This document contains a proof that the integral from 0 to 1 of x dx equals 1/2. It begins by assuming the integral exists, then provides an alternative proof that does not assume the integral exists. This is done by establishing a lemma about Riemann sums and partitions, and showing that the difference between any two Riemann sums is less than epsilon.

1. Math 441: Chapter VI, Riemann
Integration
Leonardo Di Giosia
(Problem 1)
Claim: Assuming the integral exists,
1
0
x dx =
1
2
Proof. Let > 0. Because the integral exists, we have a δ such that if S1 and S2
are any two Riemann sums of f(x) = x on the interval [0, 1] of partition width
less than or equal to δ we have that
|S1 − S2| < (Ineq. I)
Let Pn = {x0, x1, . . . , xn} be a partition of [0, 1] of width less than δ. Consider
the Riemann sum S of f using Pn,
S =
n−1
i=0
f(xi)(xi+1 − xi)
1

2. Chapter VI Riemann Integration
where xi = xi+1+xi
2 . This implies
S =
1
2
n−1
i=0
(xi+1 + xi)(xi+1 − xi) =
1
2
n−1
i=0
x2
i+1 − x2
i =
1
2
(12
− 02
) =
1
2
So if Sa is any Riemann sum of f on [0, 1] corresponding to a partition of width
less than δ, then by inequality I,
|Sa − S| = |Sa −
1
2
| <
So
1
0
x dx =
1
2
(Problem 1, Alternative)
Claim: (Without assuming the integral exists)
1
0
x dx =
1
2
Proof. Let Pn = {x0, x1 . . . xn} be any partition [0, 1] of width δ. We first
establish a lemma.
Lemma:
n−1
i=0 (xi+1 − xi)2
≤ 2δ.
Denote S = {i, 0 ≤ i ≤ n − 1|∃k, 1 ≤ k ≤ 1
δ − 1|k · δ ∈ [xi, xi+1]}. This
is the set of indices i such that a multiple of delta is somewhere in [xi, xi+1].
Let k be some integer between 0 and 1
δ − 1. Denote the set Sk = {i, 0 ≤ i ≤
n − 1|[xi, xi+1] ⊂ [k · δ, (k + 1) · δ]}. This set is the set of indices i, such that
the interval (xi, xi+1) is nested in the interval [k · δ, (k + 1) · δ]. We must have
that each interval created by Pn is either hugging a multiple of δ in [0, 1] or is
nested between two consecutive δs. We have the following:
n−1
i=0
(xi+1 − xi)2
=
i∈S
(xi+1 − xi)2
+
1
δ −1
i=0 j∈Si
(xj+1 − xj)2
(Eq. I)
For each i between 0 and 1
δ − 1, we know j∈Si
(xj+1 − xj)2
< δ2
. Why?
Consider this: if X, Y , where X < Y , are the extremities of the full interval
nested between iδ and (i+1)δ, then j∈Si
(xj+1 −xj) = Y −X ≤ δ. Because all
terms of j∈Si
(xj+1 −xj) are positive, we must have that j∈Si
(xj+1 −xj)2
<
[ j∈Si
(xj+1 − xj)]2
= [Y − X]2
≤ δ2
. So we merge this information with that
in Equation I to find
n−1
i=0
(xi+1 − xi)2
<
i∈S
(xi+1 − xi)2
+
1
δ −1
i=0
δ2
=
i∈S
(xi+1 − xi)2
+ (
1
δ
− 1)δ2
2

