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CHAPTER 7
The Riemann Integral
7.1. Riemann Integral
Definition. A partition of an interval I = [a, b] is a collection
P = {I1, I2, . . . , In}of nonoverlapping closed intervals whose union is [a, b]. We ordinarily denotethe intervals by Ii = [xi�1, xi] where
a = x0 < x1 < x2 < · · · < xi�1 < xi < · · · < xn�1 < xn = b.
The points xi (i = 0, . . . , n) are called the partition points of P .
If a point ti has been chosen from each interval Ii, then the points ti are calledtags and the set of ordered pairs
•P =
�(I1, t1), (I2, t2), . . . , (In, tn)
=n�
[xi�1, xi], ti�on
i=1
is called a tagged partition of I .
The tags may be chosen arbitrarily, and a partition point may be a tag for twoconsecutive subintervals.
116
7.1. RIEMANN INTEGRAL 117
The norm (or mesh) of P is
kPk = max�x1 � x0, x2 � x1, . . . , xn � xn�1
.
For a tagged partition•P , the Riemann sum of f : [a, b] ! R corresponding to
•P is
S(f ;•P ) =
nXi=1
f(ti)(xi � xi�1).
Definition (7.1.1). A function f : [a, b] ! R is Riemann integrable on[a, b] if
9 L 2 R 3�� 8 ✏ > 0 9 �✏ > 0 3��if
•P is any tagged partition of [a, b] with k
•Pk < �✏, then
|S(f ;•P )� L| < ✏.
The set of all Riemann functions on [a, b] is denoted R[a, b].
Note.
L =
Z b
af =
Z b
af(x) dx =
Z b
af(t) dt, etc.
118 7. THE RIEMANN INTEGRAL
Theorem (7.1.2). If f 2 R[a, b], then the value of the integral is uniquelydetermined.
Proof. Assume L0 and L00 both satisfy the definition and let ✏ > 0. Then9(1) �0✏/2 3�� if
•P1 is any tagged partition with k
•P1k < �0✏/2, then
|S(f ;•P1)� L0| <
✏
2.
(2) �00✏/2 3�� if•P2 is any tagged partition with k
•P2k < �00✏/2, then
|S(f ;•P2)� L00| <
✏
2.
Let �✏ = min{�0✏/2, �00✏/2} and let•P be a tagged partition with k
•Pk < �✏. Then
|S(f ;•P )� L0| <
✏
2and |S(f ;
•P )� L00| <
✏
2=)
|L0 � L00| = |L0 � S(f ;•P ) + S(f ;
•P )� L00|
|L0 � S(f ;•P )| + |S(f ;
•P )� L00| <
✏
2+
✏
2= ✏.
Since ✏ is arbitrary, L0 = L00. ⇤
7.1. RIEMANN INTEGRAL 119
Example.
(1) f(x) = k 2 R[a, b] 8 k 2 R.
Proof. Let•P =
n�[xi�1, xi], ti
�on
i=1be any tagged partition of [a, b]. Then
S(f ;•P ) =
nXi=1
f(ti)(xi � xi�1)
=nX
i=1
k(xi � xi�1)
= knX
i=1
(xi � xi�1)
= k(b� a).
Thus, 8 ✏ > 0, choose �✏ = 1. Then k•Pk < �✏ =)
|S(f ;•P )� k(b� a)| = 0 < ✏.
Since ✏ is arbitrary, f 2 R[a, b] and
Z b
af = k(b� a). ⇤
120 7. THE RIEMANN INTEGRAL
(2) g(x) =
(4, x 2 [1, 4)
2, x 2 [4, 6].
From the area notion of integral, we guess
Z 6
1g = 16.
Let ✏ > 0 be given. Let•P be any tagged partition of [1, 6] with k
•Pk < � (to
be determined, but as small as needed).⇥Goal: to insure |S(g;
•P )� 16| < ✏.
⇤Let
•P1 be the subset of
•P with tags in [1, 4) and
•P2 be the subset of
•P with
tags in [4, 6]. Then
(#) S(g;•P ) = S(g;
•P1) + S(g;
•P2.)
Since k•Pk < �, if u 2 [1, 4� �] and u 2 [xi�1, xi], then
xi�1 4� � =) xi < xi�1 + � 4 =) tag ti 2 [1, 4) =)
[1, 4� �] ✓[n�
[xi�1, xi], ti�
: ti 2 [1, 4)o✓ [1, 4 + �].
