1. No category

CHAPTER 8
Integration
Contrary to the impression given by most calculus courses, there are many ways
to define integration. The one given here is called the Riemann integral or the
Riemann-Darboux integral, and it is the one most commonly presented to calculus
students.
1. Partitions
A partition of the interval [a, b] is a finite set P ⇢ [a, b] such that {a, b} ⇢ P .
The set of all partitions of [a, b] is denoted part([a, b]). Basically, a partition should
be thought of as a way to divide an interval into a finite number of subintervals by
choosing some points where it is divided.
If P 2 part([a, b]), then the elements of P can be ordered in a list as a = x0 <
x1 < · · · < xn = b. The adjacent points of this partition determine n compact
intervals of the form IkP = [xk 1 , xk ], 1  k  n. If the partition is clear from the
context, we write Ik instead of IkP . It’s clear that these intervals only intersect at
their common endpoints and there is no requirement they have the same length.
Since it’s inconvenient to always list each part of a partition, we’ll use the
partition of the previous paragraph as the generic partition. Unless it’s necessary
within the context to specify some other form for a partition, assume any partition
is the generic partition. (See Figure 1.)
If I is any interval, its length is written |I|. Using the notation of the previous
paragraph, it follows that
n
X
k=1
|Ik | =
n
X
(xk
xk
1)
= xn
x0 = b
a.
k=1
The norm of a partition P is
kP k = max{|IkP | : 1  k  n}.
In other words, the norm of P is just the length of the longest subinterval determined
by P . If |Ik | = kP k for every Ik , then P is called a regular partition.
Suppose P, Q 2 part([a, b]). If P ⇢ Q, then Q is called a refinement of P . When
this happens, we write P ⌧ Q. In this case, it’s easy to see that P ⌧ Q implies
x0
I1
x1
I2
x2
I3
x3
I4
x4
I5
a
x5
b
Figure 1. The generic partition with five subintervals.
8-1
8-2
CHAPTER 8. INTEGRATION
kP k kQk. It also follows at once from the definitions that P [ Q 2 part([a, b])
with P ⌧ P [ Q and Q ⌧ P [ Q. The partition P [ Q is called the common
refinement of P and Q.
2. Riemann Sums
{x⇤k
Let f : [a, b] ! R and P 2 part([a, b]). Choose x⇤k 2 Ik for each k. The set
: 1  k  n} is called a selection from P . The expression
R (f, P, x⇤k )
=
n
X
k=1
f (x⇤k )|Ik |
is the Riemann sum for f with respect to the partition P and selection x⇤k . The
Riemann sum is the usual first step toward integration in a calculus course and can
be visualized as the sum of the areas of rectangles with height f (x⇤k ) and width |Ik |
— as long as the rectangles are allowed to have negative area when f (x⇤k ) < 0. (See
Figure 2.)
Notice that given a particular function f and partition P , there are an uncountably infinite number of di↵erent possible Riemann sums, depending on the selection
x⇤k . This sometimes makes working with Riemann sums quite complicated.
Example 8.1. Suppose f : [a, b] ! R is the constant function f (x) = c. If
P 2 part([a, b]) and {x⇤k : 1  k  n} is any selection from P , then
R (f, P, x⇤k ) =
n
X
k=1
f (x⇤k )|Ik | = c
n
X
k=1
|Ik | = c(b
a).
Example 8.2. Suppose f (x) = x on [a, b]. Choose any P 2 part([a, b]) where
kP k < 2(b a)/n. (Convince yourself this is always possible.1) Make two specific
1This is with the generic partition
y = f (x)
⇤
a x1
x0
x1
x⇤2
x2
x⇤3
x3
x⇤4
b
x4
Figure 2. The Riemann sum R (f, P, x⇤k ) is the sum of the areas of the
rectangles in this figure. Notice the right-most rectangle has negative
area because f (x⇤4 ) < 0.
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Riemann Sums
8-3
selections lk⇤ = xk 1 and rk⇤ = xk . If x⇤k is any other selection from P , then
lk⇤  x⇤k  rk⇤ and the fact that f is increasing on [a, b] gives
R (f, P, lk⇤ )  R (f, P, x⇤k )  R (f, P, rk⇤ ) .
With this in mind, consider the following calculation.
R (f, P, rk⇤ )
(68)
R (f, P, lk⇤ ) =
=
=

n
X
k=1
n
X
k=1
n
X
k=1
n
X
k=1
(rk⇤
lk⇤ )|Ik |
(xk
xk
1 )|Ik |
|Ik |2
kP k2
= nkP k2
<
a)2
4(b
n
This shows that if a partition is chosen with a small enough norm, all the Riemann
sums for f over that partition will be close to each other.
In the special case when P is a regular partition, |Ik | = (b a)/n, rk =
a + k(b a)/n and
R (f, P, rk⇤ ) =
=
n
X
rk |Ik |
k=1
n ✓
X
a+
k(b
k=1
=
b
a
na +
n
b
a
✓
a)
n
b
a
n
n
b
n
b
◆
aX
k=1
k
!
a n(n + 1)
=
na +
n
n
2
✓
◆
b a
n 1
n+1
=
a
+b
.
2
n
n
In the limit as n ! 1, this becomes the familiar formula (b2
of f (x) = x over [a, b].
◆
a2 )/2, for the integral
Definition 8.1. The function f is Riemann integrable on [a, b], if there exists a
number R(f ) such that for all " > 0 there is a > 0 so that whenever P 2 part([a, b])
with kP k < , then
|R(f ) R (f, P, x⇤k ) | < "
for any selection x⇤k from P .