3. Chapter VI Riemann Integration
Next, we describe why i∈S(xi+1 −xi)2
< ( 1
δ −1)δ2
. First, we know there are
exactly 1
δ − 1 multiples of δ in the interval (0, 1), so the cardinality of S must
be exactly 1
δ − 1. So we know i∈S(xi+1 − xi)2
≤ i∈S δ2
= ( 1
δ − 1)δ2
.
Once we place this information into the above we find that
n−1
i=0
(xi+1 − xi)2
< (
1
δ
− 1)δ2
+ (
1
δ
− 1)δ2
< 2δ
And the lemma is proven. Let S1 and S2 be two Riemann Sums of f(x) = x
corresponding to partition Pn. Because for any xi where i ∈ {0, 1, 2, . . . , n − 1},
xi ≤ f(xi) ≤ xi+1 we have the following
n−1
i=1
xi(xi+1 − xi) ≤ S1, S2 ≤
n−1
i=1
xi+1(xi+1 − xi)
This provides that (using the established lemma)
|S1 − S2| ≤
n−1
i=0
(xi+1 − xi)2
< 2δ
This implies that the integral
1
0
x dx
exists. By the result of the first problem, we know this is enough to show that
1
0
x dx =
1
2
(Problem 2)
Define f : [0, 1] → R by the following. Let x ∈ [0, 1]. If there exists some natural
number n such that 1
n = x, then f(x) = 1. Otherwise, f(x) = 0.
Claim:
1
0
f(x) dx = 0
Proof. Note that this proof will assume the existence of the integral. Let > 0.
We know that
1
0
f(x) dx =
0
f(x) dx +
1
f(x) dx
Because 1 is an upper bound on f we know 0
f(x) dx ≤ 1 · ( − 0) = .
Furthermore, we know that there are a finite number of points in [ , 1], denoted
3

4. Chapter VI Riemann Integration
x1, x2, . . . , xn, where f = 1. So we may construct n closed intervals, overlapping
only at extremites, denoted Ii for each i ∈ {1, 2, . . . , n} such that
xi ∈ Ii
and
n
i=1
Ii = [ , 1]
Hence, we have that
1
0
f(x) dx =
0
f(x) dx +
I1
f(x) dx +
I2
f(x) dx + · · · +
In
f(x) dx
By a result established in the textbook on page 114, we have that for each such
i, Ii
f(x) dx = 0. Hence,
1
0
f(x) dx =
0
f(x) dx + 0 + 0 + · · · + 0
=
0
f(x) dx
≤
Because is an arbitrary positive number we must have that
1
0
f(x) dx = 0
(Problem 3)
Let f be the function defined in problem 1 part d of chapter IV (also known as
the Thomae function).
Claim:
1
0
f(x) dx
exists
Proof. We use the lemma established on page 118 of the textbook. Let > 0.
We want to find two step functions, f1, f2 which bound f from below and above,
respectively, such that
1
0
f2(x) − f1(x) dx <
Let f1 = 0, this is a step function which bounds f from below.
Next, we work to define the upper bound step function. Let n ∈ N. Denote
S = {x ∈ [0, 1]|f(x) >
1
n
}
We claim that S is finite. This is because for each x ∈ S, x = a
b where a and b
share no factors besides ±1 and b > 0. We additionally must have that b < n
4

5. Chapter VI Riemann Integration
(Because f(x) = 1
b > 1
n ). For any b between 1 and n (a finite number of b’s),
there are a finite number of potential a’s such that a
b is in the interval [0, 1]. So
because a finite number times a finite number is finite, we must have that S is
finite.
So, say there are N points in [0, 1] such that f is greater than 2 , denoted
{x1, x2, . . . , xN }. Define a step function f2 as follows: f2(x) = 1 if x = xi for
some i ∈ {1, 2, . . . N}. Otherwise, f2(x) = 2 . We see that f2 is an upper bound
on f and
1
0
f2(x) − f1(x) dx =
2
(1 − 0) <
This implies that indeed
1
0
f(x) dx
exists.
(Problem 4)
Let [a, b] ⊂ R and let f : [a, b] → R be integrable. Let c ∈ R and define
g : [a + c, b + c] → R by g(x) = f(x − c).
Claim: Then
b+c
a+c
g(x) dx
exists and
b+c
a+c
g(x) dx =
b
a
f(x) dx
Proof. We start by asserting the following: If S is any Riemann sum of g on the
interval [a+c, b+c] of partition width less than δ for any δ > 0, then S must be
a Riemann sum for f on the interval [a, b] also of partition width less than δ. Let
S be such a Riemann sum of g. Then we have some partition {x0, x1, . . . , xn}
of [a + c, b + c] of width less than some δ > 0 and
S =
n
i=1
g(xi)(xi − xi−1) =
i=1
f(xi − c)((xi − c) − (xi−1 − c)) (Equation I)
Next, see that Pn = {x0 − c, x1 − c, . . . , xn − c} is a partition of the interval
[a, b]. Additionally, for any i ∈ {1, 2, . . . , n} we know
((xi − c) − (xi−1 − c)) = (xi − xi−1) < δ
So Pn is a partition of the interval [a, b] of width less than δ and because
xi − c ∈ [xi−1 − c, xi − c]
we have that S is a Riemann sum of f on [a, b] of partition width less than δ.
We move on to show that the basic integral
b+c
a+c
f(x) dx exists. Let > 0. Then
5

