Is the given function is Riemann integrable or not reimann integrable ??

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












Is the given function is Riemann integrable or not Riemann integrable on the interval $[0,1] $?



$$ f (x)=begincases cos x quad textifquad x in [0,frac12] \ sin x quad text if quad x in (frac12,1]endcases $$



My attempts: I know that $f$ will continuous when $cos x =sin x$ by putting $x = pi/4$. I'm confused that here $x = pi/4 $ is not given. Now how can I conclude that it is Riemann integrable or not ?



Any Hints/solution will be appreciated.



Thank you.




real-analysis




share|cite|improve this question

















  • 2




    There's a jump discontinuity but otherwise the fact that the function is continuous at every other point should tell you it's integrable
    – Spencer
    Aug 2 at 17:25










  • The function is continuous almost everywhere (you have only a point of discontinuity) so the function is riemann integrable
    – Cuoredicervo
    Aug 2 at 17:28










  • @Cuoredicervo..how the Function is continious almost everywhere..can elaboarate more
    – stupid
    Aug 2 at 17:47















up vote
0
down vote

favorite












Is the given function is Riemann integrable or not Riemann integrable on the interval $[0,1] $?



$$ f (x)=begincases cos x quad textifquad x in [0,frac12] \ sin x quad text if quad x in (frac12,1]endcases $$



My attempts: I know that $f$ will continuous when $cos x =sin x$ by putting $x = pi/4$. I'm confused that here $x = pi/4 $ is not given. Now how can I conclude that it is Riemann integrable or not ?



Any Hints/solution will be appreciated.



Thank you.




real-analysis




share|cite|improve this question

















  • 2




    There's a jump discontinuity but otherwise the fact that the function is continuous at every other point should tell you it's integrable
    – Spencer
    Aug 2 at 17:25










  • The function is continuous almost everywhere (you have only a point of discontinuity) so the function is riemann integrable
    – Cuoredicervo
    Aug 2 at 17:28










  • @Cuoredicervo..how the Function is continious almost everywhere..can elaboarate more
    – stupid
    Aug 2 at 17:47













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Is the given function is Riemann integrable or not Riemann integrable on the interval $[0,1] $?



$$ f (x)=begincases cos x quad textifquad x in [0,frac12] \ sin x quad text if quad x in (frac12,1]endcases $$



My attempts: I know that $f$ will continuous when $cos x =sin x$ by putting $x = pi/4$. I'm confused that here $x = pi/4 $ is not given. Now how can I conclude that it is Riemann integrable or not ?



Any Hints/solution will be appreciated.



Thank you.




real-analysis




share|cite|improve this question













Is the given function is Riemann integrable or not Riemann integrable on the interval $[0,1] $?



$$ f (x)=begincases cos x quad textifquad x in [0,frac12] \ sin x quad text if quad x in (frac12,1]endcases $$



My attempts: I know that $f$ will continuous when $cos x =sin x$ by putting $x = pi/4$. I'm confused that here $x = pi/4 $ is not given. Now how can I conclude that it is Riemann integrable or not ?



Any Hints/solution will be appreciated.



Thank you.





real-analysis





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 17:22
























asked Aug 2 at 17:10









stupid

50018




50018







  • 2




    There's a jump discontinuity but otherwise the fact that the function is continuous at every other point should tell you it's integrable
    – Spencer
    Aug 2 at 17:25










  • The function is continuous almost everywhere (you have only a point of discontinuity) so the function is riemann integrable
    – Cuoredicervo
    Aug 2 at 17:28










  • @Cuoredicervo..how the Function is continious almost everywhere..can elaboarate more
    – stupid
    Aug 2 at 17:47













  • 2




    There's a jump discontinuity but otherwise the fact that the function is continuous at every other point should tell you it's integrable
    – Spencer
    Aug 2 at 17:25










  • The function is continuous almost everywhere (you have only a point of discontinuity) so the function is riemann integrable
    – Cuoredicervo
    Aug 2 at 17:28










  • @Cuoredicervo..how the Function is continious almost everywhere..can elaboarate more
    – stupid
    Aug 2 at 17:47








2




2




There's a jump discontinuity but otherwise the fact that the function is continuous at every other point should tell you it's integrable
– Spencer
Aug 2 at 17:25