Since g(ti) = 4 for these tags
(⇤) 4(3� �) S(g;•P1) 4(3 + �).
7.1. RIEMANN INTEGRAL 121
Similarly, if v 2 [4 + �, 6] and v 2 [xi�1, xi], then xi � 4 + � =)xi�1 > xi � � � 4 =) tag ti 2 [4, 6] =)
[4 + �, 6] ✓[n�
[xi�1, xi], ti�
: ti 2 [4, 6]o✓ [4� �, 6].
Since g(ti) = 2 for these tags
(⇤⇤) 2(2� �) S(g;•P2) 2(2 + �).
Adding (⇤) and (⇤⇤), and using (#),
16� 6� S(g;•P ) = S(g;
•P1) + S(g;
•P2.) 16 + 6� =)
|S(g;•P )� 16| 6�.
Choosing �✏ <✏
6,
k•Pk < � =) |S(g;
•P )� 16| < ✏.
then g 2 R[1, 6] and
Z 6
1g = 16.
122 7. THE RIEMANN INTEGRAL
(3) Let 0 a < b, Q(x) = x2 for x 2 [a, b], and P =�[xi�1, xi]
n
i=1be a
partition of [a, b]. For all i, let qi =h
13
�x2
i + xixi�1 + x2i�1
�i1/2.
Since 0 xi�1 < xi, x2i�1 xixi�1 < x2
i , so
3x2i�1 < x2
i + xixi�1 + x2i�1 < 3x2
i =)
x2i�1 <
1
3
�x2
i + xixi�1 + x2i�1
�= q2
i < x2i =) xi�1 < qi < xi =) qi 2 [xi�1, xi].
Then Q(qi)(xi � xi�1) =1
3
�x2
i + xixi�1 + x2i�1
�(xi � xi�1) =
1
3(x3
i � x3i�1).
Let•Pq =
n�[xi�1, xi], qi
�o. Then
S(Q;•Pq) =
nXi=1
1
3(x3
i � x3i�1) =
1
3
nXi=1
(x3i � x3
i�1) =1
3(x3
n � x30) =
1
3(b3 � a3).
Now let ✏ > 0 be given and•P =
n�[xi�1, xi], ti
�obe an arbitrary tagged
partition of [a, b] with k•Pk < �. For each i, since ti, qi 2 [xi�1, xi], |ti�qi| < �.
Then���S(Q;•P )� 1
3(b3 � a3)
��� =���S(Q;
•P )� S(Q;
•Pq)
��� =
���nX
i=1
t2i (xi � xi�1)�nX
i=1
q2i (xi � xi�1)
��� =���
nXi=1
(t2i � q2i )(xi � xi�1)
��� nX
i=1
|(ti + qi)(ti � qi)|(xi � xi�1) < 2b�nX
i=1
(xi � xi�1) = 2b�(b� a)
So take �✏ <✏
2b(b� a). Then k
•Pk < �✏ =)
���S(Q;•P )� 1
3(b3 � a3)
��� < ✏.
Then Q 2 R[a, b] and
Z b
ax2 dx =
1
3(b3 � a3). [This is Page 207 #14.]
7.1. RIEMANN INTEGRAL 123
(4) Let A = {0, 1, 2, 3, 4, 5}, f(x) =
(3, x 2 [0, 5]\A0, x 2 A
.
We guess
Z 5
0f = 15. Let ✏ > 0 be given.
Let•P be any tagged partition of [0, 5] with k
•Pk < �. Let
•P0 be the subset of
•P with tags in A, and
•P3 be the subset of
•P with tags in [0, 5]\A.
Since 2 of the 6 points of A are endpoints of [0, 5], there are at most 10 subin-
tervals of•P with tags in A. Since
nXi=1
(xi � xi�1) = 5,nX
i=1ti 62A
(xi � xi�1) > 5� 10� =)
���S(f ;•P )� 15
��� =���S(f ;
•P3) + S(f ;
•P0)� 15
��� =���
nXi=1
ti 62A
3(xi � xi�1)� 15��� =
���3nX
i=1ti 62A
(xi � xi�1)� 15��� = 15� 3
nXi=1
ti 62A
(xi � xi�1) < 15� 3(5� 10�) = 30�.
Choosing �✏ ✏
30, if k
•Pk < �✏, then
���S(f ;•P )� 15
��� < ✏.
