Prove that $f(x)=sqrt{x}$ is Riemann integrable over $[0,1]$
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Prove that $f(x)=sqrt{x}$ is Riemann integrable over $[0,1]$
MY WORK
Consider the Riemann sum
begin{align}int^{1}_{0}f(x)dx&=limlimits_{ntoinfty}frac{1}{n}sum^{n-1}_{k=0}fleft(frac{k}{n}right)\&=limlimits_{ntoinfty}frac{1}{n}sum^{n-1}_{k=0}sqrt{frac{k}{n}}\&=limlimits_{ntoinfty}frac{1}{n^{3/2}}sum^{n-1}_{k=0}sqrt{k}end{align}
I'm stuck here, what should I do? Thanks!
calculus real-analysis limits analysis riemann-integration
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add a comment |
$begingroup$
Prove that $f(x)=sqrt{x}$ is Riemann integrable over $[0,1]$
MY WORK
Consider the Riemann sum
begin{align}int^{1}_{0}f(x)dx&=limlimits_{ntoinfty}frac{1}{n}sum^{n-1}_{k=0}fleft(frac{k}{n}right)\&=limlimits_{ntoinfty}frac{1}{n}sum^{n-1}_{k=0}sqrt{frac{k}{n}}\&=limlimits_{ntoinfty}frac{1}{n^{3/2}}sum^{n-1}_{k=0}sqrt{k}end{align}
I'm stuck here, what should I do? Thanks!
calculus real-analysis limits analysis riemann-integration
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5
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The function is continuous on a compact interval hence...
$endgroup$
– Did
Sep 11 '18 at 6:40
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Another reason: the given function is monotone and hence Riemann integrable on $[0,1]$. The answer by RRL can be used to show that any monotone function is Riemann integrable.
$endgroup$
– Paramanand Singh
Sep 11 '18 at 9:11
add a comment |
$begingroup$
Prove that $f(x)=sqrt{x}$ is Riemann integrable over $[0,1]$
MY WORK
Consider the Riemann sum
begin{align}int^{1}_{0}f(x)dx&=limlimits_{ntoinfty}frac{1}{n}sum^{n-1}_{k=0}fleft(frac{k}{n}right)\&=limlimits_{ntoinfty}frac{1}{n}sum^{n-1}_{k=0}sqrt{frac{k}{n}}\&=limlimits_{ntoinfty}frac{1}{n^{3/2}}sum^{n-1}_{k=0}sqrt{k}end{align}
I'm stuck here, what should I do? Thanks!
calculus real-analysis limits analysis riemann-integration
$endgroup$
Prove that $f(x)=sqrt{x}$ is Riemann integrable over $[0,1]$
MY WORK
Consider the Riemann sum
begin{align}int^{1}_{0}f(x)dx&=limlimits_{ntoinfty}frac{1}{n}sum^{n-1}_{k=0}fleft(frac{k}{n}right)\&=limlimits_{ntoinfty}frac{1}{n}sum^{n-1}_{k=0}sqrt{frac{k}{n}}\&=limlimits_{ntoinfty}frac{1}{n^{3/2}}sum^{n-1}_{k=0}sqrt{k}end{align}
I'm stuck here, what should I do? Thanks!
calculus real-analysis limits analysis riemann-integration
calculus real-analysis limits analysis riemann-integration
edited Dec 2 '18 at 3:02
RRL
49.5k42573
49.5k42573
asked Sep 11 '18 at 6:34
Omojola MichealOmojola Micheal
1,802324
1,802324
5
$begingroup$
The function is continuous on a compact interval hence...
$endgroup$
– Did
Sep 11 '18 at 6:40
$begingroup$
Another reason: the given function is monotone and hence Riemann integrable on $[0,1]$. The answer by RRL can be used to show that any monotone function is Riemann integrable.
$endgroup$
– Paramanand Singh
Sep 11 '18 at 9:11
add a comment |
5
$begingroup$
The function is continuous on a compact interval hence...
$endgroup$
– Did
Sep 11 '18 at 6:40
$begingroup$
Another reason: the given function is monotone and hence Riemann integrable on $[0,1]$. The answer by RRL can be used to show that any monotone function is Riemann integrable.
$endgroup$
– Paramanand Singh
Sep 11 '18 at 9:11
5
5
$begingroup$
The function is continuous on a compact interval hence...
$endgroup$
– Did
Sep 11 '18 at 6:40
$begingroup$
The function is continuous on a compact interval hence...
$endgroup$
– Did
Sep 11 '18 at 6:40
$begingroup$
Another reason: the given function is monotone and hence Riemann integrable on $[0,1]$. The answer by RRL can be used to show that any monotone function is Riemann integrable.
