Definite Integration of [arccos(x)] , [.] = GIF [closed]












-1












$begingroup$


How to find integration of [ arccos(x) ] in its domain (-1,1)? Where [.] Is Floor function.
I tried using graph method(area) but end up getting some negative terms there.






integration definite-integrals







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closed as off-topic by RRL, Carl Schildkraut, Did, Cesareo, amWhy Dec 20 '18 at 13:01


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Carl Schildkraut, Did, Cesareo, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    In what domain?
    $endgroup$
    – caverac
    Dec 19 '18 at 17:12










  • $begingroup$
    Domain of arccos x. from -1 to 1
    $endgroup$
    – user627932
    Dec 19 '18 at 17:13










  • $begingroup$
    This is a step function, so you can just add up the rectangles.
    $endgroup$
    – eyeballfrog
    Dec 19 '18 at 17:19










  • $begingroup$
    Ans is Cos1 + Cos 2 + Cos 3 + 3. So its clear that this has to be solved using graph of arccos x.
    $endgroup$
    – user627932
    Dec 19 '18 at 17:19






  • 1




    $begingroup$
    @clathratus thanks for pointing it out. I have changed it.
    $endgroup$
    – user627932
    Dec 20 '18 at 18:15
















-1












$begingroup$


How to find integration of [ arccos(x) ] in its domain (-1,1)? Where [.] Is Floor function.
I tried using graph method(area) but end up getting some negative terms there.






integration definite-integrals







share|cite|improve this question











$endgroup$



closed as off-topic by RRL, Carl Schildkraut, Did, Cesareo, amWhy Dec 20 '18 at 13:01


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Carl Schildkraut, Did, Cesareo, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    In what domain?
    $endgroup$
    – caverac
    Dec 19 '18 at 17:12










  • $begingroup$
    Domain of arccos x. from -1 to 1
    $endgroup$
    – user627932
    Dec 19 '18 at 17:13










  • $begingroup$
    This is a step function, so you can just add up the rectangles.
    $endgroup$
    – eyeballfrog
    Dec 19 '18 at 17:19










  • $begingroup$
    Ans is Cos1 + Cos 2 + Cos 3 + 3. So its clear that this has to be solved using graph of arccos x.
    $endgroup$
    – user627932
    Dec 19 '18 at 17:19






  • 1




    $begingroup$
    @clathratus thanks for pointing it out. I have changed it.
    $endgroup$
    – user627932
    Dec 20 '18 at 18:15














-1












-1








-1





$begingroup$


How to find integration of [ arccos(x) ] in its domain (-1,1)? Where [.] Is Floor function.
I tried using graph method(area) but end up getting some negative terms there.






integration definite-integrals







share|cite|improve this question











$endgroup$




How to find integration of [ arccos(x) ] in its domain (-1,1)? Where [.] Is Floor function.
I tried using graph method(area) but end up getting some negative terms there.






integration definite-integrals




integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 18:14







user627932

















asked Dec 19 '18 at 17:09









user627932user627932

73




73




closed as off-topic by RRL, Carl Schildkraut, Did, Cesareo, amWhy Dec 20 '18 at 13:01


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Carl Schildkraut, Did, Cesareo, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by RRL, Carl Schildkraut, Did, Cesareo, amWhy Dec 20 '18 at 13:01


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Carl Schildkraut, Did, Cesareo, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    In what domain?
    $endgroup$
    – caverac
    Dec 19 '18 at 17:12










  • $begingroup$
    Domain of arccos x. from -1 to 1
    $endgroup$
    – user627932
    Dec 19 '18 at 17:13










  • $begingroup$
    This is a step function, so you can just add up the rectangles.
    $endgroup$
    – eyeballfrog
    Dec 19 '18 at 17:19










  • $begingroup$
    Ans is Cos1 + Cos 2 + Cos 3 + 3. So its clear that this has to be solved using graph of arccos x.
    $endgroup$
    – user627932
    Dec 19 '18 at 17:19






  • 1




    $begingroup$
    @clathratus thanks for pointing it out. I have changed it.
    $endgroup$
    – user627932
    Dec 20 '18 at 18:15


















  • $begingroup$
    In what domain?
    $endgroup$
    – caverac
    Dec 19 '18 at 17:12










  • $begingroup$
    Domain of arccos x. from -1 to 1
    $endgroup$
    – user627932
    Dec 19 '18 at 17:13










  • $begingroup$
    This is a step function, so you can just add up the rectangles.
    $endgroup$
    – eyeballfrog
    Dec 19 '18 at 17:19










  • $begingroup$
    Ans is Cos1 + Cos 2 + Cos 3 + 3. So its clear that this has to be solved using graph of arccos x.
    $endgroup$
    – user627932
    Dec 19 '18 at 17:19






