Advanced Calculus: MATH 410

Riemann Integrals and Integrability

Professor David Levermore
6 December 2006

1. Definite Integrals
In this section we revisit the definite integral that you were intro-
duced to when you first studied calculus. You undoubtedly learned that
given a positive function f over an interval [a, b] the definite integral
Z b
f (x) dx ,
a

provided it was defined, was a number equal to the area under the
graph of f over [a, b]. You also likely learned that the definite integral
was defined as a limit of Riemann sums. The Riemann sums you most
likely used involved partitioning [a, b] into n uniform subintervals of
length (b a)/n and evaluating f at either the right-hand endpoint,
the left-hand endpoint, or the midpoint of each subinterval. At the
time your understanding of the notion of limit was likely more intu-
itive than rigorous. In this section we present the Riemann Integral,
a rigorous development of the definite integral built upon the rigorous
understanding of limit that you have studied earlier in this course.

1.1. Partitions and Darboux Sums. We will consider very general

partitions of the interval [a, b], not just those with uniform subintervals.
Definition 1.1. Let [a, b] R. A partition of the interval [a, b] is
specified by n Z+ , and {xi }ni=0 [a, b] such that
a = x0 < x1 < < xn1 < xn = b .
The partition P associated with these points is defined to be the ordered
collection of n subintervals of [a, b] given by


P = [xi1 , xi ] : i = 1, , n
This partition is denoted P = [x0 , x1 , , xn1 , xn ]. Each xi for i =
0, , n is called a partition point of P , and for each i = 1, , n
the interval [xi1 , xi ] is called a ith subinterval in P . The partition
thickness or width, denoted |P |, is defined by

|P | = max xi xi1 : i = 1, , n .
1
2

The approach to the definite integral taken here is not based on

Riemann sums, but rather on Darboux sums. This is because Darboux
sums are well-suited for analysis by the tools we have developed to
establish the existence of limits. We will be able to recover results
about Riemann sums because, as we will show, every Riemann sum is
bounded by two Darboux sums.
Let f : [a, b] R be bounded. Set
 
(1) m = inf f (x) : x [a, b] , m = sup f (x) : x [a, b] .
Because f is bounded, one knows that < m m < .
Let P = [x0 , , xn ] be a partition of [a, b]. For each i = 1, , n
set 
mi = inf f (x) : x [xi1 , xi ] ,

mi = sup f (x) : x [xi1 , xi ] .
Clearly m mi mi m for every i = 1, , n.
Definition 1.2. The lower and upper Darboux sums associated with
the function f and partition P are respectively defined by
Xn n
X
(2) L(f, P ) = mi (xi xi1 ) , U (f, P ) = mi (xi xi1 ) .
i=1 i=1

Clearly, the Darboux sums satisfy the bounds

(3) m (b a) L(f, P ) U (f, P ) m (b a) .
These inequalities will all be equalities when f is a constant.
Remark. A Riemann sum associated with the partition P is specified
by selecting a quadrature point qi [xi1 , xi ] for each i = 1, , n. Let
Q = (q1 , , qn ) be the n-tuple of quadrature points. The associated
Riemann sum is then
Xn
R(f, P, Q) = f (qi ) (xi xi1 ) .
i=1
It is easy to see that for any choice of quadrature points Q one has the
bounds
(4) L(f, P ) R(f, P, Q) U (f, P ) .
Moreover, one can show that

L(f, P ) = inf R(f, P, Q) : Q are quadrature points for P ,
(5) 
U (f, P ) = sup R(f, P, Q) : Q are quadrature points for P .
The bounds (4) are thereby sharp.
Exercise. Prove (5)
 3

1.2. Refinements. We now introduce the notion of a refinement of a

partition.

Definition 1.3. Given a partition P of an interval [a, b], a partition

P of [a, b] is called a refinement of P provided every partition point of
P is a partition point of P .

If P = [x0 , x1 , , xn1 , xn ] and P is a refinement of P then P

induces a partition of each [xi1 , xi ], which we denote by Pi . For
example, if P = [x0 , x1 , , xn 1 , xn ] with xji = xi for each i =
0, , n then Pi = [xji1 , , xji ]. Observe that
n
X n
X

(6) L(f, P ) = L(f, Pi ) ,
U (f, P ) = U (f, Pi ) .
i=1 i=1

Moreover, upon applying the bounds (3) to Pi for each i = 1, , n,

we obtain the bounds

(7) mi (xi xi1 ) L(f, Pi ) U (f, Pi ) mi (xi xi1 ) .

This observation is key to the proof of the following.

Lemma 1.1. (Refinement) Let f : [a, b] R be bounded. Let P be

a partition of [a, b] and P be a refinement of P . Then

(8) L(f, P ) L(f, P ) U (f, P ) U (f, P ) .

Proof. It follows from (2), (7), and (6) that

n
X
L(f, P ) = mi (xi xi1 )
i=1
Xn
L(f, Pi ) = L(f, P )
i=1
n
X

U (f, P ) = U (f, Pi )
i=1
n
X
mi (xi xi1 ) = U (f, P ) .
i=1


4

1.3. Comparisons. A key step in our development will be to develop

comparisons of L(f, P 1 ) and U (f, P 2 ) for any two partitions P 1 and
P 2 , of [a, b].
Definition 1.4. Given any two partitions, P 1 and P 2 , of [a, b] we
define P 1 P 2 to be the partition whose set of partition points is the
union of the partition points of P 1 and the partition points of P 2 . We
call P 1 P 2 the supremum of P 1 and P 2 .
It is easy to argue that P 1 P 2 is the smallest partition of [a, b] that
is a refinement of both P 1 and P 2 . It is therefore sometimes called the
smallest common refinement of P 1 and P 2 .
Lemma 1.2. (Comparison) Let f : [a, b] R be bounded. Let P 1
and P 2 be partitions of [a, b]. Then
(9) L(f, P 1 ) U (f, P 2 ) .
Proof. Because P 1 P 2 is a refinement of both P 1 and P 2 , it follows
from the Refinement Lemma that
L(f, P 1 ) L(f, P 1 P 2 ) U (f, P 1 P 2 ) U (f, P 2 ) .

1 2
Because the partitions P and P on either side of inequality (9) are
independent, we may obtain sharper bounds by taking the supremum
over P 1 on the left-hand side, or the infimum over P 2 on the right-hand
side. Indeed, we prove the following.
Lemma 1.3. (Sharp Comparison) Let f : [a, b] R be bounded.
Let

L(f ) = sup L(f, P ) : P is a partition of [a, b] ,
(10) 
U (f ) = inf U (f, P ) : P is a partition of [a, b] .
Let P 1 and P 2 be partitions of [a, b]. Then
(11) L(f, P 1 ) L(f ) U (f ) U (f, P 2 ) .
Moreover, if
L(f, P ) A U (f, P ) for every partition P of [a, b] ,
then A [L(f ), U (f )].
Remark. Because it is clear from (10) that L(f ) and U (f ) depend on
[a, b], strictly speaking these quantities should be denoted L(f, [a, b])
and U (f, [a, b]). This would be necessary if more than one interval was
involved in the discussion. However, that is not the case here. We
therefore embrace the less cluttered notation.
 5

Proof. If we take the infimum of the right-hand side of (9) over P 2 ,

we obtain
L(f, P1 ) U (f ) .
If we then take the supremum of the left-hand side above over P 1 , we
obtain
L(f ) U (f ) .
The bound (11) then follows.
The proof of the last assertion is left as an exercise. 
Exercise. Prove the last assertion of Lemma 1.3.

1.4. Definition of the Riemann Integral. We are now ready to

define the definite integral of Riemann.
Definition 1.5. Let f : [a, b] R be bounded. Then f is said to be
Riemann integrable over [a, b] whenever L(f ) = U (f ). In this case we
call this common value the Riemann integral of f over [a, b] and denote
Rb
it by a f :
Z b
(12) f = L(f ) = U (f ) .
a
Then f is called the integrand of the integral, a is called the lower
endpoint (or lower limit) of integration, while b is called the upper
endpoint (or upper limit) of integration.

Remark. We will call a and b the endpoints of integration rather

than the limits of integration. The word limit does enough work in
this subject. We do not need to adopt terminology that can lead to
confusion.
We begin with the following characterizations of integrability.
Theorem 1.1. (Riemann-Darboux) Let f : [a, b] R be bounded.
Then the following are equivalent:
(1) f is Riemann integrable over [a, b] (i.e. L(f ) = U (f ));
(2) for every  > 0 there exists a partition P of [a, b] such that
0 U (f, P ) L(f, P ) <  ;
(3) there exists a unique A R such that
(13) L(f, P ) A U (f, P ) for every partition P of [a, b] .
Rb
Moreover, in case (3) A = a f .
6

Remark. Characterizations (2) and (3) of the Riemann-Darboux The-

orem are useful for proving the integrability of a function f . The
Sharp Comparison Lemma shows that (13) holds if and only if A
[L(f ), U (f )]. The key thing to be established when using characteriza-
tion (3) is therefore the uniqueness of such an A.
Proof. First we show that (1) = (2). Let  > 0. By the definition
(10) of L(f ) and U (f ), we can find partitions P L and P U of [a, b] such
that

L(f ) < L(f, P L ) L(f ) ,
2

U (f ) U (f, P U ) < U (f ) + .
2
L U
Let P = P P . Because the Comparison Lemma implies that
L(f, P L ) L(f, P ) and U (f, P ) U (f, P U ), it follows from the above
inequalities that