Theorem 8.2. If f : [a, b] ! R and R(f ) exists, then R(f ) is unique.
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8-4
CHAPTER 8. INTEGRATION
Proof. Suppose R1 (f ) and R2 (f ) both satisfy the definition and " > 0. For
i = 1, 2 choose i > 0 so that whenever kP k < i , then
|Ri (f )
R (f, P, x⇤k ) | < "/2,
as in the definition above. If P 2 part([a, b]) so that kP k <
|R1 (f )
R2 (f )|  |R1 (f )
R (f, P, x⇤k ) | + |R2 (f )
1
^
2,
then
R (f, P, x⇤k ) | < "
and it follows R1 (f ) = R2 (f ).
⇤
Theorem 8.3. If f : [a, b] ! R and R(f ) exists, then f is bounded.
⇤
Proof. Left as Exercise 8.1.
3. Darboux Integration
As mentioned above, a difficulty with handling Riemann sums is there are an
uncountably so many di↵erent ways to choose partitions and selections that working
with them is unwieldy. One way to resolve this problem was shown in Example 8.2,
where it was easy to find largest and smallest Riemann sums associated with each
partition. However, that’s not always a straightforward calculation, so to use that
idea, a little more care must be taken.
Definition 8.4. Let f : [a, b] ! R be bounded and P 2 part([a, b]). For each
Ik determined by P , let
Mk = lub {f (x) : x 2 Ik }
and
mk = glb {f (x) : x 2 Ik }.
The upper and lower Darboux sums for f on [a, b] are
D(f, P ) =
n
X
k=1
Mk |Ik |
and
D(f, P ) =
n
X
k=1
mk |Ik |.
The following theorem is the fundamental relationship between Darboux sums.
Pay careful attention because it’s the linchpin holding everything together!
then
Theorem 8.5. If f : [a, b] ! R is bounded and P, Q 2 part([a, b]) with P ⌧ Q,
D(f, P )  D(f, Q)  D(f, Q)  D(f, P ).
(xk0
Proof. Let P be the generic partition and let Q = P [ {x}, where x 2
1 , xx0 ) for some k0 . Clearly, P ⌧ Q. Let
Ml = lub {f (x) : x 2 [xk0
ml = glb {f (x) : x 2 [xk0
1 , x]}
1 , x]}
Mr = lub {f (x) : x 2 [x, xk0 ]}
mr = glb {f (x) : x 2 [x, xk0 ]}
Then
mk0  ml  Ml  Mk0
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and
mk0  mr  Mr  Mk0
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Darboux Integration
8-5
so that
mk0 |Ik0 | = mk0 (|[xk0
 ml |[xk0
 Ml |[xk0
 Mk0 |[xk0
1 , x]|
+ |[x, xk0 ]|)
1 , x]| + mr |[x, xk0 ]|
1 , x]|
+ Mr |[x, xk0 ]|
1 , x]|
= Mk0 |Ik0 |.
+ Mk0 |[x, xk0 ]|
This implies
D(f, P ) =
=
n
X
k=1
mk |Ik |
X
k6=k0

X
k6=k0
mk |Ik | + mk0 |Ik0 |
mk |Ik | + ml |[xk0
1 , x]|
+ mr |[x, xk0 ]|
1 , x]|
+ Mr |[x, xk0 ]|
= D(f, Q)
 D(f, Q)
X
=
Mk |Ik | + Ml |[xk0

k6=k0
n
X
k=1
Mk |Ik |
= D(f, P )
The argument given above shows that the theorem holds if Q has one more point
than P . Using induction, this same technique also shows the theorem holds when Q
has an arbitrarily larger number of points than P .
⇤
The main lesson to be learned from Theorem 8.5 is that refining a partition
causes the lower Darboux sum to increase and the upper Darboux sum to decrease.
Moreover, if P, Q 2 part([a, b]) and f : [a, b] ! [ B, B], then,
B(b
a)  D(f, P )  D(f, P [ Q)  D(f, P [ Q)  D(f, Q)  B(b
a).
Therefore every Darboux lower sum is less than or equal to every Darboux upper
sum. Consider the following definition with this in mind.
Definition 8.6. The upper and lower Darboux integrals of a bounded function
f : [a, b] ! R are
D(f ) = glb {D(f, P ) : P 2 part([a, b])}
and
respectively.
D(f ) = lub {D(f, P ) : P 2 part([a, b])},
As a consequence of the observations preceding the definition, it follows that
D(f ) D(f ) always. In the case D(f ) = D(f ), the function is said to be Darboux
integrable on [a, b], and the common value is written D(f ).
The following is obvious.
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8-6
CHAPTER 8. INTEGRATION
Corollary 8.7. A bounded function f : [a, b] ! R is Darboux integrable if and
only if for all " > 0 there is a P 2 part([a, b]) such that D(f, P ) D(f, P ) < ".
Which functions are Darboux integrable? The following corollary gives a first
approximation to an answer.
Corollary 8.8. If f 2 C([a, b]), then D(f ) exists.
Proof. Let " > 0. According to Corollary 6.31, f is uniformly continuous, so
there is a > 0 such that whenever x, y 2 [a, b] with |x y| < , then |f (x) f (y)| <
"/(b a). Let P 2 part([a, b]) with kP k < . By Corollary 6.23, in each subinterval
Ii determined by P , there are x⇤i , yi⇤ 2 Ii such that
Since |x⇤i
f (x⇤i ) = lub {f (x) : x 2 Ii }
and
yi⇤ |  |Ii | < , we see 0  f (x⇤i )
D(f )
f (yi⇤ ) = glb {f (x) : x 2 Ii }.
f (yi⇤ ) < "/(b
D(f )  D(f, P )
=
n
X
i=1
=
n
X
D(f, P )
(f (x⇤i )
i=1
<
"
b
="
n
X
f (x⇤i )|Ii |
a
n
X
i=1
i=1
a), for 1  i  n. Then
f (yi⇤ )|Ii |
f (yi⇤ ))|Ii |
|Ii |
⇤
and the corollary follows.
