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Proving a constant function $f(x) = c$ is Riemann integrable
$begingroup$
Prove that a constant function $f(x) = c$, where $c$ is in the Real Numbers, is Riemann integrable on any interval $[a, b]$ and $int_a^bf(x) dx = c(b-a)$.
By looking at the definition, it looks like I am going to explain that it's bounded (which would be obvious since the function is constant?) Additionally, it would appear that $inf(f)$ and $sup(f)$ also obviously exist since $f$ is constant. The parts I am having trouble understanding involve explaining that $sup{L(P,f)} = inf{U(P,f)}$, as well as proving that $int_a^bf(x) dx = c(b-a)$. Let me know if what I have so far is okay, and please give me some guidance on the rest. Thanks!
real-analysis proof-writing
edited Nov 19 '13 at 16:19
mdp
11.3k12955
asked Nov 19 '13 at 16:11
Heath HuffmanHeath Huffman
174517
$endgroup$
add a comment |
$begingroup$
Prove that a constant function $f(x) = c$, where $c$ is in the Real Numbers, is Riemann integrable on any interval $[a, b]$ and $int_a^bf(x) dx = c(b-a)$.
By looking at the definition, it looks like I am going to explain that it's bounded (which would be obvious since the function is constant?) Additionally, it would appear that $inf(f)$ and $sup(f)$ also obviously exist since $f$ is constant. The parts I am having trouble understanding involve explaining that $sup{L(P,f)} = inf{U(P,f)}$, as well as proving that $int_a^bf(x) dx = c(b-a)$. Let me know if what I have so far is okay, and please give me some guidance on the rest. Thanks!
real-analysis proof-writing
edited Nov 19 '13 at 16:19
mdp
11.3k12955
asked Nov 19 '13 at 16:11
Heath HuffmanHeath Huffman
174517
$endgroup$
add a comment |
$begingroup$
Prove that a constant function $f(x) = c$, where $c$ is in the Real Numbers, is Riemann integrable on any interval $[a, b]$ and $int_a^bf(x) dx = c(b-a)$.
By looking at the definition, it looks like I am going to explain that it's bounded (which would be obvious since the function is constant?) Additionally, it would appear that $inf(f)$ and $sup(f)$ also obviously exist since $f$ is constant. The parts I am having trouble understanding involve explaining that $sup{L(P,f)} = inf{U(P,f)}$, as well as proving that $int_a^bf(x) dx = c(b-a)$. Let me know if what I have so far is okay, and please give me some guidance on the rest. Thanks!
real-analysis proof-writing
edited Nov 19 '13 at 16:19
mdp
11.3k12955
asked Nov 19 '13 at 16:11
Heath HuffmanHeath Huffman
174517
$endgroup$
Prove that a constant function $f(x) = c$, where $c$ is in the Real Numbers, is Riemann integrable on any interval $[a, b]$ and $int_a^bf(x) dx = c(b-a)$.
By looking at the definition, it looks like I am going to explain that it's bounded (which would be obvious since the function is constant?) Additionally, it would appear that $inf(f)$ and $sup(f)$ also obviously exist since $f$ is constant. The parts I am having trouble understanding involve explaining that $sup{L(P,f)} = inf{U(P,f)}$, as well as proving that $int_a^bf(x) dx = c(b-a)$. Let me know if what I have so far is okay, and please give me some guidance on the rest. Thanks!
real-analysis proof-writing
real-analysis proof-writing
edited Nov 19 '13 at 16:19
mdp
11.3k12955
asked Nov 19 '13 at 16:11
Heath HuffmanHeath Huffman
174517
edited Nov 19 '13 at 16:19
mdp
11.3k12955
asked Nov 19 '13 at 16:11
Heath HuffmanHeath Huffman
174517
edited Nov 19 '13 at 16:19
mdp
11.3k12955
edited Nov 19 '13 at 16:19
mdp
11.3k12955
edited Nov 19 '13 at 16:19
mdp
11.3k12955
11.3k12955
asked Nov 19 '13 at 16:11
Heath HuffmanHeath Huffman
174517
asked Nov 19 '13 at 16:11
Heath HuffmanHeath Huffman
174517
asked Nov 19 '13 at 16:11
Heath HuffmanHeath Huffman
174517
174517
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You're on the right track so far. For the sup=inf part, you're in a really nice position: if you take any partition and compute the max of f on each subinterval, you'll get c. So $U(P, f) = sum_i c cdot (t_{i+1} - t_i) = c cdot sum_i t_{i+1} - t_i = c(b-a)$. That means that EVERY SINGLE Upper sum turns out to be $c(b-a)$. I'll bet that you can compute the inf of a set that contains only a single number, right? Then you're on your way.
