6
Fundamental Theorems, Substitution, In
tegration by Parts, and Polar Coordinates
So far we have separately learnt the basics of integration and diﬀerentiation.
But they are not unrelated. In fact, they are
inverse operations
. This
is what we will try to explore in the ﬁrst section, via the two fundamental
theorems of Calculus. After that we will discuss the two main methods one
uses for integrating somewhat complicated functions, namely
integration
by substitution
and
integration by parts
. The ﬁnal section will discuss
integration in polar coordinates
, which comes up when there is radial
symmetry.
6.1
The fundamental theorems
Suppose
f
is an integrable function on a closed interval [
a, b
]. Then we can
consider the
signed area function A
on [
a, b
] (relative to
f
) deﬁned by the
deﬁnite integral of
f
from
a
to
x
, i.e.,
(6
.
1
.
1)
A
(
x
) =
x
∫
a
f
(
t
)
dt.
The reason for the
signed area
terminology is that
f
is not assumed to be
≥
0, so a priori
A
(
x
) could be negative.
It is extremely interesting to know how
A
(
x
) varies with
x
. What condi
tions does one need to put on
f
to make sure that
A
is continuous, or even
diﬀerentiable? The continuity part of the question is easy to answer.
Lemma 6.1
Let
f, A
be as above. Then
A
is a continuous function on
[
a, b
]
.
Proof
.
Let
c
be any point in [
a, b
]. Then
f
is continuous at
c
iﬀ we
have
lim
h
→
0
A
(
c
+
h
) =
A
(
c
)
.
Of course, in taking the limit, we consider all small enough
h
for which
c
+
h
lies in [
a, b
], and then let
h
go to zero. By the
additivity
of the integral, we
1
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View Full Documenthave (using (6.1.1)),
A
(
c
+
h
)
−
A
(
c
) =
∫
I
(
c,h
)
f
(
t
)
dt,
where
I
(
c, h
) denotes the closed interval between
c
and
c
+
h
. Clearly,
I
(
c, h
)
is [
c, c
+
h
], resp. [
c
+
h, c
], if
h
is positive, resp. negative. When
h
goes to
zero,
I
(
c, h
) shrinks to the point
{
c
}
, and so
lim
h
→
0
A
(
c
+
h
)
−
A
(
c
) = 0
,
which is what we needed to show.
2
Remark 6.1.2
:
The general moral to remember is that, just as in real life,
Integration is good and Diﬀerentiation is bad
!
Indeed, as seen in this Lemma,
integration
makes functions better; here it
takes an integrable, but not necessarily a continuous, function
f
, and from
it obtains a continuous function. The Theorem below says that if
f
is in
addition continuous, then its integral is even diﬀerentiable. Diﬀerentiation,
on the other hand, makes functions worse. The derivative of a diﬀerentiable
function
f
is often not diﬀerentiable (think of
f
(
x
) = sign(
x
)
x
2
at
x
= 0); in
fact,
f
′
need not even be continuous (think of
f
(
x
) =
x
2
sin(1
/x
) when
x
̸
= 0
and = 0 when
x
= 0).
A satisfactory answer to the question of diﬀerentiability of the integral is
given by the following important result, which comes with an appropriately
honoriﬁc title:
Theorem 6.2
(
The ﬁrst fundamental theorem of Calculus
) Let
f
be
an integrable function on
[
a, b
]
, and let
A
be the function deﬁned by (6.1.1).
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 Fall '08
 Borodin,A
 Calculus, Integration By Parts, Polar Coordinates, dx, Mj

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