6. Chapter VI Riemann Integration
there exists a δ1 such that if S1 and S2 are any two Riemann sums for f on the
interval [a, b] of partition width less that δ1, then |S1 −S2| < (by a proposition
in the textbook on page 118). Let Sa and Sb be two Riemann sums of g on the
interval [a + c, b + c] of partition width less than δ1. Then by our proven result,
we must have that |Sa − Sb| < and using the definition established on page
118, we must have that g is Riemann integrable on the interval [a+c, b+c]. We
now demonstrate that the two integrals are equal. By using the trick of taking
the minimum of two existent deltas, we must have that there exists a δ2 such
that if R and P are any two Riemann sums of g and f respectively, partitioned
on their respective domains, of partition width less than δ2 then
R −
b+c
a+c
g(x) dx <
2
(Ineq. I)
and
P −
b
a
f(x) dx <
2
(Ineq. II)
So let A be some such Riemann sum of g on [a + c, b + c]. We know A then is
also such a Riemann sum of f on [a, b], by our result earlier in the proof. So
both inequalities I and II hold for Riemann sum A and we may combine them
to form
b+c
a+c
g(x) dx −
b
a
f(x) dx <
Because is an arbitrary positive number we have that the quantity on the left
in the above is zero. Finally, this implies that
b+c
a+c
g(x) dx =
b
a
f(x) dx
(Problem 5)
Let [a, b] ⊂ R and let f : [a, b] → R be continuous.
Claim: Then f is integrable on [a, b].
Proof. Because f is defined on a compact metric space and is continuous, it
must be uniformly continuous. Let > 0. Then (b−a) > 0 and there must exist
some δ > 0 such that if x, y are any elements of [a, b] such than |x − y| < δ we
have that |f(x) − f(y)| < 2(b−a) . Let SX, SY be Riemann sums with respective
partitions Xn = {x0, x1, . . . , xn} and Ym = {y0, y1, . . . , ym} of [a, b] of width less
than δ
2 . Denote Zp = {z0, z1, . . . , zp} = Xn ∪ Ym. Then Zp is also a partition of
of [a, b] of width less than δ
2 .
We want to define terms Ci where i ∈ {1, 2, . . . , p} such that
SX − SY =
n
i=1
f(xi)(xi − xi−1) −
m
i=1
f(yi)(yi − yi−1) =
p
i=1
Ci(zi − zi−1)
6

7. Chapter VI Riemann Integration
So, let i ∈ {1, 2, . . . , p}. There exists some unique k based on this i, denoted
k(i), such that the interval [zi−1, zi] ⊂ [xk(i)−1, xk(i)] , and a similar unique s(i)
such that [zi−1, zi] ⊂ [ys(i)−1, ys(i)]. So define Ci = f(xk(i)) − f(ys(i)). It is easy
to see that indeed
n
i=1
f(xi)(xi − xi−1) −
m
i=1
f(yi)(yi − yi−1) =
p
i=1
Ci(zi − zi−1)
We know by the Triangle inequality that
p
i=1
Ci(zi − zi−1) ≤
p
i=1
|Ci(zi − zi−1)| (Ineq. I)
We know that for any i, xk(i) and ys(i) are in two intervals of width less than δ
2
which intersect at some point z ∈ [zi−1, zi]. So we know
|xk(i) − z| <
δ
2
and
|ys(i) − z| <
δ
2
So by the triangle inequality, we have that
|xk(i) − ys(i)| < δ
Implying that for any i ∈ {1, 2, . . . , p}, |Ci| < (b−a) . So we combine this infor-
mation with inequality I to find
SX − SY ≤
p
i=1
|Ci(zi − zi−1)| <
b − a
p
i=1
(zi − zi−1) =
Hence, we must have that f is integrable on [a, b].
(Problem 7)
Let f : [a, b] → R be bounded and continuous everywhere except at a finite
number of points, xi, x2 . . . , xn ∈ (a, b). Claim: Then
b
a
f(x) dx
exists.
Proof. Let > 0. There exists some M such that |f(x)| < M for all x ∈ [a, b].
Denote δ = min({xi+1 − xi|1 < i < n − 1}). Define
δ = min({x1 − a, b − xn,
δ
2
,
8M · n
})
7