answered Nov 19 '13 at 16:16
John HughesJohn Hughes
64.1k24191
$endgroup$
$begingroup$
I got it! This worked out very well. Thank you!
$endgroup$
– Heath Huffman
Nov 19 '13 at 20:31
add a comment |
$begingroup$
Constant function is a continoues function and we know that every continoues function is a Riemann integrable.Since Constant function is a Riemann Integrable.
answered Jan 5 '17 at 17:54
Shekhar KumarShekhar Kumar
1
$endgroup$
add a comment |
$begingroup$
Sorry, this answer is coming in after five years.
Consider, for each $n$, a uniform partition, $P_n$ on $[a,b]$ which is a finite sequence of real numbers such that
$$a=x_0<x_<cdots<x_n=b.$$
Define $I_j=[x_{j-1},x_j)$ for each $1leq jleq n-1,$ and $,I_n=[x_{n-1},x_n]$. Clearly, $f$ is bounded on $[a,b]$, so $m:=inf f$ and $M:=sup f$ both exist. Fix $jin {1,2,cdots,n}$ and $,xin I_j,$ then
$$m_jleq f(t_j)leq M_j,;;text{where};;t_jin I_j,$$
begin{align}m_j=inf{f(x):;xin I_j }=c ;;text{and};; M_j=sup{f(x):;xin I_j }=c.end{align}
Summing up to $n,$ results to
begin{align}L(f,P_n)&=c(b-a)=c(x_{n}-x_{0})=csum^{n}_{j=1}(x_{j}-x_{j-1})=sum^{n}_{j=1}m_j(x_{j}-x_{j-1})leq sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})\&leq sum^{n}_{j=1}M_j(x_{j}-x_{j-1})=csum^{n}_{j=1}(x_{j}-x_{j-1})=c(x_{n}-x_{0})=c(b-a)=U(f,P_n),end{align}
Thus,
begin{align}suplimits_{P_n} L(f,P_n)&=limlimits_{nto infty}L(f,P_n)=c(b-a)leq limlimits_{nto infty} sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})\&leq c(b-a)=limlimits_{nto infty}U(f,P_n)=inflimits_{P_n} U(f,P_n),end{align}
and
begin{align}int^{a}_{b}f(x)dx= limlimits_{nto infty}sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})=suplimits_{P_n} L(f,P_n)=inflimits_{P_n} U(f,P_n)=c(b-a),end{align}
as required.
answered Jan 15 at 17:38
Omojola MichealOmojola Micheal
1,889324
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
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active
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$begingroup$
You're on the right track so far. For the sup=inf part, you're in a really nice position: if you take any partition and compute the max of f on each subinterval, you'll get c. So $U(P, f) = sum_i c cdot (t_{i+1} - t_i) = c cdot sum_i t_{i+1} - t_i = c(b-a)$. That means that EVERY SINGLE Upper sum turns out to be $c(b-a)$. I'll bet that you can compute the inf of a set that contains only a single number, right? Then you're on your way.
answered Nov 19 '13 at 16:16
John HughesJohn Hughes
64.1k24191
$endgroup$
$begingroup$
I got it! This worked out very well. Thank you!