Then f 2 R[0, 5] and
Z 5
0f = 15.
124 7. THE RIEMANN INTEGRAL
With an argument similar to that of example (4), one can prove the followingtheorem.
Theorem (7.1.3). If g is Riemann integrable on [a, b] and if f(x) = g(x)except for a finite number of points in [a, b], then f is Riemann integrable
and
Z b
af =
Z b
ag.
Theorem (7.1.5). Suppose f, g 2 R[a, b]. Then
(a) if k 2 R, kf 2 R[a, b] and
Z b
akf = k
Z b
af.
(b) f + g 2 R[a, b] and
Z b
a(f + g) =
Z b
af +
Z b
ag.
(c) If f(x) g(x) 8x 2 [a, b], then
Z b
af
Z b
ag.
Proof.
(a) Given ✏ > 0, since f 2 R[a, b], if k 6= 0, 9 �✏/|k| > 0 3�� 8•P with
k•Pk < �✏/|k|, ���S(f ;
•P )�
Z b
af��� <
✏
|k|.
Let �✏ = �✏/|k|. Then 8•P with k
•Pk < �✏,
���S(kf ;•P )� k
Z b
af��� =
���nX
i=1
(kf)(ti)(xi � xi�1)� k
Z b
af��� =
���knX
i=1
f(ti)(xi � xi�1)� k
Z b
af��� = |k|
���S(f ;•P )�
Z b
af��� < |k| · ✏
|k| = ✏.
Thus
Z b
akf = k
Z b
af. The result is clearly true for k = 0.
7.1. RIEMANN INTEGRAL 125
(b) Let ✏ > 0 be given.
f 2 R[a, b] =) 9 �0 > 0 3�� 8•P 0 with k
•P 0k < �0,
���S(f ;•P 0)�
Z b
af��� <
✏
2.
g 2 R[a, b] =) 9 �00 > 0 3�� 8•
P 00 with k•
P 00k < �00,���S(g;
•P 00)�
Z b
af��� <
✏
2.
Let �✏ = min{�0, �00}. Then, for k•Pk < �✏,
(⇤)���S(f ;
•P )�
Z b
af��� <
✏
2and
���S(g;•P )�
Z b
af��� <
✏
2,
so���S(f + g;•P )�
⇣Z b
af +
Z b
ag⌘��� =
���S(f ;•P ) + S(g;
•P )�
Z b
af �
Z b
ag���
���S(f ;•P )�
Z b
af��� +
���S(g;•P )�
Z b
ag��� <
✏
2+
✏
2= ✏,
and thus
Z b
a(f + g) =
Z b
af +
Z b
ag.
(c) From (⇤),
�✏
2< S(f ;
•P )�
Z b
af <
✏
2and � ✏
2< S(g;
•P )�
Z b
ag <
✏
2,
so Z b
af < S(f ;
•P ) +
✏
2 S(g;
•P ) +
✏
2<
Z b
ag + ✏ =)
Z b
af
Z b
ag
since ✏ is arbitrary. ⇤
126 7. THE RIEMANN INTEGRAL
Theorem (7.1.6). If f 2 R[a, b], then f is bounded on [a, b].
Proof. Suppose f is unbounded on [a, b] withR b
a f = L.
Then 9 � > 0 3�� 8•P with k
•Pk < �,���S(f ;
•P )� L
��� < 1 =)���S(f ;
•P )��� < L + 1.
Let Q be a partition of [a, b] with kQk < �.
There exists at least one subinterval, say [xk�1, xk], on which f is not bounded.
Tag Q by ti = xi for i 6= k and pick tk 2 [xk�1, xk] 3��
|f(tk)(xk � xk�1)| > |L| + 1 +���X
i6=k
f(ti)(ti � ti�1)
���.Then���S(f ;
•Q)��� =
���f(tk)(xk � xk�1) +Xi6=k
f(ti)(ti � ti�1)
��� ����f(tk)(xk � xk�1)
���� ���Xi6=k
f(ti)(ti � ti�1)
��� > |L| + 1,
a contradiction. ⇤
7.1. RIEMANN INTEGRAL 127
Example.
(5)g(x) =
8<:
0, x 2 [0, 1] \Q1
x, x 2 [0, 1]\Q
. Show g 62 R[0, 1].
Proof.
(1) Since g is unbounded on [0, 1], g 62 R[0, 1].
