MTH 122 — Calculus II
Essex County College — Division of Mathematics and Physics
1
Lecture Notes #3 — Sakai Web Project Material
1
Integration by Parts
Let’s start with an example where we’re asked to verify:
Z
xe
x
d
x
=
xe
x

e
x
+
C.
As you know, we basically just have to check that:
d
d
x
(
xe
x

e
x
+
C
) =
xe
x
.
Here goes:
d
d
x
(
xe
x

e
x
+
C
)
=
d
d
x
(
xe
x
)

d
d
x
(
e
x
) +
d
d
x
(
C
)
=
(
xe
x
+
e
x
)

(
e
x
) + (0)
=
xe
x
The main problem here is not the checking, but it is the actual process of finding the antideriva
tive.
So far we’ve only used one real method, that was
u
substitution, and although it is an
important method it would fail miserably in this particular case. Another technique known as
integration by parts
uses the
product rule
. As you recall, the product rule is:
d
d
x
(
uv
) =
u
0
v
+
uv
0
,
where
u
and
v
are functions of
x
and
u
0
and
v
0
are the derivatives with respect to
x
. We can
rewrite this product rule as follows:
uv
0
=
d
d
x
(
uv
)

u
0
v,
and then integrate both sides,
Z
uv
0
d
x
=
Z
d
d
x
(
uv
) d
x

Z
u
0
v
d
x.
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 Spring '10
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 Calculus, Division, Derivative, Integration By Parts, dx

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