8. Chapter VI Riemann Integration
We know that because we chose our δ small enough that the intervals [a, x1 −
δ ], [x1 + δ , x2 − δ ], [x2 + δ , x3 − δ ] . . . , [xn + δ , b] are well defined and that f
is continuous on each of these intervals. Denote them I0, I1, . . . , In respectively.
Hence, for each interval we have step functions Li, Gi : Ii → R such that for all
x ∈ Ii,
Li(x) ≤ f(x) ≤ Gi(x)
and
Ii
Gi(x) − Li(x) dx <
2 · n
Our next step is two create to mega step functions defined on all of [a, b]. Let
x ∈ [a, b]. Then define G(x) by



G1(x) x ∈ I1
G2(x) x ∈ I2
. . .
Gn(x) x ∈ In
M else
And L(x) by 


L1(x) x ∈ I1
L2(x) x ∈ I2
. . .
Ln(x) x ∈ In
−M else
Note that L and G are indeed step functions on [a, b] such that for any x ∈ [a, b]
we have L(x) ≤ f(x) ≤ G(x). All we must show is the the integral of their
difference is less than . Consider
b
a
G(x) − L(x) dx =
x1−δ
a
G(x) − L(x) dx +
x1+δ
x1−δ
G(x) − L(x) dx + · · ·
+
xn+δ
xn−δ
G(x) − L(x) dx +
b
xn+δ
G(x) − L(x) dx
=
n
i=1 Ii
Gi(x) − Fi(x) dx +
n
i=1
xi+δ
xi−δ
G(x) − F(x) dx
<
n
i=1
2 · n
+
n
i=1
(2δ ) · (2M)
<
2
+
n
i=1
4 · M · N
· (2M)
=
2
+
2
=
8

9. Chapter VI Riemann Integration
So f is integrable on [a, b].
(Problem 8)
Let f : [a, b] → R be nondecreasing.
Claim: Then
b
a
f(x) dx
exists.
Proof. Let > 0. There exists some natural number N such that
b − a
N
<
f(b) − f(a)
Denote this quantity on the above left as δ. We define two step functions,
A, B : [a, b] → R. Let x ∈ R. Define A(x) as



f(a) x ∈ [a, a + δ)
f(a + δ) x ∈ [a + δ, a + 2δ)
· · ·
f(a + (N − 1)δ) x ∈ [a + (N − 1)δ, b]
and define B(x) as



f(a + δ) x ∈ [a, a + δ)
f(a + 2δ) x ∈ [a + δ, a + 2δ)
· · ·
f(b) x ∈ [a + (N − 1)δ, b]
Because f is nondecreasing, we must have that for all x ∈ [a, b],
A(x) ≤ f(x) ≤ B(x)
Now consider the following computations
b
a
B(x) − A(x) dx =
N−1
i=0
(f(a + (i + 1)δ) − f(a + iδ)) · δ
= (f(b) − f(a))δ
<
So f is integrable on [a, b].
9