$endgroup$
– Heath Huffman
Nov 19 '13 at 20:31
add a comment |
$begingroup$
You're on the right track so far. For the sup=inf part, you're in a really nice position: if you take any partition and compute the max of f on each subinterval, you'll get c. So $U(P, f) = sum_i c cdot (t_{i+1} - t_i) = c cdot sum_i t_{i+1} - t_i = c(b-a)$. That means that EVERY SINGLE Upper sum turns out to be $c(b-a)$. I'll bet that you can compute the inf of a set that contains only a single number, right? Then you're on your way.
answered Nov 19 '13 at 16:16
John HughesJohn Hughes
64.1k24191
$endgroup$
$begingroup$
I got it! This worked out very well. Thank you!
$endgroup$
– Heath Huffman
Nov 19 '13 at 20:31
add a comment |
$begingroup$
You're on the right track so far. For the sup=inf part, you're in a really nice position: if you take any partition and compute the max of f on each subinterval, you'll get c. So $U(P, f) = sum_i c cdot (t_{i+1} - t_i) = c cdot sum_i t_{i+1} - t_i = c(b-a)$. That means that EVERY SINGLE Upper sum turns out to be $c(b-a)$. I'll bet that you can compute the inf of a set that contains only a single number, right? Then you're on your way.
answered Nov 19 '13 at 16:16
John HughesJohn Hughes
64.1k24191
$endgroup$
You're on the right track so far. For the sup=inf part, you're in a really nice position: if you take any partition and compute the max of f on each subinterval, you'll get c. So $U(P, f) = sum_i c cdot (t_{i+1} - t_i) = c cdot sum_i t_{i+1} - t_i = c(b-a)$. That means that EVERY SINGLE Upper sum turns out to be $c(b-a)$. I'll bet that you can compute the inf of a set that contains only a single number, right? Then you're on your way.
answered Nov 19 '13 at 16:16
John HughesJohn Hughes
64.1k24191
answered Nov 19 '13 at 16:16
John HughesJohn Hughes
64.1k24191
answered Nov 19 '13 at 16:16
John HughesJohn Hughes
64.1k24191
answered Nov 19 '13 at 16:16
John HughesJohn Hughes
64.1k24191
64.1k24191
$begingroup$
I got it! This worked out very well. Thank you!
$endgroup$
– Heath Huffman
Nov 19 '13 at 20:31
add a comment |
$begingroup$
I got it! This worked out very well. Thank you!
$endgroup$
– Heath Huffman
Nov 19 '13 at 20:31
$begingroup$
I got it! This worked out very well. Thank you!
$endgroup$
– Heath Huffman
Nov 19 '13 at 20:31
$begingroup$
I got it! This worked out very well. Thank you!
$endgroup$
– Heath Huffman
Nov 19 '13 at 20:31
add a comment |
$begingroup$
Constant function is a continoues function and we know that every continoues function is a Riemann integrable.Since Constant function is a Riemann Integrable.
answered Jan 5 '17 at 17:54
Shekhar KumarShekhar Kumar
1
$endgroup$
add a comment |
$begingroup$
Constant function is a continoues function and we know that every continoues function is a Riemann integrable.Since Constant function is a Riemann Integrable.
answered Jan 5 '17 at 17:54
Shekhar KumarShekhar Kumar
1
$endgroup$
add a comment |
$begingroup$
Constant function is a continoues function and we know that every continoues function is a Riemann integrable.Since Constant function is a Riemann Integrable.
answered Jan 5 '17 at 17:54
Shekhar KumarShekhar Kumar
1
$endgroup$
Constant function is a continoues function and we know that every continoues function is a Riemann integrable.Since Constant function is a Riemann Integrable.
answered Jan 5 '17 at 17:54
Shekhar KumarShekhar Kumar
1
answered Jan 5 '17 at 17:54
Shekhar KumarShekhar Kumar
1
answered Jan 5 '17 at 17:54
Shekhar KumarShekhar Kumar
1
answered Jan 5 '17 at 17:54
Shekhar KumarShekhar Kumar
1
1
add a comment |
add a comment |
$begingroup$
Sorry, this answer is coming in after five years.
