This preview shows pages 1–3. Sign up to view the full content.
Math021, week 10
Integration by Substitution
Theorem 10.1
If
f
and
g
are functions,
f
(
g
(
x
))
is an antiderivative of
f
0
(
g
(
x
))
g
0
(
x
)
.
That is,
Z
f
0
(
g
(
x
))
g
0
(
x
)
dx
=
f
(
g
(
x
)) +
C.
proof: nothing diﬀerent from the orinary chain rule.
Example 10.2
Evaluate
Z
2
x
(1 +
x
2
)
8
dx.
solution:
Let
g
(
x
) = 1 +
x
2
and
h
(
x
) =
x
8
. We are asking for an antiderivative of
g
0
(
x
)
h
(
g
(
x
)). Now if
f
(
x
) =
1
9
x
9
,
f
is an antiderivative of
h
. In other words,
we are asking for an antiderivative of
g
0
(
x
)
f
0
(
g
(
x
)). According to the version of
chain rule above, an antiderivative of it is
f
(
g
(
x
)) =
1
9
(1 +
x
2
)
8
. Thus,
Z
2
x
(1 +
x
2
)
8
dx
=
1
9
(1 +
x
2
)
9
+
C.
Remark 10.3
The previous procedure can easily be memorized by putting
u
=
x
2
+ 1
,
du
= 2
xdx
. Then,
Z
2
x
(1 +
x
2
)
8
dx
=
Z
u
8
du
=
1
9
u
9
+
C
=
1
9
(1 +
x
2
)
9
+
C.
Example 10.4
Evaluate
Z
e
4
x
dx.
Let
u
= 4
x
so that
du
= 4
dx
. Then,
R
e
4
x
dx
=
R
e
u
1
4
du
=
1
4
e
u
+
C
=
1
4
e
4
x
+
C.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentExample 10.5
Evaluate
Z
e
sin
x
cos
xdx.
solution:
Let
u
= sin
x
so that
du
= cos
xdx
. Then,
R
e
sin
x
cos
xdx
=
R
e
u
du
=
e
u
+
C
=
e
sin
x
+
C.
Example 10.6
Evaluate the area of the region bounded by the graph of
f
(
x
) =
e
sin
x
cos
x
, the xaxis, the lines
x
= 0
and
x
=
π/
2
.
solution:
Note that
f
(
x
)
≥
0 for 0
≤
x
≤
π/
2. Hence, the area of the given region is
Z
π/
2
0
e
sin
x
cos
xdx.
We saw that
g
(
x
) =
e
sin
x
is an antiderivative of
f
(
x
) =
e
sin
x
cos
x
. By the
fundamental theorem of calculus,
Z
π/
2
0
e
sin
x
cos
xdx
=
e
sin
x

π/
2
0
=
e
1

e
0
=
e

1
which is the area of the given region.
Remark 10.7
This is the end of the preview. Sign up
to
access the rest of the document.