L(f ) < L(f, P ) L(f ) ,
2

U (f ) U (f, P ) < U (f ) + .
2
Hence, if L(f ) = U (f ) one concludes that
   
 
0 U (f, P ) L(f, P ) < U (f ) + L(f ) = .
2 2
This shows that (1) = (2).
Next we show that (2) = (3). Suppose that (3) is false. The Sharp
Comparison Lemma shows that (13) holds for every A [L(f ), U (f )],
and that this interval is nonempty. So the only way (3) can be false is
if uniqueness fails. In that case there exists A1 and A2 such that
L(f, P ) A1 < A2 U (f, P ) for every partition P of [a, b] .
One thereby has that
U (f, P ) L(f, P ) A2 A1 > 0 for every partition P of [a, b] .
Hence, (2) must be false. It follows that (2) = (3).
Rb
Finally, we show that (3) = (1) and that (3) implies A = a f .
The Sharp Comparison Lemma shows that (13) holds if and only if
A [L(f ), U (f )]. But (3) states that such an A is unique. Hence,
Rb
A = L(f ) = U (f ), which implies (1) and A = a f . 
 7

1.5. Convergence of Riemann and Darboux Sums. We now make

a connection with the notion of a definite integral as the limit of a se-
quence of Riemann sums.
Recall for any given f : [a, b] R a Riemann sum associated with a
partition P = [x0 , x1 , xn ] of [a, b] is specified by selecting a quadra-
ture point qi [xi1 , xi ] for each i = 1, , n. Let Q = (q1 , , qn ) be
the n-tuple of quadrature points. The associated Riemann sum is then
Xn
(14) R(f, P, Q) = f (qi ) (xi xi1 ) .
i=1

If f : [a, b] R is bounded (so that the Darboux sums L(f, P ) and

U (f, P ) are defined) then for any choice of quadrature points Q one
has the bounds
(15) L(f, P ) R(f, P, Q) U (f, P ) .
A sequence of Riemann sums for any given f : [a, b] R is therefore
specified by a sequence {P n }
n=1 of partitions of [a, b] and a sequence
n
{Q }n=1 of associated quadrature points. The sequence of partitions
cannot be arbitrary.
Definition 1.6. Let f : [a, b] R be bounded. A sequence {P n }
n=1 of
partitions of [a, b] is said to be Archimedean for f provided
lim U (f, P n ) L(f, P n ) = 0 .


(16)
n

Our main theorem is the following.

Theorem 1.2. (Archimedes-Riemann) Let f : [a, b] R be bounded.
Then f is Riemann integrable over [a, b] if and only if there exists a
sequence of partitions of [a, b] that is Archimedean for f . If {P n }
n=1
is any such sequence then
Z b Z b
n n
(17) lim L(f, P ) = f , and lim U (f, P ) = f.
n a n a
Moreover, if {Qn }
n=1 is any sequence of associated quadrature points
then
Z b
n n
(18) lim R(f, P , Q ) = f,
n a
where the Riemann sums R(f, P, Q) are defined by (14).
Remark. The content of this theorem is that once one has found Rb a
sequence of partitions P n such that (16) holds, then the integral a f
exists and may be evaluated as the limit of any associated sequence
of Darboux sums (17) or Riemann sums (18). This theorem thereby
8

splits the task of evaluating a definite integrals into two steps. The
first step is by far the easier. It is a rare integrand f for which one can
find a sequence of Darboux or Riemann sums that allows one of the
limits (17) or (18) to be evaluated directly.
Proof. If f is Riemann integrable over [a, b] then one can use character-
ization (2) of the Riemann-Darboux Theorem to construct a sequence
of partitions that satisfies (16), and is thereby Archimedean for f . Con-
versely, given a sequence of partitions of [a, b] that is Archimedean for
f , then the fact that f is integrable over [a, b] follows directly from
characterization (2) of the Riemann-Darboux Theorem. The details of
these arguments are left as an exercise.
We now establish the limits (17) and (18). Let {P n } n=1 be a sequence
of partitions of [a, b] that is Archimedean for f and let {Qn } n=1 be a
sequence of associated quadrature points. The bounds on Riemann
sums given by (15) yield the inequalities
L(f, P n ) R(f, P n , Qn ) U (f, P n ) ,
while, because f is Riemann integrable, we also have the inequalities
Z b
n
L(f, P ) f U (f, P n ) .
a
It follows from these inequalities that
Z b
n n n
L(f, P ) U (f, P ) L(f, P ) f
a
Z b
n n
R(f, P , Q ) f
a
Z b
n
U (f, P ) f U (f, P n ) L(f, P n ) ,
a
which implies that
 Z b 
 
L(f, P n ) f  U (f, P n ) L(f, P n ) ,
 
a
 Z b 
 
R(f, P n , Qn ) f  U (f, P n ) L(f, P n ) ,
 
a
 Z b 
 
U (f, P n ) f  U (f, P n ) L(f, P n ) .

a

Because {P n } is Archimedean for f , it satisfies (16), whereby the

n=1
right-hand sides above vanish as n tends to . The limits (17) and
(18) follow. 
 9

1.6. Partitions Lemma. We now prove a Lemma that will subse-

quently provide us with a simple criterion for a sequence of partitions
to be Archimedean for any Riemann integrable function.
Lemma 1.4. (Partitions) Let f : [a, b] R be bounded. Then for
every  > 0 there exists a > 0 such that for every partition P of [a, b]
one has
(
0 L(f ) L(f, P ) <  ,
(19) |P | < =
0 U (f, P ) U (f ) <  .

Proof. Let  > 0. There exist partitions PL and PU of [a, b] such that
 
0 L(f ) L(f, PL ) < , 0 U (f, PU ) U (f ) < .
2 2
Let P = PL PU . Let n be the number of subintervals in P  . Pick
   

> 0 such that


n 2M < , where M = sup |f (x)| : x [a, b] .

2
We will show that (19) holds for this .
Now let P = [x0 , x1 , , xn ] be an arbitrary partition of [a, b] such
that |P | < . Set P = P P  . We consider
0 L(f ) L(f, P ) = L(f ) L(f, P )

+ L(f, P ) L(f, P ) ,


(20)
0 U (f, P ) U (f ) = U (f, P ) U (f )

+ U (f, P ) U (f, P ) .

We will prove the theorem by showing that each of the four terms in
parentheses on the right-hand sides above is less than /2.
Because P is a refinement of P  , which is a refinement of both PL
and PU , the Refinement Lemma implies that
0 L(f ) L(f, P ) L(f ) L(f, P  )

L(f ) L(f, PL ) < ,
2
0 U (f, P ) U (f ) U (f, P  ) U (f )

U (f, PU ) U (f ) < .
2
Thus, the first terms on the right-hand sides of (20) are less than /2.
Because P is a refinement of P , for each i = 1, , n we let Pi de-
note the partition of [xi1 , xi ] induced by P . The Refinement Lemma
10

then yields
n
X
0 L(f, P ) L(f, P ) = L(f, Pi ) mi (xi xi1 ) ,
 
i=1
n
X
0 U (f, P ) U (f, P ) = mi (xi xi1 ) U (f, Pi ) .
 
i=1
 
Because P has at most n 1 partition points that are not partition
points of P , there are at most n 1 indices i for which [xi1 , xi ] contains
at least one partition point of Pi within (xi1 , xi ). For all other indices
the terms in the above sums are zero. Each of the nonzero terms in
the above sums satisfy the bounds
0 mi (xi xi1 ) U (f, Pi ) 2M (xi xi1 ) < 2M ,
0 L(f, Pi ) mi (xi xi1 ) 2M (xi xi1 ) < 2M .
Because there are at most n 1 such terms, we obtain the bounds

0 U (f, P ) U (f, P ) < n 2M < ,
2
 
0 L(f, P ) L(f, P ) < n 2M < .
2
This shows the second terms on the right-hand sides of (20) are each
less than /2, thereby completing the proof of the lemma. 
An immediate consequence of the Partitions Lemma is the following
characterization of Riemann integrable functions.
Theorem 1.3. (Darboux) Let f : [a, b] R be bounded. Then f
Riemann integrable over [a, b] if and only if for every  > 0 there exists
a > 0 such that for every partition P of [a, b] one has
|P | < = 0 U (f, P ) L(f, P ) <  .
Proof. The direction (=) follows from the Riemann-Darboux The-
orem. The direction (=) follows from the definition of the Riemann
integral and the Partitions Lemma. The details of the proof are left as
an exercise. 
An immediate consequence of the Darboux Theorem is that there a
simple criterion for a sequence of partitions to be Archimedean for any
Riemann integrable function.
Theorem 1.4. (Archimedean Sequence) Every sequence {P n } n=1
of partitions of [a, b] such that |P n | 0 as n is Archimedean for
every function f : [a, b] R that is Riemann integrable over [a, b].
 11

Proof. Exercise. 
Remark. The condition that |P n | 0 as n is not necessary for a
sequence {P n }
n=1 of partitions to be Archimedean. For example, every
sequence of partitions is Archimedean for every constant function.
The Partitions Lemma also leads to a useful characterization of Rie-
mann integrable functions in terms of limits of Riemann sums. The
key step towards this characterization is taken by the following lemma.
Lemma 1.5. Let f : [a, b] R be bounded. Let {P n } n=1 be any
sequence of partitions of [a, b] such that |P n | 0 as n . Then
(21) lim L(f, P n ) = L(f ) , lim U (f, P n ) = U (f ) .
n n

Moreover, there exist sequences {QnL } n

n=1 and {QU }n=1 of associated
quadrature points such that
(22) lim R(f, P n , QnL ) = L(f ) , lim R(f, P n , QnU ) = U (f ) .
n n

Proof. The proof of (21) follows from the Partitions Lemma. To prove
(22), it follows from (5) that for each partition P n there exist sets of
associated quadrature points QnL and QnU such that
1
0 R(f, P n , QnL ) L(f, P n ) < n ,
2
1
0 U (f, P n ) R(f, P n , QnU ) < n .
2
The details of the proof are left as an exercise. 
An immediate consequence of the previous lemma is the following
characterization of Riemann integrable functions in terms of limits of
Riemann sums.
Theorem 1.5. Let f : [a, b] R be bounded. Let {P n } n=1 be any
n
sequence of partitions of [a, b] such that |P | 0 as n . Suppose
that there exists a unique A R such that for every sequence {Q n }
n=1
of associated quadrature points one has
(23) lim R(f, P n , Qn ) = A .
n

Then f is Riemann integrable over [a, b] and

Z b
(24) f = A.
a

Proof. Exercise. 
12

2. Riemann Integrable Functions

In the previous section we defined the Riemann integral and gave
characterizations of Riemann integrable functions. However, we did
not use those characterizations to identify a large class of Riemann
integrable functions. That is what we will do in this section. Before
beginning that task, we remark that there are many functions that are
not Riemann integrable.
Exercise. Let f be the function
(
1 if x Q ,
f (x) =
0 if x
/ Q.
Show that the restriction of f to any closed bounded interval [a, b] with
a < b is not Riemann integrable.

