4 Integration
The theory of infinite series addresses the problem of summing infinitely many finite quantities. By
contrast, integration is the business of summing infinitely many infinitesimal quantities. Mathemati-
cians have attempted to do both for well over 2000 years, and the philosophical objections are just as
old.
1
The development and increased application of calculus from the late 1600s spurred mathemati-
cians to try to put the theory on a firmer footing, though from Newton and Leibniz it took another
150 years before Bernhard Riemann (1856) provided a thorough development of the integral.
32 The Riemann Integral
The basic idea behind Riemann integration is to approximate area using a sequence of rectangles
whose width tends to zero. The following discussion is hopefully familiar.
Example 4.1. Consider f (x) = x
2
defined on [0, 1].
For each n ∈ N, let ∆x =
1
n
and define x
i
= i∆x.
Above each subinterval [x
i−1
, x
i
], raise a rectangle of height
f (x
i
) = x
2
i
.
The sum of the areas of these rectangles is the Riemann sum
with right-endpoints
2
R
n
=
n
∑
i=1
f (x
i
) ∆x =
n
∑
i=1
i
2
n
3
=
n(n + 1)(2n + 1)
6n
3
=
1
3
+
3n + 1
6n
2
The Riemann sum with left-endpoints is defined similarly:
L
n
=
n
∑
i=1
f (x
i−1
) ∆x =
n
∑
i=1
(i −1)
2
n
3
=
1
3
−
3n −1
6n
2
Since f is increasing, the area A under the curve satisfies
L
n
≤ A ≤ R
n
and the squeeze theorem allows us to conclude that A =
1
3
.
0
1
0 1
n =
16
R
n
=
0.365234
0
1
0 1
n =
16
L
n
=
0.302734
The example contains the essential idea, but more flexibility is needed. To get further, we must
properly define the concepts of partition and Riemann sum.
1
Two of Zeno’s ancient paradoxes are relevant here: Achilles and the Tortoise concerns a convergent infinite series,
while the Arrow Paradox discusses a difficulty with integration by questioning whether time can be considered as a sum
of instants. Perhaps the most famous contemporary criticism comes from Bishop George Berkeley, who gave his name
to the Californian city and thus the first UC campus: in The Analyst (1734), Berkeley savaged the foundations of calcu-
lus, describing the infinitesimal increments required in Newton’s theory of fluxions (derivatives) as merely the “ghosts of
departed quantities.”
2
Now is a good time to review some identities:
n
∑
i=1
i =
1
2
n(n + 1),
n
∑
i=1
i
2
=
1
6
n(n + 1) (2n + 1),
n
∑
i=1
i
3
=
1
4
n
2
(n + 1)
2
Definition 4.2. A partition P = {x
0
, . . . , x
n
} of an interval [a, b] is a finite sequence such that
a = x
0
< x
1
< ··· < x
n−1
< x
n
= b
For each 1 ≤ i ≤ n, define ∆x
i
= x
i
− x
i−1
. The mesh of the partition is mesh(P) := max ∆x
i
.
Choose a sample point x
∗
i
in each subinterval [x
i−1
, x
i
].
If f : [a, b] → R, the Riemann sum
n
∑
i=1
f (x
∗
i
) ∆x
i
computes the area of a family of n rectangles.
f (x)
x
x
1
x
∗
1
x
2
x
∗
2
x
3
x
∗
3
x
4
x
∗
4
x
5
x
∗
5
x
6
x
∗
6
b = x
7
x
∗
7
a = x
0
b = x
7
In elementary calculus, one typically computes Riemann sums for equally-spaced partitions with left,
right or middle sample points. The double freedom of partition & sample points makes applying the
definition a challenge, so instead we consider two special families of rectangles.
Definition 4.3. Given a partition P of [a, b] and a bounded function f on [a, b], define
M
i
= sup
x∈[x
i−1
,x
i
]
f (x) U( f , P) =
n
∑
i=1
M
i
∆x
i
m
i
= inf
x∈[x
i−1
,x
i
]
f (x) L( f, P) =
n
∑
i=1
m
i
∆x
i
U( f , P) and L( f , P) are the upper and lower Darboux sums for
f with respect to P. The upper and lower Darboux integrals are
U( f ) = inf U( f , P) L( f ) = sup L( f , P)
where the supremum and infimum are over all partitions.
We say that f is (Riemann) integrable on [a, b] if U( f ) = L( f )
and denote this value by
Z
b
a
f or
Z
b
a
f (x) dx
If the interval is understood or irrelevant, it is common just
to say that f is integrable and write
R
f .
a x
1
x
2
x
3
x
4
x
5
x
6
x
7
b
Upper Darboux sum U( f , P)
a x
1
x
2
x
3
x
4
x
5
x
6
x
7
b
Lower Darboux sum L( f , P)
Intuitively, L( f , P) is the sum of the areas of rectangles built on P which just fit under the graph of
f . It is also the infimum of all Riemann sums on P. If f is discontinuous, then L( f , P) need not be a
Riemann sum; there might not be suitable sample points!
2
Examples 4.4. 1. We revisit Example 4.1 in this language.
Given a partition Q = {x
0
, . . . , x
n
} of [ 0, 1] and sample points x
∗
i
∈ [x
i−1
, x
i
], we compute the
Riemann sum for f (x) = x
2
n
∑
i=1
f (x
∗
i
) ∆x
i
=
n
∑
i=1
(x
∗
i
)
2
(x
i
− x
i−1
)
Since f is increasing, we have x
2
i−1
≤ (x
∗
i
)
2
≤ x
2
i
on each interval, whence
L( f , Q) =
n
∑
i=1
(x
i−1
)
2
(x
i
− x
i−1
) ≤
n
∑
i=1
(x
∗
i
)
2
(x
i
− x
i−1
) ≤
n
∑
i=1
(x
i
)
2
(x
i
− x
i−1
) = U( f , Q)
The Darboux sums are therefore the Riemann sums for right and left endpoints.
If we take Q
n
to be the partition with subintervals of equal width ∆x =
1
n
, then
U( f ) = inf
P
U( f , P) ≤ U( f , Q
n
) =
n
∑
i=1
i
n
2
∆x = R
n
is the right Riemann sum discussed originally. Similarly L( f ) ≥ L
n
. Since L
n
and R
n
both
converge to
1
3
as n → ∞, the squeeze theorem forces
L
n
≤ L( f ) ≤ U( f ) ≤ R
n
=⇒ L( f ) = U( f ) =
1
3
whence f is Riemann integrable on [0, 1] with
R
1
0
x
2
dx =
1
3
.
2. Suppose f (x) = kx + c on the interval [a, b] where
k > 0. Take the evenly spaced partition P
n
where
x
i
= a +
b−a
n
i with ∆x
i
=
b−a
n
. Since f is increasing,
the upper Darboux sum is again the Riemann sum
with right-endpoints:
U( f , P
n
) = R
n
=
n
∑
i=1
f (x
i
) ∆x
=
b − a
n
n
∑
i=1
k(b −a)
n
i + ak + c
0
ba
c
ak + c
bk + c
U( f , P
n
)
=
b − a
n
k(b −a)
n
·
1
2
n(n + 1) + (ak + c)n
−−−→
n→∞
1
2
k(b −a)
2
+ (b − a)(ak + c) =
k
2
( b
2
− a
2
) + c(b − a)
We similarly see that the lower Darboux sum is given by the Riemann sum with left endpoints,
and that
L( f , P
n
) = L
n
=
b − a
n
k(b −a)
n
·
1
2
n(n −1) + (ak + c)n
−−−→
n→∞
k
2
( b
2
− a
2
) + c(b − a)
By the same argument as above, L
n
≤ L( f ) ≤ U( f ) ≤ R
n
and the squeeze theorem show that
f is integrable with
R
b
a
f =
k
2
( b
2
− a
2
) + c(b − a).
