Ch7/MATH0211/YMC/200910
1
Chapter 7.
Integration
7.1.
Indefinite Integrals
We have studied how to differentiate a given function
f
(
x
)
in the last two chapters. Starting with
this section, we would like to know what function we diffferentiate to get
f
(
x
)
, or equivalently, we
would like to find a function
F
whose derivative is the given function
f
. If such a function
F
exists, it
is called the
antiderivative
of
f
:
Definition 7.1
An
antiderivative
of the function
f
is a function
F
such that
F
0
(
x
) =
f
(
x
)
For example,
x
2
is the antiderivative of
2
x
as
d
dx
x
2
= 2
x
. However, it is not the only antiderivative
of
2
x
:
d
dx
(
x
2
+ 1) = 2
x
and
d
dx
(
x
2

3) = 2
x.
In fact, any function of the form
x
2
+
C
, for any constant
C
, gives an antiderivative of
2
x
. It follows
that
2
x
has infinitely many antiderivatives, and any two of them differ only by a constant. We can refer
to
x
2
+
C
the most general antiderivative of
2
x
and denote it by
R
2
x dx
, that is,
Z
2
x dx
=
x
2
+
C.
This is the called an
indefinite integral
of
2
x
. The symbol
R
is called the
integral sign
, the function
2
x
is called the
integrand
and
C
is the
constant of integration
. The
dx
is part of the integral notation
and indicates the variable involved.
If
F
is any antiderivative of
f
, then
Z
f
(
x
)
dx
=
F
(
x
) +
C
where
C
is a constant
Here are some important properties of indefinite integrals:
Properties of the Indefinite Integration
1.
Z
k dx
=
kx
+
C
where
k
is constant.
2.
Z
kf
(
x
)
dx
=
k
Z
f
(
x
)
dx
where
k
is constant.
3.
Z
(
f
(
x
)
±
g
(
x
))
dx
=
Z
f
(
x
)
dx
±
Z
g
(
x
))
dx
.
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Ch7/MATH0211/YMC/200910
2
To actually compute indefinite integrals we begin with some of the basic integration formulas:
Basic Integration Formulas
1.
Z
x
n
dx
=
x
n
+1
n
+ 1
+
C
for
n
6
=

1
.
2.
Z
1
x
dx
=
Z
dx
x
= ln

x

+
C
.
3.
Z
e
x
dx
=
e
x
+
C
Example 7.1
Find the following indefinite integrals:
(
a
)
Z
(3 + 5
x

x
4
)
dx
(
b
)
Z
1
√
t
dt
(
c
)
Z
x
3

1
x
+ 4
e
x
dx
Solution
.
(a)
Z
(3 + 5
x

x
4
)
dx
=
Z
3
dx
+
Z
5
x dx

Z
x
4
dx
= (3
x
+
C
1
) + 5
x
2
2
+
C
2

x
5
5
+
C
3
= 3
x
+
5
2
x
2
+
1
5
x
5
+ (
C
1
+
C
2

C
3
)
For convenience, we will replace the constant
C
1
+
C
2

C
3
by a single constant
C
. We then
have
Z
(3 + 5
x

x
4
)
dx
= 3
x
+
5
2
x
2
+
1
5
x
5
+
C.
From now on when we integrate an expression involving more than one term, we will not write
the constant of integration for each term. Only one constant is needed.
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 Spring '10
 CHAN
 Math, Integrals, dx

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