Prove that $f(x)=sqrt{x}$ is Riemann integrable over $[0,1]$












1












$begingroup$


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




    $begingroup$
    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


















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















1












1








1


3



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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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
















  • 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












2 Answers
2






active

oldest

votes


















7












$begingroup$

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 $epsilon$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















1












$begingroup$

Recall that a continuous function $f:[a,b] to mathbb{R}$ is integrable on the interval $[a,b]$.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    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 $epsilon$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
















    7












    $begingroup$

    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 $epsilon$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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














    7












    7








    7





    $begingroup$

    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 $epsilon$.






    share|cite|improve this answer











    $endgroup$



    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 $epsilon$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    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


















    • $begingroup$
      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











    1












    $begingroup$

    Recall that a continuous function $f:[a,b] to mathbb{R}$ is integrable on the interval $[a,b]$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Recall that a continuous function $f:[a,b] to mathbb{R}$ is integrable on the interval $[a,b]$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Recall that a continuous function $f:[a,b] to mathbb{R}$ is integrable on the interval $[a,b]$.






        share|cite|improve this answer









        $endgroup$



        Recall that a continuous function $f:[a,b] to mathbb{R}$ is integrable on the interval $[a,b]$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 11 '18 at 6:49









        gimusigimusi

        1




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