(2) Suppose
Z 1
0g = L.
8 ✏ > 0 9 �✏ > 0 3�� 8•P with k
•Pk < �✏,
���S(g;•P )� L
��� < ✏ =)
L� ✏ < S(g;•P ) < L + ✏
Now L � 0.
If L > 0, take a partition•P 0 with all rational tags 3�� k
•P 0k < �✏. Taking
✏ = L,
0 = L� ✏ <nX
i=1
g(ti)(xi � xi�1) = 0,
a contradiction. If L = 0, take ✏ = 1 and
a partition•
P 00 with all irrational tags 3�� k•
P 00k < �✏. ThennX
i=1
g(ti)(xi � xi�1) >nX
i=1
(xi � xi�1) = 1 = ✏,
a contradiction. Thus g 62 R[0, 1]. ⇤
[This is part of Page 207 # 10.]
128 7. THE RIEMANN INTEGRAL
(6) The Ruler (Thomae’s) function on [0, 1].
h(x) =
8<:
0, if x 2 [0, 1]\Q1
n, if x 2 [0, 1] \Q with x =
m
n, n 2 N,
m
nin lowest terms
Guess
Z b
ah = 0. Lt ✏ > 0 be given.
E✏ =nx 2 [0, 1] : h(x) � ✏
2
ois a finite set.
Let n✏ = # of elements in E✏ and �✏ =✏
4n✏if n✏ 6= 0, �✏ = 1 if n✏ = 0.
Let•P be any partition of [0, 1] with k
•Pk < �✏.
Let•P1 be the subset of
•P with tags in E✏ and
•P2 the subset with tags elsewhere.
Now•P1 has at most 2n✏ intervals withX
ti2E✏
(xi � xi�1) 2n✏�e =✏
2and 0 < h(ti) 1 8 tags in
•P1.
Also, Xti 62E✏
(xi � xi�1) 1 and h(ti) <✏
28 tags in
•P2.
Then ���S(h;•P )� 0
��� = S(h;•P1) + S(h;
•P2) < 1 · ✏
2+
✏
2· 1 = ✏.
Since ✏ is arbitrary,
Z b
ah = 0.
7.2. RIEMANN INTEGRABLE FUNCTIONS 129
Homework
Pages 206-07 # 8
(E1) Let f(x) =
(0, x 2 [0, 2], x 6= 1
1, x = 1. Show
Z 2
0f = 0.
(E2) Let g(x) =
(0, x 2 [0, 1]\Q1, x 2 [0, 1] \Q
. Show g 62 R[0, 1].
7.2. Riemann Integrable Functions
Theorem (7.2.1 — Cauchy Criterion). f : [a, b] ! R is in R[a, b] ()8 ✏ > 0 9 ⌘✏ > 0 3�� if
•P and
•Q are any tagged partitions of [a, b] with
k•Pk < ⌘✏ and k
•Qk < ⌘✏, then
���S(f ;•P )� S(f ;
•Q)��� < ✏.
Proof.
(=)) Let f 2 R[a, b] with
Z b
af = L. Given ✏ > 0,
9 ⌘✏ = �✏/2 3�� if•P and
•Q are any tagged partitions of [a, b] with k
•Pk < ⌘✏
and k•Qk < ⌘✏, then���S(f ;
•P )� L
��� <✏
2and
���S(f ;•Q)� L
��� <✏
2.
Then���S(f ;•P )� S(f ;
•Q)��� =
���S(f ;•P )� L + L� S(f ;
•Q)��� ���S(f ;
•P )� L
��� +���S(f ;
•Q)� L
��� <✏
2+
✏
2= ✏.
130 7. THE RIEMANN INTEGRAL
((=) 8n 2 N, let �n > 0 be such that if•P and
•Q have k
•Pk < �n and
k•Qk < �n, then ���S(f ;
•P )� S(f ;
•Q)��� <
1
n.
8n 2 N, we may assume �n � �n+1; otherwise replace �n+1 by min{�1, . . . , �n}.
8n 2 N, let•
Pn be a tagged partition of [a, b] with k•
Pnk < �n.