3
Following the examples, a few remarks are in order.
Riemann versus Darboux Definition 4.3 is really that of the Darboux integral. Riemann’s definition is
as follows: for f [a, b] → R to be integrable with integral
R
b
a
f means
∀ϵ > 0, ∃δ such that ∀P, x
∗
i
, mesh(P) < δ =⇒
n
∑
i=1
f (x
∗
i
) ∆x
i
−
Z
b
a
f
< ϵ
It can be shown that this is equivalent to the Darboux integral. We won’t pursue Riemann’s
formulation further, except to observe that if a function is integrable and mesh(P
n
) → 0, then
R
b
a
f = lim
n→∞
n
∑
i=1
f (x
∗
i
) ∆x
i
: this allows us to approximate integrals using any sample points we
choose, hence why right endpoints (x
∗
i
= x
i
) are so common in Freshman calculus.
Monotone Functions Darboux sums are particularly easy to compute for monotone functions. As in
the examples, if f is increasing, then each M
i
= f (x
i
), from which U( f , P) is the Riemann sum
with right-endpoints. Similarly, L( f , P) is the Riemann sum with left-endpoints. The roles reverse
if f is decreasing.
Area If f is positive and continuous,
3
the Riemann integral
R
b
a
f serves as a definition for the area
under the curve y = f (x). This should make intuitive sense:
1. In the second example where we have a straight line, we obtain the same value for the
area by computing directly as the sum of a rectangle and a triangle!
2. If the area under the curve is to make sense, then, for any partition P, it plainly satisfies
the inequalities
L( f , P) ≤ Area ≤ U( f , P)
But these are exactly the same as those satisfied by the integral itself:
L( f , P) ≤ L( f ) =
Z
b
a
f = U( f ) ≤ U( f , P)
In the examples we exhibited a sequence of partitions (P
n
) where U( f , P
n
) and L( f , P
n
) both con-
verged to the same limit. The next results develop some basic properties of partitions and make this
process rigorous.
Lemma 4.5. Suppose f : [a, b] → R is bounded and suppose P, Q are partitions of [a, b].
1. If Q is a refinement of P, that is P ⊆ Q, then
L( f , P) ≤ L( f , Q) ≤ U( f , Q) ≤ U( f , P)
2. For any partitions P, Q, we have L( f , P) ≤ U( f , Q)
3. L( f ) ≤ U( f )
3
We’ll see later (Theorem 4.16) that every continuous function is integrable.
4
Proof. 1. We prove inductively. First suppose that Q = P ∪{t} contains exactly one additional point
t ∈ (x
k−1
, x
k
). Write
m
1
= inf{f (x) : x ∈ [x
k−1
, t]}
m
2
= inf{f (x) : x ∈ [t, x
k−1
]}
m = inf{f (x) : x ∈ [x
k−1
, x
k
]} = min{m
1
, m
2
}
The Darboux sums L( f , P) and L( f , Q) are identical ex-
cept for the terms involving t; indeed
x
k−1
x
k
t
Extra area!
··· ···
m
1
m
2
L( f , Q) − L( f , P) = m
1
( t − x
k−1
) + m
2
(x
k
−t) −m(x
k
− x
k−1
)
= (m
1
−m)(t −x
k−1
) + (m
2
−m)(x
k
−t) ≥ 0
Since partitions are finite sets, by induction we see that P ⊆ Q =⇒ L( f , P) ≤ L( f , Q).
The argument for U( f , Q) ≤ U( f , P) is similar, and the middle inequality is trivial.
2. If P and Q are partitions, then P ∪ Q is a refinement of both P and Q. By part 1,
L( f , P) ≤ L( f , P ∪Q) ≤ U( f , P ∪ Q) ≤ U( f , Q) (∗)
3. We leave this as an exercise.
Theorem 4.6 (Cauchy criterion for integrability). Suppose f : [a, b] → R is bounded.
1. f is integrable ⇐⇒ ∀ϵ > 0, ∃P such that U( f , P) − L( f , P) < ϵ
2. f is integrable ⇐⇒ ∃(P
n
)
n∈N
such that U( f , P
n
) − L( f , P
n
) → 0. Moreover, in such a case
both sequences U( f , P
n
) and L( f , P
n
) converge to
R
b
a
f .
We call this a Cauchy criterion since integrability is demonstrated without mention of the integral!
Proof. 1. (⇒) Suppose f is integrable. Since inf U( f , Q) =
R
f = sup L( f , R), ∃Q, R such that
U( f , Q) −
Z
f <
ϵ
2
and
Z
f − L( f , R) <
ϵ
2
Let P = Q ∪ R and apply (∗): L( f , R) ≤ L( f , P) ≤ U( f , P) ≤ U( f , Q). But then
U( f , P) − L( f , P) ≤ U( f , Q) − L( f , R) = U( f , Q) −
Z
f +
Z
f − L( f , R) < ϵ
(⇐) For every partition, L( f , P) ≤ L( f ) ≤ U( f ) ≤ U( f , P). Thus
0 ≤ U( f ) − L( f ) ≤ U( f , P) − L( f , P) < ϵ
Since this holds for all ϵ > 0, we see that U( f ) = L( f ).
2. This is an exercise.
5
Examples 4.7. 1. The freedom to choose a partition can be very useful. Consider f (x) =
√
x on the
interval [0, b]. We choose a partition that evaluates nicely when fed to this function:
P
n
= {x
0
, . . . , x
n
} where x
i
=
i
n
2
b
=⇒ ∆x
i
= x
i
− x
i−1
=
b
n
2
i
2
−(i −1)
2
=
(2i −1)b
n
2
Since f is increasing on [0, b], we see that
U( f , P
n
) =
n
∑
i=1
f (x
i
) ∆x
i
=
n
∑
i=1
i
√
b
n
·
(2i −1)b
n
2
=
b
3/2
n
3
n
∑
i=1
2i
2
−i
=
b
3/2
n
3
1
3
n(n + 1)(2n + 1) −
1
2
n(n + 1)
−−−→
n→∞
2
3
b
3/2
Similarly
L( f , P
n
) =
n
∑
i=1
f (x
i−1
) ∆x
i
=
n
∑
i=1
(i −1)
√
b
n
·
(2i −1)b
n
2
=
b
3/2
n
3
n
∑
i=1
2i
2
−3i + 1
=
b
3/2
n
3
1
3
n(n + 1)(2n + 1) −
3
2
n(n + 1) + n
−−−→
n→∞
2
3
b
3/2
Since these limits are equal, we conclude that f is integrable and that
R
b
0
√
x dx =
2
3
b
3/2
.
0
0 b
√
b
Upper Sum U( f , P
n
)
0
0 b
√
b
Lower Sum L( f , P
n
)
2. We finish this section with the classic example of a non-integrable function. Let f : [a, b] → R
to be the indicator function of the irrational numbers,
f (x) =
(
1 if x ∈ Q
0 if x ∈ Q
Since any interval of positive length contains both rational and irrational numbers, we see that
sup
f (x) : x ∈ [x
i−1
, x
i
]
= 1 and inf
f (x) : x ∈ [x
i−1
, x
i
]
= 0
for any partition P = {x
0
, . . . , x
n
}. We conclude that
U( f , P) =
n
∑
i=1
(x
i
− x
i−1
) = b − a =⇒ U( f ) = b − a and
L( f , P) = 0 =⇒ L( f ) = 0
Since the upper and lower integrals are unequal, f is not Riemann integrable.