$endgroup$
– Paramanand Singh
Sep 11 '18 at 9:11
$begingroup$
Another reason: the given function is monotone and hence Riemann integrable on $[0,1]$. The answer by RRL can be used to show that any monotone function is Riemann integrable.
$endgroup$
– Paramanand Singh
Sep 11 '18 at 9:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There are a number of possible approaches.
If you are trying to show that the integral exists with Riemann (Darboux) sums, then since $f(x) = sqrt{x}$ is increasing, the upper and lower sums for a uniform partition $P = (0, 1/n, 2/n, ldots, 1)$ are
$$U(P,f) = frac1{n}sum_{k=1}^nsqrt{k/n}= frac1{n^{3/2}}sum_{k=1}^nsqrt{k},\L(P,f) = frac1{n}sum_{k=1}^nsqrt{(k-1)/n}= frac1{n^{3/2}}sum_{k=1}^nsqrt{k-1}$$
Hence,
$$U(P,f)-L(P,f) = frac1{n^{3/2}}sum_{k=1}^nleft(sqrt{k}-sqrt{k-1}right)$$
Since the sum is telescoping, we have for $n > 1/epsilon$,
$$U(P,f)-L(P,f) = frac{sqrt{n}}{n^{3/2}}= frac1{n}< epsilon$$
Therefore, $f$ is integrable by the Riemann criterion -- since for any $epsilon >0$ there exists a partition for which the difference between upper and lower sums is less than $epsilon$.
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Thanks very much, I am grateful!
$endgroup$
– Omojola Micheal
Sep 11 '18 at 7:22
$begingroup$
@Mike: You're welcome. Glad this helped.
$endgroup$
– RRL
Sep 11 '18 at 7:40
add a comment |
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Recall that a continuous function $f:[a,b] to mathbb{R}$ is integrable on the interval $[a,b]$.
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add a comment |
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2 Answers
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active
oldest
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2 Answers
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active
oldest
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active
oldest
votes
$begingroup$
There are a number of possible approaches.
If you are trying to show that the integral exists with Riemann (Darboux) sums, then since $f(x) = sqrt{x}$ is increasing, the upper and lower sums for a uniform partition $P = (0, 1/n, 2/n, ldots, 1)$ are
$$U(P,f) = frac1{n}sum_{k=1}^nsqrt{k/n}= frac1{n^{3/2}}sum_{k=1}^nsqrt{k},\L(P,f) = frac1{n}sum_{k=1}^nsqrt{(k-1)/n}= frac1{n^{3/2}}sum_{k=1}^nsqrt{k-1}$$
Hence,
$$U(P,f)-L(P,f) = frac1{n^{3/2}}sum_{k=1}^nleft(sqrt{k}-sqrt{k-1}right)$$
Since the sum is telescoping, we have for $n > 1/epsilon$,
$$U(P,f)-L(P,f) = frac{sqrt{n}}{n^{3/2}}= frac1{n}< epsilon$$
Therefore, $f$ is integrable by the Riemann criterion -- since for any $epsilon >0$ there exists a partition for which the difference between upper and lower sums is less than $epsilon$.
$endgroup$
$begingroup$
Thanks very much, I am grateful!
$endgroup$
– Omojola Micheal
Sep 11 '18 at 7:22
$begingroup$
@Mike: You're welcome. Glad this helped.
$endgroup$
– RRL
Sep 11 '18 at 7:40
add a comment |
$begingroup$
There are a number of possible approaches.
If you are trying to show that the integral exists with Riemann (Darboux) sums, then since $f(x) = sqrt{x}$ is increasing, the upper and lower sums for a uniform partition $P = (0, 1/n, 2/n, ldots, 1)$ are
$$U(P,f) = frac1{n}sum_{k=1}^nsqrt{k/n}= frac1{n^{3/2}}sum_{k=1}^nsqrt{k},\L(P,f) = frac1{n}sum_{k=1}^nsqrt{(k-1)/n}= frac1{n^{3/2}}sum_{k=1}^nsqrt{k-1}$$
Hence,
$$U(P,f)-L(P,f) = frac1{n^{3/2}}sum_{k=1}^nleft(sqrt{k}-sqrt{k-1}right)$$
Since the sum is telescoping, we have for $n > 1/epsilon$,
$$U(P,f)-L(P,f) = frac{sqrt{n}}{n^{3/2}}= frac1{n}< epsilon$$
Therefore, $f$ is integrable by the Riemann criterion -- since for any $epsilon >0$ there exists a partition for which the difference between upper and lower sums is less than $epsilon$.
$endgroup$
$begingroup$
Thanks very much, I am grateful!
$endgroup$
– Omojola Micheal
Sep 11 '18 at 7:22
$begingroup$
@Mike: You're welcome. Glad this helped.