For m > n, k•
Pmk < �n also. Thus, for m > n,
(⇤)���S(f ;
•Pn)� S(f ;
•Pm)
��� <1
n=)
⇣S(f ;
•Pm)
⌘1m=1
is a Cauchy sequence in R =)
limm!1
S(f ;•
Pm) = A for some A 2 R. Then, 8n 2 N and ↵ > 0, 9 m 2 N 3�����S(f ;
•Pn)�A
��� =���S(f ;
•Pn)� S(f ;
•Pm) + S(f ;
•Pm)�A
��� ���S(f ;•
Pn)� S(f ;•
Pm)��� +
���S(f ;•
Pm)�A��� <
1
n+ ↵ =)
���S(f ;•
Pn)�A��� 1
n.
Now, given ✏ > 0, let K 2 N 3�� K >2
✏. If
•Q is any tagged partition with
k•Qk < �K,���S(f ;
•Q)�A
��� =���S(f ;
•Q)� S(f ;
•PK) + S(f ;
•PK)�A
��� ���S(f ;•Q)� S(f ;
•PK)
��� +���S(f ;
•PK)�A
��� 1
K+
1
K=
2
K< ✏.
Since ✏ is arbitrary,
Z b
af = A. ⇤
7.2. RIEMANN INTEGRABLE FUNCTIONS 131
Example. (2) (redone) (2) g(x) =
(4, x 2 [1, 4)
2, x 2 [4, 6].
We showed if•P was any tagged partition with k
•Pk < �,
16� 6� S(g;•P ) 16 + 6�.
If•Q were another such partition,
16� 6� S(g;•Q) 16 + 6�.
Then
(16� 6� � (16 + 6�) S(g;•P )� S(g;
•Q) (16 + 6�)� (16� 6�) =)
�12� S(g;•P )� S(g;
•Q) 12� =)���S(g;
•P )� S(g;
•Q)��� 12�.
Thus, for a given ✏ > 0, choose ⌘✏ = � <✏
12.
Then, by the Cauchy Criterion, g 2 R[a, b].
132 7. THE RIEMANN INTEGRAL
Note. f : [a, b] ! R is not in R[a, b] () 9 ✏0 > 0 3�� 8 ⌘ > 0 9 tagged
partitions•P and
•Q with k
•Pk < ⌘ and k
•Qk < ⌘ 3�����S(f ;
•P )� S(f ;
•Q)��� � ✏0.
Example. (5) (redone) g(x) =
8<:
0, x 2 [0, 1] \Q1
x, x 2 [0, 1]\Q
.
Take ✏0 =1
2. If
•P is any partition with all irrational tags,
S(g;•P ) =
nXi=1
f(ti)(xi � xi�1) >nX
i=1
(xi � xi�1) = 1.
If•Qis any partition with all rational tags,
S(g;•Q) = 0.
Then ���S(g;•P )� S(g;
•Q)��� > 1 >
1
2= ✏0.
Thus g 62 R[0, 1].
7.2. RIEMANN INTEGRABLE FUNCTIONS 133
Theorem (7.2.3 — Squeeze Theorem). Let f : [a, b] ! R. Then f 2R[a, b] () 8✏ > 0 9 functions ↵✏,!✏ 2 R[a, b] 3��
(#) ↵✏(x) f(x) !✏(x)
8 x 2 [a, b] and
Z b
a(!✏ � ↵✏) < ✏.
Proof. (=)) Take ↵✏ = !✏ = f 8 ✏ > 0.
((=) Let ✏ > 0 be given. Since ↵✏,!✏ 2 R[a, b], 9 �✏ > 0 3�� for k•Pk < �✏,���S(↵✏;
•P )�
Z b
a↵✏
��� < ✏ and���S(!✏;
•P )�
Z b
a!✏
��� < ✏ =)Z b
a↵✏ � ✏ < S(↵✏;
•P ) and S(!✏;
•P ) <
Z b
a!✏ + ✏
From (#),
S(↵✏;•P ) S(f ;
•P ) S(!✏;
•P ) =)Z b
a↵✏ � ✏ < S(f ;
•P ) <
Z b
a!✏ + ✏.
If also k•Qk < �✏, Z b
a↵✏ � ✏ < S(f ;
•Q) <
Z b
a!✏ + ✏.
so ���S(f ;•P )� S(f ;
•Q)��� <
Z b
a!✏ �
Z b
a↵✏ + 2✏ =
Z b
a(!✏ � ↵✏) + 2✏ < 3✏.
Since ✏ is arbitrary, f 2 R[a, b] by the Cauchy Criterion. ⇤
134 7. THE RIEMANN INTEGRAL
Definition. � : [a, b] ! R is a step function if it has only a finite numberof distinct values, each value being assumed on one or more subintervals of[a, b].