There's a jump discontinuity but otherwise the fact that the function is continuous at every other point should tell you it's integrable
– Spencer
Aug 2 at 17:25












The function is continuous almost everywhere (you have only a point of discontinuity) so the function is riemann integrable
– Cuoredicervo
Aug 2 at 17:28




The function is continuous almost everywhere (you have only a point of discontinuity) so the function is riemann integrable
– Cuoredicervo
Aug 2 at 17:28












@Cuoredicervo..how the Function is continious almost everywhere..can elaboarate more
– stupid
Aug 2 at 17:47





@Cuoredicervo..how the Function is continious almost everywhere..can elaboarate more
– stupid
Aug 2 at 17:47











2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










The measure theory approach here is a bit of overkill I think. Let $epsilon>0.$ Let $n>2.$ Choose partitions $P_n$ of $[0,1/2-1/n]$ and $Q_n$ of $[1/2+1/n,1]$ such that



$$U(P_n,f)-L(P_n,f)<epsilon,,,U(Q_n,f)-L(Q_n,f)<epsilon.$$



Then $R_n=P_ncup Q_n$ is a partition of $[0,1],$ and



$$tag 1 U(R_n,f)-L(R_n,f) < epsilon + (C_n-c_n)frac2n + epsilon.$$



Here $C_n,c_n$ are the $sup, inf$ of $f$ over $[1/2-1/n,1/2+1/n].$ The Riemann integrability of $f$ on $[0,1]$ follows.






share|cite|improve this answer






























    up vote
    1
    down vote













    I assume that you've learned about Darboux theory. Then $f$ is integrable on $[a,b]$ iff there exists a partition of $[a,b]$ whose sum of oscillation $sum_1^n omega_j varDelta x_j < varepsilon$. Then for the $f$ in the post, the integrability of $f$ on $[0, 1/2]$ and $[1/2, 1]$ implies the existence of partitions $mathcal P_1, mathcal P_2$ of each interval where the sum $sum_mathcal P_k omega_j varDelta x_j < varepsilon/2 [k = 1,2]$. Then the partition of $[0,1]$ formed by union of the partitions above, whose sum $sum omega_j varDelta x_j = sum_mathcal P_1 + sum_mathcal P_2 < varepsilon$.



    The comments has referred to following facts:



    Definition. $f$ is continuous almost everywhere on the interval $[a, b]$ if for each $varepsilon >0$ the discontinuity of $f$ could be covered by an at most countable collection of open intervals $(I_j)_1^infty$ with total length less than $varepsilon$.



    Theorem. $fboxLebesgue criterion$ $f$ is Riemann integrable on $[a,b]$ iff $f$ is bounded and $f$ is almost everywhere continuous on $[a,b ]$ [Proof omitted].



    Now $f$ only has one discontinuity, which could be covered by one open interval with length $varepsilon $ as small as possible. Hence the integrability.






    share|cite|improve this answer























      Your Answer




      StackExchange.ifUsing("editor", function ()
      return StackExchange.using("mathjaxEditing", function ()
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      );
      );
      , "mathjax-editing");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "69"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      convertImagesToLinks: true,
      noModals: false,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );








       

      draft saved


      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870292%2fis-the-given-function-is-riemann-integrable-or-not-reimann-integrable%23new-answer', 'question_page');

      );

      Post as a guest

















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      The measure theory approach here is a bit of overkill I think. Let $epsilon>0.$ Let $n>2.$ Choose partitions $P_n$ of $[0,1/2-1/n]$ and $Q_n$ of $[1/2+1/n,1]$ such that



      $$U(P_n,f)-L(P_n,f)<epsilon,,,U(Q_n,f)-L(Q_n,f)<epsilon.$$



      Then $R_n=P_ncup Q_n$ is a partition of $[0,1],$ and



      $$tag 1 U(R_n,f)-L(R_n,f) < epsilon + (C_n-c_n)frac2n + epsilon.$$



      Here $C_n,c_n$ are the $sup, inf$ of $f$ over $[1/2-1/n,1/2+1/n].$ The Riemann integrability of $f$ on $[0,1]$ follows.