6
As any freshman calculus student can attest, if you can find an anti-derivative, then the fundamental
theorem of calculus (Section 34) makes evaluating integrals far easier. For instance, you are probably
desperate to write
d
dx
2
3
x
3/2
= x
1/2
=⇒
Z
b
0
√
x dx =
2
3
x
3/2
b
0
=
2
3
b
3/2
rather than computing Riemann/Darboux sums as in the previous example! In most practical cases,
however, no easy-to-compute anti-derivative exists, so the best we can do is approximate integrals
by evaluating Riemann sums for progressively finer partitions. Thankfully computers excel at such
tedious work!
Exercises 32 1. For each function on the given interval, use partitions to find the upper and lower
Darboux integrals. Hence prove that the function is integrable and compute its integral.
(a) f (x) = x
3
on [0, b] for any b > 0.
(b) g(x) =
3
√
x on [0, b].
(Hint: mimic Example 4.7.1)
2. Repeat question 1 for the following two functions. You cannot simply compute Riemann sums
for left and right endpoints and take limits: why not?
(a) h(x) = x(2 − x) on [ 0, 2]
(Hint: choose a partition with 2n points such that x
n
= 1 and observe that h(2 − x) = h(x))
(b) k(x) =
(
2x if x ≤ 1
5 − x if x > 1
on [0, 3].
(Hint: this time try a partition with 3n points. . . )
3. Let f (x) = x for rational x and f (x) = 0 for irrational x.
(a) Calculate the upper and lower Darboux integrals for f on the interval [0, b].
(b) Is f integrable on [0, b]?
4. Prove part 3 of Lemma 4.5: L( f ) ≤ U( f ).
5. Prove part 2 of Theorem 4.6.
f is integrable ⇐⇒ ∃(P
n
)
n∈N
such that U( f , P
n
) − L( f , P
n
) → 0.
Moreover, both U( f , P
n
) and L( f , P
n
) converge to
R
f .
6. (a) Reread Definition 4.3. What happens if we allow f : [a, b] → R to be unbounded?
(b) Read “Riemann versus Darboux” on page 4. Explain why being Riemann integrable also
forces f to be bounded.
7. (If you like coding) Write a short program to estimate
R
b
a
f (x) dx using Riemann sums. This
can be very simple (equal partitions with right endpoints), or more complex (random partition
and sample points given a mesh). Apply your program to estimate
R
5
0
sin(x
2
e
−
√
x
) dx.
7
33 Properties of the Riemann Integral
The rough take-away of this long section is that everything you think is integrable probably is! There
will not be many examples since we have not established many explicit values for integrals.
Theorem 4.8 (Linearity). If f , g are integrable and k, l are constant, then k f + lg is integrable and
Z
k f + lg = k
Z
f + l
Z
g
Example 4.9. Thanks to examples in the previous section, we can now calculate, for instance
Z
2
0
5x
3
−3
√
x dx = 5 ·
1
4
·2
4
−3 ·
2
3
·2
3/2
= 20 −4
√
2
Proof. Suppose ϵ > 0 is given. By Theorem 4.6 part 3, there exist partitions R, S such that
U( f , R) − L( f , R) <
ϵ
2
and U(g, S) − L(g, S) <
ϵ
2
By Theorem 4.6 part 1, if P := R ∪ S, then both inequalities are satisfied by P. On each subinterval,
inf f (x) + inf g(x) ≤ inf( f (x) + g(x)) and sup( f (x) + g(x)) ≤ sup f (x) + sup g(x)
since the individual suprema/infima could be ‘evaluated’ at different places. Thus
L( f , P) + L(g, P) ≤ L( f + g, P) ≤ U( f + g, P) ≤ U( f , P) + U(g, P)
whence U( f + g, P) − L( f + g, P) < ϵ and f + g is integrable. Moreover,
Z
( f + g) −
Z
f −
Z
g ≤
U( f , P) −
Z
f
+
U(g, P) −
Z
g
< ϵ
Using lower Darboux integrals similarly, we see that
−ϵ <
Z
( f + g) −
Z
f −
Z
g < ϵ
Since this holds for all ϵ > 0, we conclude that
R
( f + g) =
R
f +
R
g.
That k f is integrable with
R
k f = k
R
f is an exercise. Put these together for the result.
Corollary 4.10 (Changing endvalues). Suppose g is integrable on [a, b] and that f : [ a, b] → R
satisfies f (x) = g(x) on (a, b). Then f is integrable on [a, b] and
R
b
a
f =
R
b
a
g.
Definition 4.11 (Integration on an open interval). A bounded function f : (a, b) → R is integrable if
it has an integrable extension g : [a, b] → R where f (x) = g(x) on (a, b). In such a case, we define
R
b
a
f :=
R
b
a
g.
The Corollary (its proof is an exercise) shows that the choice of extension is irrelevant.
8
Theorem 4.12 (Basic Comparisons). Suppose f and g are integrable on [a, b].
1. If f (x) ≤ g(x), then
R
f ≤
R
g.
2. If m ≤ f (x) ≤ M then m( b − a) ≤
R
b
a
f ≤ M(b − a).
3. f g is integrable.
4.
|
f
|
is integrable and
R
f
≤
R
|
f
|
.
5. max( f , g) and min( f , g) are integrable.
Part 3 is not integration by parts and does not tell us how
R
f g relates to
R
f and
R
g!
Proof. 1. Since g( x) − f (x) ≥ 0 is integrable, L(g − f , P) ≥ 0 for all partitions P, and so
0 ≤ L(g − f ) =
Z
g − f =
Z
g −
Z
f
2. Apply part 1 twice.
3. This is an exercise.
4. The integrability is an exercise. For the comparison, apply part 1 to −
|
f
|
≤ f ≤
|
f
|
.
5. max( f , g) =
1
2
( f + g) +
1
2
|
f − g
|
, etc.
Theorem 4.13 (Domain splitting). Suppose that f : [a, b] → R
and let c ∈ (a, b). If f is integrable on both [a, c] and [c, b], then it
is integrable on [a, b] and
Z
b
a
f =
Z
c
a
f +
Z
b
c
f
f (x)
a c b x
R
c
a
f
R
b
c
f
In light of this result, it is conventional to allow integral limits to be reversed:
Z
a
b
f := −
Z
b
a
f is consistent with
Z
a
a
f = 0
Proof. Let ϵ > 0 be given, then ∃R, S partitions of [a, c], [c, b] such that
U( f , R) − L( f , R) <
ϵ
2
, U( f , S) − L( f , S) <
ϵ
2
Choose P = R ∪S to partition [a, b], then
U( f , P) − L( f , P) = U( f , R) + U( f , S) − L( f , R) −L( f , S) < ϵ
Moreover
f (x)
a c b x
R
z
} | {
S
z
} | {
Z
b
a
f −
Z
c
a
f −
Z
b
c
f ≤ U( f , P) − L( f , R) − L( f , S) = U( f , P) − L( f , P) < ϵ
The other side is similar.
9
Example 4.14. If f (x) =
√
x on [0, 1] and f (x) = 1 on [1, 2], then
Z
2
0
f =
Z
1
0
√
x dx +
Z
2
1
1 dx =
2
3
+ 1 =
5
3
Monotonic & Continuous Functions
We are now in a position to establish the integrability of two large classes of functions.
Definition 4.15. A function f : [a, b] → R is:
Monotonic if it is either increasing (x < y =⇒ f (x) ≤ f (y)) or decreasing.
Piecewise monotonic if there is a partition P = {x
0
, . . . , x
n
} of [a, b] such that f is monotonic on each
open subinterval (x
k−1
, x
k
).
Piecewise continuous if there is a partition such that f is uniformly continuous on each (x
k−1
, x
k
).
Theorem 4.16. If f is monotonic or continuous on [a, b], then it is integrable.