10. Chapter VI Riemann Integration
(Problem 9)
Let f : [a, b] → R be integrable. Claim: Then |f| : [a, b] → R is integrable and
b
a
f(x) dx ≤
b
a
|f(x)| dx
Proof. Let > 0. Because f is integrable, we know that there exists step
functions L, U : [a, b] → R such that for all x in [a, b],
L(x) ≤ f(x) ≤ U(x)
and
b
a
U(x) − L(x) dx <
Next, we define two new functions, M, m : [a, b] → R. Let x ∈ [a, b]. Define
M(x) = max({|U(x)|, |L(x)|})
m(x) = min({|U(x)|, |L(x)|})
We know both |U| and |L| are step functions, so the functions M and m are
also step functions. If x ∈ [a, b] we know m(x) ≤ |f(x)| ≤ M(x). Additionally,
by the reverse triangle inequality,
M(x)−m(x) ≤ |M(x)−m(x)| = ||U(x)|−|L(x)|| ≤ |U(x)−L(x)| = U(x)−L(x)
So because the function M(x)−m(x) is bounded above by U(x)−L(x) on [a, b],
we have that
b
a
M(x) − m(x) dx ≤
b
a
U(x) − L(x) dx <
So we must have that |f| is integrable on [a, b]. Furthermore, because f, −f ≤ |f|
on the interval [a, b], we must have that
b
a
f(x) dx ≤
b
a
|f(x)| dx
and
−
b
a
f(x) dx ≤
b
a
|f(x)| dx
So
b
a
f(x) dx ≤
b
a
|f(x)| dx
10

11. Chapter VI Riemann Integration
(Problem 11)
Let f : [a, b] → R be a continuous real valued nonnegative function such that
there exists some x0 ∈ [a, b] such thatf(x0) > 0. Claim: Then
0 <
b
a
f(x) dx
Proof. Because f(x0)
2 > 0 and because f is continuos, there exists some δ > 0
such that if x ∈ [a, b] such that |x − x0| < δ, we have
|f(x) − f(x0)| <
f(x0)
2
f(x0)
2
< f(x)
That is, there exists some x1, x2 ∈ [a, b] such that f(x0)
2 is a lower bound on f
on the interval [x1, x2]. Everywhere else, 0 is a lower bound on f. So we have
0 ≤
x1
a
f(x) dx
f(x0)
2(x2 − x1)
<
x2
x1
f(x) dx
0 ≤
b
x2
f(x) dx
If we sum these inequalities, we find that
0 <
f(x0)
2(x2 − x1)
<
x1
a
f(x) dx +
x1
x1
f(x) dx +
b
x2
f(x) dx =
b
a
f(x) dx
(Problem 12)
Let f : [a, b] → R be continuous. Claim: Then there exists some c ∈ [a, b] such
that
f(c)(b − a) =
b
a
f(x) dx
Proof. Denote F : [a, b] → R by
F(x) =
x
a
f(t) dt
then F is a continuous function on [a, b] and differentiable on (a, b). So there
must exist (by the Mean Value theorem) a c ∈ [a, b] such that
F (c) =
F(b) − F(a)
b − a
11

12. Chapter VI Riemann Integration
By the Fundamental Theorem of Calculus, we must have that F (c) = f(c) so
f(c)(b − a) =
b
a
f(x) dx
(Problem 14)
Let f : [0, ∞) → R be a continuous function such that limx→+∞ f(x) = c.
Claim: Then
lim
x→+∞
x
0
f(t) dt
x
= c
Proof. By the Fundamental Theorem of Calculus, we know the numerator func-
tion is differentiable and that its derivative is f(x). By L’Hospital’s rule, we
know
lim
x→+∞
x
0
f(t) dt
x
= lim
x→+∞
f(x)
1
= c
(Problem 15)
Let [a, b], [c, d] ⊂ R and let g : [a, b] × [c, d] → R be continuous. Define h :
[a, b] → R by
h(x) =
d
c
g(x, t) dt
for all x ∈ [a, b].
Claim: Then h is continuous.
Proof. Let x0 ∈ [a, b]. Let > 0. Because g is continuous on a compact metric
space we know that it must be uniformly continuous. So there exists some δ > 0
such that if (x1, x2) and (y1, y2) are any two points of [a, b] × [c, d] such that
(x1 − y1)2 + (x2 − y2)2 < δ
we have that
|g(x1, x2) − g(y1, y2)| <
d − c
Let x ∈ [a, b] such that |x − x0| < δ. For any y ∈ [c, d] we know
(x0 − x)2 + (y − y)2 = |x − x0| < δ
so
|g(x, y) − g(x0, y)| <
d − c
12