$endgroup$
– RRL
Sep 11 '18 at 7:40
add a comment |
$begingroup$
There are a number of possible approaches.
If you are trying to show that the integral exists with Riemann (Darboux) sums, then since $f(x) = sqrt{x}$ is increasing, the upper and lower sums for a uniform partition $P = (0, 1/n, 2/n, ldots, 1)$ are
$$U(P,f) = frac1{n}sum_{k=1}^nsqrt{k/n}= frac1{n^{3/2}}sum_{k=1}^nsqrt{k},\L(P,f) = frac1{n}sum_{k=1}^nsqrt{(k-1)/n}= frac1{n^{3/2}}sum_{k=1}^nsqrt{k-1}$$
Hence,
$$U(P,f)-L(P,f) = frac1{n^{3/2}}sum_{k=1}^nleft(sqrt{k}-sqrt{k-1}right)$$
Since the sum is telescoping, we have for $n > 1/epsilon$,
$$U(P,f)-L(P,f) = frac{sqrt{n}}{n^{3/2}}= frac1{n}< epsilon$$
Therefore, $f$ is integrable by the Riemann criterion -- since for any $epsilon >0$ there exists a partition for which the difference between upper and lower sums is less than $epsilon$.
$endgroup$
There are a number of possible approaches.
If you are trying to show that the integral exists with Riemann (Darboux) sums, then since $f(x) = sqrt{x}$ is increasing, the upper and lower sums for a uniform partition $P = (0, 1/n, 2/n, ldots, 1)$ are
$$U(P,f) = frac1{n}sum_{k=1}^nsqrt{k/n}= frac1{n^{3/2}}sum_{k=1}^nsqrt{k},\L(P,f) = frac1{n}sum_{k=1}^nsqrt{(k-1)/n}= frac1{n^{3/2}}sum_{k=1}^nsqrt{k-1}$$
Hence,
$$U(P,f)-L(P,f) = frac1{n^{3/2}}sum_{k=1}^nleft(sqrt{k}-sqrt{k-1}right)$$
Since the sum is telescoping, we have for $n > 1/epsilon$,
$$U(P,f)-L(P,f) = frac{sqrt{n}}{n^{3/2}}= frac1{n}< epsilon$$
Therefore, $f$ is integrable by the Riemann criterion -- since for any $epsilon >0$ there exists a partition for which the difference between upper and lower sums is less than $epsilon$.
edited Sep 11 '18 at 7:01
answered Sep 11 '18 at 6:48
RRLRRL
49.5k42573
49.5k42573
$begingroup$
Thanks very much, I am grateful!
$endgroup$
– Omojola Micheal
Sep 11 '18 at 7:22
$begingroup$
@Mike: You're welcome. Glad this helped.
$endgroup$
– RRL
Sep 11 '18 at 7:40
add a comment |
$begingroup$
Thanks very much, I am grateful!
$endgroup$
– Omojola Micheal
Sep 11 '18 at 7:22
$begingroup$
@Mike: You're welcome. Glad this helped.
$endgroup$
– RRL
Sep 11 '18 at 7:40
$begingroup$
Thanks very much, I am grateful!
$endgroup$
– Omojola Micheal
Sep 11 '18 at 7:22
$begingroup$
Thanks very much, I am grateful!
$endgroup$
– Omojola Micheal
Sep 11 '18 at 7:22
$begingroup$
@Mike: You're welcome. Glad this helped.
$endgroup$
– RRL
Sep 11 '18 at 7:40
$begingroup$
@Mike: You're welcome. Glad this helped.
$endgroup$
– RRL
Sep 11 '18 at 7:40
add a comment |
$begingroup$
Recall that a continuous function $f:[a,b] to mathbb{R}$ is integrable on the interval $[a,b]$.
$endgroup$
add a comment |
$begingroup$
Recall that a continuous function $f:[a,b] to mathbb{R}$ is integrable on the interval $[a,b]$.
$endgroup$
add a comment |
$begingroup$
Recall that a continuous function $f:[a,b] to mathbb{R}$ is integrable on the interval $[a,b]$.
$endgroup$
Recall that a continuous function $f:[a,b] to mathbb{R}$ is integrable on the interval $[a,b]$.
answered Sep 11 '18 at 6:49
gimusigimusi
1
1
add a comment |
add a comment |
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5
$begingroup$
The function is continuous on a compact interval hence...
$endgroup$
– Did
Sep 11 '18 at 6:40
$begingroup$
Another reason: the given function is monotone and hence Riemann integrable on $[0,1]$. The answer by RRL can be used to show that any monotone function is Riemann integrable.
$endgroup$
– Paramanand Singh
Sep 11 '18 at 9:11