This corollary should not be construed to imply that only continuous functions
are Darboux integrable. In fact, the set of integrable functions is much more
extensive than only the continuous functions. Consider the following example.
Example 8.3. Let f be the salt and pepper function of Example 6.15. It was
shown that C(f ) = Qc . We claim that f is Darboux integrable over any compact
interval [a, b].
To see this, let " > 0 and N 2 N so that 1/N < "/2(b a). Let
{qki : 1  i  m} = {qk : 1  k  N } \ [a, b]
and choose P 2 part([a, b]) such that kP k < "/2m. Then
n
X
D(f, P ) =
lub {f (x) : x 2 I` }|I` |
`=1
=
X
qki 2I
/ `
lub {f (x) : x 2 I` }|I` | +
1
(b a) + mkP k
N
"
"
<
(b a) + m
2(b a)
2m
= ".
X
qki 2I`
lub {f (x) : x 2 I` }|I` |

Since f (x) = 0 whenever x 2 Qc , it follows that D(f, P ) = 0. Therefore, D(f ) =
D(f ) = 0 and D(f ) = 0.
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The Integral
8-7
4. The Integral
There are now two di↵erent definitions for the integral. It would be embarassing,
if they gave di↵erent answers. The following theorem shows they’re really di↵erent
sides of the same coin.2
Theorem 8.9. Let f : [a, b] ! R.
(a) R(f ) exists i↵ D(f ) exists.
(b) If R(f ) exists, then R(f ) = D(f ).
Proof. (a) (=)) Suppose R(f ) exists and " > 0. By Theorem 8.3, f is
bounded. Choose P 2 part([a, b]) such that
|R(f )
R (f, P, x⇤k ) | < "/4
for all selections x⇤k from P . From each Ik , choose xk and xk so that
"
"
Mk f (xk ) <
and f (xk ) mk <
.
4(b a)
4(b a)
Then
D(f, P )
R (f, P, xk ) =
=
n
X
k=1
n
X
Mk |Ik |
"
4(b
a)
k=1
f (xk )|Ik |
f (xk ))|Ik |
(Mk
k=1
<
n
X
(b
a) =
"
.
4
In the same way,
R (f, P, xk )
Therefore,
D(f )
D(f, P ) < "/4.
D(f )
= glb {D(f, Q) : Q 2 part([a, b])}
 D(f, P ) D(f, P )
⇣
"⌘
< R (f, P, xk ) +
4
⇣
lub {D(f, Q) : Q 2 part([a, b])}
R (f, P, xk )
 |R (f, P, xk )
R (f, P, xk )| +
< |R (f, P, xk )
R(f )| + |R(f )
"
2
"⌘
4
R (f, P, xk ) | +
"
2
<"
Since " is an arbitrary positive number, this shows D(f ) exists and equals R(f ),
which is part (b) of the theorem.
2Theorem 8.9 shows that the two integrals presented here are the same. But, there are many
other integrals, and not all of them are equivalent. For example, the well-known Lebesgue integral
includes all Riemann integrable functions, but not all Lebesgue integrable functions are Riemann
integrable. The Denjoy integral is another extension of the Riemann integral which is not the same
as the Lebesgue integral. For more discussion of this, see [9].
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8-8
CHAPTER 8. INTEGRATION
((=) Suppose f : [a, b] ! [ B, B], D(f ) exists and " > 0. Since D(f ) exists,
there is a P1 2 part([a, b]), with points a = p0 < · · · < pm = b, such that
"
D(f, P1 ) D(f, P1 ) < .
2
Set = "/8mB. Choose P 2 part([a, b]) with kP k < and let P2 = P [ P1 . Since
P1 ⌧ P2 , according to Theorem 8.5,
"
D(f, P2 ) D(f, P2 ) < .
2
Thinking of P as the generic partition, the interiors of its intervals (xi 1 , xi )
may or may not contain points of P1 . For 1  i  n, let
Qi = {xi
1 , xi }
[ (P1 \ (xi
1 , xi ))
2 part(Ii ).
If P1 \(xi 1 , xi ) = ;, then D(f, P ) and D(f, P2 ) have the term Mi |Ii | in common
because Qi = {xi 1 , xi }.
Otherwise, P1 \ (xi 1 , xi ) 6= ; and
D(f, Qi )
BkP2 k
BkP k >
B .
Since P1 has m 1 points in (a, b), there are at most m 1 of the Qi not contained
in P .
This leads to the estimate
n
X
"
D(f, P ) D(f, P2 ) = D(f, P )
D(f, Qi ) < (m 1)2B < .
4
i=1
In the same way,
D(f, P2 )
D(f, P ) < (m
1)2B <
"
.
4
Putting these estimates together yields
D(f, P )
D(f, P ) =
D(f, P )
D(f, P2 ) + D(f, P2 )
This shows that, given " > 0, there is a
D(f, P )
Since
D(f, P2 ) + (D(f, P2 )
> 0 so that kP k <
D(f, P ))
" " "
< + + ="
4 2 4
implies
D(f, P ) < ".
D(f, P )  D(f )  D(f, P ) and D(f, P )  R(f, P, x⇤i )  D(f, P )
for every selection x⇤i from P , it follows that |R(f, P, x⇤i ) D(f )| < " when kP k < .