Examples 4.17. 1. Since sine is continuous, we can approximate via a sequence of Riemann sums
Z
π
0
sin x dx =
π
n
lim
n→∞
n
∑
i=1
sin
πi
n
Evaluating this limit is another matter entirely, one best handled in the next section...
2. Similarly, e
√
x
can be integrated and therefore approximated via Riemann sums:
Z
1
0
e
√
x
dx =
1
n
lim
n→∞
n
∑
i=1
exp
r
i
n
= lim
n→∞
n
∑
i=1
2j −1
n
exp
j
n
Both sums use right endpoints; the first has equal subintervals and the second is analogous to
Example 4.7.1. These limits would typically be estimated using a computer.
Proof. Suppose f : [a, b] → R is continuous. Since [a, b] is closed and bounded, f is uniformly contin-
uous. Let ϵ > 0 be given:
∃δ > 0 such that ∀x, y ∈ [a, b],
|
x −y
|
< δ =⇒
|
f (x) − f (y)
|
<
ϵ
b − a
Let P be a partition with mesh P < δ. Since f attains its bounds on each [x
i−1
, x
i
] (extreme value
theorem),
∃x
∗
i
, y
∗
i
∈ [x
i−1
, x
i
] such that M
i
−m
i
= f (x
∗
i
) − f (y
∗
i
) <
ϵ
b − a
from which
U( f , P) − L( f , P) <
n
∑
i=1
ϵ
b − a
(x
i
− x
i−1
) = ϵ
The monotonicity argument is an exercise.
10
Corollary 4.18. Piecewise continuous and bounded piecewise monotonic functions are integrable.
Proof. If f is piecewise continuous, then the restriction of f to (x
k−1
, x
k
) has a continuous extension
g
k
: [x
k−1
, x
k
] → R; integrable by Theorem 4.16. By Corollary 4.10, f is integrable on [x
k−1
, x
k
] with
R
x
k
x
k−1
f =
R
x
k
x
k−1
g
k
. Several applications of Theorem 4.13 finish things off:
Z
b
a
f =
n
∑
k=1
Z
x
k
x
k−1
f
The argument for piecewise monotonicity is similar.
Example 4.19. The ‘fractional part’ function f (x) = x − ⌊x⌋
is both piecewise continuous and piecewise monotone on any
bounded interval. It is therefore integrable on any [a, b].
0
1
0 1 2 3 4 5
We finish with the final incarnation of the intermediate value theorem.
Corollary 4.20 (IVT for integrals). If f is continuous on [a, b], then ∃ξ ∈ (a, b) for which
f ( ξ) =
1
b − a
Z
b
a
f
Proof. Since f is continuous, it is integrable on [a, b]. By the extreme value theorem it is also bounded
and attains its bounds: ∃p, q ∈ [a, b] such that
f (p) := inf
x∈[a,b]
f (x), f ( q) = sup
x∈[a,b]
f (x)
Applying Theorem 4.12, part 2, with m = f (p) and
M = f (q) , we see that
( b − a) f (p) ≤
Z
b
a
f ≤ (b − a) f (q)
f
av
M
m
ξa bp q
Now divide by b −a and apply the usual intermediate value theorem for f to see that the required ξ
exists between p and q.
In the picture, when f is positive and continuous, the grey area equals that under the curve; imagine
levelling off the blue hill with a bulldozer. . . The notation f
av
=
1
b−a
R
b
a
f is short for the average value
of f on [a, b]: to see why this interpretation is sensible, approach
R
f via a sequence of Riemann sums
on equally-spaced partitions P
n
, then
1
b − a
Z
b
a
f = lim
n→∞
n
∑
i=1
f (x
∗
i
) ∆x = lim
n→∞
f (x
∗
1
) + ··· + f (x
∗
n
)
n
is the limit of a sequence of averages of equally-spaced samples f (x
∗
i
).
11
What can/can’t be integrated? (non-examinable)
We now know a great many examples of integrable functions: essentially
• Piecewise continuous & monotonic functions are integrable.
• Linear combinations, products, absolute values, maximums and minimums of (already) inte-
grable functions.
After so many positive integrability conditions, it is reasonable to ask precisely which functions are
Riemann integrable. There is a precise answer, though it is quite tricky to understand.
Theorem 4.21 (Lebesgue). Suppose f : [a, b] → R is bounded. Then
f is Riemann integrable ⇐⇒ it is continuous except on a set of measure zero
Na
¨
ıvely, the measure of a set is the sum of the lengths of its maximal subintervals; though unfor-
tunately this doesn’t make for a very useful definition.
4
Any countable subset has measure zero;
Lebesgue’s result is almost as if we can extend Corollary 4.18 to allow for infinite sums. Indeed
you might have encountered a function which is continuous only on the irrationals; such a function
is Riemann integrable. There are also some uncountable sets with measure zero such as Cantor’s
middle-third set: if f is the indicator function of Cantor’s set
f (x) =
(
1 if x ∈ C
0 otherwise
then f is continuous except on C, and is Riemann integrable with
R
1
0
f (x) dx = 0.
Exercises 33 1. Explain why
R
2π
0
x
2
sin
8
( e
x
) dx ≤
8
3
π
3
2. If f is integrable on [a, b] prove that it is integrable on any interval [c, d] ⊆ [a, b].
3. We complete the proof of Theorem 4.8 (linearity of integration).
(a) Suppose k > 0, let A ⊆ R and define kA := {kx : x ∈ A}. Prove that sup kA = k sup A
and inf kA = k inf A.
(b) If k > 0 prove that k f is integrable on any interval and that
R
k f = k
R
f .
(c) How should you modify your argument if k < 0?
4. Give an example of an integrable but discontinuous function on a closed bounded interval [a, b]
for which the conclusion of the Intermediate Value Theorem for Integrals is false.
4
Formally, the length of an open interval (a, b) is b − a and a set A ⊆ R has measure zero if
∀ϵ > 0, ∃ open intervals I
n
such that A ⊆
∞
[
n=1
I
n
and
∞
∑
i=1
length(I
n
) < ϵ
More generally, the measure of a set (subject to a technical condition) is the infimum of the sum of the lengths of any
countable collection of open covering intervals. A rigorous discussion of measure theory is properly a matter for graduate
analysis. Somewhat surprisingly, there exist sets with positive measure that contain no subintervals, and even sets which
are non-measurable!
12
5. Explicitly compute the value of the integral
R
15/2
1/2
x −⌊x⌋dx (recall Example 4.19).
6. We prove and extend Corollary 4.10. Suppose f is integrable on [a, b].
(a) If g : [a, b] → R satisfies f (x) = g(x) for all x ∈ (a, b), prove that g is integrable and
R
b
a
g =
R
b
a
f .
(Hint: consider h = f − g and show that
R
h = 0)
(b) Now suppose g : [a, b] → R satisfies f (x) = g(x) for all x ∈ [ a, b] except at finitely many
points. Prove that g is integrable and
R
b
a
g =
R
b
a
f .
7. Show that an increasing function on [a, b] is integrable and thus complete Theorem 4.16.
(Hint: Choose a partition P with mesh P <
ϵ
f (b)−f (a)
)
8. Suppose f and g are integrable on [a, b].
(a) Define h(x) = ( f (x))
2
. We know:
• f is bounded: ∃K such that
|
f (x)
|
≤ K on [a, b].
• Given ϵ > 0, ∃P such that U( f , P) − L( f , P) <
ϵ
2K
. For each subinterval [x
i−1
, x
i
], let
M
i
= sup f (x), m
i
= inf f (x), M
i
= sup h(x), m
i
= inf h(x)
Prove that M
i
−m
i
≤ 2(M
i
−m
i
)K and use this to conclude that h is integrable.
(b) Prove that f g is integrable.