  • 1




    $begingroup$
    @clathratus thanks for pointing it out. I have changed it.
    $endgroup$
    – user627932
    Dec 20 '18 at 18:15
















$begingroup$
In what domain?
$endgroup$
– caverac
Dec 19 '18 at 17:12




$begingroup$
In what domain?
$endgroup$
– caverac
Dec 19 '18 at 17:12












$begingroup$
Domain of arccos x. from -1 to 1
$endgroup$
– user627932
Dec 19 '18 at 17:13




$begingroup$
Domain of arccos x. from -1 to 1
$endgroup$
– user627932
Dec 19 '18 at 17:13












$begingroup$
This is a step function, so you can just add up the rectangles.
$endgroup$
– eyeballfrog
Dec 19 '18 at 17:19




$begingroup$
This is a step function, so you can just add up the rectangles.
$endgroup$
– eyeballfrog
Dec 19 '18 at 17:19












$begingroup$
Ans is Cos1 + Cos 2 + Cos 3 + 3. So its clear that this has to be solved using graph of arccos x.
$endgroup$
– user627932
Dec 19 '18 at 17:19




$begingroup$
Ans is Cos1 + Cos 2 + Cos 3 + 3. So its clear that this has to be solved using graph of arccos x.
$endgroup$
– user627932
Dec 19 '18 at 17:19




1




1




$begingroup$
@clathratus thanks for pointing it out. I have changed it.
$endgroup$
– user627932
Dec 20 '18 at 18:15




$begingroup$
@clathratus thanks for pointing it out. I have changed it.
$endgroup$
– user627932
Dec 20 '18 at 18:15










2 Answers
2






active

oldest

votes


















0












$begingroup$

$cos^{-1}x$ is a monotonically decreasing function, such that:





  • $3lecos^{-1}xlepi$ for $-1le xlecos3$


  • $2lecos^{-1}xle3$ for $cos3le xlecos2$


  • $1lecos^{-1}xle2$ for $cos2le xlecos1$


  • $0lecos^{-1}xle1$ for $cos1le xle 1$


$I=displaystyleint_{-1}^1lfloorcos^{-1}xrfloor dx\=displaystyleint_{-1}^{cos3}3dx+int_{cos3}^{cos2}2dx+int_{cos2}^{cos1}1dx+int_{cos1}^10dx\=cos1+cos2+cos3+3$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Maybe a small sketch will help you



    enter image description here



    It is just a series of steps, so the integral is simple. The trick is to find the transitions, for example, to find the left-most consider $3 = arccos(x)$, whose solution is $x = cos(3)$, so in the region $-1 leq x leq cos(3)$ the function is just three, and the area is



    $$
    int_{-1}^{cos(3)} lfloor arccos(x) rfloor {rm d}x = 3 [cos(3) + 1]
    $$



    You can continue with the other steps, the result is



    $$
    int_{-1}^{1} lfloor arccos(x) rfloor {rm d}x = cos 1 + cos 2 + cos 3 + 3
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks i just found what i was doing wrong
      $endgroup$
      – user627932
      Dec 19 '18 at 17:29


















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    $cos^{-1}x$ is a monotonically decreasing function, such that:





    • $3lecos^{-1}xlepi$ for $-1le xlecos3$


    • $2lecos^{-1}xle3$ for $cos3le xlecos2$


    • $1lecos^{-1}xle2$ for $cos2le xlecos1$


    • $0lecos^{-1}xle1$ for $cos1le xle 1$


    $I=displaystyleint_{-1}^1lfloorcos^{-1}xrfloor dx\=displaystyleint_{-1}^{cos3}3dx+int_{cos3}^{cos2}2dx+int_{cos2}^{cos1}1dx+int_{cos1}^10dx\=cos1+cos2+cos3+3$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $cos^{-1}x$ is a monotonically decreasing function, such that:





      • $3lecos^{-1}xlepi$ for $-1le xlecos3$


      • $2lecos^{-1}xle3$ for $cos3le xlecos2$


      • $1lecos^{-1}xle2$ for $cos2le xlecos1$


      • $0lecos^{-1}xle1$ for $cos1le xle 1$


      $I=displaystyleint_{-1}^1lfloorcos^{-1}xrfloor dx\=displaystyleint_{-1}^{cos3}3dx+int_{cos3}^{cos2}2dx+int_{cos2}^{cos1}1dx+int_{cos1}^10dx\=cos1+cos2+cos3+3$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $cos^{-1}x$ is a monotonically decreasing function, such that:





        • $3lecos^{-1}xlepi$ for $-1le xlecos3$


        • $2lecos^{-1}xle3$ for $cos3le xlecos2$


        • $1lecos^{-1}xle2$ for $cos2le xlecos1$


        • $0lecos^{-1}xle1$ for $cos1le xle 1$


        $I=displaystyleint_{-1}^1lfloorcos^{-1}xrfloor dx\=displaystyleint_{-1}^{cos3}3dx+int_{cos3}^{cos2}2dx+int_{cos2}^{cos1}1dx+int_{cos1}^10dx\=cos1+cos2+cos3+3$






        share|cite|improve this answer









        $endgroup$



        $cos^{-1}x$ is a monotonically decreasing function, such that:





        • $3lecos^{-1}xlepi$ for $-1le xlecos3$


        • $2lecos^{-1}xle3$ for $cos3le xlecos2$


        • $1lecos^{-1}xle2$ for $cos2le xlecos1$


        • $0lecos^{-1}xle1$ for $cos1le xle 1$


        $I=displaystyleint_{-1}^1lfloorcos^{-1}xrfloor dx\=displaystyleint_{-1}^{cos3}3dx+int_{cos3}^{cos2}2dx+int_{cos2}^{cos1}1dx+int_{cos1}^10dx\=cos1+cos2+cos3+3$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 17:25









        Shubham JohriShubham Johri

        5,192717




        5,192717























            0












            $begingroup$

            Maybe a small sketch will help you



            enter image description here



            It is just a series of steps, so the integral is simple. The trick is to find the transitions, for example, to find the left-most consider $3 = arccos(x)$, whose solution is $x = cos(3)$, so in the region $-1 leq x leq cos(3)$ the function is just three, and the area is



            $$
            int_{-1}^{cos(3)} lfloor arccos(x) rfloor {rm d}x = 3 [cos(3) + 1]
            $$



            You can continue with the other steps, the result is



            $$
            int_{-1}^{1} lfloor arccos(x) rfloor {rm d}x = cos 1 + cos 2 + cos 3 + 3
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks i just found what i was doing wrong
              $endgroup$
              – user627932
              Dec 19 '18 at 17:29
















            0












            $begingroup$

            Maybe a small sketch will help you



            enter image description here



            It is just a series of steps, so the integral is simple. The trick is to find the transitions, for example, to find the left-most consider $3 = arccos(x)$, whose solution is $x = cos(3)$, so in the region $-1 leq x leq cos(3)$ the function is just three, and the area is



            $$
            int_{-1}^{cos(3)} lfloor arccos(x) rfloor {rm d}x = 3 [cos(3) + 1]
            $$



            You can continue with the other steps, the result is



            $$
            int_{-1}^{1} lfloor arccos(x) rfloor {rm d}x = cos 1 + cos 2 + cos 3 + 3
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks i just found what i was doing wrong
              $endgroup$
              – user627932
              Dec 19 '18 at 17:29














            0












            0








            0





            $begingroup$

            Maybe a small sketch will help you



            enter image description here



            It is just a series of steps, so the integral is simple. The trick is to find the transitions, for example, to find the left-most consider $3 = arccos(x)$, whose solution is $x = cos(3)$, so in the region $-1 leq x leq cos(3)$ the function is just three, and the area is



            $$
            int_{-1}^{cos(3)} lfloor arccos(x) rfloor {rm d}x = 3 [cos(3) + 1]
            $$



            You can continue with the other steps, the result is



            $$
            int_{-1}^{1} lfloor arccos(x) rfloor {rm d}x = cos 1 + cos 2 + cos 3 + 3
            $$






            share|cite|improve this answer









            $endgroup$



            Maybe a small sketch will help you



            enter image description here



            It is just a series of steps, so the integral is simple. The trick is to find the transitions, for example, to find the left-most consider $3 = arccos(x)$, whose solution is $x = cos(3)$, so in the region $-1 leq x leq cos(3)$ the function is just three, and the area is



            $$
            int_{-1}^{cos(3)} lfloor arccos(x) rfloor {rm d}x = 3 [cos(3) + 1]
            $$



            You can continue with the other steps, the result is



            $$
            int_{-1}^{1} lfloor arccos(x) rfloor {rm d}x = cos 1 + cos 2 + cos 3 + 3
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 19 '18 at 17:27









            caveraccaverac

            14.6k31130




            14.6k31130












            • $begingroup$
              Thanks i just found what i was doing wrong
              $endgroup$
              – user627932
              Dec 19 '18 at 17:29


















            • $begingroup$
              Thanks i just found what i was doing wrong
              $endgroup$
              – user627932
              Dec 19 '18 at 17:29
















            $begingroup$
            Thanks i just found what i was doing wrong
            $endgroup$
            – user627932
            Dec 19 '18 at 17:29




            $begingroup$
            Thanks i just found what i was doing wrong
            $endgroup$
            – user627932
            Dec 19 '18 at 17:29



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