We conclude f is Riemann integrable and R(f ) = D(f ).
⇤
From Theorem 8.9, we are justified in using a single notation for both R(f ) and
Rb
Rb
D(f ). The obvious choice is the familiar a f (x) dx, or, more simply, a f .
When proving statements about the integral, it’s convenient to switch back and
forth between the Riemann and Darboux formulations. Given f : [a, b] ! R the
following three facts summarize much of what we know.
Rb
(1) a f exists i↵ for all " > 0 there is a > 0 and an ↵ 2 R such that whenever
P 2 part([a, b]) and x⇤i is a selection from P , then |R (f, P, x⇤i ) ↵| < ".
Rb
In this case a f = ↵.
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The Cauchy Criterion
8-9
Rb
(2) a f exists i↵ 8" > 09P 2 part([a, b]) D(f, P ) D(f, P ) < "
(3) For any P 2 part([a, b]) and selection x⇤i from P ,
D(f, P )  R (f, P, x⇤i )  D(f, P ).
5. The Cauchy Criterion
Rb
We now face a conundrum. In order to show that a f exists, we must know
its value. It’s often very hard to determine the value of an integral, even if the
integral exists. We’ve faced this same situation before with sequences. The basic
definition of convergence for a sequence, Definition 3.2, requires the limit of the
sequence be known. The path out of the dilemma in the case of sequences was the
Cauchy criterion for convergence, Theorem 3.22. The solution is the same here,
with a Cauchy criterion for the existence of the integral.
Theorem 8.10 (Cauchy Criterion). Let f : [a, b] ! R. The following statements
are equivalent.
Rb
(a) a f exists.
(b) Given " > 0 there exists P 2 part([a, b]) such that if P ⌧ Q1 and
P ⌧ Q2 , then
|R (f, Q1 , x⇤k )
(69)
R (f, Q2 , yk⇤ )| < "
for any selections from Q1 and Q2 .
Rb
Proof. (=)) Assume a f exists. According to Definition 8.1, there is a > 0
Rb
such that whenever P 2 part([a, b]) with kP k < , then | a f R (f, P, x⇤i ) | < "/2
for every selection. If P ⌧ Q1 and P ⌧ Q2 , then kQ1 k < , kQ2 k < and a simple
application of the triangle inequality shows
|R (f, Q1 , x⇤k )
R (f, Q2 , yk⇤ )|

R (f, Q1 , x⇤k )
Z
b
f +
a
Z
b
f
a
R (f, Q2 , yk⇤ ) < ".
((=) Let " > 0 and choose P 2 part([a, b]) satisfying (b) with "/2 in place of ".
We first claim that f is bounded. To see this, suppose it is not. Then it must be
unbounded on an interval Ik0 determined by P . Fix a selection {x⇤k 2 Ik : 1  k  n}
and let yk⇤ = x⇤k for k 6= k0 with yk⇤0 any element of Ik0 . Then
"
> |R (f, P, x⇤k )
2
R (f, P, yk⇤ )| = f (x⇤k0 )
f (yk⇤0 ) |Ik0 | .
But, the right-hand side can be made bigger than "/2 with an appropriate choice
of yk⇤0 because of the assumption that f is unbounded on Ik0 . This contradiction
forces the conclusion that f is bounded.
Thinking of P as the generic partition and using mk and Mk as usual with
Darboux sums, for each k, choose x⇤k , yk⇤ 2 Ik such that
Mk
January 19, 2015
f (x⇤k ) <
"
and f (yk⇤ )
4n|Ik |
mk <
"
.
4n|Ik |
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8-10
CHAPTER 8. INTEGRATION
With these selections,
D(f, P )
D(f, P )
= D(f, P )
=

n
X
|Mk
<
R (f, P, yk⇤ ) + R (f, P, yk⇤ )
f (x⇤k ))|Ik | + R (f, P, x⇤k )
R (f, P, yk⇤ ) +
f (x⇤k )| |Ik | + |R (f, P, x⇤k )
R (f, P, yk⇤ )| +
(Mk
k=1
n
X
k=1
R (f, P, x⇤k ) + R (f, P, x⇤k )
n
X
k=1
"
|Ik | + |R (f, P, x⇤k )
4n|Ik |
n
X
(f (yk⇤ )
k=1
n
X
R (f, P, yk⇤ )| +
D(f, P )
mk )|Ik |
(f (yk⇤ )
k=1
n
X
mk )|Ik |
"
|Ik |
4n|Ik |
k=1
" " "
< + + <"
4 2 4
Corollary 8.7 implies D(f ) exists and Theorem 8.9 finishes the proof.
Corollary 8.11. If
Rb
a
f exists and [c, d] ⇢ [a, b], then
Rd
c
⇤
f exists.
Proof. Let P0 = {a, b, c, d} 2 part([a, b]) and " > 0. Using Theorem 8.10,
choose a partition P" such that P0 ⌧ P" and whenever P" ⌧ P and P" ⌧ P 0 , then
|R (f, P, x⇤k )
R (f, P 0 , yk⇤ ) | < ".
Let P"1 = P" \ [a, c], P"2 = P" \ [c, d] and P"3 = P" \ [d, b]. Suppose P"2 ⌧ Q1 and
P"2 ⌧ Q2 . Then P"1 [ Qi [ P"3 for i = 1, 2 are refinements of P" and
|R (f, Q1 , x⇤k )
R (f, Q2 , yk⇤ ) | =
|R f, P"1 [ Q1 [ P"3 , x⇤k
R f, P"1 [ Q2 [ P"3 , yk⇤ | < "
for any selections. An application of Theorem 8.10 shows
Rb
a
f exists.
