Harvey Mudd College Math Tutorial:
Integration by Parts
We will use the
Product Rule
for derivatives to derive a powerful integration formula:
•
Start with (
f
(
x
)
g
(
x
))
0
=
f
(
x
)
g
0
(
x
) +
f
0
(
x
)
g
(
x
).
•
Integrate both sides to get
f
(
x
)
g
(
x
) =
Z
f
(
x
)
g
0
(
x
)
dx
+
Z
f
0
(
x
)
g
(
x
)
dx
. (We need not
include a constant of integration on the left, since the integrals on the right will also
have integration constants.)
•
Solve for
Z
f
(
x
)
g
0
(
x
)
dx
, obtaining
Z
f
(
x
)
g
0
(
x
)
dx
=
f
(
x
)
g
(
x
)

Z
f
0
(
x
)
g
(
x
)
dx.
This formula frequently allows us to compute a difficult integral by computing a much simpler
integral. We often express the Integration by Parts formula as follows:
Let
u
=
f
(
x
)
dv
=
g
0
(
x
)
dx
du
=
f
0
(
x
)
dx
v
=
g
(
x
)
Then the formula becomes
Z
u dv
=
uv

Z
v du.
To integrate by parts, strategically choose
u
,
dv
and then apply the formula.
Example
Let’s evaluate
Z
xe
x
dx
.
Let
u
=
x
dv
=
e
x
dx
du
=
dx
v
=
e
x
Then by integration by parts,
Z
xe
x
=
xe
x

Z
e
x
dx
=
xe
x

e
x
+
C.
A Faulty Choice
A Reduction Formula
Integration by parts “works” on definite integrals as well:
Z
b
a
u dv
=
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 HONG
 Integration By Parts, dx, ex cos

Click to edit the document details