      share|cite|improve this answer



























        up vote
        1
        down vote



        accepted










        The measure theory approach here is a bit of overkill I think. Let $epsilon>0.$ Let $n>2.$ Choose partitions $P_n$ of $[0,1/2-1/n]$ and $Q_n$ of $[1/2+1/n,1]$ such that



        $$U(P_n,f)-L(P_n,f)<epsilon,,,U(Q_n,f)-L(Q_n,f)<epsilon.$$



        Then $R_n=P_ncup Q_n$ is a partition of $[0,1],$ and



        $$tag 1 U(R_n,f)-L(R_n,f) < epsilon + (C_n-c_n)frac2n + epsilon.$$



        Here $C_n,c_n$ are the $sup, inf$ of $f$ over $[1/2-1/n,1/2+1/n].$ The Riemann integrability of $f$ on $[0,1]$ follows.






        share|cite|improve this answer

























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The measure theory approach here is a bit of overkill I think. Let $epsilon>0.$ Let $n>2.$ Choose partitions $P_n$ of $[0,1/2-1/n]$ and $Q_n$ of $[1/2+1/n,1]$ such that



          $$U(P_n,f)-L(P_n,f)<epsilon,,,U(Q_n,f)-L(Q_n,f)<epsilon.$$



          Then $R_n=P_ncup Q_n$ is a partition of $[0,1],$ and



          $$tag 1 U(R_n,f)-L(R_n,f) < epsilon + (C_n-c_n)frac2n + epsilon.$$



          Here $C_n,c_n$ are the $sup, inf$ of $f$ over $[1/2-1/n,1/2+1/n].$ The Riemann integrability of $f$ on $[0,1]$ follows.






          share|cite|improve this answer















          The measure theory approach here is a bit of overkill I think. Let $epsilon>0.$ Let $n>2.$ Choose partitions $P_n$ of $[0,1/2-1/n]$ and $Q_n$ of $[1/2+1/n,1]$ such that



          $$U(P_n,f)-L(P_n,f)<epsilon,,,U(Q_n,f)-L(Q_n,f)<epsilon.$$



          Then $R_n=P_ncup Q_n$ is a partition of $[0,1],$ and



          $$tag 1 U(R_n,f)-L(R_n,f) < epsilon + (C_n-c_n)frac2n + epsilon.$$



          Here $C_n,c_n$ are the $sup, inf$ of $f$ over $[1/2-1/n,1/2+1/n].$ The Riemann integrability of $f$ on $[0,1]$ follows.







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 3 at 16:59


























          answered Aug 3 at 16:40









          zhw.

          65.1k42769




          65.1k42769




















              up vote
              1
              down vote













              I assume that you've learned about Darboux theory. Then $f$ is integrable on $[a,b]$ iff there exists a partition of $[a,b]$ whose sum of oscillation $sum_1^n omega_j varDelta x_j < varepsilon$. Then for the $f$ in the post, the integrability of $f$ on $[0, 1/2]$ and $[1/2, 1]$ implies the existence of partitions $mathcal P_1, mathcal P_2$ of each interval where the sum $sum_mathcal P_k omega_j varDelta x_j < varepsilon/2 [k = 1,2]$. Then the partition of $[0,1]$ formed by union of the partitions above, whose sum $sum omega_j varDelta x_j = sum_mathcal P_1 + sum_mathcal P_2 < varepsilon$.



              The comments has referred to following facts:



              Definition. $f$ is continuous almost everywhere on the interval $[a, b]$ if for each $varepsilon >0$ the discontinuity of $f$ could be covered by an at most countable collection of open intervals $(I_j)_1^infty$ with total length less than $varepsilon$.



              Theorem. $fboxLebesgue criterion$ $f$ is Riemann integrable on $[a,b]$ iff $f$ is bounded and $f$ is almost everywhere continuous on $[a,b ]$ [Proof omitted].



              Now $f$ only has one discontinuity, which could be covered by one open interval with length $varepsilon $ as small as possible. Hence the integrability.






              share|cite|improve this answer



























                up vote
                1
                down vote













                I assume that you've learned about Darboux theory. Then $f$ is integrable on $[a,b]$ iff there exists a partition of $[a,b]$ whose sum of oscillation $sum_1^n omega_j varDelta x_j < varepsilon$. Then for the $f$ in the post, the integrability of $f$ on $[0, 1/2]$ and $[1/2, 1]$ implies the existence of partitions $mathcal P_1, mathcal P_2$ of each interval where the sum $sum_mathcal P_k omega_j varDelta x_j < varepsilon/2 [k = 1,2]$. Then the partition of $[0,1]$ formed by union of the partitions above, whose sum $sum omega_j varDelta x_j = sum_mathcal P_1 + sum_mathcal P_2 < varepsilon$.