Lemma (7.2.4). If J is a subinterval of [a, b] with endpoints c < d and if
�J(x) =
(1, x 2 J
0, x 2 [a, b]\J ,
then �J 2 R[a, b] and
Z b
a�J = d� c.
Proof. Let ✏ > 0 be given. Let �✏ be chosen 3���✏ <
✏
4and c + �✏ < d� �✏.
Then, for k•Pk < �✏,
[c + �✏, d� �✏] ✓[
ti2[c,d]
n�[xi�1, xi], ti
�o✓ [c� �✏, d + �✏].
Thus, for k•Pk < �✏,
d� c� 2�✏ S(�J ;•P ) d� c + 2�✏ =)���S(�j;
•P )� (d� c)
��� < 2�✏ < 2 · ✏
4=
✏
2< ✏.
Therefore,
Z b
a�J = d� c. ⇤
7.2. RIEMANN INTEGRABLE FUNCTIONS 135
Theorem (7.2.5). If � : [a, b] ! R is a step function, then � 2 R[a, b].
Proof. � is a linear combination of “elementary step functions” (as in thelemma):
� =nX
j=1
kj�Jj where Jj has endpoints cj < dj.
From the Lemma and Theorem 7.1.4,Z b
a� =
nXj=1
kj(dj � cj).
⇤
Theorem (7.2.7). If f is continuous on [a, b], then f 2 R[a, b].
Proof. f continuous on [a, b] =) f is u.c. on [a, b].
Thus, given ✏ > 0, 9 �✏ > 0 3��u, v 2 [a, b] and |u� v| < �✏ =) |f(u)� f(v)| <
✏
b� a.
Let P =�Ii
n
i=1be a partition 3�� kPk < �✏, ui a point where f attains its
minimum on Ii and vi a point where f attains its maximum on Ii. Let
↵✏(x) =
(f(ui), x 2 [xi�1, xi], i = 1, . . . , n� 1
f(un), x 2 [xn�1, xn],
!✏(x) =
(f(vi), x 2 [xi�1, xi], i = 1, . . . , n� 1
f(vn), x 2 [xn�1, xn].
Then ↵✏(x) f(x) !✏(x) 8 x 2 [a, b]. Thus
0 Z b
a(!✏�↵✏) =
nXi=1
⇥f(vi)�f(ui)
⇤(xi�xi�1) <
nXi=1
⇣ ✏
b� a
⌘(xi�xi�1) = ✏.
Then f 2 R[a, b] by the Squeeze Theorem. ⇤
136 7. THE RIEMANN INTEGRAL
Theorem (7.2.8). If f : [a, b] ! R is monotone on [a, b], then f 2R[a, b].
Proof. WLOG, suppose f is increasing on [a, b]. Given ✏ > 0,
9 q 2 N 3�� h =f(b)� f(a)
q<
✏
b� a.
Let yk = f(a) + kh for k = 1, . . . , q. Consider the sets
Ak = f�1�[yk�1, yk)
�for k = 0, 1, . . . , q and Aq = f�1
�[yq�1, yq)
�.
Toss out any empty Ak’s and relabel the others consecutively. Add endpointsif necessary to the remaining Ak’s to get closed intervals Ik’s. Then the Ik’soverlap only at endpoints, [a, b] =
Sqk=1 Ik, and f(x) 2 [yk�1, yk] 8 x 2 Ik.
Define step functions ↵✏,!✏ : [a, b] ! R by ↵✏(x) = yk�1 and !✏(x) = yk forx 2 Ak. 8x 2 [a, b],
↵✏(x) f(x) !✏(x) =)Z b
a(!✏�↵✏) =
qXk=1
(yk� yk�1)(xk�xk�1) =qX
k=1
h(xk�xk�1) = h(b� a) < ✏.
Since ✏ is arbitrary, f 2 R[a, b] by the Squeeze Theorem. ⇤
Theorem (7.2.9 — Additivity). Let f : [a, b] ! R and let c 2 (a, b).Then f 2 R[a, b] () its restrictions to [a, c] and [c, b] are both Riemannintegrable. In this case, Z b
af =
Z c
af +
Z b
cf.
Definition. If f 2 R[a, b] and if ↵,� 2 [a, b] with ↵ < �, defineZ ↵
�f = �
Z �
↵f and
Z ↵
↵f = 0.
