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16.1
Definition (Nice functions.)
I will say that a real valued function
defined on an interval
is a
nice function on
, if
is continuous on
and integrable on every subinterval of
.
Remark: We know that piecewise monotonic continuous functions on
are
nice. It turns out that every continuous function on is nice, but we
are not
going to prove this. The theorems stated in this chapter for nice functions
are usually stated for continuous functions. You can find a proof that every
continuous
function on an interval is integrable on (and hence that every
continuous function on is nice on ) in [44, page 246]
or in
[1, page 153]. However both of these sources use a slightly
different
definition of continuity and of integral than we do, so you will need to do
some work
to translate the proofs in these references into proofs in our terms. You
might try
to prove the result yourself, but the proof is rather tricky. For all the
applications we will make in this course, the functions examined will be
continuous
and piecewise monotonic so the theorems as we prove them are good enough.
16.2
Exercise.
A
Can you give an example of a continuous function on a closed interval
that is
not piecewise monotonic?
You may describe your example rather loosely, and you do not need to
prove that it is continuous.
Proof: By the definition of antiderivative, is continuous on and
on . Let be arbitrary points in . I will
suppose
. (Note that if (16.4) holds
when , then it holds when
,
since both sides of the equation change sign when and are interchanged.
Also
note that the theorem clearly holds for .)
Let
be any partition of , and let be an
integer
with . If we can apply the mean value theorem to
on
to find a number
such that
If , let . Then
is a sample for
such that
We have shown that for every partition of there is a sample for
such that
Let be a sequence of partitions for such that
,
and for each
let be a sample for such that
Then, since is integrable on ,
16.5
Example.
The fundamental theorem will allow us to evaluate many integrals easily. For
example, we know that
.
Hence, by the fundamental theorem,
This says that the two sets
and
have the same area - a rather remarkable result.
Proof: Since is continuous on we can find numbers
such
that
By the inequality theorem for integrals
(here and denote constant functions) i.e.,
i.e.,
We can now apply the intermediate value property to on the interval whose
endpoints are and to get a number between and such that
The number is in the interval , so we are done.
16.8
Exercise.
A
Explain how corollary
16.7 follows
from theorem
16.6. (There is nothing to show unless
)
Proof: Let . I will show that is continuous at .
Since is integrable on there is a number such that
By the corollary to the inequality theorem for integrals (8.17), it
follows
that
for all . Thus, for all ,
Now
, so by the squeezing rule
for limits of functions,
.
It follows that
.
Proof: Let
and let be a point in . Let be any sequence in
such that . Then
By the mean value theorem for integrals, there is a number between
and
such that
Now
and since
,
we
have
, by the squeezing rule for sequences.
Since is continuous, we conclude that
; i.e.,
i.e.,
This proves that
for . In addition is
continuous
on by lemma 16.9. Hence is an antiderivative for on
.
Remark Leibnitz's
statement of the fundamental principle of the calculus was
the following:
Differences and sums are the inverses of one another, that is
to say, the sum of the differences of a series is a term of the series,
and the difference of the sums of a series is a term of the
series; and I enunciate the former thus, , and the
latter thus, [34, page 142].
To see the relation between Leibnitz's formulas and ours, in the equation
, write to get
, or
. This corresponds to equation
(16.11). Equation (16.4) can be written as
If we cancel the 's (in the next chapter we will show that this
is actually justified!) we get
. This is
not quite the same as . However if you choose the
origin of coordinates to be , then the two formulas
coincide.
To emphasize the inverse-like relation between differentiation and
integration, I will restate our formulas for both parts of the
the fundamental theorem, ignoring all hypotheses:
By exploiting the ambiguous notation for indefinite integrals, we
can get a form almost identical with Leibniz's:
16.12
Example.
Let
We will calculate the derivatives of
, and
. By the fundamental
theorem,
Now
, so by the chain rule,
We have
, so by the product rule,
16.13
Exercise.
Calculate the derivatives of the following functions. Simplify your
answers as
much as you can.
- a)
-
- b)
-
- c)
-
- d)
-
- e)
-
.
(We defined
and
in exercise
14.56.) Find simple
formulas
(not involving any integrals) for
and for
.
A
16.15
Exercise.
Let
and
be the functions whose graphs are shown below:
Let
for
. Sketch the graphs of
and
. Include some discussion about why your answer is correct.
Next: 17. Antidifferentiation Techniques
Up: Math 111 Calculus I
Previous: 15.3 Convexity
  Index
Ray Mayer
2007-09-07