⇤
6. Properties of the Integral
Rb
Rb
Theorem 8.12. If a f and a g both exist, then
Rb
Rb
Rb
Rb
(a) If ↵, 2 R, then a (↵f + g) exists and a (↵f + g) = ↵ a f + a g.
Rb
(b) a f g exists.
Rb
(c) a |f | exists.
Proof. (a) Let " > 0. If ↵ = 0, in light of Example 8.1, it is clear ↵f is
integrable. So, assume ↵ 6= 0, and choose a partition Pf 2 part([a, b]) such that
whenever Pf ⌧ P , then
R (f, P, x⇤k )
January 19, 2015
Z
b
f <
a
"
.
2|↵|
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Properties of the Integral
8-11
Then
R (↵f, P, x⇤k )
↵
Z
b
f =
a
n
X
k=1
n
X
= |↵|
k=1
=
↵
f (x⇤k )|Ik |
R (f, P, x⇤k )
= |↵|
< |↵|
↵f (x⇤k )|Ik |
"
2|↵|
Z
Z
Z
b
f
a
b
f
a
b
f
a
"
.
2
Rb
Rb
This shows ↵f is integrable and a ↵f = ↵ a f .
Assuming 6= 0, in the same way, we can choose a Pg 2 part([a, b]) such that
when Pg ⌧ P , then
Z
R (g, P, x⇤k )
b
g <
a
"
.
2| |
Let P" = Pf [ Pg be the common refinement of Pf and Pg , and suppose P" ⌧ P .
Then
|R (↵f + g, P, x⇤k )
↵
Z
b
f+
a
Z
b
a
 |↵| R (f, P, x⇤k )
!
g |
Z
b
a
f + | | R (g, P, x⇤k )
Z
b
g <"
a
Rb
Rb
Rb
for any selection. This shows ↵f + g is integrable and a (↵f + g) = ↵ a f + a g.
Rb
Rb 2
(b) Claim: If a h exists, then so does a h
To see this, suppose first that 0  h(x)  M on [a, b]. If M = 0, the claim is
trivially true, so suppose M > 0. Let " > 0 and choose P 2 part([a, b]) such that
D(h, P )
D(h, P ) 
"
.
2M
For each 1  k  n, let
mk = glb {h(x) : x 2 Ik }  lub {h(x) : x 2 Ik } = Mk .
Since h
0,
m2k = glb {h(x)2 : x 2 Ik }  lub {h(x)2 : x 2 Ik } = Mk2 .
January 19, 2015
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8-12
CHAPTER 8. INTEGRATION
Using this, we see
D(h2 , P )
D(h2 , P ) =
=
n
X
k=1
n
X
(Mk2
m2k )|Ik |
(Mk + mk )(Mk
k=1
 2M
n
X
mk )|Ik |
(Mk
mk )|Ik |
= 2M D(h, P )
< ".
D(h, P )
k=1
!
Therefore, h2 is integrable when h 0.
If h is not nonnegative, let m = glb {h(x) : a  x  b}. Then h m
h m is integrable by (a). From the claim, (h m)2 is integrable. Since
h2 = (h
m)2 + 2mh
0, and
m2 ,
it follows from (a) that h2 is integrable.
Finally, f g = 14 ((f + g)2 (f g)2 ) is integrable by the claim and (a).
p
(c) Claim: If h 0 is integrable, then so is h.
To see this, let " > 0 and choose P 2 part([a, b]) such that
For each 1  k  n, let
and define
mk = glb {
Then
D(h, P )
p
p
h(x) : x 2 Ik }  lub { h(x) : x 2 Ik } = Mk .
A = {k : Mk
(70)
mk < "}
X
k2A
Using the fact that mk
X
(71)
(Mk
k2B
D(h, P ) < "2 .
(Mk
and
B = {k : Mk
mk )|Ik | < "(b
mk
"}.
a).
0, we see that Mk mk  Mk + mk , and
1X
mk )|Ik | 
(Mk + mk )(Mk mk )|Ik |
"
k2B
1X
=
(Mk2 m2k )|Ik |
"
k2B
1

D(h, P )
"
<"
D(h, P )
Combining (70) and (71), it follows that
p
p
D( h, P ) D( h, P ) < "(b a) + " = "((b a) + 1)
p
can be made arbitrarily
small. Therefore, h is integrable.
p
Since |f | = f 2 an application of (b) and the claim suffice to prove (c).
Rb
Theorem 8.13. If a f exists, then
January 19, 2015
⇤
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The Fundamental Theorem of Calculus
8-13
Rb
(a) If f 0 on [a, b], then a f 0.
Rb
Rb
(b) | a f |  a |f |
Rb
Rc
Rb
(c) If a  c  b, then a f = a f + c f .
Proof. (a) Since all the Riemann sums are nonnegative, this follows at once.
Rb
(b) It is always true that |f | ± f 0 and |f | f 0, so by (a), a (|f | + f ) 0
Rb
Rb
Rb
Rb
Rb
and a (|f | f ) 0. Rearranging these shows
f  a |f | and a f  a |f |.
a
Rb
Rb
Therefore, | a f |  a |f |, which is (b).
(c) By Corollary 8.11, all the integrals exist. Let " > 0 and choose Pl 2
part([a, c]) and Pr 2 part([c, b]) such that whenever Pl ⌧ Ql and Pr ⌧ Qr , then,
R (f, Ql , x⇤k )
Z
c
f <
a
"
2
R (f, Qr , yk⇤ )
and
Z
b
f <
c
"
.