2.1. Integrability of Monotonic Functions. We first show that the

class of Riemann integrable functions includes the class of monotonic
functions. Recall that this class is defined as follows.
Definition 2.1. Let D R. A function f : D R is said to be
nondecreasing over D provided that
x < y = f (x) f (y) for every x, y D .
A function f : D R is said to be nonincreasing over D provided that
x < y = f (x) f (y) for every x, y D .
A function that is either nondecreasing or nonincreasing is said to be
monotonic over D.
A function that is monotonic over a closed bounded interval [a, b] is
clearly bounded by its endpoint values.
Theorem 2.1. (Monotonic Integrability) Let f : [a, b] R be
monotonic. Then f is Riemann integrable over [a, b]. Moreover, for
every partition P of [a, b] one has
 
(25) 0 U (f, P ) L(f, P ) |P | f (b) f (a) .
Proof. Given (25), it is easy to prove that f is Riemann integrable
over [a, b]. Indeed, let  > 0. Let P be any partition of [a, b] such that
|P | |f (b) f (a)| < . Then by (25) one has
 
0 U (f, P ) L(f, P ) |P | f (b) f (a) <  .
Hence, f is Riemann integrable by characterization (2) of the Riemann-
Darboux Theorem.
 13

All that remains to be done is to prove (25). For any partition

P = [x0 , , xn ] we have the following basic estimate. Because f is
monotonic, over each subinterval [xi1 , xi ] one has that
 
mi mi = f (xi ) f (xi1 ) .

We thereby obtain
n
X
0 U (f, P ) L(f, P ) = (mi mi ) (xi xi1 )
i=1
n
X n
X  
|P | (mi mi ) = |P | f (xi ) f (xi1 ) ,
i=1 i=1

where |P | = max{xi xi1 : i = 1, , n} is the thickness of P .

Because f is monotonic, the terms f (xi ) f (xi1 ) are either all non-
negative, or all nonpositive. We may therefore pass the absolute value
outside the last sum above, which then telescopes. We thereby obtain
the bound
n
X  
0 U (f, P ) L(f, P ) |P | f (xi ) f (xi1 )
i=1n 
X

= |P |  f (xi ) f (xi1 ) 
 
 
 i=1 
= |P | f (b) f (a) .

This establishes (25), and thereby proves the theorem. 

Remark. It is a classical fact that a monotonic function over [a, b] is
continuous at all but at most a countable number of points where it
has a jump discontinuity. One example of such a function defined over
the interval [0, 1] is

1 1 1
for k+1 < x k ,
f (x) = 2 k 2 2
0 for x = 0 .

One can show that

1
2
Z
f= .
0 3
14

2.2. Integrability of Continuous Functions. The class of Riemann

integrable functions also includes the class of continuous functions.
Theorem 2.2. (Continuous Integrability) Let f : [a, b] R be
continuous. Then f is Riemann integrable over [a, b].
Remark. The fact that f is continuous over [a, b] implies that it is
bounded over [a, b] and that it is uniformly continuous over [a, b]. The
fact f is bounded is needed to know that the Darboux sums L(f, P )
and U (f, P ) make sense for any partition P . The fact f is uniformly
continuous will play the central role in our proof.
Proof. Let  > 0. Because f is uniformly continuous over [a, b], there
exists a > 0 such that

|x y| < = |f (x) f (y)| < for every x, y [a, b] .
ba
Let P = [x0 , x1 , , xn ] be any partition of [a, b] such that |P | < .
Because f is continuous, it takes on extreme values over each subin-
terval [xi1 , xi ] of P . Hence, for every i = 1, , n there exist points
xi and xi in [xi1 , xi ] such that mi = f (xi ) and mi = f (xi ). Because
|P | < it follows that |xi xi | < , whereby

mi mi = f (xi ) f (xi ) < .
ba
We thereby obtain
Xn
0 U (f, P ) L(f, P ) = (mi mi ) (xi xi1 )
i=1
n
 X 
(xi xi1 ) = (b a) =  .
ba i=1
ba
This shows that for every partition P of [a, b] one has
|P | < = 0 U (f, P ) L(f, P ) <  .
But  > 0 was arbitrary. It follows that f is Riemann integrable by the
Darboux Theorem (Theorem 1.3). 
Exercise. A function f : [a, b] R is said to be Holder continuous of
order (0, 1] if there exists a C R+ such that for every x, y [a, b]
one has
|f (x) f (y)| C |x y| .
Show that for every partition P of [a, b] one has
0 U (f, P ) L(f, P ) |P | C (b a) .
 15

2.3. Linearity and Order for Riemann Integrals. Linear combi-

nations of Riemann integrable functions are again Riemann integrable.
Riemann integrals respect linearity and order.

2.3.1. Linearity. One basic fact about Riemann integrals is that they
depend linearly on the integrand. This fact is not completely trivial
because we defined the Riemann integral through Darboux sums, which
do not depend linearly on the integrand.

Proposition 2.1. (Linearity) Let f : [a, b] R and g : [a, b] R

be Riemann integrable over [a, b]. Let R. Then f + g and f are
Riemann integrable over [a, b] with
Z b Z b Z b Z b Z b
(f + g) = f+ g, (f ) = f.
a a a a a

Proof. A key step towards establishing the additivity is to prove that

if P is any partition of [a, b] then



L(f, P )+L(g, P ) L (f +g), P U (f +g), P U (f, P )+U (g, P ) .

A key step towards establishing the scalar multiplicity is to prove that

if > 0 and P is any partition of [a, b] then

L(f, P ) = L(f, P ) , U (f, P ) = U (f, P ) .

The proof is left as an exercise. 

Remark. It follows immediately from the above proposition that every
linear combination of Riemann integrable functions is also Riemann
integrable, and that its integral is the same linear combination of the
respective integrals. More precisely, if fk : [a, b] R is Riemann
integrable over [a, b] for every k = 1, 2, , n then for every {k }nk=1
R one knows that
n
X
k f k is Riemann integrable over [a, b] ,
k=1

with
n
! n
Z b X X Z b
k f k = k fk .
a k=1 k=1 a
16

2.3.2. Nonnegativity. Another basic fact about definite integrals is that

they respect nonnegativity of the integrand.

Proposition 2.2. (Nonnegativity) Let f : [a, b] R be nonnegative

and Riemann integrable over [a, b]. Then
Z b
0 f.
a

Proof. Exercise. 

2.3.3. Order. The basic comparison property of definite integrals now

follows from Propositions 2.1 and 2.2.

Proposition 2.3. (Order) Let f : [a, b] R and g : [a, b] R be

Riemann integrable over [a, b]. Let f (x) g(x) for every x [a, b].
Then
Z b Z b
f g.
a a

Proof. Exercise. 

2.3.4. Bounds. The basic bounds on definite integrals now follows from
Proposition 2.3.

Proposition 2.4. (Bounds) Let f : [a, b] R be Riemann integrable

(and hence, bounded) over [a, b]. Suppose that Range(f ) [m, m].
Then
Z b
m (b a) f m (b a) .
a

Moreover,
Z b 
 

 f  M (b a) ,
a

where M = sup |f (x)| : x [a, b] .

Proof. Exercise. 
 17

2.4. Nonlinearity. More general combinations of Riemann integrable

functions are again Riemann integrable.
Theorem 2.3. (Continuous Compositions) Let f : [a, b] R be
Riemann integrable over [a, b]. Suppose that Range(f ) [m, m]. Let
G : [m, m] R be continuous. Then G(f ) : [a, b] R is Riemann
integrable over [a, b].
Proof. Because G : [m, m] R is continuous, it is bounded. Suppose
that Range(G) [m , m ] where m < m .
Let  > 0. Because G is uniformly continuous over [m, m], there
exists a > 0 such that for every y, z [m, m] one has

|y z| < = |G(y) G(z)| < .
2(b a)
Because f is Riemann integrable over [a, b] there exists a partition P
such that

0 U (f, P ) L(f, P ) < .
2(m m )
Let P = [x0 , x1 , , xn ]. For every i = 1, .n define mi , mi , mi , and
mi by
mi = inf{f (x) : x [xi1 , xi ]} ,
mi = sup{f (x) : x [xi1 , xi ]} ,
mi = inf{G(f (x)) : x [xi1 , xi ]} ,
mi = sup{G(f (x)) : x [xi1 , xi ]} .
We want to show that
n
X
0 U (G(f ), P ) L(G(f ), P ) = (mi mi )(xi xi1 ) <  .
i=1

The key step is to decompose the indices i = 1, , n into two sets:

I< = {i : mi mi < } , I = {i : mi mi } .
We analyze the above sum over each of these sets separately.
We first show that the sum over the good set I< is small because
each (mi mi ) is sufficiently small. The values of f over [xi1 , xi ] lie in
[mi , mi ]. Because G is continuous, the Extreme-Value Theorem implies
that G takes on its inf and sup over [mi , mi ], say at the points y i and
y i respectively. Because |y i y i | < for every i I< , one has
mi mi sup{G(y) : y [mi , mi ]} inf{G(y) : y [mi , mi ]}

= G(y i ) G(y i ) < ,
2(b a)
18

whereby the sum over I< satisfies

X  X 
(26) (mi mi )(xi xi1 ) < (xi xi1 ) .
iI<
2(b a) iI 2
<

We now show that the sum over the bad set I is small because
the partition P is refined enough to make the set I sufficiently small.
Indeed, because mi mi for every i I , we have
X X
(xi xi1 ) (mi mi )(xi xi1 )
iI iI


U (f, P ) L(f, P ) < ,
2(m m )
whereby the set I is small in the sense that
X 
(xi xi1 ) < .
iI
2(m m )

Hence, the sum over I satisfies

X X 
(27) (mi mi )(xi xi1 ) < (m m ) (xi xi1 ) < .
iI iI
2

Upon combining bounds (26) and (27), we obtain

0 U (G(f ), P ) L(G(f ), P )
X X
= (mi mi )(xi xi1 ) + (mi mi )(xi xi1 )
iI< iI
 
< + = .
2 2
Because  was arbitrary, G(f ) is Riemann integrable by characteriza-
tion (2) of the Riemann-Darboux Theorem. 
An important consequence of the Composition Theorem is that the
product of Riemann integrable functions is also Riemann integrable.
Proposition 2.5. (Product) Let f : [a, b] R and g : [a, b] R
be Riemann integrable over [a, b]. Then the product f g : [a, b] R is
Riemann integrable over [a, b].
Remark. Taken together the Linearity and Product Propositions show
that the class of Riemann integrable functions is an algebra.
Proof. The proof is based on the algebraic identity
f g = 14 (f + g)2 (f g)2 .


