This preview shows pages 1–5. Sign up to view the full content.
CALCULUS II
Spring 2009
LECTURE 4
This lecture provides eight examples of improper integrals.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document Recall
We recall from Lecture 3
•
every polynomial can (theoretically) be written as a product of linear
and quadratic factors
•
integration of rational functions can be eﬀected by partial fraction
decomposition of the integrand
End of Recall
2
Mathematics Interlude I
Improper Integrals
We recall brieﬂy the deﬁnition of the Riemann Integral.
Deﬁnition
. Let
f
denote a realvalued function deﬁned on the closed and
bounded interval [
a,b
]. Let
L
denote a real number such that for each
± >
0 there exists
δ >
0 such that if
{
x
i
}
n
i
=0
is a partition of [
a,b
] with
max
1
≤
i
≤
n
Δ
x
i
< δ
, then
±
±
±
±
±
n
X
i
=1
f
(
ξ
i
)Δ
x
i

L
±
±
±
±
±
< ±
for every choice of the
ξ
i
’s constrained by
x
i

1
≤
ξ
i
≤
x
i
,
1
≤
i
≤
n
. The
number
L
is called the
integral
of
f
, or the
Riemann integral
of
f
. One
writes the integral of
f
by
Z
b
a
f
(
x
)
dx.
If a realvalued function deﬁned on [
a,b
] has a Riemann integral then it is
said to be
integrable
or
Riemann integrable
. Most functions that we deal
with are Riemann integrable, but not all. The function
f
(
x
) =
(
1
√
x
0
< x
≤
1
0
x
= 0
is not Riemann integrable. But it has an integral in another sense, called
an improper integral. (This is our ﬁrst example below.)
Deﬁnition
. For a realvalued function,
f
, deﬁned on [
a,b
], a
Riemann sum
is an expression of the form
n
X
i
=1
f
(
ξ
i
)Δ
x
i
3
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
n
is a positive integer,
{
x
i
}
n
i
=0
is a partition of [
a,b
],
x
i

1
≤
ξ
i
≤
x
i
,
1
≤
i
≤
n
and Δ
x
i
=
x
i

x
i

1
,
1
≤
i
≤
n
.
Roughly speaking, one says that the integral of
f
is the limit, as
n
→ ∞
, of
its Riemann sums. Abusing our deﬁnition of a limit we write
Z
b
a
f
(
x
)
dx
=
lim
maxΔ
x
i
→
0
n
X
i
=1
f
(
ξ
i
)Δ
x
i
.
The next example is not that of a Riemann integral but of an
improper Rie
mann integral
. The integrand,
1
√
x
, is not Riemann integrable on [0
,
1], since
it is not bounded there. So an additional limiting process is imposed. We
work through the example and then look back to see what happened.
EXAMPLE.
Evaluate the improper integral
Z
1
0
1
√
x
dx.
It is understood that the function
1
√
x
is not deﬁned at
x
=
0. But as is seen below, the expression
R
1
0
1
√
x
dx
is still
meaningful.
Solution. One writes
Z
1
0
1
√
x
dx
=
lim
t
→
0
+
Z
1
t
t

1
2
dx
=
lim
t
→
0
+
2
x
1
2
±
±
±
±
1
t
=
lim
t
→
0
+
²
2

2
t
1
2
³
= 2

2 lim
t
→
0
+
t
1
2
= 2
.
Thus, the value of the improper integral is given by
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 02/09/2009 for the course MATH 1020 taught by Professor Ecker during the Spring '08 term at Rensselaer Polytechnic Institute.
 Spring '08
 ECKER
 Factors, Improper Integrals, Integrals

Click to edit the document details