the riemann integral - christian brothers universityfacstaff.cbu.edu/wschrein/media/m414...

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. ⇤


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). ⇤


(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


•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 + �).


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.


(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 =


�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.]


(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


•P with tags in A, and


•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.


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.


(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. ⇤


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(tk)(xk � xk�1)

�� Xi6=k


�� > |L| + 1,

a contradiction. ⇤


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.]


(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= ✏.


((=) 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< ✏.


Z b

af = A. ⇤


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].


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].


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. ⇤


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]


�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. ⇤


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. ⇤


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.”


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�� < ✏.


Z b

af = F (b)� F (a). ⇤


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.


(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.


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]. ⇤


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].


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).


(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.

the riemann integral - christian brothers universityfacstaff.cbu.edu/wschrein/media/m414...

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