 19

By the Linearity Proposition the functions f +g and f g are Riemann

integrable over [a, b]. By Composition Theorem (applied to G(z) = z 2 )
the functions (f + g)2 and (f g)2 are Riemann integrable over [a, b].
Hence, by applying the Linearity Proposition to the above identity, one
sees that f g is Riemann integrable over [a, b]. 
Remark. We could just as well have built a proof of the Product
Lemma based on the identity
f g = 21 (f + g)2 f 2 g 2 ,

or the identity
1
f 2 + g 2 (f g)2 .


fg = 2

The Composition Theorem also implies that the absolute-value of a

Riemann integrable function is also Riemann integrable. When com-
bined with the Order, Bounds, and Product Propositions 2.3, 2.4, and
2.5, this leads to the following useful bound.
Proposition 2.6. (Absolute-Value) Let f : [a, b] R and g :
[a, b] R be Riemann integrable over [a, b]. Suppose that g is non-
negative. Then f g : [a, b] R and |f |g : [a, b] R are Riemann
integrable over [a, b] and satisfy
Z b  Z b Z b
 

 f g  |f |g M g,
a a a

where M = sup{|f (x)| : x [a, b]}.

Proof. Exercise. 

2.5. Restrictions and Interval Additivity. A property of the def-

inite integral that you learned when you first studied integration is
interval additivity. In its simplest form this property states that, pro-
vided all the integrals exist, for every a, b, c R such that a < b < c
one has
Z c Z b Z c
(28) f= f+ f.
a a b
In elementary calculus courses this formula is often stated without
much emphasis on implicit integrability assumptions. As we will see
below, Riemann integrals have this property. In that setting this for-
mula assumes that f is Riemann integrable over [a, c], and that the
restrictions of f to [a, b] and [b, c] are Riemann integrable over those
intervals. As the next lemma shows, these last two assumptions follow
from the first.
20

Lemma 2.1. (Restriction) Let f : [a, d] R be Riemann integrable

over [a, d]. Then for every [b, c] [a, d] with b < c the restriction of f
to [b, c] is Riemann integrable over [b, c].
Proof. Let [b, c] [a, d]. Let  > 0. Because f is Riemann integrable
over [a, d] by characterization (2) of the Riemann-Darboux Theorem
there exists a partition P of [a, d] such that
0 U (f, P ) L(f, P ) <  .
By the Refinement Lemma we may assume that b and c are partition
points of P , otherwise we can simply replace P by P [a, b, c, d].
Let P be the partition of [b, c] induced by P . Then
0 U (f, P ) L(f, P ) U (f, P ) L(f, P ) <  .
Hence, f is Riemann integrable over [b, c] by characterization (2) of the
Riemann-Darboux Theorem. 
Now return to the interval additivity formula (28). More interest-
ing from the viewpoint of building up the class of Riemann integrable
functions is the fact that if the restrictions of f to [a, b] and [b, c] are
Riemann integrable over those intervals then f is Riemann integrable
over [a, c]. More generally, we have the following.
Proposition 2.7. (Interval Additivity) Let P = [p0 , , pk ] be any
partition of [a, b]. Then f is Riemann integrable over [a, b] if and only
if the restriction of f to [pi1 , pi ] is Riemann integrable for every i =
1, , k. Moerover, in that case one has
Z b X k Z pi
(29) f= f.
a i=1 pi1

Proof. ( = ) This follows from the Restriction Lemma.

(=) Because f is Riemann integrable over [pi1 , pi ] for every i =
1, , k there exists a partition Pi of [pi1 , pi ] such that

0 U (f, Pi ) L(f, Pi ) < .
k
Let P be the refinement of P such that Pi is the induced partition of
[pi1 , pi ]. One then sees that
0 U (f, P ) L(f, P )
k k
X X 
U (f, Pi ) L(f, Pi )


= < = .
i=1 i=1
k
Hence, by characterization (2) of the Riemann-Darboux Theorem, f is
Riemann integrable over [a, b].
 21

One can use Riemann sums to establish (29). This part of the proof
is left as an exercise. 
The restriction a < b < c in the interval additivity formula (28) can
be dropped provided one adopts the following convention.
Definition 2.2. Let f : [a, b] R be Riemann integrable over [a, b].
Define Z a Z b
f = f.
b a

Exercise. Show that (28) holds for every a, b, and c R, provided

that we adopt Definition 2.2 and f is Riemann integrable over all the
intervals involved.

2.6. Extensions and Piecewise Integrability. Now we will use in-

terval additivity to build up the class of Riemann integrable functions.
This will require a lemma regarding extensions. To motivate the need
for this lemma, let us consider the function f : [1, 1] R defined by

x + 2 for x [1, 0) ,

f (x) = 1 for x = 0 ,

x for x (0, 1] .
It is easy to use the Riemann-Darboux Theorem to verify that this
function is Riemann integrable with
Z 1
f = 2,
1
yet this fact does not follow directly from other theorems we have
proved. For example, f restricted to either [1, 0] or [0, 1] is neither
monotonic nor continuous because of its behavior at x = 0. However,
our intuition tells us (correctly) that the value of f at 0 should not
effect whether or not it is Riemann integrable. The following lemma
shows this to be the case if the points in questions are the endpoints
of the interval of integration.
Lemma 2.2. (Extension) Let f : (a, b) R be bounded. Suppose
that for every [c, d] (a, b) the restriction of f to [c, d] is Riemann
integrable over [c, d]. Let f : [a, b] R be any extension of f to [a, b].
Then f is Riemann integrable over [a, b]. Moreover, if f1 and f2 are
two such extensions of f then
Z b Z b

f1 = f2 .
a a
22

Proof. Let  > 0. Let Range(f) [m, m]. Let > 0 such that
 ba
(m m) < and < .
3 2
Because the restriction of f to [a + , b ] is Riemann integrable, there
exists a partition P of [a + , b ] such that

0 U (f, P ) L(f, P ) < .
3
Let P be the extension of P to [a, b] obtained by adding a and b as
partition points. Then
0 U (f, P ) L(f, P )
= U (f, [a, a + ]) L(f, [a, a + ]) + U (f, P ) L(f, P )
   

+ U (f, [b , b]) L(f, [b , b])

 


(m m) + + (m m) <  .
3
Hence, the extension f is Riemann integrable over [a, b] by characteri-
zation (2) of the Riemann-Darboux Theorem.
Now let f1 and f2 be two extensions of f to [a, b]. Let {P n } n=1 be
any sequence of partitions of [a, b] such that |P n | 0 as n .
This sequence is Archimedean for both f1 and f2 by Theorem 1.4. Let
{Qn }
n=1 be any sequence of associated quadrature points such that
neither a nor b are quadrature points. Because f1 (x) = f2 (x) for every
x (a, b), we have R(f1 , P n , Qn ) = R(f2 , P n , Qn ) for every n Z+ .
Therefore the Archimedes-Riemann Theorem yields
Z b Z b
f1 = lim R(f1 , P , Q ) = lim R(f2 , P , Q ) =
n n n n
f2 .
a n n a


It is a consequence of the Extension Lemma and interval additivity
that two functions that differ at only a finite number of points are the
same when is comes to Riemann integrals.
Theorem 2.4. Let f : [a, b] R be Riemann integrable over [a, b].
Let g : [a, b] R such that g(x) = f (x) at all but a finite number of
points in [a, b]. Then g is Riemann integrable over [a, b] and
Z b Z b
g= f.
a a
 23

Proof. Exercise. 
Remark. The same cannot be said of two functions that differ at a
countable number of points. Indeed, consider the function
(
1 if x Q ,
g(x) =
0 if x / Q.
Its restriction to any closed bounded interval [a, b] is not Riemann
integrable, yet it differs from f = 0 at a countable number of points.
We can now show that all functions that are piecewise monotonic
over [a, b] are also Riemann intergrable over [a, b]. We first recall the
definition of piecewise monotonic function.
Definition 2.3. A function f : [a, b] R is said to be piecewise
monotonic if it is bounded and there exists a partition P = [x0 , , xn ]
of [a, b] such that f is monotonic over (xi1 , xi ) for every i = 1, , n.
Theorem 2.5. Let f : [a, b] R be piecewise monotonic. Then f is
Riemann integrable over [a, b].
Proof. This follows from Proposition 2.1, the Extension Lemma, and
the Interval Additivity Proposition. The details are left as an exercise.

We can also show that all functions that are piecewise continuous
over [a, b] are also intergrable over [a, b]. We first recall the definition
of piecewise continuous function.
Definition 2.4. A function f : [a, b] R is said to be piecewise
continuous if it is bounded and there exists a partition P = [x0 , , xn ]
of [a, b] such that f is continuous over (xi1 , xi ) for every i = 1, , n.
We remark the class of piecewise continuous functions includes some
fairly wild functions. For example, it contains f : [1, 1] R given by

1 + sin(1/x)
if x (0.1] ,
f (x) = 4 if x = 0 ,

1 + sin(1/x) if x [1, 0) .