                The comments has referred to following facts:



                Definition. $f$ is continuous almost everywhere on the interval $[a, b]$ if for each $varepsilon >0$ the discontinuity of $f$ could be covered by an at most countable collection of open intervals $(I_j)_1^infty$ with total length less than $varepsilon$.



                Theorem. $fboxLebesgue criterion$ $f$ is Riemann integrable on $[a,b]$ iff $f$ is bounded and $f$ is almost everywhere continuous on $[a,b ]$ [Proof omitted].



                Now $f$ only has one discontinuity, which could be covered by one open interval with length $varepsilon $ as small as possible. Hence the integrability.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  I assume that you've learned about Darboux theory. Then $f$ is integrable on $[a,b]$ iff there exists a partition of $[a,b]$ whose sum of oscillation $sum_1^n omega_j varDelta x_j < varepsilon$. Then for the $f$ in the post, the integrability of $f$ on $[0, 1/2]$ and $[1/2, 1]$ implies the existence of partitions $mathcal P_1, mathcal P_2$ of each interval where the sum $sum_mathcal P_k omega_j varDelta x_j < varepsilon/2 [k = 1,2]$. Then the partition of $[0,1]$ formed by union of the partitions above, whose sum $sum omega_j varDelta x_j = sum_mathcal P_1 + sum_mathcal P_2 < varepsilon$.



                  The comments has referred to following facts:



                  Definition. $f$ is continuous almost everywhere on the interval $[a, b]$ if for each $varepsilon >0$ the discontinuity of $f$ could be covered by an at most countable collection of open intervals $(I_j)_1^infty$ with total length less than $varepsilon$.



                  Theorem. $fboxLebesgue criterion$ $f$ is Riemann integrable on $[a,b]$ iff $f$ is bounded and $f$ is almost everywhere continuous on $[a,b ]$ [Proof omitted].



                  Now $f$ only has one discontinuity, which could be covered by one open interval with length $varepsilon $ as small as possible. Hence the integrability.






                  share|cite|improve this answer















                  I assume that you've learned about Darboux theory. Then $f$ is integrable on $[a,b]$ iff there exists a partition of $[a,b]$ whose sum of oscillation $sum_1^n omega_j varDelta x_j < varepsilon$. Then for the $f$ in the post, the integrability of $f$ on $[0, 1/2]$ and $[1/2, 1]$ implies the existence of partitions $mathcal P_1, mathcal P_2$ of each interval where the sum $sum_mathcal P_k omega_j varDelta x_j < varepsilon/2 [k = 1,2]$. Then the partition of $[0,1]$ formed by union of the partitions above, whose sum $sum omega_j varDelta x_j = sum_mathcal P_1 + sum_mathcal P_2 < varepsilon$.



                  The comments has referred to following facts:



                  Definition. $f$ is continuous almost everywhere on the interval $[a, b]$ if for each $varepsilon >0$ the discontinuity of $f$ could be covered by an at most countable collection of open intervals $(I_j)_1^infty$ with total length less than $varepsilon$.



                  Theorem. $fboxLebesgue criterion$ $f$ is Riemann integrable on $[a,b]$ iff $f$ is bounded and $f$ is almost everywhere continuous on $[a,b ]$ [Proof omitted].



                  Now $f$ only has one discontinuity, which could be covered by one open interval with length $varepsilon $ as small as possible. Hence the integrability.







                  share|cite|improve this answer















                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 3 at 2:19


























                  answered Aug 3 at 2:03









                  xbh

                  1,0006




                  1,0006






















                       

                      draft saved


                      draft discarded


























                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870292%2fis-the-given-function-is-riemann-integrable-or-not-reimann-integrable%23new-answer', 'question_page');

                      );

                      Post as a guest
































































                      Comments

                      Post a Comment

                      Popular posts from this blog

                      What is the equation of a 3D cone with generalised tilt?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
                      Read more

                      Color the edges and diagonals of a regular polygon

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
                      Read more

                      Relationship between determinant of matrix and determinant of adjoint?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
                      Read more