4.6 copyright © 2014 pearson education, inc. integration techniques: integration by parts objective...
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4.6
Copyright © 2014 Pearson Education, Inc.
Integration Techniques: Integration by Parts
OBJECTIVE• Evaluate integrals using the formula for integration by parts.
• Solve applied problems involving integration by parts.
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THEOREM 7
The Integration-by-Parts Formula
u dv uv v du
4.6 Integration Techniques: Integration by Parts
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Tips on Using Integration by Parts:1. If you have had no success using substitution, try
integration by parts.2. Use integration by parts when an integral is of the
form
Match it with an integral of the form
by choosing a function to be u = f (x), where f (x) can be differentiated, and the remaining factor to be dv = g (x) dx, where g (x) can be integrated.
f (x)g(x) dx .
u dv
4.6 Integration Techniques: Integration by Parts
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3. Find du by differentiating and v by integrating.4. If the resulting integral is more complicated than the
original, make some other choice for u = f (x) and dv = g (x) dx.
5. To check your solution, differentiate.
4.6 Integration Techniques: Integration by Parts
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Example 1: Evaluate: ln x dx.
4.6 Integration Techniques: Integration by Parts
1
and .du dx v xx
Then,
Let ln and .u x dv dx
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4.6 Integration Techniques: Integration by Parts
Example 1 (concluded): Then, the Integration-by-Parts Formula gives
lnx x x C
lnx x dx
ln x dx
1(ln )x x x dx
x
u dv uv v du
1ln and .u x and du dx v x
x
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4.6 Integration Techniques: Integration by Parts
Quick Check 1
Evaluate: 3 .xxe dx3Let and .xu x dv e dx
31 and .
3xdu dx v e Then,
Then, the integration by parts formula gives
3 3 31
3 3x x xx
xe e e dx
3 31
3 9x xx
e e C
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Example 2: Evaluate:
Let’s examine several choices for u and dv.
Attempt 1: This will not work because we do not know how to integrate
x ln x dx.
4.6 Integration Techniques: Integration by Parts
ln .dv x dx
Let and ln .u x dv x dx
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Example 2 (continued): Attempt 2:
Using the Integration-by-Parts Formula, we have
4.6 Integration Techniques: Integration by Parts
Let ln and .u x dv x dx
21
Then and .2x
du dx vx
( ln )x x dx
2 2 1ln
2 2
x xx dx
x
2 1
ln2 2
xx xdx
2 2
ln2 4
x xx C
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Example 3: Evaluate:
Using the Integration by Parts Formula gives us
x 5x 1 dx.
4.6 Integration Techniques: Integration by Parts
3 22Then 1 and 5 1 .
15du dx v x
1 2Let and (5 1) .u x dv x dx
( 5 1)x x dx 3 2 3 22 2(5 1) (5 1)
15 15x x x dx
3 2 5 22 2 2(5 1) (5 1)
15 25 15x x x C
3 2 5 22 4(5 1) (5 1)
15 375x x x C
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4.6 Integration Techniques: Integration by Parts
Quick Check 2
Evaluate: 2 3 2 .x x dxLet 2 and 3 2 .u x dv x dx
Then, 3/222 and 3 2 .
9du dx v x
Using the integration by parts formula, we get:
3/2 3/24 22 3 2 3 2 3 2 2
9 9x x dx x x x dx
3/2 5/24 43 2 3 2
9 135x x x C
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Example 4: Evaluate:
Note that we already found the indefinite integral in Example 1. Now we evaluate it from 1 to 2.
ln x dx1
2
4.6 Integration Techniques: Integration by Parts
2ln2 1
2ln2 2 0 1
(2 ln2 2) (1 ln1 1)
2
1ln x dx
2
1lnx x x
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Example 5: Evaluate to find the area of the shaded region shown below.
x2e xdx0
7
4.6 Integration Techniques: Integration by Parts
Then 2 and .xdu x dx v e
2Let and .xu x dv e dx
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Example 5 (continued):Using the Integration-by-Parts Formula gives us
To evaluate the integral on the right, we can apply integration by parts again, as follows.
4.6 Integration Techniques: Integration by Parts
2( )xx e dx 2( ) (2 )x xx e e x dx
2 2x xx e xe dx
Then 2 and .xdu dx v e
Let 2 and .xu x dv e dx
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Example 5 (continued):Using the Integration-by-Parts Formula again gives us
Then we can substitute this solution into the formula on the last slide.
4.6 Integration Techniques: Integration by Parts
2 2x xxe e C
2 ( )xx e dx 2 ( ) (2 )x xx e e dx
2( 2 2)xe x x C
2 2 2x x xx e xe e C 2 xx e dx
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Example 5 (concluded):Then, we can evaluate the definite integral.
4.6 Integration Techniques: Integration by Parts
1.94
765 2e
7 2 0 2(7 2 7 2) (0 2 0 2)e e
7 72 2
00( 2 2)x xx e dx e x x C
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4.6 Integration Techniques: Integration by Parts
Quick Check 3
Evaluate:3
0.
1
xdx
x 1
Let and .1
u x dv dxx
Then, 1/2 and 2 1 .du dx v x
Using the integration by parts formula, we get:
1/2 1/22 1 2 1
1
xdx x x x dx
x
1/2 3/242 1 1
3x x x C
Slide 4- 18Copyright © 2014 Pearson Education, Inc.
4.6 Integration Techniques: Integration by Parts
Quick Check 3 Concluded
Then we can evaluate the definite integral.
3
3 1/2 3/2
00
42 1 1
31
xdx x x x
x
1/2 3/2 1/2 3/24 42 3 3 1 3 1 2 0 0 1 0 1
3 3
1/2 3/2 3/24 46 4 4 1
3 3
4 8 412
3 3
36 32 4
3 3 3
8 2 or 2
3 3
Slide 4- 19Copyright © 2014 Pearson Education, Inc.
4.6 Integration Techniques: Integration by Parts
Section Summary
• The Integration-by-Parts Formula is the reverse of the Product Rule for differentiation:
• The choices for u and dv should be such that the integral is simpler than the original integral. If this does not turn out to be the case, other choices should be made.
v du
.u dv uv v du