2
If P = Pl [ Pr and Q = Ql [ Qr , then P, Q 2 part([a, b]) and P ⌧ Q. The triangle
inequality gives
Z c
Z b
R (f, Q, x⇤k )
f
f < ".
a
c
Since every refinement of P has the form Ql [ Qr , part (c) follows.
⇤
Rb
There’s some notational trickery that can be played here. If a f exists, then
Ra
Rb
Rb
Rc
Rb
we define b f =
f . With this convention, it can be shown a f = a f + c f
a
no matter the order of a, b and c, as long as at least two of the integrals exist. (See
Problem 8.4.)
7. The Fundamental Theorem of Calculus
Theorem 8.14 (Fundamental Theorem of Calculus 1). Suppose f, F : [a, b] ! R
satisfy
Rb
(a) a f exists
(b) F 2 C([a, b]) \ D((a, b))
(c) F 0 (x) = f (x), 8x 2 (a, b)
Rb
Then a f = F (b) F (a).
Proof. Let " > 0 and choose P" 2 part([a, b]) such that whenever P" ⌧ P ,
then
Z b
R (f, P, x⇤k )
f < ".
a
for every selection from P . On each interval [xk 1 , xk ] determined by P , the
function F satisfies the conditions of the Mean Value Theorem. (See Corollary 7.13.)
Therefore, for each k, there is an ck 2 (xk 1 , xk ) such that
F (xk )
January 19, 2015
F (xk
1)
= F 0 (ck )(xk
xk
1)
= f (ck )|Ik |.
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8-14
CHAPTER 8. INTEGRATION
So,
Z
b
f
(F (b)
F (a)) =
a
=
=
Z
Z
Z
b
f
a
b
f
a
n
X
k=1
n
X
k=1
b
f
a
(F (xk )
F (xk
1)
f (ck )|Ik |
R (f, P, ck )
<"
⇤
and the theorem follows.
Example 8.4. Suppose f and its first n derivatives are all continuous on [a, b].
There is a function Rn (x, t) such that
Rn (x, t) = f (x)
n
X1
k=0
f (k) (t)
(x
k!
t)k
for a  t  b. Di↵erentiating both sides of the equation with respect to t, note that
the right-hand side telescopes, so the result is
d
Rn (x, t) =
dt
(x t)n 1 (n)
f (t).
(n 1)!
Using Theorem 8.14 and the fact that Rn (x, x) = 0 gives
Rn (x, c) = Rn (x, c) Rn (x, x)
Z c
d
=
Rn (x, t) dt
x dt
Z x
(x t)n 1 (n)
=
f (t) dt,
(n 1)!
c
which is the integral form of the remainder from Taylor’s formula.
Corollary 8.15 (Integration by Parts). If f, g 2 C([a, b]) \ D((a, b)) and both
f 0 g and f g 0 are integrable on [a, b], then
Z b
Z b
0
fg +
f 0 g = f (b)g(b) f (a)g(a).
a
a
Proof. Use Theorems 7.3(c) and 8.14.
⇤
Rb
Suppose a f exists. By Corollary 8.11, f is integrable on every interval [a, x],
Rx
for x 2 [a, b]. This allows us to define a function F : [a, b] ! R as F (x) = a f ,
called the indefinite integral of f on [a, b].
Theorem 8.16 (Fundamental Theorem of Calculus 2). Let f be integrable on
[a, b] and F be the indefinite integral of f . Then F 2 C([a, b]) and F 0 (x) = f (x)
whenever x 2 C(f ) \ (a, b).
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The Fundamental Theorem of Calculus
8-15
M
m
x+h
x
Figure
Z 3.
1
lim
h!0 h
This figure illustrates a “box” argument showing
x+h
f = f (x).
x
Rb
Proof. To show F 2 C([a, b]), let x0 2 [a, b] and " > 0. Since a f exists,
there is an M > lub {|f (x)| : a  x  b}. Choose 0 <
< "/M and x 2
(x0
, x0 + ) \ [a, b]. Then
Z x
|F (x) F (x0 )| =
f  M |x x0 | < M < "
x0
and x0 2 C(F ).
Let x0 2 C(f )\(a, b) and " > 0. There is a > 0 such that x 2 (x0
(a, b) implies |f (x) f (x0 )| < ". If 0 < h < , then
Z
F (x0 + h) F (x0 )
1 x0 +h
f (x0 ) =
f f (x0 )
h
h x0
Z
1 x0 +h
=
(f (t) f (x0 )) dt
h x0
Z
1 x0 +h

|f (t) f (x0 )| dt
h x0
Z
1 x0 +h
<
" dt
h x0
= ".
, x0 + ) ⇢
This shows F+0 (x0 ) = f (x0 ). It can be shown in the same way that F 0 (x0 ) = f (x0 ).
Therefore F 0 (x0 ) = f (x0 ).
⇤
The right picture makes Theorem 8.16 almost obvious. Consider Figure 3.
Suppose x 2 C(f ) and " > 0. There is a > 0 such that
f ((x
d, x + d) \ [a, b]) ⇢ (f (x)
"/2, f (x) + "/2).
Let
m = glb {f y : |x
January 19, 2015
y| < }  lub {f y : |x
y| < } = M.
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8-16
CHAPTER 8. INTEGRATION
Apparently M
mh 
Since M
m < " and for 0 < h < ,
Z
x+h
x
f  M h =) m 
F (x + h)
h
F (x)
 M.
m ! 0 as h ! 0, a “squeezing” argument shows
lim
h#0
F (x + h)
h
F (x)
= f (x).
