a
continuous
derivative on the
π‘Ž, 𝑏 and 𝑓 is
continuous on
the range of 𝑔,
then
𝑏
𝑔 𝑏
𝑓 𝑔 π‘₯ 𝑔′ π‘₯ 𝑑π‘₯ =
π‘Ž
𝑓 𝑒 𝑑𝑒
𝑔 π‘Ž
Exercise 11
Evaluate 𝐴 =
4
0
2π‘₯ + 1 𝑑π‘₯.
Objective 2
You will be able to use
symmetry as an aid in
integration
Exercise 12
Evaluate each definite integral.
1.
2 6
π‘₯
βˆ’2
+ 1 𝑑π‘₯
2.
2 6
π‘₯
0
+ 1 𝑑π‘₯
Exercise 13
Evaluate
πœ‹/3
sin
3π‘₯
𝑑π‘₯.
βˆ’πœ‹/3
Symmetry
Recall that the graphs of even and odd functions
have a useful kind of symmetry:
Even = 𝑦-axis Symmetry
𝑓 βˆ’π‘₯ = 𝑓 π‘₯
Odd = Origin Symmetry
𝑓 βˆ’π‘₯ = βˆ’π‘“ π‘₯
Integration of Even/Odd Functions
Let 𝑓 be integrable on βˆ’π‘Ž, π‘Ž .
If 𝑓 is an even function, then
π‘Ž
π‘Ž
𝑓
π‘₯
𝑑π‘₯
=
2
𝑓 π‘₯ 𝑑π‘₯.
βˆ’π‘Ž
0
If 𝑓 is an odd function, then
π‘Ž
𝑓 π‘₯ 𝑑π‘₯ = 0.
βˆ’π‘Ž
Exercise 14
Evaluate
πœ‹/2
3
sin
π‘₯
cos
π‘₯
βˆ’πœ‹/2
+ sin π‘₯ cos π‘₯ 𝑑π‘₯.
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