13. Chapter VI Riemann Integration
Finally, this implies that
|h(x)−h(x0)| =
d
c
g(x, t)−g(x0, t) dt ≤
d
c
|g(x, t)−g(x0, t)| dt <
d − c
·(d−c) =
So h is continuous at x0 and is thus continuous on [a, b].
(Problem 16)
Let [a, b] ⊂ R. Let g : C[a, b] → R be defined as if f ∈ C[a, b],
g(f) =
b
a
f(x) dx
Claim: Then g is uniformly continuous.
Proof. Let > 0. let δ = b−a . Let f1 and f2 be elements of C[a, b] such that
d(f1, f2) < δ
Max({|f1(x) − f2(x)||x ∈ [a, b]}) <
epsilon
b − a
Because b−a is an upper bound on the function |f1 − f2| on [a, b], we have that
b
a
|f1(x) − f2(x)| dx <
b − a
(b − a) =
b
a
f1(x) − f2(x) dx <
b
a
f1(x) dx −
b
a
f2(x) dx <
g(f1) − g(f2) <
So we must have that g is uniformly continuos on C[a, b].
(Problem 17)
Let [a, b] ⊂ R and let U be some open subset of R containing [a, b]. Let u, v :
U → R be differentiable with continuous derivatives on U. Claim: Then
b
a
u (x)v(x) dx = u(b)v(b) − u(a)v(a) −
b
a
u(x)v (x) dx
Proof. We know that the function F : U → R given by F(x) = u(x)v(x) for all
x ∈ U is differentiable on U where
F (x) = u (x)v(x) + u(x)v (x)
13

14. Chapter VI Riemann Integration
for all x ∈ U. Denote F (x) = f(x). Because u and v are continuous on U by
hypothesis, we must have that f, the derivative of F, is continuous as well on
U. Because a, b ∈ U, we have established all prerequisites of Corollary 2 from
page 127 of the textbook. So we have that
b
a
f(x) dx = F(b) − F(a)
b
a
u (x)v(x) + u(x)v (x) dx = u(b)v(b) − u(a)v(a)
b
a
u (x)v(x) dx = u(b)v(b) − u(a)v(a) −
b
a
u(x)v (x) dx
(Problem 18)
Let x0 ∈ R
Claim: Then
x0/
√
1+x2
0
0
dt
√
1 − t2
=
1
0
du
1 + u2
Proof. First, define Φ : R → (−1, 1) by
Φ(x) =
x
√
1 + x2
for all x ∈ R (do take note that Φ truly is defined for all x and that for any
x, −1 < Φ(x) < 1). Additionally, we see that Φ is differentiable on R with
derivative given by
d
dx
Φ(x) =
1
(1 + x2)
3
2
The function a− 3
2 is continuous on [0, ∞) and 1 + x2
is continuous on all of R
with range of [1, ∞). this implies the composition
(1 + x2
)− 3
2 =
d
dx
Φ(x)
is continuous. So we have that Φ : R → (−1, 1) is differentiable with continuous
derivative and that f : (−1, 1) → R where f(t) = 1√
1−t2
is continuous. Finally,
because 0, x0 ∈ R, we have established all the prerequisites of the change of
variable theorem established on page 128 of the textbook. Hence
Φ(x0)
Φ(0)
f(t) dt =
x0
0
f(Φ(u)) · Φ (u) du
x0/
√
1+x2
0
0
dt
√
1 − t2
=
x0
0
1
1 + u2
14