(Hint: f g =
1
4
( f + g)
2
−
1
4
( f − g)
2
)
(c) Prove that U(
|
f
|
, P) − L(
|
f
|
, P) ≤ U( f , P) − L( f , P) for any partition P. Hence conclude
that
|
f
|
is integrable.
(One can extend these arguments—it’s a bit harder!—to show that if j is continuous, then j ◦ f is
integrable. Parts (a) and (c) correspond, respectively, to j(x) = x
2
and j(x) =
|
x
|
.)
9. (Hard) Let f (x) =
x if x = 0 and sin
1
x
> 0
−x if x = 0 and sin
1
x
< 0
0 if x = 0
(a) Show that f is not piecewise continuous on [0, 1].
(b) Show that f is not piecewise monotonic on [0, 1].
(c) Show that f is integrable on [0, 1].
(Hint: given ϵ, hunt for a suitable partition to make U( f , P) − L( f , P) < ϵ by considering [0, x
1
]
differently to the other subintervals)
(d) Make a similar argument to show that g = sin
1
x
is integrable on ( 0, 1], where
g(x) =
(
sin
1
x
if x = 0
0 if x = 0
Note that neither argument evaluates the integrals!
13
34 The Fundamental Theorem of Calculus
The key result linking integration and differentiation is usually presented in two parts.
5
While there
are significant subtleties, the rough statements are as follows:
Part I Differentiation reverses integration:
d
dx
R
x
a
f ( t) dt = f (x)
Part II Integration reverses differentiation:
R
b
a
F
′
(x) dx = F(b) − F(a)
These facts seemed intuitively obvious to early practitioners of calcu-
lus: indeed, given a continuous positive function f :
• Let F(x) denote the area under the curve between 0 and x.
• A small increase ∆x results in the area increasing by ∆F.
• ∆F ≈ f (x)∆x is approximately the area of a rectangle, whence
∆F
∆x
≈ f (x). This is part I.
• F(b) − F(a) ≈
∑
∆F
i
≈
∑
f (x
i
) ∆x
i
. Since F
′
= f , this is part II.
∆F
∆x
x
f (x)
In fact when Leibniz introduced the symbols
R
and d in the late 1600’s, it was partly to reflect the
fundamental theorem.
6
If you’re happy with non-rigorous notions of limit, rate of change, area, and
(infinite) sums, the above is all you need!
Of course, we are very much concerned with the details: What must we assume regarding f and F,
and how are these properties used in the proof?
Theorem 4.22 (FTC, part I). Suppose f is integrable on [a, b]. For any x ∈ [a, b], define
F(x) :=
Z
x
a
f ( t) dt
Then:
1. F is uniformly continuous on [a, b];
2. If f is continuous at c ∈ [a, b], then F is differentiable at c with F
′
( c) = f (c).
As ever, the condition at c = a should be right-continuous and the conclusion right-differentiable, etc.
Compare this with the na
¨
ıve version above where we assumed f was continuous. We now require
only the integrability of f , and its continuity at one point for the full result.
5
We follow the traditional numbering; some authors reverse these.
6
R
is a stylized S for sum, while d stands for difference. Given a sequence F = (F
0
, F
1
, F
2
, . . . , F
n
), construct a new
sequence of differences
dF = (F
1
− F
0
, F
2
− F
1
, . . . , F
n
− F
n−1
)
which can then be summed:
Z
dF = (F
1
− F
0
) + (F
2
− F
1
) + ···(F
n
− F
n−1
) = F
n
− F
0
(∗)
Viewing a function as an ‘infinite sequence’ of values spaced along an interval, dF becomes a sequence of infinitesimals and
(∗) is essentially the fundamental theorem:
R
dF = F(b) − F(a). It is the conception of a function that is suspect here, not
the essential relationship between sums and differences.
14
Examples 4.23. You should have seen many examples in an elementary calculus course.
1. Since f (x) = sin
2
(x
3
−7) is continuous on any bounded interval, we conclude that
d
dx
Z
x
4
sin
2
( t
3
−7) dt = sin
2
(x
3
−7)
If one follows Theorem 4.13 and its resulting conventions, then this is valid for all x ∈ R.
2. The chain rule permits more complicated examples. For instance, since f (t) = sin
√
t is contin-
uous on its domain [0, ∞) and y(x) = x
2
+ 3 has range [3, ∞) ⊆ dom( f ), we have
d
dx
Z
x
2
+3
0
sin
√
t dt =
dy
dx
d
dy
Z
y
0
sin
√
t dt = 2x sin
p
x
2
+ 3
3. For a final positive example, observe that
d
dx
Z
e
x
sin x
tan(t
2
) dt = e
x
tan(e
2x
) − cos x tan(sin
2
x)
To evaluate this, one first chooses any constant a and writes
Z
e
x
sin x
=
Z
e
x
a
+
Z
a
sin x
=
Z
e
x
a
−
Z
sin x
a
before differentiating. This is valid provided sin x, e
x
and a all lie in the same subinterval of
dom tan(t
2
) = R \{±
q
π
2
, ±
q
3π
2
, ±
q
5π
2
, . . .}
Since
|
sin x
|
≤ 1 <
p
π
2
, this requires
e
2x
<
π
2
⇐⇒ x <
1
2
ln
π
2
Choosing a = 1 would certainly suffice.
4. Now consider why the theorem requires continuity. The piecewise
continuous function
f : [0, 2] → R : x 7→
(
2x if x ≤ 1
1
2
if x > 1
has a jump discontinuity at x = 1. We can still compute
F(x) =
(
R
x
0
2t dt = x
2
if x ≤ 1
R
1
0
2t dt +
R
x
1
1
2
dt =
1
2
(x + 1) if x > 1
This is continuous, indeed uniformly so. However the discontinu-
ity of f results in F having a corner and thus being non-differentiable
at x = 1. Indeed, F
′
(x) = f (x) for all x = 1; that is, at all values of
x where f is continuous.
0
1
2
f (x)
0 1 2
x
0
1
F(x )
0 1 2
x
15
Proving FTC I Neither half of the theorem is particularly difficult once you write down what you
know and what you need to prove. Here are the key ingredients:
1. F uniformly continuous means controlling the size of
|
F(y) − F(x)
|
=
Z
y
a
f ( t) dt −
Z
x
a
f ( t) dt
=
Z
y
x
f ( t) dt
≤
Z
y
x
|
f ( t)
|
dt
But the boundedness of f allows us to bound this last integral. . .
2. F
′
( c) = f (c) means showing that lim
x→c
F(x)−F(c)
x−c
= f (c), which means controlling the size of
F(x) − F(c)
x −c
− f (c)
=
1
x −c
Z
x
c
f ( t) dt − f (c)
The trick here will be to bring the constant f (c) inside the integral as
1
x−c
R
x
c
f ( c) dt so that what
we really have to control is the size of
1
|
x−c
|
R
x
c
|
f ( t) − f (c)
|
dt. This is where the continuity of
f comes in. . .
Proof. 1. Since f is integrable, it is bounded: ∃M > 0 such that
|
f (x)
|
≤ M for all x.
Let ϵ > 0 be given and define δ =
ϵ
M
. Then, for any x, y ∈ [a, b],
0 < y − x < δ =⇒
|
F(y) − F(x)
|
=
Z
y
x
f ( t) dt
≤
Z
y
x
|
f ( t)
|
dt (Theorem 4.12, part 4)
≤ M(y − x) (Theorem 4.12, part 2)
< Mδ = ϵ
We conclude that F is uniformly continuous on [a, b].