Consider, for each $n$, a uniform partition, $P_n$ on $[a,b]$ which is a finite sequence of real numbers such that
$$a=x_0<x_<cdots<x_n=b.$$
Define $I_j=[x_{j-1},x_j)$ for each $1leq jleq n-1,$ and $,I_n=[x_{n-1},x_n]$. Clearly, $f$ is bounded on $[a,b]$, so $m:=inf f$ and $M:=sup f$ both exist. Fix $jin {1,2,cdots,n}$ and $,xin I_j,$ then
$$m_jleq f(t_j)leq M_j,;;text{where};;t_jin I_j,$$
begin{align}m_j=inf{f(x):;xin I_j }=c ;;text{and};; M_j=sup{f(x):;xin I_j }=c.end{align}
Summing up to $n,$ results to
begin{align}L(f,P_n)&=c(b-a)=c(x_{n}-x_{0})=csum^{n}_{j=1}(x_{j}-x_{j-1})=sum^{n}_{j=1}m_j(x_{j}-x_{j-1})leq sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})\&leq sum^{n}_{j=1}M_j(x_{j}-x_{j-1})=csum^{n}_{j=1}(x_{j}-x_{j-1})=c(x_{n}-x_{0})=c(b-a)=U(f,P_n),end{align}
Thus,
begin{align}suplimits_{P_n} L(f,P_n)&=limlimits_{nto infty}L(f,P_n)=c(b-a)leq limlimits_{nto infty} sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})\&leq c(b-a)=limlimits_{nto infty}U(f,P_n)=inflimits_{P_n} U(f,P_n),end{align}
and
begin{align}int^{a}_{b}f(x)dx= limlimits_{nto infty}sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})=suplimits_{P_n} L(f,P_n)=inflimits_{P_n} U(f,P_n)=c(b-a),end{align}
as required.
answered Jan 15 at 17:38
Omojola MichealOmojola Micheal
1,889324
$endgroup$
add a comment |
$begingroup$
Sorry, this answer is coming in after five years.
Consider, for each $n$, a uniform partition, $P_n$ on $[a,b]$ which is a finite sequence of real numbers such that
$$a=x_0<x_<cdots<x_n=b.$$
Define $I_j=[x_{j-1},x_j)$ for each $1leq jleq n-1,$ and $,I_n=[x_{n-1},x_n]$. Clearly, $f$ is bounded on $[a,b]$, so $m:=inf f$ and $M:=sup f$ both exist. Fix $jin {1,2,cdots,n}$ and $,xin I_j,$ then
$$m_jleq f(t_j)leq M_j,;;text{where};;t_jin I_j,$$
begin{align}m_j=inf{f(x):;xin I_j }=c ;;text{and};; M_j=sup{f(x):;xin I_j }=c.end{align}
Summing up to $n,$ results to
begin{align}L(f,P_n)&=c(b-a)=c(x_{n}-x_{0})=csum^{n}_{j=1}(x_{j}-x_{j-1})=sum^{n}_{j=1}m_j(x_{j}-x_{j-1})leq sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})\&leq sum^{n}_{j=1}M_j(x_{j}-x_{j-1})=csum^{n}_{j=1}(x_{j}-x_{j-1})=c(x_{n}-x_{0})=c(b-a)=U(f,P_n),end{align}
Thus,
begin{align}suplimits_{P_n} L(f,P_n)&=limlimits_{nto infty}L(f,P_n)=c(b-a)leq limlimits_{nto infty} sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})\&leq c(b-a)=limlimits_{nto infty}U(f,P_n)=inflimits_{P_n} U(f,P_n),end{align}
and
begin{align}int^{a}_{b}f(x)dx= limlimits_{nto infty}sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})=suplimits_{P_n} L(f,P_n)=inflimits_{P_n} U(f,P_n)=c(b-a),end{align}
as required.
answered Jan 15 at 17:38
Omojola MichealOmojola Micheal
1,889324
$endgroup$
add a comment |
$begingroup$
Sorry, this answer is coming in after five years.
