have (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).