2. Let ϵ > 0 be given. Since f is continuous at c, ∃δ > 0 such that, for all t ∈ [a, b],
|
t −c
|
< δ =⇒
|
f ( t) − f (c)
|
<
ϵ
2
Now for all x ∈ [a, b] (except c),
0 <
|
x −c
|
< δ =⇒
F(x) − F(c)
x −c
− f (c)
=
1
x −c
Z
x
c
f ( t) − f (c) dt
(Theorem 4.8)
≤
1
|
x −c
|
Z
x
c
|
f ( t) − f (c)
|
dt (Theorem 4.12)
≤
1
|
x −c
|
ϵ
2
|
x −c
|
=
ϵ
2
< ϵ
Clearly lim
x→c
F(x)−F(c)
x−c
= f (c), and so F is differentiable at c with F
′
( c) = f (c).
16
The Fundamental Theorem, part II As with part I, the formulaic part of the result should be familiar,
though we are more interested in the assumptions and where they are needed.
Theorem 4.24 (FTC, part II). Suppose g is continuous on [a, b], differentiable on (a, b), and that g
′
is integrable
7
on (a, b). Then,
Z
b
a
g
′
= g(b) − g(a)
Part II is often expressed in terms of anti-derivatives: F being an anti-derivative of f if F
′
= f . Com-
bined with FTC I, we recover the familiar ‘+c’ result and a simpler version of the fundamental theo-
rem often seen in elementary calculus.
Corollary 4.25. Let f be continuous on [a, b].
• If F is an anti-derivative of f , then
R
b
a
f = F(b) − F(a).
• Every anti-derivative has the form F(x) =
R
x
a
f ( t) dt + c for some constant c.
Examples 4.26. Again, basic examples should be familiar.
1. Plainly g(x) = x
2
+ 2x
3/2
is continuous on [1, 4] and differentiable on (1, 4) with derivative
g
′
(x) = 2x + 3
√
x; this last is continuous (and thus integrable) on (1, 4). We conclude that
Z
4
1
2x + 3
√
x dx = x
2
+ 2x
3/2
4
1
= (16 + 16) − (1 + 2) = 29
2. If g(x) = sin(3x
2
), then g
′
(x) = 6x cos(3x
2
). Certainly g satisfies the hypotheses of the theorem
on any bounded interval [a, b]. We conclude
Z
b
a
6x cos(3x
2
) dx = sin(3b
2
) − sin(3a
2
)
Moreover, every anti-derivative of f (x) = 6x cos(3x
2
) has the form F(x) = sin(3x
2
) + c.
3. Recall Example 4.23.4 where we saw that the discontinuity of f led to the non-differentiability
of F(x) =
R
x
0
f ( t) dt at x = 1. The function F therefore fails the hypotheses of FTC II on the
interval [0, 2].
However, except at x = 1, F is an anti-derivative of f and moreover
R
2
0
f (x) dx = F(2) − F(0),
so we appear to have the formulaic conclusion of FTC II, though this is tautological given the
definition of F!
The way out of this conundrum is to note that other anti-derivatives
ˆ
F of f exist (except at
x = 1), and which fail to satisfy the conclusion. For instance
ˆ
F(x) =
(
x
2
if x < 1
1
2
x if x > 1
=⇒
ˆ
F(2) −
ˆ
F(0) = 1 =
3
2
=
Z
2
0
f (x) dx
7
See Definition 4.11 if you’re unsure what it means for g
′
to be integrable on a bounded open interval.
17
Proving FTC II See Exercise 10 for a relatively easy proof when g
′
= f is continuous. For the real
McCoy, we can only rely on the integrability of g
′
: the trick is to use the mean value theorem to write
g(b) − g(a) as a Riemann sum over a suitable partition.
Proof. Let ϵ > 0 be given and choose a partition P such that U(g
′
, P) − L(g
′
, P) < ϵ. Since g satisfies
the mean value theorem on each subinterval of the partition P, we see that
∃ξ
i
∈ (x
i−1
, x
i
) such that g
′
( ξ
i
) =
g(x
i
) − g(x
i−1
)
x
i
− x
i−1
from which
g(b) − g(a) =
n
∑
i=1
g(x
i
) − g(x
i−1
) =
n
∑
i=1
g
′
( ξ
i
)( x
i
− x
i−1
)
This is a Riemann sum for g
′
associated to the partition P, hence,
L(g
′
, P) ≤ g(b) − g(a) ≤ U(g
′
, P)
However we also have L(g
′
, P) ≤
R
b
a
g
′
≤ U(g
′
, P). Since these hold for all ϵ, the proof is complete.
While we certainly used the integrability of g
′
in the proof, it might seem strange that we assumed it
at all: shouldn’t every derivative be integrable? Perhaps surprisingly, the answer is no! If you want
a challenge, look up the Volterra function, which is differentiable everywhere, but whose derivative is
non-integrable (on, for instance, [0, 1])!
The Rules of Integration
If one wants to evaluate an integral, rather than merely show it exists, there are really only two options:
1. Evaluate Riemann sums and take limits: often difficult if not impossible to do explicitly.
2. Use FTC II. The problem now becomes the finding of anti-derivatives, for which the core method
is essentially guess and differentiate. To obtain general rules, we attempt to reverse the rules of
differentiation.
Integration by Parts First consider the product rule: the product g = uv of two differentiable
functions is differentiable with g
′
= u
′
v + uv
′
. Now apply Theorems 4.8, 4.12 and FTC II.
Corollary 4.27 (Integration by Parts). Suppose u, v are continuous on [a, b], differentiable on (a, b),
and that u
′
, v
′
are integrable on (a, b). Then
Z
b
a
u
′
(x)v(x)dx = u(b)v(b) − u(a)v(a) −
Z
b
a
u(x)v
′
(x)dx
This is significantly less useful than the product rule since it is only capable of transforming the
integral of a product into another such integral.
18
Examples 4.28. You should have seen myriad examples in a previous course. With practice, there
is no need to explicitly state u and v.
1. Let u(x) = x and v
′
(x) = cos x. Then u
′
(x) = 1 and v(x) = sin x, whence
Z
π/2
0
x cos x dx =
[
x sin x
]
π/2
0
−
Z
π/2
0
sin x dx =
π
2
sin
π
2
−0 −
[
−cos x
]
π/2
0
=
π
2
+ cos
π
2
−cos 0 =
π
2
−1
2. Let u(x) = ln x and v
′
(x) = 1. Then u
′
(x) =
1
x
and v(x) = x, whence
Z
e
2
e
ln x dx =
[
x ln x
]
e
2
e
−
Z
e
2
e
x
x
dx = e
2
ln e
2
−e ln e −
[
x
]
e
2
e
= 2e
2
−e −e
2
+ e = e
2
Change of Variables/Substitution We now turn our attention to the chain rule. If g(x) = F
u(x)
,
where F and u are differentiable, then g is differentiable with
g
′
(x) =
dg
dx
=
dF
du
du
dx
= F
′
u(x)
u
′
(x)
Now integrate both sides; the only issue is what assumptions are needed to invoke FTC II.
Theorem 4.29 (Substitution Rule). Suppose we have two continuous functions: u : [a, b] → R and
f : range(u) → R. Suppose also that u is differentiable on (a, b) with integrable derivative u
′
. Then
Z
b
a
f
u(x)
u
′
(x) dx =
Z
u(b)
u(a)
f ( u) du
This is the famous ‘u-sub’ formula from elementary calculus.
Proof. We leave as an exercise the verification that both integrals exist. We may also assume that
range(u) is an interval of positive length
8
for otherwise both integrals are trivially zero.
Choose any c ∈ range(u) and define
F : range(u) → R by F(v) :=
Z
v
c
f ( t) dt
Since f is continuous, by FTC I we see that F is differentiable with F
′
( u) = f (u). But now
Z
b
a
f
u(x)
u
′
(x) dx =
Z
b
a
d
dx
F
u(x)
dx (chain rule)
= F
u(b)
− F
u( a)
(FTC II)
=
Z
u(b)
u(a)
f ( u) du
8
By the intermediate and extreme value theorems, range(u) is already a closed bounded interval.