Consider, for each $n$, a uniform partition, $P_n$ on $[a,b]$ which is a finite sequence of real numbers such that
$$a=x_0<x_<cdots<x_n=b.$$
Define $I_j=[x_{j-1},x_j)$ for each $1leq jleq n-1,$ and $,I_n=[x_{n-1},x_n]$. Clearly, $f$ is bounded on $[a,b]$, so $m:=inf f$ and $M:=sup f$ both exist. Fix $jin {1,2,cdots,n}$ and $,xin I_j,$ then
$$m_jleq f(t_j)leq M_j,;;text{where};;t_jin I_j,$$
begin{align}m_j=inf{f(x):;xin I_j }=c ;;text{and};; M_j=sup{f(x):;xin I_j }=c.end{align}
Summing up to $n,$ results to
begin{align}L(f,P_n)&=c(b-a)=c(x_{n}-x_{0})=csum^{n}_{j=1}(x_{j}-x_{j-1})=sum^{n}_{j=1}m_j(x_{j}-x_{j-1})leq sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})\&leq sum^{n}_{j=1}M_j(x_{j}-x_{j-1})=csum^{n}_{j=1}(x_{j}-x_{j-1})=c(x_{n}-x_{0})=c(b-a)=U(f,P_n),end{align}
Thus,
begin{align}suplimits_{P_n} L(f,P_n)&=limlimits_{nto infty}L(f,P_n)=c(b-a)leq limlimits_{nto infty} sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})\&leq c(b-a)=limlimits_{nto infty}U(f,P_n)=inflimits_{P_n} U(f,P_n),end{align}
and
begin{align}int^{a}_{b}f(x)dx= limlimits_{nto infty}sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})=suplimits_{P_n} L(f,P_n)=inflimits_{P_n} U(f,P_n)=c(b-a),end{align}
as required.
answered Jan 15 at 17:38
Omojola MichealOmojola Micheal
1,889324
$endgroup$
Sorry, this answer is coming in after five years.
Consider, for each $n$, a uniform partition, $P_n$ on $[a,b]$ which is a finite sequence of real numbers such that
$$a=x_0<x_<cdots<x_n=b.$$
Define $I_j=[x_{j-1},x_j)$ for each $1leq jleq n-1,$ and $,I_n=[x_{n-1},x_n]$. Clearly, $f$ is bounded on $[a,b]$, so $m:=inf f$ and $M:=sup f$ both exist. Fix $jin {1,2,cdots,n}$ and $,xin I_j,$ then
$$m_jleq f(t_j)leq M_j,;;text{where};;t_jin I_j,$$
begin{align}m_j=inf{f(x):;xin I_j }=c ;;text{and};; M_j=sup{f(x):;xin I_j }=c.end{align}
Summing up to $n,$ results to
begin{align}L(f,P_n)&=c(b-a)=c(x_{n}-x_{0})=csum^{n}_{j=1}(x_{j}-x_{j-1})=sum^{n}_{j=1}m_j(x_{j}-x_{j-1})leq sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})\&leq sum^{n}_{j=1}M_j(x_{j}-x_{j-1})=csum^{n}_{j=1}(x_{j}-x_{j-1})=c(x_{n}-x_{0})=c(b-a)=U(f,P_n),end{align}
Thus,
begin{align}suplimits_{P_n} L(f,P_n)&=limlimits_{nto infty}L(f,P_n)=c(b-a)leq limlimits_{nto infty} sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})\&leq c(b-a)=limlimits_{nto infty}U(f,P_n)=inflimits_{P_n} U(f,P_n),end{align}
and
begin{align}int^{a}_{b}f(x)dx= limlimits_{nto infty}sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})=suplimits_{P_n} L(f,P_n)=inflimits_{P_n} U(f,P_n)=c(b-a),end{align}
as required.
answered Jan 15 at 17:38
Omojola MichealOmojola Micheal
1,889324
answered Jan 15 at 17:38
Omojola MichealOmojola Micheal
1,889324
answered Jan 15 at 17:38
Omojola MichealOmojola Micheal
1,889324
answered Jan 15 at 17:38
Omojola MichealOmojola Micheal
1,889324
1,889324
add a comment |
add a comment |
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