Theorem 2.6. Let f : [a, b] R be piecewise continuous. Then f is

Riemann integrable over [a, b].
Proof. This follows from Proposition 2.2, the Extension Lemma, and
the Interval Additivity Proposition. The details are left as an exercise.

24

2.7. Lebesgue Theorem. In this section we state a beautiful theo-

rem of Lebesgue that characterizes those functions that are Riemann
integrable. In order to do this we need to introduce the following notion
of very small subsets of R.
Definition 2.5. A set A R is said to have measure zero if for every
 > 0 there exists a countable collection of open intervals {(a i , bi )}
i=1
such that

[ X
A (ai , bi ) , and (bi ai ) <  .
i=1 i=1

In other words, a set has measure zero if it can be covered by an

arbitrarily small countable collection of open intervals, where the size
of the countable collection of intervals is defined by the above sum.
Example. Every finite or countable subset of R has measure zero. In
particular, Q has measure zero. Indeed, consider a countable set A =
1
{xi } i i
i=1 R. Let  > 0. Let r < 3 and set (ai , bi ) = (xi r , xi + r )

for every i Z+ . The collection of open intervals {(ai , bi )}i=1 clearly
covers A. Moreover,

X X 2r
(bi ai ) = 2r i  = < .
i=1 i=1
1 r

The fact that measure zero is a reasonable concept of very small

is confirmed by the following facts.
Proposition 2.8. If {An }n=1 is a collection of subsets of R each of
which has measure zero then
[
A= An has measure zero .
n=1

If B R has measure zero and A B then A has measure zero.

Proof. Exercise. 
Of course, one has to show that many sets do not have measure zero.
Exercise. Show that every nonempty open interval (a, b) does not
have measure zero.
Remark. When the above exercise is combined with the last assertion
of Proposition 2.8, we see that a set does not have measure zero if it
contains a nonempty open interval. There are many such sets.
The following example shows that there are some very interesting
sets that have measure zero.
 25

Example. The Cantor set is an uncountable set that has measure

zero. The Cantor set is the subset C of the interval [0, 1] obtained by
sequentially removing middle thirds as follows. Define the sequence
of sets {Cn }
n=1 as follows

C1 = [0, 1] ( 31 , 23 ) = [0, 13 ] [ 23 , 1] ,
C2 = C1 ( 19 , 29 ) ( 97 , 98 )
= [0, 19 ] [ 29 , 31 ] [ 23 , 97 ] [ 89 , 1] ,
1 2 7 8
C3 = C2 ( 27 , 27 ) ( 27 , 27 ) ( 19 , 20 ) ( 27
27 27
25 26
, 27 )
1 2 1
= [0, 27 ] [ 27 , 9 ] [ 29 , 27
7 8 1
] [ 27 , 3]
[ 23 , 27
19 20 7
] [ 27 , 9 ] [ 98 , 25
27
] [ 26
27
, 1] ,
..
.
In general one has
[
Cn = Cn1 ( 2k1
3n
, 32kn ) for n > 3 .
2k<3n

One can show by induction that each Cn is the union of 2n closed inter-
vals each of which have length 1/3n . Each Cn is therefore sequentially
compact. Moreover, these sets are nested as
C1 C2 Cn Cn+1 .
The Cantor set C is then defined to be the intersection of these sets:
\
C= Cn .
n=1

Being the intersection of nested sequentially compact sets, this set is

nonempty by the Cantor Theorem. It is harder to show that C is
uncountable. We will not do so here. However, from the information
given above you should be able to show that C has measure zero.
Exercise. Show the Cantor set has measure zero.
We need one more definition.
Definition 2.6. Let S R. Let A(x) be any assertion about a point x.
Then we say A(x) for almost every x S or A almost everywhere
in S provided

x S : A(x) is false has measure zero .
Roughly speaking, a property holds almost everywhere if it fails to hold
on a set of measure zero.
26

Example. Let f : R R be given by

(
1 if x Q ,
f (x) =
0 if x
/ Q.
Then f = 0 almost everywhere.
We are now ready to state the Lebesgue Theorem.
Theorem 2.7. (Lebesgue) Let f : [a, b] R be bounded. Then f
is Riemann integrable over [a, b] if and only if it is continuous almost
everywhere in [a, b].
Proof. The proof is omitted. It is quite involved. One can be found
in Principles of Analysis by Walter Rudin. 
The Lebesgue Theorem allows us to sharpen our Nonnegativity and
Order Propositions for Riemann integrals (Propositions 2.2 and 2.3).
Proposition 2.9. (Positivity) Let f : [a, b] R be Riemann inte-
grable. Suppose that f 0 and that f > 0 almost everywhere over a
nonempty (c, d) [a, b]. Then
Z b
f > 0.
a

Proof. The key step is to show that there exists a p (c, d) such that
f (p) > 0 and f is continuous at p. Indeed, consider the sets
Y = {x (c, d) : f (x) = 0} ,
Z = {x (c, d) : f is not continous at x} .
The set Y has measure zero by hypothesis. The set Z has measure
zero by the Lebesgue Theorem. The set Y Z therefore has measure
zero by the first assertion of Proposition 2.8. But then Y Z cannot
contain (c, d) because otherwise the last assertion of Proposition 2.8
would imply (c, d) has measure zero, which it does not. Hence, the set
(c, d) Y Z is nonempty. The rest of the proof is left as an exercise.

Proposition 2.10. (Strict Order) Let f : [a, b] R and g : [a, b]
R be Riemann integrable. Suppose that f g and that f (x) < g(x)
almost everywhere over a nonempty (c, d) [a, b]. Then
Z b Z b
f< g.
a a

Proof. Exercise. 
 27

2.8. Power Rule. We have now built up a large class of Riemann

integrable functions. However, we have not computed many definite
integrals. In this section we will derive the so-called power rule for
definite integrals specifically, that for any p R and any [a, b] R+
one has
p+1
b ap+1
for p 6= 1 ,
Z b

p+1

(30) xp dx =
b
a
log for p = 1 .


a
Of course, you should be familiar with this rule from your previous
study of calculus. You should recall that it follows easily from the
Fundamental Theorem of Calculus. Here however we will derive it by
taking limits of Riemann sums.
We begin with the observation that for any p R the power func-
tion x 7 xp is both monotonic and continuous over R+ . It is there-
fore Riemann integrable over [a, b] either by Theorem 2.1 or by The-
orem 2.2. Moreover, a sequence {P n } n=1 of partitions of [a, b] will be
n
Archimedean whenever |P | 0 as n . The problem therefore re-
duces to finding such a sequence of partitions and a sequence {Qn } n=1
of associated quadrature sets for which one can show that
p+1
b ap+1
for p 6= 1 ,


p+ 1

lim R(xp , P n , Qn ) =
n b
log for p = 1 .


a
We will take two approaches to this problem.

2.8.1. Uniform Partitions. Whenever p 0 it is clear that the function

x 7 xp is Riemann integrable over [0, b]. If one uses the uniform
partitions over [0, b] given by
ib
P n = [x0 , x1 , , xn ] , , xi =
n
and the right-hand rule quadrature sets Qn = (x1 , , xn ) then
n  p
p n n b X ib bp+1
R(x , P , Q ) = = p+1 S p (n) ,
n i=1 n n

where
n
X
p
S (n) = ip .
i=1
28

One must therefore show that for every p 0 one has

b
br+1 p bp+1
Z
(31) xp dx = lim S (n) = .
0 n np+1 p+1

Once this is done then for every [a, b] (0, ) one has

b b a
bp+1 ap+1
Z Z Z
p p
x dx = x dx xp dx = ,
a 0 0 p+1

which agrees with (30).

In order to prove (31) one must establish the limit

1 1
(32) lim S p (n) = .
n np+1 p+1

The details of proving (32) are presented in the book for the cases
p = 0, 1, 2 with b = 1. Most calculus books prove this limit for cases no
higher than p = 3. They usually proceed by first establishing formulas
for S p (n) like

n(n + 1)
S 0 (n) = n , S 1 (n) = ,
2
n(n + 1)(2n + 1) n2 (n + 1)2
S 2 (n) = , S 3 (n) = .
6 4

The first of these formulas is trivial. The others are typically verified by
an induction argument on n. Given such an explicit formula for S p (n),
establishing (32) is easy. However, this approach does not give any
insight into how to obtain these formulas, which grow in complexity as
p increases.
Here we will take a different approach that allows us to prove (32)
for every p N. We will first find a relation that expresses S p (n) in
terms of all the S j (n) with j = 0, , p 1. Then, instead of using
this relation to generate complicated explicit formulas for S p (n), we
will use it to prove (32) via an induction argument on p.
Proof. Clearly S 0 (n) = n, so that limit (32) holds for p = 0. Now
assume that for some q 1 limit (32) holds for every p < q. By a
telescoping sum, the binomial formula, and the definition of S p (n), one
 29

obtains the identity

n
X
q+1
(i + 1)q+1 iq+1
 
(n + 1) 1=
i=1
n Xq
X (q + 1)!
= ip
i=1 p=0
p!(q p + 1)!
q
X (q + 1)!
= S p (n)
p=0
p!(q p + 1)!
q1
X (q + 1)!
= (q + 1) S q (n) + S p (n) .
p=0
p!(q p + 1)!

Upon solving for S q (n) and dividing by nq+1 , we obtain the relation
"
1 q 1 (n + 1)q+1 1
S (n) = q+1
n q+1 q+1 n q+1 n
(33) q1
#
X (q + 1)! 1
q+1
S p (n) .
p=0
p!(q p + 1)! n

Because we know
(n + 1)q+1 1
lim = 1 , lim = 0,
n nq+1 n nq+1

and because, by the induction hypothesis, we know

1
lim S p (n) = 0 for every p < q ,
n nq+1
we can pass to the n limit in relation (33). We thereby establish
that limit (32) holds for p = q. 
Remark. The place in our proof that required p to be a natrual
number was the point were we used of the binomial formula.
Exercise. Relation (33) can be recast as
p1
" #
1 X (p + 1)!
S p (n) = (n + 1)p+1 1 S j (n) .
p+1 j=0
j!(p j + 1)!