A similar argument establishes the limit from the left and F 0 (x) = f (x).
It’s easy to read too much into the Fundamental Theorem of Calculus. We
are tempted to start thinking of integration and di↵erentiation as opposites of
each other. But, this is far from the truth. The operations of integration and
antidi↵erentiation are di↵erent operations, that happen to sometimes be tied together
by the Fundamental Theorem of Calculus. Consider the following examples.
Example 8.5. Let
f (x) =
(
|x|/x,
0,
x 6= 0
x=0
It’s
R x easy to prove that f is integrable over any compact interval, and that F (x) =
f = |x| 1 is an indefinite integral of f . But, F is not di↵erentiable at x = 0
1
and f is not a derivative, according to Theorem 7.16.
Example 8.6. Let
f (x) =
(
x2 sin x12 ,
0,
x 6= 0
x=0
It’s straightforward to show that f is di↵erentiable and
(
2
1
2x sin x12
0
x cos x2 , x 6= 0
f (x) =
0,
x=0
Since f 0 is unbounded near x = 0, it follows from Theorem 8.3 that f 0 is not
integrable over any interval containing 0.
Example 8.7. Let f be the salt and pepper function of Example 6.15. It was
Rb
Rx
shown in Example 8.3 that a f = 0 on any interval [a, b]. If F (x) = 0 f , then
F (x) = 0 for all x and F 0 = f only on C(f ) = Qc .
8. Change of Variables
Integration by substitution works side-by-side with the Fundamental Theorem of
Calculus in the integration section of any calculus course. Most of the time calculus
books require all functions in sight to be continuous. In that case, a substitution
theorem is an easy consequence of the Fundamental Theorem and the Chain Rule.
(See Exercise 8.7.) More general statements are true, but they are harder to prove.
Theorem 8.17. If f and g are functions such that
(a) g is strictly monotone on [a, b],
(b) g is continuous on [a, b],
R g(b)
Rb
(c) both g(a) f and a (f g)g 0 exist,
January 19, 2015
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8. CHANGE OF VARIABLES
then
Z
(72)
8-17
g(b)
f=
g(a)
Z
b
(f
g)g 0 .
a
Proof. Suppositions (a) and (b) show g is a bijection from [a, b] to an interval
[c, d]. The correspondence between the endpoints depends on whether g is increasing
or decreasing.
Let " > 0.
From (d) and Definition 8.1, there is a 1 > 0 such that whenever P 2 part([a, b])
with kP k < 1 , then
Z b
"
(73)
R ((f g)g 0 , P, x⇤i )
(f g)g 0 <
2
a
for any selection from P . Choose P1 2 part([a, b]) such that kP1 k < 1 .
Using the same argument, there is a 2 > 0 such that whenever Q 2 part([c, d])
with kQk < 2 , then
Z d
"
(74)
R (f, Q, x⇤i )
f <
2
c
for any selection from Q. As above, choose Q1 2 part([c, d]) such that kQ1 k < 2 .
Setting P2 = P1 [ {g 1 (x) : x 2 Q1 } and Q2 = P1 [ {g(x) : x 2 P1 }, it is
apparent that P1 ⌧ P2 , Q1 ⌧ Q2 , kP2 k  kP1 k < 1 , kQ2 k  kQ1 k < 2 and
Q2 = {g(x) : x 2 P2 }. From (73) and (74), it follows that
Z b
"
(f g)g 0 R ((f g)g 0 , P2 , x⇤i ) <
2
a
(75)
and
Z
d
f
c
R (f, Q2 , yi⇤ ) <
"
2
for any selections from P2 and Q2 .
Label the points of P2 as a = x1 < x2 < · · · < xn = b and those of Q2 as
c = y0 < y1 < · · · < yn = d. From (b), (c) and the Mean Value Theorem, for each i,
choose ci 2 (xi 1 , xi ) such that
(76)
g(xi )
g(xi
1)
= g 0 (ci )(xi
xi
1 ).
Notice that {ci : 1  i  n} is a selection from P2 .
First, assume g is strictly increasing. In this case g(xi ) = yi for 0  i  n and
g(ci ) 2 (yi 1 , yi ) for 0 < i  n, so g(ci ) is a selection from Q2 .
Z
Z
g(b)
f
g(a)
=
b
(f
Z
g(b)
f
g(a)

January 19, 2015
g)g 0
a
Z
R (f, Q2 , g(ci )) + R (f, Q2 , g(ci ))
Z
b
(f
g(b)
f
g(a)
g)g 0
a
R (f, Q2 , g(ci )) + R (f, Q2 , g(ci ))
Z
b
(f
g)g 0
a
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8-18
CHAPTER 8. INTEGRATION
Use the triangle inequality and (75). Expand the second Riemann sum.
<
n
X
"
+
f (g(ci )) (g(xi )
2
i=1
g(xi
Z
1 ))
b
g)g 0
(f
a
Apply the Mean Value Theorem, as in (76), and then use (75).
=
n
X
"
+
f (g(ci ))g 0 (ci ) (xi
2
i=1
"
= + R ((f
2
" "
< +
2 2
="
0
g)g , P2 , ci )
xi
Z
Z
1)
b
b
(f
g)g 0
a
g)g 0
(f
a
and (72) follows.