15. Chapter VI Riemann Integration
(Problem 19)
Let U ⊂ R be some open interval in R and let f : U → R have a continuous
(n + 1)th
derivative on U. Let a, b ∈ U.
Claim: Then
f(b) = f(a) +
f (a)(b − a)
1!
+ · · · +
f(n)
(a)(b − a)n
n!
+
b
a
f(n+1)
(x)(b − x)n
n!
dx
Proof. We utilize the terms Rn(b, a) and Rn(b, x) where x is between a and b
defined on page 106 of the textbook. So
f(b) = f(a) +
f (a)(b − a)
1!
+
f (a)(b − a)2
2!
+ · · · +
f(n)
(a)(b − a)n
n!
+ Rn(b, a)
(Eq. I)
and
f(b) = f(x) +
f (x)(b − x)
1!
+
f (x)(b − x)2
2!
+ · · · +
f(n)
(x)(b − x)n
n!
+ Rn(b, x)
Because f is (n + 1) times differentiable, we use the result established on page
107. So Rn(b, x) is a differentiable function on U where
d
dx
Rn(b, x) = −f(n+1)
(x)
(b − x)n
n!
(Eq. II)
Additionally, because f(n+1)
is a continuous function on U by hypothesis, we
have that d
dx Rn(b, x) too must be a continuous function on U. So
intb
a
d
dx
Rn(b, x) dx
Exists. By the Fundamental Theorem of Calculus, we have that
b
a
d
dx
Rn(b, x) dx = Rn(b, b) − Rn(b − a) = −Rn(b, a) (Eq. III)
Additionally, utilizing equation II, we find
b
a
d
dx
Rn(b, x) dx = −
b
a
f(n+1)
(x)
(b − x)n
n!
dx
So combining this result with equation III provides
Rn(b, a) =
b
a
f(n+1)
(x)
(b − x)n
n!
dx
Finally, substituting the above information into equation I provides
[f(b) = f(a) +
f (a)(b − a)
1!
+ · · · +
f(n)
(a)(b − a)n
n!
+
b
a
f(n+1)
(x)(b − x)n
n!
dx
15

16. Chapter VI Riemann Integration
(Problem 21)
(a)
Let k ∈ (0, ∞). Denote {sn}∞
n=1 ⊂ R by the following
sn =
1k
+ 2k
+ · · · + nk
nk+1
for any n ∈ N. Claim:
lim
n→∞
sn =
1
k + 1
Proof. Note that for any natural number n
sn =
n
i=1
1
n
·
i
n
k
So sn is a Riemann sum of the function g(x) = xk
of partition width exactly 1
n
on [0, 1]. Because g is continuous on [0, 1] we know it is integrable so
1
0
g(x) dx
exists and
1
0
g(x) dx =
1
k + 1
Let > 0. Because this integral exists, there exists some δ such that if S is any
Riemann sum of g on [0, 1] of partition width less than δ then
S −
1
k + 1
<
So let N > 1
δ . Let n be some natural number greater than N. Then 1
n < δ.
This implies that sn is such a Riemann sum of partition width less that δ so we
have that
sn −
1
k + 1
<
(b)
Denote {pn}∞
n=1 ⊂ R by the following
pn =
1
n + 1
+
1
n + 2
+ · · · +
1
2n
for any n ∈ N.
Claim:
16

17. Chapter VI Riemann Integration
Proof.
lim
n→∞
pn = log (2)
Note that for any natural number n,
pn =
n
i=1
1
n
·
1
1 + ( i
n )
So pn is a partition of the function f(x) = 1
1+x on the interval [0, 1] of partition
with exactly 1
n . We know f is continuous on the interval [0, 1] so it is integrable
and
1
0
f(x) dx = log (2)
Let > 0. Then there exists some δ > 0 such that for any Riemann sum S of f
on [0, 1] of width less than δ, we have that
S − log (2) < .
So let N > 1
δ . Let n be some natural number greater than N. Then 1
n < δ, so
pn is a Riemann sum of f on [0, 1] of partition width less than δ so we have that
pn − log (2) <
(Problem 22)
Denote {sn}∞
n=1 ⊂ R by
sn = 1 +
1
2
+ · · · +
1
n
for all natural numbers n and denote {pn}∞
n=1 ⊂ R by
pn = sn − log (n)
for all natural numbers n.
Claim: Then for any natural number n, 1 ≥ pn > 0, pn ≥ pn+1, and thus,
limn→∞ pn = p exists and p ∈ [0, 1].
Proof. Let n be some natural number. We define step Fn : [1, n + 1] → R by
the following:
Fn(x) =
1
x
for all x ∈ [1, n + 1]. We see that Fn is actually a step function on [1, n + 1] and
hence it is integrable where
n+1
1
Fn(x) dx = sn
17