This can be used to generate explicit formulas for S p (n) for any p 1.
To get an idea of how complicated these explict formulas become, start
with the fact S 0 (n) = n and use the above relation to generate explicit
formulas for S 1 (n), S 2 (n), S 3 (n), and S 4 (n).
30

2.8.2. Nonuniform Partitions. The difficulty with the approach using

uniform partitions was that the resulting Riemann sums could not be
evaluated easily. Fermat saw that this difficulty can be elegantly over-
come by using the nonuniform partitions over [a, b] (0, ) given by
  ni
n b
P = [x0 , x1 , , xn ] , xi = a .
a
By introducing
  n1
b
rn = ,
a
the partition points can be expressed as xi = a rni . If one uses the
left-hand rule quadrature sets Qn = (x0 , , xn1 ) then
n1 n1
p n n
X
p X
arni arni+1 arni p+1
rni(p+1) .


R(x , P , Q ) = =a (rn 1)
i=0 i=0
(p+1)
Notice that the last sum is a finite geometric series with ratio rn .
It can therefore be evaluated as

n(p+1)
n1 rn
1
X
i(p+1) for p 6= 1 ,
rn = rn (p+1)
1

n
i=0 for p = 1 .
When p 6= 1 the Riemann sums are thereby evaluated as
n(p+1)
p n n p+1 rn 1 rn 1
= bp+1 ap+1


R(x , P , Q ) = a (rn 1) (p+1) (p+1)
.
rn 1 rn 1
n
Here we have used the fact that rn = b/a to see that
ap+1 rnn(p+1) 1 = bp+1 ap+1 .

Given the above explicit formula for R(xp , P n , Qn ), one only needs
to show that
rn 1 1
(34) lim (p+1) = .
n r
n 1 p+1
Then
rn 1 bp+1 ap+1
lim R(xp , P n , Qn ) = bp+1 ap+1 lim (p+1)


= ,
n n r
n 1 p+1
which yields (30) for the case p 6= 1. The case p = 1 is left as an
exercise. 
Exercise. Prove (34).
Exercise. Prove (30) for the case p = 1.
 31

Exercise. By taking limits of Riemann sums, show for every positive

a and b that Z b
ab 1
ax dx = .
0 log(a)
Hint: Use uniform partitions.
Remark. Fermat discovered his beautiful derivation of the power rule
(30) before Newton and Leibniz developed the fundamental theorems
of calculus. In other words, there was no easy way to do the problem
when Fermat discovered the power rule. It took a genius like Fermat to
solve a problem that the easy way makes routine. In fact, Fermats
power rule provided an essential clue that led to the development of
the easy way by Newton and Leibniz.
32

3. Relating Integration with Differentiation

Both integration and differentiation predate Newton and Leibniz.
The definite integral has roots that go back at least as far as Eudoxos
and Archimedes, some two thousand years earlier. The derivative goes
back at least as far as Fermat. The fact they are connected in some
instances was understood by Fermat, who worked out special cases, and
by Barrow, who extended Fermats work. Barrow was one of Newtons
teachers and his work was known to Leibniz. The big breakthrough
of Newton and Leibniz was the understanding that this connection is
general. This realization made the job of computing definite intergrals
much easier, which enabled major advances in science, engineering, and
mathematics. This connection takes form in what we now call the first
and second fundamental theorems of calculus.

3.1. First Fundamental Theorem of Calculus. The business of

evaluating integrals by taking limits of Riemann sums is usually either
difficult or impossible. However, as you have known since you first
studied integration, for many integrands there is a much easier way.
We begin with a definition.
Definition 3.1. Let f : [a, b] R. A function F : [a, b] R is said
to be a primitive or antiderivative of f over [a, b] provided
the function F is continuous over [a, b],
there exists a partition [p0 , , pn ] of [a, b] such that for each i =
1, , n the function F restricted to (pi1 , pi ) is differentiable
and satisfies
(35) F 0 (x) = f (x) for every x (pi1 , pi ) .
Remark. Definition 3.1 states that F is continuous and piecewise
differentiable over [a, b]. There are at most a finite number of points in
[a, b] at which either F 0 is not defined or F 0 is defined but F 0 6= f . For
example, consider the function

1
for x (0, 1] ,
f (x) = 0 for x = 0 ,

1 for x [1, 0) .

The function F (x) = |x| is a primitive of f over [1, 1], yet is not
differentiable at x = 0. Similarly, consider the function
(
1 for x [1, 0) (0, 1] ,
f (x) =
0 for x = 0 ,
 33

The differentiable function F (x) = x is a primitive of this f over [1, 1],

yet F 0 (0) 6= f (0). In both examples there is clearly no function F that
is differentiable over [1, 1] such that F 0 = f because f does not have
the intermediate-value property.
Exercise. Let f : [a, b] R. Let F : [a, b] R be a primitive of f
over [a, b]. Let g : [a, b] R such that g(x) = f (x) at all but a finite
number of points of [a, b]. Show that F is also a primitive of g over
[a, b].
It is clear that if F is a primitive of a function f over [a, b] then so
is F + c for any constant c. It is a basic fact that a primitive is unique
up to this arbitrary additive constant.
Lemma 3.1. Let f : [a, b] R. Let F1 : [a, b] R and F2 : [a, b] R
be primitives of f over [a, b]. Then there exists a constant c such that
F2 (x) = F1 (x) + c for every x [a, b].

Proof. Let G = F2 F1 . We must show that this function is a constant

over [a, b]. Let P 1 and P 2 be the partitions associated with F1 and F2
respectively. Set P = P 1 P 2 . Express P in terms of its partition
points as P = [p0 , , pn ]. For each i = 1, , n the restriction of G
to [pi1 , pi ] is continuous over [pi1 , pi ] and differentiable over (pi1 , pi )
with
G0 (x) = F20 (x) F10 (x) = f (x) f (x) = 0 for every x (pi1 , pi ) .
It follows from the Lagrange Mean-Value Theorem that restriction of
G to each [pi1 , pi ] is constant ci over that subinterval. But for each
i = 1, , n1 the point pi is in the subintervals [pi1 , pi ] and [pi , pi+1 ],
whereby ci = G(pi ) = ci+1 . Hence, G must be a constant over [a, b]. 

Corollary 3.1. Let f : [a, b] R have a primitive over [a, b]. Let
xo [a, b] and yo R. Then f has a unique primitive F such that
F (xo ) = yo .

Proof. Exercise. 
Exercise. Let f : [0, 3] R be defined by

x
for 0 x < 1 ,
f (x) = x for 1 x < 2 ,

1 for 2 x 3 .
Find F , the primitive of f over [0, 3] specified by F (0) = 1.
We are now ready to for the big theorem.
34

Theorem 3.1. (First Fundamental Theorem of Calculus) Let

f : [a, b] R be Riemann integrable and have a primitive F over [a, b].
Then Z b
f = F (b) F (a) .
a

Remark. This theorem essentially reduces the problem of evaluating

definite integrals to that of finding an explicit primitive of f . While
such an explicit primitive cannot always be found, it can be found for
a wide class of elementary integrands f .
Proof. We must show that for every partition P of [a, b] one has
(36) L(f, P ) F (b) F (a) U (f, P ) .
Let P be an arbitrary partition of [a, b] and let P denote the re-
finement P [p0 , , pn ]. Express P in terms of its partition points
as P = [x0 , , xn ]. Then for every i = 1, , n one knows that
F : [xi1 , xi ] R is continuous, and that F : (xi1 , xi ) R is dif-
ferentiable. Then by the Lagrange Mean-Value Theorem there exists
qi (xi1 , xi ) such that
F (xi ) F (xi1 ) = F 0 (qi ) (xi xi1 ) = f (qi ) (xi xi1 ) .
Because mi f (qi ) mi , we see from the above that
mi (xi xi1 ) F (xi ) F (xi1 ) mi (xi xi1 ) .
Upon summing these inequalities we obtain
n
X
L(f, P ) F (xi ) F (xi1 ) U (f, P ) .


i=1

Because the above sum telescopes, we see that

n
X

F (xi ) F (xi1 ) = F (b) F (a) .
i=1

The Refinement Lemma therefore yields

L(f, P ) L(f, P ) F (b) F (a) U (f, P ) U (f, P ) ,
from which (36) follows. 
Remark. Notice that the First Fundamental Theorem of Calculus
does not require f to be continuous, or even piecewise continuous. It
only requires f to be Riemann integrable and to have a primitive. Also
notice how Definition 3.1 of primitives allows the use of the Lagrange
Mean-Value Theorem in the above proof.
 35

The following is an immediate corollary of the First Fundamental

Theorem of Calculus.
Corollary 3.2. Let F : [a, b] R be continuous over [a, b] and differ-
entiable over (a, b). Suppose F 0 : (a, b) R is bounded over (a, b) and
Riemann integrable over every [c, d] (a, b). Let f be any extension of
F 0 to [a, b]. Then f is Riemann integrable over [a, b] and
Z b
f = F (b) F (a) .
a

Example. Let F be defined over [1, 1] by

(
x cos(log(1/|x|)) if x 6= 0 ,
F (x) =
0 if x = 0 .
Then F is continuous over [1, 1] and continuously differentiable over
[1, 0) (0, 1] with
F 0 (x) = cos(log(1/|x|)) + sin(log(1/|x|)) .
As this function is bounded, we have
Z 1
 
cos(log(1/|x|)) + sin(log(1/|x|)) dx = F (1) F (1) = 2 .
1

Here the integrand can be assigned any value at x = 0.