Now assume g is strictly decreasing on [a, b]. The proof is much the same as above,
except the bookkeeping is trickier because order is reversed instead of preserved
by g. This means g(xk ) = yn k when 0  k  n and g(cn k+1 ) 2 (yk 1 , yk ) for
0 < k  n. Therefore, g(cn k+1 ) is a selection from Q2 . From the Mean Value
Theorem,
yk
yk
(77)
1
= g(xn
=
k)
(g(xn
k+1 )
0
=
g(xn
g (cn
k+1 )
g(xn
k ))
k+1 )(xn k+1
xn
k ),
where ck 2 (xk 1 , xk ) is as above. The rest of the proof is much like the case when
g is increasing.
Z
Z
g(b)
f
g(a)
=

Z
Z
b
(f
g)g 0
a
Z
g(a)
g(b)
f + R (f, Q2 , g(cn
k+1 ))
R (f, Q2 , g(cn
k+1 ))
g(b)
g(a)
f + R (f, Q2 , g(cn
k+1 )) +
R (f, Q2 , g(cn
k+1 ))
b
(f
g)g 0
a
Z
b
(f
g)g 0
a
Use (75), expand the second Riemann sum and apply (77).
<
=
"
+
2
"
+
2
January 19, 2015
n
X
k=1
n
X
f (g(cn
f (g(cn
k=1
k+1 ))(yk
k+1 ))g
0
(cn
yk
1)
Z
b
(f
g)g 0
a
k+1 )(xn k+1
xn
k)
Z
b
(f
a
g)|g 0 |
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Integral Mean Value Theorems
8-19
Reverse the order of the sum and use (75).
n
X
"
= +
f (g(ck ))g 0 (ck )(xk
2
k=1
"
+ R ((f
2
" "
< +
2 2
="
g)g 0 , P2 , ck )
=
xk
Z
Z
1)
b
(f
b
(f
g)g 0
a
g)g 0
a
⇤
The theorem has been proved.
R1 p
Example 8.8. Suppose we want to calculate 1 1 x2 dx. Using the notation
p
of Theorem 8.17, let f (x) = 1 x2 , g(x) = sin x and [a, b] = [ ⇡/2, ⇡/2]. In this
case, g is an increasing function. Then (72) becomes
Z 1p
Z sin(⇡/2) p
1 x2 dx =
1 x2 dx
1
Z
=
Z
=
sin( ⇡/2)
⇡/2
⇡/2
⇡/2
p
1
sin2 x cos x dx
cos2 d dx
⇡/2
⇡
.
2
On the other hand, it can also be done with a decreasing function. If g(x) = cos x
and [a, b] = [0, ⇡], then
Z 1p
Z cos 0 p
2
1 x dx =
1 x2 dx
1
cos ⇡
Z cos ⇡ p
=
1 x2 dx
cos 0
Z ⇡p
=
1 cos2 x( sin x) dx
0
Z ⇡p
=
1 cos2 x sin x dx
0
Z ⇡
=
sin2 x dx
=
0
⇡
=
2
9. Integral Mean Value Theorems
Theorem 8.18. Suppose f, g : [a, b] ! R are such that
(a) g(x) 0 on [a, b],
(b) f is bounded and m  f (x)  M for all x 2 [a, b], and
Rb
Rb
(c) a f and a f g both exist.
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8-20
CHAPTER 8. INTEGRATION
There is a c 2 [m, M ] such that
Proof. Obviously,
(78)
If
m
Rb
a
Z
Z
b
fg = c
a
b
a
g
Z
Z
b
g.
a
b
fg  M
a
Z
b
g.
a
g = 0, we’re done. Otherwise, let
Then
Rb
a
fg = c
Rb
a
Rb
fg
c = Ra b .
g
a
g and from (78), it follows that m  c  M .
⇤
Corollary 8.19. Let f and g be as in Theorem 8.18, but additionally assume
f is continuous. Then there is a c 2 (a, b) such that
Z b
Z b
f g = f (c)
g.
a
a
Proof. This follows from Theorem 8.18 and Corollaries 6.23 and 6.26.
⇤
Theorem 8.20. Suppose f, g : [a, b] ! R are such that
(a) g(x) 0 on [a, b],
(b) f is bounded and m  f (x)  M for all x 2 [a, b], and
Rb
Rb
(c) a f and a f g both exist.
There is a c 2 [a, b] such that
Z b
Z c
Z b
fg = m
g+M
g.
a
a
Proof. For a  x  b let
G(x) = m
Z
By Theorem 8.16, G 2 C([a, b]) and
Z b
Z
glb G  G(b) = m
g
a
c
x
g+M
a
b
a
fg  M
Z
g.
x
Z
Now, apply Corollary 6.26 to find c where G(c) =
10. Exercises
b
b
a
Rb
a
g = G(a)  lub G.
f g.
⇤
8.1. If f : [a, b] ! R and R(f ) exists, then f is bounded.
8.2. Let
f (x) =
(
1,
0,
x2Q
.
x2
/Q
(a) Use Definition 8.1 to show f is not integrable on any interval.
(b) Use Definition 8.6 to show f is not integrable on any interval.
January 19, 2015
http://math.louisville.edu/⇠lee/ira
10. EXERCISES
8.3. Calculate
R5
2
8-21
x2 using the definition of integration.
8.4. If at least two of the integrals exist, then
Z b
Z c
Z
f=
f+
a
a
b
f
c
no matter the order of a, b and c.
8.5. If ↵ > 0, f : [a, b] ! [↵, ] and
Rb
a
f exists, then
Rb
a
1/f exists.
8.6. If f : [a, b] ! [0, 1) is continuous and D(f ) = 0, then f (x) = 0 for all
x 2 [a, b].
8.7. In the statement of Theorem 8.17, make the additional assumption that f is
continuous. Use the Fundamental Theorem of Calculus to give an easier proof.
January 19, 2015
http://math.louisville.edu/⇠lee/ira