19
Examples 4.30. Reading the theorem is bad enough; its application often requires significant creativ-
ity in order to recognize a suitable substitution.
9
1. To evaluate the integral
R
√
π
0
2x sin x
2
dx, consider the substitution u(x) = x
2
defined on
[0,
√
π]. Certainly u is continuous, and its derivative u
′
(x) = 2x is integrable on (0,
√
π).
Finally f ( u) = sin u is continuous on range(u) = [0, π]. The hypotheses are satisfied, whence
Z
√
π
0
2x sin x
2
dx =
Z
√
π
0
f
u(x)
u
′
(x) dx =
Z
π
0
sin u du = −cos u
π
0
= 2
2. For the following integral with f (u) =
1
u
2
+1
, we make the substitution u(x) = x
2
− 2. Note
that u : [
√
2,
√
3] → [0, 1] and that u
′
(x) = 2x is integrable; moreover, f (u) is continuous on
range(u) = [0, 1]. We conclude that
Z
√
3
√
2
2x
x
4
−4x
2
+ 5
dx =
Z
√
3
√
2
2x
(x
2
−2)
2
+ 1
dx =
Z
1
0
1
u
2
+ 1
du = arctan u
1
0
=
π
4
3. The hypotheses on u really are all that is necessary. In particular, u doesn’t need to be left-
/right-differentiable at the endpoints of [a, b]! For instance, with f (u) = u
2
and u(x) =
√
x on
[0, 4], we easily verify
8
3
=
Z
4
0
1
2
√
x dx =
Z
4
0
x
2
√
x
dx =
Z
4
0
f
u(x)
u
′
(x) dx =
Z
2
0
f ( u) du =
Z
2
0
u
2
du =
8
3
4. Sloppy use of the substitution rule might lead to utter nonsense. For instance, consider the
‘substitution’ u = x
2
in the following:
Z
2
−1
1
x
dx =
Z
2
−1
1
2x
2
2x dx =
Z
4
1
1
2u
du =
1
2
(ln 4 −ln 1) = ln 2
Of course the left hand integral does not exist since
1
x
is undefined at 0 ∈ (−1, 2), so the
conclusion is false. In the language of the substitution rule, f (u) =
1
2u
is not continuous on
range(u) = [0, 4]: it is not even defined at u = 0! You are very unlikely to make precisely this
mistake since the first integral is so clearly undefined, but for more complicated functions. . .
9
Hence the old adage, “Differentiation is a science; integration an art.” To illustrate via an example, consider the function
f (x) = tan(e
x
cos(3x
2
) + 4x
3
). The product and chain rules allow one to explicitly compute the derivative
d f
dx
=
1
1 + (e
x
cos(3x
2
) + 4x
3
)
2
e
x
cos(3x
2
) − 6xe
x
sin(3x
2
) + 12x
2
By contrast, the integration analogues (integration by parts/substitution) are essentially useless in attempting to find an
explicit anti-derivative facilitating the integration of the same function via FTC II; for instance, the integral
Z
1
0
tan(e
x
cos(3x
2
) + 4x
3
) dx
is likely impossible to evaluate explicitly and can only be approximated (e.g. via Riemann sums).
20
Exercises 34 1. Calculate the following limits:
(a) lim
x→0
1
x
Z
x
0
e
t
2
dt (b) lim
h→0
1
h
Z
3+h
3
e
t
2
dt
2. Let f ( t) =
0 if t < 0
t if 0 ≤ t ≤ 1
4 if t > 1
(a) Determine the function F(x) =
R
x
0
f ( t) dt.
(b) Sketch F. Where is F continuous?
(c) Where is F differentiable? Calculate F
′
at the points of differentiability.
3. Let f be a continuous function on R.
(a) Define F(x) =
R
x+1
x−1
f ( t) dt. Carefully show that F is differentiable on R and compute F
′
.
(b) Repeat for the function G(x) =
R
sin x
0
f ( t) dt.
4. Recall Examples 4.23.4 and 4.26.3. Find all anti-derivatives F of f on [ 0, 1) ∪ (1, 2]. How many
satisfy
R
2
0
f (x) dx = F(2) − F(0)?
5. Consider integration by parts. Plainly
R
x
a
u
′
( t)v(t) dt is an anti-derivative of u
′
(x)v(x) by FTC I:
what does integration by parts say is another?
6. Use change of variables to integrate
R
1
0
x
√
1 − x
2
dx
7. Use integration by parts and the substitution rule to evaluate
R
b
0
arcsin x dx for any b < 1.
8. Use integration by parts to evaluate
R
b
0
x arctan x dxfor any b > 0
9. Check that the assumptions of int by subs guarantee that both integrals are well-defined (i.e.
that ( f ◦u)u
′
and f are integrable on the required intervals.
10. We prove a simpler version of the fundamental theorem of calculus.
(a) Suppose f is continuous on [a, b] and define F(x) =
R
x
a
f ( t) dt. For any c, x ∈ [a, b] where
c = x, prove that
m ≤
F(x) − F(c)
x −c
≤ M
where m, M are the maximum and minimum values of f (t) on the closed interval bounded
by c, x. Make sure to explain why m, M exist, and use this to deduce that F
′
( c) = f (c).
(b) Suppose f is continuous on [a, b] and that F is any anti-derivative of f on a, b (that is,
F
′
= f ). Use part (a) and the mean value theorem to prove that
R
b
a
f ( t) dt = F(b) − F(a).
21
36 Improper Integrals
The Riemann integral has several limitations. Even allowing for functions to be integrable on open
intervals (Exercise 32.6), the definition of
R
b
a
f (x) dx requires the following:
• That (a, b) be a bounded interval.
• That f be bounded on (a, b).
There is a natural way to extend the Riemann integral to unbounded intervals and functions: limits.
Definition 4.31. Suppose f : [a, b) → R satisfies the following properties:
• f is integrable on every closed bounded subinterval [a, t] ⊆ [a, b).
• Either b = ∞, or b is finite and f is unbounded at b,
The improper integral of f on [a, b) is defined to be
Z
b
a
f (x) dx := lim
t→b
−
Z
t
a
f (x) dx
This is convergent or divergent in the same manner as the limit.
If an integral is improper at its lower limit then
R
b
a
f (x) dx := lim
s→a
+
R
b
s
f (x) dx.
If an integral is improper at both ends, choose any c ∈ (a, b) and define
Z
b
a
f (x) dx = lim
s→a
+
Z
c
s
f (x) dx + lim
t→b
−
Z
t
c
f (x) dx
provided both one-sided improper integrals exist and the limit sum makes sense.
Theorem 4.13 says that the choice of c for a doubly-improper integral is irrelevant.
Many properties of the Riemann integral transfer to improper integrals, though not all. For example,
part 1 of Theorem 4.12 extends:
Theorem 4.32. If 0 ≤ f (x) ≤ g(x) on [a, b), then
R
b
a
f ≤
R
b
a
g, whenever the integrals exist (standard
or improper). In particular:
•
R
b
a
f = ∞ =⇒
R
b
a
g = ∞
•
R
b
a
g converges =⇒
R
b
a
f converges to a value ≤
R
b
a
g.
We leave some of the detail to Exercise 36.7.
22
Examples 4.33. 1.
R
t
0
x
2
dx =
1
3
t
3
for any t > 0. Clearly
Z
∞
0
x
2
dx = lim
t→∞
1
3
t
3
= ∞
More formally, the improper integral
R
∞
0
x
2
dx diverges to infinity.