3.2. Second Fundamental Theorem of Calculus. It is natural to

ask if every Riemann integrable function has a primitive. It is clear
from the First Fundamental Theorem that if f is Riemann integrable
over [a, b] and has a primitive F that one must have
Z x
F (x) = F (a) + f.
a

So given a function f that is Riemann integrable over [a, b], we can

define F by the above formula. One then checks if F 0 (x) = f (x) except
at a finite number of points. In general this will not be the case. For
example, if f : [0, 1] R is the Riemann function given by
(
1
q
if x Q with x = pq in lowest terms .
f (x) =
0 otherwise .
This function is continuous at all the irrationals, and so is Riemann
integrable by the Lebesgue Theorem. Moreover, one can show that for
36

every x [0, 1] one has

Z x
F (x) = f = 0.
0
Hence, F is differentiable but F 0 (x) 6= f (x) at every rational. Therefore
F is not a primitive of f . Therfore f has no primitives.
The Second Fundamental Theorem of Calculus shows that the above
construction does yield a primitive for a large classes of functions.
Theorem 3.2. (Second Fundamental Theorem of Calculus) Let
f : [a, b] R be Riemann integrable. Define F : [a, b] R by
Z x
F (x) = f for every x [a, b] .
a
Then F (a) = 0, F is Lipschitz continuous over [a, b], and if f is con-
tinuous at c [a, b] then F is differentiable at c with F 0 (c) = f (c).
In particular, if f is continuous over [a, b] then F is continuously
differentiable over [a, b] with F 0 = f . If f is piecewise continuous
over [a, b] then F is piecewise continuously differentiable over [a, b] with
F 0 = f at all but a finite number of points in [a, b].
Proof. The fact that F (a) = 0 is obvious. Next, we show that F is
Lipschitz continuous over [a, b]. Let M = sup{|f (x)| : x [a, b]}. For
every x, y [a, b] one has
Z y Z x  Z x 
   
|F (y) F (x)| =  f (t) dt f (t) dt = 
  f (t) dt
a a y
Z y  Z y 
   
 |f (t)| dt M  dt = M |y x| .
x x
This shows that F is Lipschitz continuous over [a, b].
Now let f be continuous at c [a, b]. Let  > 0. Because f is
continuous at c there exists a > 0 such that for every z [a, b] one
has  
|z c| < = f (z) f (c) <  .
Because f (c) is a constant, for every x [a, b] such that x 6= c one has
Z x
1
f (c) = f (c) dz .
xc c
It follows that
Z x Z x
F (x) F (c) 1 1
f (c) = f (z) dz f (c) dz
xc xc c xc c
Z x
1

= f (z) f (c) dz .
xc c
 37

Therefore for every x [a, b] one has

0 < |x c| < =
   Z x 
 F (x) F (c)   1

 f (c) =  f (z) f (c) dz 
 xc xc c
1  x 
Z 
 
f (z) f (c) dz 
|x c|  c
  x 
Z 

< dz  = |x c| =  .
|x c| c
 |x c|
But this is the - characterization of
F (x) F (c)
lim = f (c) .
xc xc
Hence, F is differentiable at c with F 0 (c) = f (c).
The remainder of the proof is left as an exercise. 
Remark. Roughly speaking, the First and Second Fundamental The-
orems of Calculus respectively state that
Z x Z x
0 d
F (x) = F (a) + F (t) dt , f (x) = f (t) dt .
a dx a
In words, the first states that integration undoes differentiation (up
to a constant), while the second states that differentiation undoes in-
tegration. In other words, integration and differentiation are (nearly)
inverses of each other. This is the realization that Newton and Leibniz
had.
Remark. Newton and Leibniz were influenced by Barrow. He had
proved the Second Fundamental Theorem for the special case where f
was continuous and monotonic. This generalized Fermats observation
that the Second Fundamental Theorem holds for the power functions
xp , which are continuous and monotonic over x > 0. Of course, neither
Barrows statement nor his proof of this theorem were given in the
notation we use today. Rather, they were given in a highly geometric
setting that was commonly used at the time. This made it harder to
see that his result could be generalized further. You can get an idea of
what he did by assuming that f is nondecreasing and continuous over
[a, b] and drawing the picture that goes with the inequality
F (y) F (x)
a x < y b = f (x) f (y) ,
yx
Rx
where F (x) = a f . By letting y x while using the continuity of f ,
one obtains F 0 (x) = f (x).
38

3.3. Integration by Parts. An important consequence of the First

Fundamental Theorem of Calculus and the Product Rule for derivatives
is the following lemma regarding integration by parts.
Lemma 3.2. (Integration by Parts) Let f : [a, b] R and g :
[a, b] R be Riemann integrable and have primitives F and G respec-
tively over [a, b]. Then F g and Gf are Riemann integrable over [a, b]
and
Z b Z b
(37) F g = F (b)G(b) F (a)G(a) Gf .
a a

Proof. The functions F and G are Riemann integrable over [a, b]

because they are continuous. The functions F g and Gf are therefore
Riemann integrable over [a, b] by the Product Lemma.
The function F G is continuous over [a, b]. Let P and Q be the
partitions of [a, b] associated with F and G respectively. Let R = P Q.
Express R in terms of its partition points as R = [r0 , , rn ]. Then for
every i = 1, , n the function F G is differentiable over (ri1 , ri ) with
(by the Product Rule)
(F G)0 (x) = F (x)G0 (x) + G(x)F 0 (x)
= F (x)g(x) + G(x)f (x)
= (F g + Gf )(x) for every x (ri1 , ri ) .
Therefore F G is a primitive of F g + Gf over [a, b]. Equation (37)
then follows by the First Fundamental Theorem of Calculus and the
Additivity Lemma. 
In the case where f and g are continuous over [a, b] then the Second
Fundamental Theorem of Calculus implies that f and g have primitives
F and G that are piecewise continuously differentiable over [a, b]. In
that case integration by parts reduces to the following.
Corollary 3.3. Let F : [a, b] R and G : [a, b] R be continuously
differentiable over [a, b]. Then
Z b Z b
0
F G = F (b)G(b) F (a)G(a) GF 0 .
a a

3.4. Substitution. An important consequence of the First Fundamen-

tal Theorem of Calculus and the Chain Rule for derivatives is the
following lemma regarding changing the variable of integration in a
definite integral by monotonic substitution y = G(x).
 39

Proposition 3.1. (Monotonic Substitution) Let g : [a, b] R be

Riemann integrable and have a primitive G that is increasing over [a, b].
Let f : [G(a), G(b)] R be Riemann integrable and have a primitive
F over [G(a), G(b)] such that f (G)g is Riemann integrable over [a, b].
Then one has the change of variable formula
Z G(b) Z b
(38) f= f (G)g .
G(a) a

Remark. If we show the variables of integration explicitly then the

change of variable formula (38) takes the form
Z G(b) Z b
f (y) dy = f (G(x))g(x) dx .
G(a) a

Remark. The assumption that G is increasing over [a, b] could have

equivalently been stated as g is positive almost everywhere over [a, b].
Because G is a primitive, it is continuous as well as increasing. Its
range is therefore the interval [G(a), G(b)], the interval overwhich f
and F are assumed to be defined. This insures the compositions f (G)
and F (G) are defined over [a, b].
Proof. Let P = [p0 , , pl ] be the partition of [G(a), G(b)] associated
with the primitive F . Let Q = [q0 , , qm ] be the partition of [a, b]
associated with the primitive G. Because G : [a, b] [G(a), G(b)]
is increasing, G1 (P ) = [G1 (p0 ), , G1 (pl )] is a partition of [a, b].
Consider the partition R = Q G1 (P ) of [a, b]. Express R in terms
of its partition points as R = [r0 , , rn ].
The function F (G) : [a, b] R is continuous over [a, b]. Then for
every i = 1, , n the function F (G) is differentiable over (ri1 , ri )
with (by the Chain Rule)
F (G)0 (x) = F 0 (G(x)) G0 (x) = f (G(x)) g(x) for every x (ri1 , ri ) .
Therefore F (G)G is a primitive of f (G)g over [a, b]. Because f (G)g
is Riemann integrable over [a, b], the First Fundamental Theorem of
Calculus implies
Z b
f (G)g = F (G)(b) F (G)(a) = F (G(b)) F (G(a)) .
a
On the other hand, because f is Riemann integrable and F is a primi-
tive of f over [G(a), G(b)], the First Fundamental Theorem of Calculus
also implies
Z G(b)
f = F (G(b)) F (G(a)) .
G(a)
40

The change of variable formula (38) immediately follows from the last
two equations. 
Remark. The assumption that G is increasing over [a, b] could have
been replaced by the assumption that G is decreasing over [a, b]. In
that case the interval [G(b), G(a)] replaces the interval [G(a), G(b)] in
the hypotheses regarding f and F , but the change of variable formula
(38) remains unchanged.
Exercise. The assumption that G is increasing over [a, b] in Proposi-
tion 3.1 can be weakened to the assumption that G is nondecreasing
over [a, b]. Prove this slightly strengthend lemma. The proof can be
very similar to the one given above, however you will have to work a
bit harder to show that F (G) is a primitive of f (G)g over [a, b].
It is natural to ask whether one needs a hypothesis like G is mono-
tonic over [a, b] in order to establish the change of variable formula
(38). Indeed, one does not. However, without it one must take care
to insure the compositions f (G) and F (G) are defined over [a, b], to
insure that F (G) is a primitive of f (G)g over [a, b], and to insure that
f (G)g is Riemann integrable over [a, b]. Here we do this by assuming
that f is continuous over an interval containing Range(G).
Proposition 3.2. (Nonmonotonic Substitution) Let g : [a, b] R
be Riemann integrable and have a primitive G over [a, b]. Suppose that
Range(G) [m, m] and let f : [m, m] R be continuous over [m, m].
Then the change of variable formula (38) holds.

