(b) Fix r and limit m to infinity to conclude ak+¹(1 − r)rk - 1 1-guk+1 ≤ L(ƒ) ≤U(ƒ) ≤ a*+¹(1 − r); (c) Prove that fro all 0 < r < 1 we have ak+1pk ; ≤L(S) ≤U(S) ≤; 1+r+r²+ + puk (d) Prove that f is Riemann integrable and that rdr = ak+1 1+r+r²+ ak+1 k+1
(b) Fix r and limit m to infinity to conclude ak+¹(1 − r)rk - 1 1-guk+1 ≤ L(ƒ) ≤U(ƒ) ≤ a*+¹(1 − r); (c) Prove that fro all 0 < r < 1 we have ak+1pk ; ≤L(S) ≤U(S) ≤; 1+r+r²+ + puk (d) Prove that f is Riemann integrable and that rdr = ak+1 1+r+r²+ ak+1 k+1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
parts b, c, d
![2. RIEMANN SUMS
(a) Let a > 0 be a real number, let k be a natural number, and consider the function f : [0, a] → R
defined by f(x) = x*. For any natural number m and any 0 <r <1 consider the partition
Pm,r = {0, arm, arm-1,...ar², ar, a}
of [0, a]. Prove the formulas for L(f, Pm,r) and U(f, Pm,r) given by
L(f, Pm,r)= ah+1(1 − p) pk -
1-p(k+1)m
1-pk+1
and
1-p(k+1)m
= ak+1₂(k+1)m +ak+¹(1-r). 1-pk+1
U(f, Pm,r) = ak
(b) Fix r and limit m to infinity to conclude
1
*1 — pk+1 ≤ L(f) ≤ U (f) ≤ a^+¹ (1 − r);
ak+¹(1 — r)rk-
(c) Prove that fro all 0 <r< 1 we have
ak+1 gk
<L(f) ≤U(f) ≤
1+r+p² +...+ juht
(d) Prove that f is Riemann integrable and that
S² =
rdx
=
ak+1
k+1
1
- pk+1'
ak+1
1+r+p²+...+goti"](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F04625ac1-43ff-4999-b93a-55388fc0c5e2%2Fd1575885-ce8f-4888-8810-cb80cba9449a%2Fo4ubvil_processed.png&w=3840&q=75)
Transcribed Image Text:2. RIEMANN SUMS
(a) Let a > 0 be a real number, let k be a natural number, and consider the function f : [0, a] → R
defined by f(x) = x*. For any natural number m and any 0 <r <1 consider the partition
Pm,r = {0, arm, arm-1,...ar², ar, a}
of [0, a]. Prove the formulas for L(f, Pm,r) and U(f, Pm,r) given by
L(f, Pm,r)= ah+1(1 − p) pk -
1-p(k+1)m
1-pk+1
and
1-p(k+1)m
= ak+1₂(k+1)m +ak+¹(1-r). 1-pk+1
U(f, Pm,r) = ak
(b) Fix r and limit m to infinity to conclude
1
*1 — pk+1 ≤ L(f) ≤ U (f) ≤ a^+¹ (1 − r);
ak+¹(1 — r)rk-
(c) Prove that fro all 0 <r< 1 we have
ak+1 gk
<L(f) ≤U(f) ≤
1+r+p² +...+ juht
(d) Prove that f is Riemann integrable and that
S² =
rdx
=
ak+1
k+1
1
- pk+1'
ak+1
1+r+p²+...+goti"
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it looks like part c and part d were cut off (just says U( and we dont see what L(f)< in part d
also part b looks incomplete, so can u give a more thorough answer for that too
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