18. Chapter VI Riemann Integration
Next, we show that f(x) = 1
x bounds Fn(x) from below on [1, n + 1]. Let
x ∈ [1, n + 1]. We know x ≤ x, so
1
x
≤
1
x
f(x) ≤ Fn(x)
So we have that
n+1
1
f(x) dx ≤
n+1
1
Fn(x) dx
log (n + 1) ≤ sn
log
n + 1
n
≤ pn
Because 0 < log n+1
n we have that 0 < pn. We now show that pn is non
increasing.
Consider f on the interval [1, n+1
n ]. We know f must be bounded below by n
n+1
so
n
n + 1
·
n + 1
n
− 1 ≤
n+1
n
1
1
x
dx
1
n + 1
≤ log
n + 1
n
1 +
1
2
+ · · · +
1
n
+
1
n + 1
− log(n + 1) ≤ 1 +
1
2
+ · · · +
1
n
− log(n)
pn+1 ≤ pn
So {pn}∞
n=1 is non increasing and because it is bounded below, it must converge.
Furthermore, consider p1 = 1−log 1 ≤ 1. Because {pn}∞
n=1 is non increasing, pn
is bounded above by 1 for all natural numbers n. In total, because 0 < pn ≤ 1
for all n ∈ N, we must have that
lim
n→∞
pn ∈ [0, 1]
(Problem 23)
Let f : R → R be differentiable such that f = f on R and f(0) = 1.
Claim: Then for all x ∈ R.
f(x) = ex
Proof. Let x0 ∈ R. If x0 = 0 then both f(x0) = ex0
= 1. So assume x0 = 0.
In order to show that f(x0) = ex0
we will take the time to show the sequence
{sn}∞
n=1 ⊂ R defined by
sn =
|x0|n
n!
18

19. Chapter VI Riemann Integration
converges to 0. Let > 0. By the Archimedean principle, there exists some
natural number N such that N > |x0|. We know that
|x0|N
is a fixed positive number. We know that |x0|
N is a fixed positive number between
0 and 1. We know that the sequence of functions fn = xn
converges pointwise
to 0 on the interval [0, 1). Because |x0|
N is in this interval, there exists some N
such that
n > N →
|x0|
N
n
<
|x0|N
(Implication I)
Additionally, if n > N we know 1
n , 1
n−1 . . . 1
N+1 < 1
N so
n > N →
|x0|n−N
n · (n − 1) · · · (N + 1)
<
|x0|
N
n
|x0|n−N
· N!
n!
<
|x0|
N
n
|x0|n−N
n!
<
|x0|
N
n
(Implication II)
So, if n > max{(N, N }) both the two above implications hold and we have that
|x0|n−N
n!
<
|x0|N
|x0|n
n!
<
|sn − 0| <
And the result is proven. We know that both ex
and f(x) are infinitely differ-
entiable, so their nth
derivatives are always defined. By the integral version of
Taylor’s remainder theorem (Problem 19 of this chapter, we have both that
f(x0) = f(0) + f (0)|x0| + · · · +
f(n)
(0)|x0|n
n!
+
x0
0
f(n+1)
(x) ·
(x0 − x)
n!
dx
= 1 + |x0| + · · · +
|x0|n
n!
+
x0
0
f(x) ·
(x0 − x)
n!
dx
and
ex0
= e0
+ e0
|x0| + · · · +
e0
|x0|n
n!
+
x0
0
ex
·
(x0 − x)n
n!
dx
= 1 + |x0| + · · · +
|x0|n
n!
+
x0
0
f(x) ·
(x0 − x)
n!
dx
19

20. Chapter VI Riemann Integration
Which implies
f(x0) − ex0
=
x0
0
(f(x) − ex
) ·
(x0 − x)n
n!
dx
but we know |x0|n
is an upper bound on (x0 − x)n
where x is an element of
the closed interval with extremities x0 and 0, and we must have some positive
number M which bounds |f − ex
| from above on the same interval. Hence, we
must have that
f(x0) − ex0
=
x0
0
(f(x) − ex
) ·
(x0 − x)n
n!
dx ≤
M · |x0|n+1
n!
= sn · M · |x0|
for any natural number n. But, because we showed that {sn}∞
n=1 converges to
zero, there must exist some n such that
sn <
M|x0|
So
f(x0) − ex0
<
Because is an arbitrary positive number, we must have that f(x0) = ex0
so
f(x) = ex
for all x ∈ R.
20