Problem 3. Define f [0, 1] → Ras : f(x) = { 1 x = [0,1/2) x € [1/1, 1] Using the Integrability Criteria (Theorem 7.2.8) prove that f is Riemann Integrable.
Problem 3. Define f [0, 1] → Ras : f(x) = { 1 x = [0,1/2) x € [1/1, 1] Using the Integrability Criteria (Theorem 7.2.8) prove that f is Riemann Integrable.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![Problem 3.
Define f : [0, 1] → R as
S1 z€ (0, })
f(x) =
10 zE,1] *
x € [3,1]
Using the Integrability Criteria (Theorem 7.2.8) prove that f is Riemann Integrable.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d4bcd37-5435-406e-9dcc-959e9ecc5bc8%2F03845361-5ab5-445f-8ce3-e72b7c7b32b8%2Flup3e28_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem 3.
Define f : [0, 1] → R as
S1 z€ (0, })
f(x) =
10 zE,1] *
x € [3,1]
Using the Integrability Criteria (Theorem 7.2.8) prove that f is Riemann Integrable.
![Personal Hotspot : Used 100 MB
65%
abbott-secon...
7.2. The Definition of the Riemann Integral
221
of integrable functions. The preceding inequality reveals that integrability is
really equivalent to the existence of partitions whose upper and lower sums are
arbitrarily close together.
Theorem 7.2.8 (Integrability Criterion). A bounded function ƒ is inte-
grable on [a, b] if and only if, for every e > 0, there erists a partition P. of [a, b]
such that
U(f, P.) – L(f, P.) < e.
Proof. Let e > 0. If such a partition P exists, then
U(f) – L(f) < U(f, P.) – L(f, P.) < e.
Because e is arbitrary, it must be that U(f) = L(), so ƒ is integrable. (To be
absolutely precise here, we could throw in a reference to Theorem 1.2.6.)
The proof of the converse statement is a familiar triangle inequality argument
with parentheses in place of absolute value bars because, in each case, we know
which quantity is larger. Because U(f) is the greatest lower bound of the upper
sums, we know that, given some e > 0, there must exist a partition P such that
U(f, P1) < U(f) +
Likewise, there exists a partition P2 satisfying
L(f, P2) > L(f) –
Now, let P. = P, U P2 be the common refinement. Keeping in mind that the
integrability of ƒ means U(f) = L(f), we can write
U(f, P.) – L(f, P.) < U(f,Pi) – L(f, P2)
(U) +5) - (LU) – 5)
In the discussion at the beginning of this chapter, it became clear that inte-
grability is closely tied to the concept of continuity. To make this observation
more precise, let P = {ro, ¤1,72; . .., In} be an arbitrary partition of [a, b], and
define Ark = xk - Ik-1. Then,
U(f, P) – L(f, P) =E(Mk – mx)Ark,
k=1
where M and m, are the supremum and infimum of the function on the interval
[Ik-1, Ik], respectively. Our ability to control the size of U (f, P)–L(S, P) hinges
on the differences Mg – mk, which we can interpret as the variation in the range
of the function over the interval [rk-1, Tx]. Restricting the variation of ƒ over
arbitrarily small intervals in [a, b] is precisely what it means to say that f is
uniformly continuous on this set.
222
Chapter 7. The Riemann Integral](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d4bcd37-5435-406e-9dcc-959e9ecc5bc8%2F03845361-5ab5-445f-8ce3-e72b7c7b32b8%2Fj0d6vv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Personal Hotspot : Used 100 MB
65%
abbott-secon...
7.2. The Definition of the Riemann Integral
221
of integrable functions. The preceding inequality reveals that integrability is
really equivalent to the existence of partitions whose upper and lower sums are
arbitrarily close together.
Theorem 7.2.8 (Integrability Criterion). A bounded function ƒ is inte-
grable on [a, b] if and only if, for every e > 0, there erists a partition P. of [a, b]
such that
U(f, P.) – L(f, P.) < e.
Proof. Let e > 0. If such a partition P exists, then
U(f) – L(f) < U(f, P.) – L(f, P.) < e.
Because e is arbitrary, it must be that U(f) = L(), so ƒ is integrable. (To be
absolutely precise here, we could throw in a reference to Theorem 1.2.6.)
The proof of the converse statement is a familiar triangle inequality argument
with parentheses in place of absolute value bars because, in each case, we know
which quantity is larger. Because U(f) is the greatest lower bound of the upper
sums, we know that, given some e > 0, there must exist a partition P such that
U(f, P1) < U(f) +
Likewise, there exists a partition P2 satisfying
L(f, P2) > L(f) –
Now, let P. = P, U P2 be the common refinement. Keeping in mind that the
integrability of ƒ means U(f) = L(f), we can write
U(f, P.) – L(f, P.) < U(f,Pi) – L(f, P2)
(U) +5) - (LU) – 5)
In the discussion at the beginning of this chapter, it became clear that inte-
grability is closely tied to the concept of continuity. To make this observation
more precise, let P = {ro, ¤1,72; . .., In} be an arbitrary partition of [a, b], and
define Ark = xk - Ik-1. Then,
U(f, P) – L(f, P) =E(Mk – mx)Ark,
k=1
where M and m, are the supremum and infimum of the function on the interval
[Ik-1, Ik], respectively. Our ability to control the size of U (f, P)–L(S, P) hinges
on the differences Mg – mk, which we can interpret as the variation in the range
of the function over the interval [rk-1, Tx]. Restricting the variation of ƒ over
arbitrarily small intervals in [a, b] is precisely what it means to say that f is
uniformly continuous on this set.
222
Chapter 7. The Riemann Integral
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