MAT 1235 Calculus II Section 7.1 Integration By Parts .
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MAT 1235 Calculus II Section 7.1 Integration By Parts http://myhome.spu.edu/lauw
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Transcript of MAT 1235 Calculus II Section 7.1 Integration By Parts .
MAT 1235Calculus II
Section 7.1
Integration By Parts
http://myhome.spu.edu/lauw
Homework and …
WebAssign 7.1(17 problems, 38 min.)
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Exam II
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Preview
The reverse of the product rule
Product Rule
dxxgxfxgxfdxxgxf
xgxfdxxgxfdxxgxf
xgxfdxxgxfxgxf
xgxfxgxfxgxfdx
d
)()()()()()(
)()()()()()(
)()()()()()(
)()()()()()(
Product Rule
dxxgxfxgxfdxxgxf
xgxfdxxgxfdxxgxf
xgxfdxxgxfxgxf
xgxfxgxfxgxfdx
d
)()()()()()(
)()()()()()(
)()()()()()(
)()()()()()(
Definition of Antiderivatives
dxxgxfxgxfdxxgxf
xgxfdxxgxfdxxgxf
xgxfdxxgxfxgxf
xgxfxgxfxgxfdx
d
)()()()()()(
)()()()()()(
)()()()()()(
)()()()()()(
Rationale
dxxgxfxgxfdxxgxf )()()()()()(
The integral on the right hand side is easier to evaluate than the one on the left hand side
Easier
Difficult
Rationale
dxxgxfxgxfdxxgxf )()()()()()(
The integral on the right hand side is easier to evaluate than the one on the left hand side
Easier
Difficult
xdxxxxdxx sinsincos
Alternative Form
dxxgxfxgxfdxxgxf )()()()()()(
vduuvudv
dxxgdvdxxfdu
xgvxfu
have We
)( ,)(Then
)( ),(Let
Integration By Parts
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )b b
b
aa a
f x g x dx f x g x f x g x dx
udv uv vdu
f x g x dx f x g x f x g x dx
Example 1
cosx xdx
Example 1(Analysis) vduuvudv
In order to use IBP, we need to choose and
One possible choice is xdx dvxu cos ,
cosx xdx
Example 1(Analysis)
In order to use IBP, we need to choose and
One possible choice is
xdx dvxu cos ,Let
cosx xdx
vduuvudv
Example 1(Analysis)
Now, we need to find and
xdxx cos
x
xdxvdxdu
xdx dvxu
sin
cos , Then,
cos , Let
vduuvudv
Example 1(Analysis)
Now, we need to find and
xdxx cos
x
xdxvdxdu
xdx dvxu
sin
cos , Then,
cos , Let
vduuvudv
Example 1
cos
sin sin
x xdx
x x xdx
x
xdxvdxdu
xdx dvxu
sin
cos , Then,
cos , Let
vduuvudv
Example 1
cos
sin sin
x xdx
x x xdx
vduuvudv
x
xdxvdxdu
xdx dvxu
sin
cos , Then,
cos , Let
What if?
B
Example 1 vduuvudv
What if?
2 ,sin Then,
,cos Let 2x
xdxvxdxdu
xdx dvxu
cos
cos
x xdx
x xdx
Example 2xxe dx
vduuvudv
xxe dx
Expectations
All supporting steps are required and done on the right side.
Example 32
1
ln xdx
vduuvudv
2
1
ln xdx
Example 4
cosxe xdx
vduuvudv
cosxe xdx xe