7.3. THE FUNDAMENTAL THEOREM 137
Homework
(1) Give an example of an integrable function h : [0, 1] ! R with h(x) >
0 8 x 2 [0, 1], but such that k(x) =1
h(x)is not integrable on [0, 1].
(2) Give an example of a function f : [0, 1] ! R that is not integrable on [0, 1],but is such that |f | 2 R[0, 1].
(3) Give an example of an integrable function f : [0, 1] ! R and a nonintegrablefunction g : [0, 1] ! R such that fg 2 R[0, 1].
7.3. The Fundamental Theorem
We now explore the inverse relationship between di↵erentiation and integration.We begin by looking at “integrating a derivative.”
138 7. THE RIEMANN INTEGRAL
Theorem (7.3.1 — Fundamental Theorem of Calculus (First Form)).
Suppose 9 a finite set E in [a, b] and functions f, F : [a, b] ! R 3��:
(a) F is continuous on [a, b],
(b) F 0(x) = f(x) 8 x 2 [a, b]\E,
(c) f 2 R[a, b].
Then
Z b
af = F (b)� F (a).
Proof. We prove the theorem for the case where E = {a, b}. The generalcase can be obtained by breaking the interval into the union of a finite number
of intervals. Let ✏ > 0 be given. Since f 2 R[a, b], 9 �✏ > 0 3�� if•P is any
tagged partition with k•Pk < �✏, then���S(f ;
•P )�
Z b
af��� < ✏.
Applying the MVT to F on each subinterval [xi�1, xi] (we do not need di↵er-entiability at a and b for this), =) 9 ui 2 (xi�1, xi) 3��
F (xi)� F (xi�1) = F 0(ui)(xi � xi�1).
Using F 0(ui) = f(ui),
F (b)� F (a) =nX
i=1
⇥F (xi)� F (xi�1)
⇤=
nXi=1
f(ui)(xi � xi�1).
Now let•
Pu =n�
[xi�1, xi], ui
�on
i=1. Then
S⇣f ;
•Pu
⌘=
nXi=1
f(ui)(xi�xi�1) = F (b)�F (a) =)����F (b)�F (a)
��Z b
af��� < ✏.
Since ✏ is arbitrary,
Z b
af = F (b)� F (a). ⇤
7.3. THE FUNDAMENTAL THEOREM 139
Note.
(1) If E = ;, hypothesis (a) is automatically satisfied.
(2) If f is not defined at c 2 E, take f(c) = 0.
(3) Even if E = ;, hypothesis (c) is not automatically satisfied since 9 functionsF 3�� F 0 62 R[a, b] (see Example (4) below)
Example.
(1) If F (x) =1
3x3 on [a, b], then f(x) = F 0(x) = x2 is continuous on [a, b] =)
f 2 R[a, b]. With E = ; here,Z b
ax2 dx = F (b)� F (a) =
1
3(b3 � a3).
(2) F (x) = |x� 1| for x 2 [�10, 10].
f(x) = F 0(x) =
(�1, x 2 [�10, 1)
1, x 2 (1, 10].
Here E = {1}. Defining f(1) = 0, f is a step function =) f 2 R[�10, 10].Thus Z 10
�10f(x) dx = F (10)� F (�10) = 9� 11 = �2.
(3) F (x) = 4x1/4 for x 2 [0, b].
F is continuous on [0, b] and F 0(x) = x�3/4 on (0, b].
Since f(x) = F 0(x) is not bounded on [0, b], f 62 R[0, b]
and so the FTC does not apply.
140 7. THE RIEMANN INTEGRAL
(4) K(x) =
8<:
x2 cos1
x2, x 2 (0, 1]
0, x = 0. Then
K 0(x) =
8<:
2x cos1
x2+
2
xsin
1
x2, x 2 (0, 1]
0, x = 0
Thus K is continuous and di↵erentiable at every point of [0, 1].
Since the first term of K 0 is continuous on [0, 1], it belongs to R[0, 1].
But the second term is unbounded, so does not belong to R[0, 1].
Thus K 0 62 R[0, 1] and the FTC does not apply (see note (3) above).
We now look at the case of “di↵erentiating an integral.”
Definition. A function F such that F 0(x) = f(x) 8 x 2 [a, b] is called anantiderivative or primitive of f on [a, b].