2. With f (x) = x
−4/3
defined on [1, ∞),
Z
∞
1
x
−4/3
dx = lim
t→∞
Z
t
1
x
−4/3
dx = lim
t→∞
h
−3x
−1/3
i
t
1
= lim
t→∞
3 −3t
−1/3
= 3
3. Consider f (x) =
|
x
|
e
−x
2
/2
on (−∞, ∞). On a bounded interval [0, t), we have
Z
t
0
f (x) dx =
Z
t
0
xe
−x
2
/2
dx =
h
−e
−x
2
/2
i
t
0
= 1 − e
−t
2
/2
−−→
t→∞
1
By symmetry, we conclude that
Z
∞
−∞
|
x
|
e
−x
2
/2
dx = 1 + 1 = 2
This example is important in probability: multiplying by
1
√
2π
, we have computed the the ex-
pectation of
|
X
|
when X is a normally-distributed random variable
E(
|
X
|
) =
Z
∞
−∞
1
√
2π
|
x
|
e
−x
2
/2
dx =
r
2
π
4. If t ∈ [ 0, 1), we can use our knowledge of derivatives
d
dx
sin
−1
x =
1
√
1−x
2
to evaluate
Z
1
0
1
√
1 − x
2
dx = lim
t→1
−
Z
t
0
1
√
1 − x
2
dx = lim
t→1
−
sin
−1
t =
π
2
and that, moreover
R
1
−1
1
√
1−x
2
dx = π. By comparison, we see that
1
√
1 − x
4
≤
1
√
1 − x
2
=⇒
Z
1
−1
1
√
1 − x
4
dx ≤
Z
1
−1
1
√
1 − x
2
dx = π
5. Improper integrals need not exist. For instance,
lim
t→∞
Z
t
0
sin x dx = lim
t→∞
1 −cos t
diverges by oscillation.
23
Exercises 36 1. Use your answers from the previous section to decide whether the improper inte-
grals
R
1
0
arcsin x dx and
R
∞
0
x arctan x dx exist. If so, what are their values?
2. Let p be a positive constant. Prove the following:
Z
1
0
1
x
p
dx =
(
1
1−p
if p < 1
∞ if p ≥ 1
Z
∞
1
1
x
p
dx =
(
1
p−1
if p > 1
∞ if p ≤ 1
3. Explain why
R
b
a
f (x) dx = lim
t→b
−
R
t
a
f (x) dx holds, even when f is integrable on [a, b].
4. State a version of integration by parts modified for when
R
b
a
u
′
(x)v(x) dx is an improper inte-
gral. Now evaluate
R
∞
0
xe
−4x
dx.
5. What is wrong with the following calculation?
Z
∞
−∞
x dx = lim
t→∞
1
2
x
2
t
−t
= lim
t→∞
1
2
( t
2
−t
2
) = lim
t→∞
0 = 0
6. Prove or disprove: if
R
f and
R
g are convergent improper integrals, so is
R
f g.
7. Prove part of Theorem 4.32. Suppose 0 ≤ f (x) ≤ g(x) for all x ∈ [a, b), and that
R
b
a
g is a
convergent improper integral. Prove that
R
b
a
f converges and that
R
b
a
f ≤
R
b
a
g.
24
Generalizing the Riemann Integral (non-examinable)
In the 1890’s, Thomas Stieltjes
10
offered a generalization of the Riemann integral.
Definition 4.34. Let α be a monotonically increasing function on an interval [a, b]. Given a partition
P = {x
0
, . . . , x
n
} of [a, b] and a function f , define the differences
∆α
i
= α(x
i
) − α(x
i−1
)
The upper/lower Darboux–Stieltjes sums/integrals are defined analogously to the pure Riemann case:
U( f , P, α) =
n
∑
i=1
M
i
∆α
i
L( f , P, α ) =
n
∑
i=1
m
i
∆α
i
U( f , α) = inf U( f , P, α) L( f , α) = sup L( f , P, α)
f is Riemann–Stieltjes integrable of class R(α) if U( f , α) = L( f , α): we denote this value
R
b
a
f (x) dα.
The standard Riemann integral corresponds to α(x) = x. It is the ability to choose other functions α
that makes the Riemann–Stieltjes integral both powerful and applicable.
Standard Properties Most results in sections 32 and 33 hold with suitable modifications, as does the
discussion of improper integrals. For instance,
f ∈ R(α) ⇐⇒ ∃P such that U( f , P, α) − L( f , P, α) < ϵ
The result regarding piecewise continuity of f is a notable exception: if f and α are simultane-
ously piecewise continuous then f might not lie in R(α).
Weighted integrals If α is differentiable, then we obtain a standard Riemann integral
Z
b
a
f (x) dα =
Z
b
a
f (x)α
′
(x) dx
weighted so that f (x) contributes more when α is increasing rapidly.
Probability If α(a) = 0 and α(b) = 1, then α may be viewed as a probability distribution function.Its
derivative α
′
is the corresponding probability density function. For example:
1. The uniform distribution on [a, b] has α =
1
b−a
(x −a) so that
Z
b
a
f (x) dα =
1
b − a
Z
b
a
f (x) dx
Since α
′
is constant, the integrals weigh all values of x uniformly.
2. The standard normal distribution has α(x) =
R
x
−∞
1
√
2π
e
−t
2
/2
dt. The fact that α
′
=
1
√
2π
e
−x
2
/2
is maximal when x = 0 reflects the fact that a normally distributed variable is clustered
near its mean.
In all cases,
R
f (x) dα = E( f (X)) computes an expectation (see, for instance, Example 4.33.3).
10
Stieltjes was Dutch; for the pronunciation try ‘steelchez.’
25
Non-differentiable α A major flexibility comes when we allow α to be non-differentiable, or even dis-
continuous! For example, given a partition Q = {s
0
, . . . , s
n
} of [a, b], and a positive sequence
( c
k
)
n
k=1
, define
α(x) =
0 if x = a
k
∑
i=1
c
i
if x ∈ (s
k−1
, s
k
]
This is an increasing step function on [a, b]. The Riemann–Stieltjes integral becomes a weighted
sum
Z
b
a
f (x) dα =
n
∑
i=1
c
i
f ( s
i
)
Taking instead an infinite sequence (s
n
) ⊆ [a, b] results in an infinite series, which helps explain
why so many results for series and integrals look similar!
This also touches on probability. For example, let p ∈ [0, 1], n ∈ N, and s
k
= k on the interval
[0, n]. If c
k
=
(
n
k
)
p
k
(1 − p)
n−k
, then
Z
f (x) dα =
n
∑
k=0
n
k
p
k
(1 − p)
n−k
f (x) = E( f (X))
is the expectation of f (X) when X ∼ B(n, p) is a binomially distributed random variable.
Integrals and Convergence
The Lebesgue integral is another common generalization. Its main purpose is to permit the transfer
of integrability to the limit of a sequence of integrable functions.
11
To see the problem, consider the
sequence
f
n
: [0, 1] → R : x 7→
(
1 if x =
p
q
∈ Q with q ≤ n
0 otherwise
Each f
n
is piecewise continuous and thus Riemann integrable with
R
1
0
f
n
(x) dx = 0. However, the
pointwise limit of f
n
is the function
f (x) =
(
1 if x ∈ Q
0 if x ∈ Q
which is not Riemann integrable. In the Lebesgue theory, the limit f turns out to be integrable with
integral 0, so that
lim
n→∞
Z
1
0
f
n
(x) dx =
Z
1
0
lim
n→∞
f
n
(x) dx
Recall that the interchange of limits and integrals would be automatic if the convergence f
n
→ f
were uniform: of course the convergence isn’t uniform here.
11
Recall how uniform convergence does this for continuity.
26