Proof. By the Second Fundamental Theorem of Calculus f has a con-

tinuously differentiable primitive F over [m, m]. It is then easy to show
that F (G) is a primitive of f (G)g over [a, b]. Because f (G) is contin-
uous (hence, Riemann integrable) while g is Riemann integrable over
[a, b], it follows from the Product Lemma that f (G)g is Riemann inte-
grable over [a, b]. The rest of the proof proceeds as that of Proposition
3.1, except here the partitions P , Q, and R are trivial. 

3.5. Integral Mean-Value Theorem. We will now give a useful the-

orem that a first glance does not seem to have a connection either with
the Fundamental Theorems of Calculus or with a Mean-Value Theorem
for differentiable functions. However, as will be explained later, there
is a connection.
Theorem 3.3. (Integral Mean-Value) Let f : [a, b] R be con-
tinuous. Let g : [a, b] R be Riemann integrable and positive almost
 41

everywhere over [a, b]. Then there exists a point p (a, b) such that
Z b Z b
(39) f g = f (p) g
a a

Proof. Because f is continuous over [a, b], the Extreme-Value Theorem

there exists points x and x [a, b] such that
 
f (x) = inf f (x) : x [a, b] , f (x) = sup f (x) : x [a, b] .
Then
f (x) f (x) f (x) for every x [a, b] ,
which, because g is nonnegative, implies that
Z b Z b Z b
f (x) g f g f (x) g.
a a a
If f (x) = f (x) then f is constant and (39) holds for every p (a, b).
So suppose f (x) < f (x).
Because f (x) < f (x) and because f is continuous there exists [c, d]
[a, b] such that x [c, d] and that f (x) < 21 (f (x) + f (x)). Then
f (x) f (x) > 12 f (x) f (x) > 0 for every x (c, d) .

Because (f (x) f )g 0, and because (f (x) f (x))g(x) > 0 almost

everywhere over the nonempty interval (c, d), the Positivity Proposition
implies
Z b Z b Z b


0< f (x) f g = f (x) g fg .
a a a
In a similar manner we can argue that
Z b Z b
0< f g f (x) g.
a a
Because g is positive almost everywhere over [a, b], the Positivity Propo-
Rb
sition also implies that a g > 0. Therefore, we see that
Rb
fg
f (x) < Ra b < f (x) .
a
g
Because f is continuous, the Intermediate-Value Theorem implies there
exists a p between x and x such that (39) holds. 
Remark. The connection of this theorem to both the First and Sec-
ond Fundamental Theorem of Calculus and to the Cauchy Mean-Value
Theorem for differentiable functions is seen when both f and g are
continuous. Then by the Second Fundamental Theorem of Calculus f g
and g have continuously differentiable primitives F and G. The Cauchy
42

Mean-Value Theorem applied to F and G then yields a p (a, b) such

that
F 0 (p)


F (b) F (a) = 0 G(b) G(a) = f (p) G(b) G(a) .
G (p)
By the First Fundamental Theorem of Calculus we therefore have
Z b Z b


f g = F (b) F (a) = f (p) G(b) G(a) = f (p) g.
a a

In other words, when both f and g are continuous the Integral Mean-
Value Theorem is just the Cauchy Mean-Value Theorem for differen-
tiable functions applied to primitives of f g and g.

3.6. Cauchy Riemainder Theorem. Recall that if f is n-times dif-

ferentiable over an interval (a, b) and c (a, b) then the nth Taylor
polynomial approximation of f at c is given by
n
X
(k) (x c)k
(40) Tcn f (x) = f (c) .
k=0
k!

Recall too that if f is (n + 1)-times differentiable over the interval (a, b)

then the Lagrange Remainder Theorem states that for every x (a, b)
there exists a point p between c and x such that
(x c)n+1
(41) f (x) = Tcn f (x) + f (n+1) (p) .
(n + 1)!
Our proof of the Lagrange Remainder Theorem was based on a direct
application of the Lagrange Mean-Value Theorem.
Here we give an alternative representation of the remainder due to
Cauchy. Its proof is based on a direct application of the First Fun-
demental Theorem of Calculus, the proof of which also rests on the
Lagrange Mean-Value Theorem. We will see that the resulting repre-
sentation contains more information than that of Lagrange.
Theorem 3.4. (Cauchy Remainder) Let f : (a, b) R be (n + 1)-
times differentiable over (a, b) and let f (n+1) be Riemann integrable
over every closed subinterval of (a, b). Let c (a, b). Then for every
x (a, b) one has
Z x
(x t)n
(42) n
f (x) = Tc f (x) + f (n+1) (t) dt .
c n!
 43

Proof. Let x (a, b). Then define F : (a, b) R by

n
X (x t)k
(43) F (t) = Ttn f (x) = f (t)+ f (k) (t) for every t (a, b) .
k=1
k!
Clearly F is differentiable over (a, b) with (notice the telescoping sum)
n 
(x t)k (x t)k1
X 
0 0 (k+1) (k)
F (t) = f (t) + f (t) f (t)
k=1
k! (k 1)!
(x t)n
= f (n+1) (t) .
n!
Because c and x are in (a, b) and because f (n+1) (and hence F 0 ) is
Riemann integrable over every closed subinterval of (a, b), the First
Fundamental Theoren of Calculus yields
Z x Z x
0 (x t)n
(44) F (x) F (c) = F (t) dt = f (n+1) (t) dt .
c c n!
However, it is clear from definition (43) of F (t) that
F (x) = f (x) , while F (c) = Tcn f (x) .
Formula (42) therefore follows from (44). 
Remark. The Lagrange Remainder Theorem can be derived from
Cauchys if one assumes that f (n+1) is continuous over (a, b). In that
case the Integral Mean-Value Theorem implies that for each x (a, b)
there exists a point p between c and x such that
Z x Z x
(n+1) (x t)n (n+1) (x t)n
f (t) dt = f (p) dt .
c n! c n!
A direct calculation then shows that
Z x
(x t)n (x c)n+1
dt = ,
c n! (n + 1)!
whereby
Z x
(x t)n (x c)n+1
(45) f (n+1) (t) dt = f (n+1) (p) .
c n! (n + 1)!
When this is placed into Cauchys formula (42) one obtains Lagranges
formula (41).
Remark. One cannot derive the Cauchy Remainder Theorem from
that of Lagrange. This is because the Lagrange Remainder Theorem
only tells you that the point p appearing in (41) lies between c and x
while the Cauchy theorem provides you with the explicit formula (42)
for the remainder.
44

Remark. The only way to bound the Taylor remainder using the
Lagrange form (41) is to use uniform bounds on f (n+1) (p) over all p
that lie between c and x. While this approach is sufficient for some
tasks (like showing that the formal Taylor series of ex , cos(x), and
sin(x) converge to those functions for every x R), it fails for other
tasks. However, if you are able to obtain suitable pointwise bounds on
its intrgrand then the Cauchy form (42) can sometimes yield bounds on
the Taylor remainder that are sufficient for those tasks. This remark
is illustrated by the following example.
Example. Let f (x) = log(1 + x) for every x > 1. Then
(k 1)!
f (k) (x) = (1)k1 for every x > 1 and k Z+ .
(1 + x)k
The formal Taylor aeries of f about 0 is therefore

X (1)k1
xk .
k=1
k
The Absolute Ratio Test shows that this series converges absolutely for
|x| < 1 and diverges for |x| > 1. For x = 1 the series is the negative of
the harmonic series, and therefore diverges. For x = 1 the Alternating
Series Test shows the series converges. We therefore conclude that the
series converges if and only if x (1, 1]. These arguments do not
show however that the sum of the series is f (x). This requires showing
that for every x (1, 1] the Taylor remainder f (x) T0n f (x) vanishes
as n .
First let us approach this problem using the Lagrange form of the
remainder (41): there exists a p between 0 and x such that
n+1
xn+1 (1)n

n (n+1) x
f (x) T0 f (x) = f (p) = .
(n + 1)! n+1 1+p
If x (0, 1] then p (0, x) and we obtain the bound
f (x) T0n f (x) < 1 xn+1 .
 
n+1
This bound clearly vanishes as n for every x (0, 1]. On the
other hand, if x (1, 0) then p (|x|, 0) and we obtain the bound
 n+1
f (x) T n f (x) < 1 |x|
 
0 .
n + 1 1 |x|
This bound will only vanish as n for x [ 21 , 0). This approach
leaves the question open for x (1, 21 ).
 45

Now let us approach this problem using the Cauchy form of the
remainder (42):
Z x
(x t)n
n
f (x) T0 f (x) = f (n+1) (t) dt
0 n!
Z x Z x n
n (x t)n tx dt
= (1) n+1
dt = .
0 (1 + t) 0 1+t 1+t
Let us only consider the case x (1, 0). Consider the substitution
tx 1+x 1+x 1+x
s= =1 , t+1= , dt = ds .
1+t 1+t 1s (1 s)2
Notice that s decreases from x (= |x|) to 0 as t goes from 0 to x.
Hence, because
1 1 1
< = for every s (0, |x|) ,
1s 1 |x| 1+x
we obtain the bound
Z |x| n
 
f (x) T0n f (x) = s
ds
0 1s
Z |x|
1 1 |x|n+1
< sn ds = .
1+x 0 1+x n+1
This bound clearly vanishes as n for every x (1, 0).
Collecting all of our results, we have shown that

X (1)k1 k
log(1 + x) = x for every x (1, 1] ,
k=1
k
and that the series diverges for all other values of x.
Exercise. When q N the binomial expansion yields
n q
q
X q! k
X q(q 1) (q k + 1) k
(1 + x) = x =1+ x .
k=0
k!(q k)! k=1
k!
Now let q R N. Let f (x) = (1 + x)q for every x > 1. Then
f (k) (x) = q(q1) (qk+1)(1+x)qk for every x > 1 and k Z+ .
The formal Taylor series of f about 0 is therefore

X q(q 1) (q k + 1) k
1+ x .
k=1
k!
Show that this series converges absolutely to (1 + x)q when |x| < 1
and diverges when |x| > 1. (This formula is Newtons extension of the
binomial expansion to powers q that are real.)