Definition (7.3.3). If f 2 R[a, b], then
F (z) =
Z z
af for z 2 [a, b]
is called the indefinite integral of f with basepoint a.
7.3. THE FUNDAMENTAL THEOREM 141
Theorem (7.3.4). Let F (z) =
Z z
af for z 2 [a, b]. Then F is continu-
ous (actually, u.c.) on [a, b].
Proof. (We show F is Lipschitz on [a, b].)
By additivity, if a w z b, then
F (z) =
Z z
af =
Z w
af +
Z z
wf = F (w) +
Z z
wf =)
F (z)� F (w) =
Z z
wf.
Since f is bounded on [a, b], 9 M 2 R 3���M f(x) M 8x 2 [a, b] =)
�M(z � w) Z z
wf M(z � w) =)
|F (z)� F (w)| =���Z z
wf��� M |z � w| =)
F is Lipschitz on [a, b] =) F is u.c. on [a, b]. ⇤
142 7. THE RIEMANN INTEGRAL
Theorem (7.3.5 — Fundamental Theorem of Calculus (Second Form)).
Let f 2 R[a, b] and let f be continuous at c 2 [a, b]. Then F (z) =
Z z
af is
di↵erentiable at c and F 0(c) = f(c).
Proof. (for c 2 (a, b)) Since f is continuous at c, given ✏ > 0, 9 �✏ > 0 3��|z � c| < �✏ and z 2 [a, b] =) |f(z)� f(c)| < ✏.
Now1
z � c
Z z
c1 = 1, so
���F (z)� F (c)
z � c� f(c)
��� =
���� 1
z � c
h Z z
af �
Z c
afi� f(c)
���� ���� 1
z � c
h Z z
af +
Z a
cfi� f(c)
���� =
���� 1
z � c
Z z
cf � f(c)
z � c
Z z
c1
���� =
1
|z � c|
���Z z
c[f(x)� f(c)] dx
��� 1
|z � c|(✏)|z � c| = ✏.
Since ✏ is arbitrary, limz!c
F (z)� F (c)
z � c= f(c) =) F 0(c) = f(c). ⇤
Theorem (7.3.6). If f is continuous on [a, b], then F (z) =
Z z
af is
di↵erentiable on [a, b] and F 0(z) = f(z) 8 z 2 [a, b].
7.3. THE FUNDAMENTAL THEOREM 143
Example.
(1) There is no linear combination of elementary functions equal to
Ze�x2
dx.
But for any [a, b] ✓ R, define F (z) =
Z z
ae�x2
dx. Then, by FTC2,
F 0(z) = e�z2, so F (z) =
Z z
ae�x2
dx is an antiderivative of e�z2.
(2) Suppose G(z) =
Z sin z
aarctanx2 dx.
We need the chain rule to find G0(z).
Thus G0(z) = cos z arctan(sin2 z).
144 7. THE RIEMANN INTEGRAL
(3) f(x) =
8><>:�1, x 2 [�10, 1)
0, x = 1
1, x 2 (1, 10]
is in R[0, 1]. Then
F (z) =
Z z
�10f =
8>><>>:
Z z
�10(�1) dx =
⇥� x
⇤z�10
= �z � 10, z 2 [�10, 1]
�11 +
Z z
11 dx = �11 +
⇥x⇤z1
= z � 12, z 2 [1, 10],
So
F (z) =
Z z
�10f = |z � 1|� 11
is an indefinite integral with basepoint �10 of f , but is not an antiderivativeon [�10, 10] since F 0(1) DNE.
(4) The Ruler (Thomae’s) function on [0, 1].
f(x) =
8<:
0, if x 2 [0, 1]\Q1
n, if x 2 [0, 1] \Q with x =
m
n, n 2 N,
m
nin lowest terms
is in R[0, 1]. Now
F (z) =
Z z
0f = 0 8 z 2 [0, 1],
so F 0(z) = 0 8 z 2 [0, 1]. Thus F is an indefinite integral of f with basepoint 0,but F is not an antiderivative of f on [0, 1] since F 0(z) 6= f(z) 8 z 2 [0, 1]\Q.
Homework
Pages 223-25 # 2, 3
(E1) Find F 0(z) for F (z) =
Z z
0ln⇣1 +
x
x2 + 2
⌘dx.
(E2) Find F 0(z) for F (z) =
Z z2+4z
5cos x dx.