Integrating minimum of two periodic functions












0












$begingroup$


I am trying to see what is the simplest way to calculate the definite integral of a function of the form



$$f(x) = min(sin(x), sin(2x))$$



I can of course check when $sin(x)<sin(2x)$ and combine the definite integrals for each section. I can imagine how I would do this when $x$ is in units of $2pi$, as I can just multiply the result for the first period. But let's say $x=6.5pi$, would the correct approach be to have the simplified calculation for the first $6pi$ and then for the remaining section separately?






calculus







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$endgroup$












  • $begingroup$
    Visualisation: $f$ is periodic with period $2pi$.
    $endgroup$
    – TheSimpliFire
    Dec 27 '18 at 8:34
















0












$begingroup$


I am trying to see what is the simplest way to calculate the definite integral of a function of the form



$$f(x) = min(sin(x), sin(2x))$$



I can of course check when $sin(x)<sin(2x)$ and combine the definite integrals for each section. I can imagine how I would do this when $x$ is in units of $2pi$, as I can just multiply the result for the first period. But let's say $x=6.5pi$, would the correct approach be to have the simplified calculation for the first $6pi$ and then for the remaining section separately?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Visualisation: $f$ is periodic with period $2pi$.
    $endgroup$
    – TheSimpliFire
    Dec 27 '18 at 8:34














0












0








0


1



$begingroup$


I am trying to see what is the simplest way to calculate the definite integral of a function of the form



$$f(x) = min(sin(x), sin(2x))$$



I can of course check when $sin(x)<sin(2x)$ and combine the definite integrals for each section. I can imagine how I would do this when $x$ is in units of $2pi$, as I can just multiply the result for the first period. But let's say $x=6.5pi$, would the correct approach be to have the simplified calculation for the first $6pi$ and then for the remaining section separately?






calculus







share|cite|improve this question











$endgroup$




I am trying to see what is the simplest way to calculate the definite integral of a function of the form



$$f(x) = min(sin(x), sin(2x))$$



I can of course check when $sin(x)<sin(2x)$ and combine the definite integrals for each section. I can imagine how I would do this when $x$ is in units of $2pi$, as I can just multiply the result for the first period. But let's say $x=6.5pi$, would the correct approach be to have the simplified calculation for the first $6pi$ and then for the remaining section separately?






calculus




calculus






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 8:39









Robert Z

98.6k1068139




98.6k1068139










asked Dec 27 '18 at 8:28









CaharpukaCaharpuka

1012




1012












  • $begingroup$
    Visualisation: $f$ is periodic with period $2pi$.
    $endgroup$
    – TheSimpliFire
    Dec 27 '18 at 8:34


















  • $begingroup$
    Visualisation: $f$ is periodic with period $2pi$.
    $endgroup$
    – TheSimpliFire
    Dec 27 '18 at 8:34
















$begingroup$
Visualisation: $f$ is periodic with period $2pi$.
$endgroup$
– TheSimpliFire
Dec 27 '18 at 8:34




$begingroup$
Visualisation: $f$ is periodic with period $2pi$.
$endgroup$
– TheSimpliFire
Dec 27 '18 at 8:34










2 Answers
2






active

oldest

votes


















2












$begingroup$

Hint. Use the fact that
$$
begin{align}min(sin(x),sin(2x))&=frac{1}{2}left(sin(x)+sin(2x)-|sin(x)-sin(2x)|right).
end{align}$$
Hence if we integrate over a period,
$$begin{align}
int_0^{2pi}min(sin(x),sin(2x)),dx&=
-frac{1}{2}int_0^{2pi}|sin(x)-sin(2x)|,dx\
&=-int_0^{pi}|sin(x)-sin(2x)|,dx\
&=int_0^{pi/3}(sin(x)-sin(2x)),dx\
&quad-int_{pi/3}^pi(sin(x)-sin(2x)),dx
end{align}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
    $endgroup$
    – Caharpuka
    Dec 27 '18 at 8:41



















0












$begingroup$

$$f(x)=sin2x-sin x=2sindfrac x2cosdfrac{3x}2$$



As the $f(x+2pi)=f(x)$



we can safely focus on $0le xle2pi$ where $sindfrac x2not<0$



$f(x)$ will be $>0$



if $cosdfrac{3x}2>0$



$(i)0<dfrac{3x}2<dfracpi2$ or $(ii)dfrac{3pi}2<dfrac{3x}2<2pi$



$(i)0<x<dfracpi3$ or $(ii)pi<x<dfrac{4pi}3$



Consequently, $displaystyleint_0^{2pi}$min$(sin x,sin2x) dx$



$$=int_0^{pi/3}sin x dx+int_{pi/3}^pisin2x dx+int_pi^{4pi/3}sin x dx+int_{4pi/3}^{2pi }sin2x dx=?$$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    2












    $begingroup$

    Hint. Use the fact that
    $$
    begin{align}min(sin(x),sin(2x))&=frac{1}{2}left(sin(x)+sin(2x)-|sin(x)-sin(2x)|right).
    end{align}$$
    Hence if we integrate over a period,
    $$begin{align}
    int_0^{2pi}min(sin(x),sin(2x)),dx&=
    -frac{1}{2}int_0^{2pi}|sin(x)-sin(2x)|,dx\
    &=-int_0^{pi}|sin(x)-sin(2x)|,dx\
    &=int_0^{pi/3}(sin(x)-sin(2x)),dx\
    &quad-int_{pi/3}^pi(sin(x)-sin(2x)),dx
    end{align}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
      $endgroup$
      – Caharpuka
      Dec 27 '18 at 8:41
















    2












    $begingroup$

    Hint. Use the fact that
    $$
    begin{align}min(sin(x),sin(2x))&=frac{1}{2}left(sin(x)+sin(2x)-|sin(x)-sin(2x)|right).
    end{align}$$
    Hence if we integrate over a period,
    $$begin{align}
    int_0^{2pi}min(sin(x),sin(2x)),dx&=
    -frac{1}{2}int_0^{2pi}|sin(x)-sin(2x)|,dx\
    &=-int_0^{pi}|sin(x)-sin(2x)|,dx\
    &=int_0^{pi/3}(sin(x)-sin(2x)),dx\
    &quad-int_{pi/3}^pi(sin(x)-sin(2x)),dx
    end{align}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
      $endgroup$
      – Caharpuka
      Dec 27 '18 at 8:41














    2












    2








    2





    $begingroup$

    Hint. Use the fact that
    $$
    begin{align}min(sin(x),sin(2x))&=frac{1}{2}left(sin(x)+sin(2x)-|sin(x)-sin(2x)|right).
    end{align}$$
    Hence if we integrate over a period,
    $$begin{align}
    int_0^{2pi}min(sin(x),sin(2x)),dx&=
    -frac{1}{2}int_0^{2pi}|sin(x)-sin(2x)|,dx\
    &=-int_0^{pi}|sin(x)-sin(2x)|,dx\
    &=int_0^{pi/3}(sin(x)-sin(2x)),dx\
    &quad-int_{pi/3}^pi(sin(x)-sin(2x)),dx
    end{align}$$






    share|cite|improve this answer











    $endgroup$



    Hint. Use the fact that
    $$
    begin{align}min(sin(x),sin(2x))&=frac{1}{2}left(sin(x)+sin(2x)-|sin(x)-sin(2x)|right).
    end{align}$$
    Hence if we integrate over a period,
    $$begin{align}
    int_0^{2pi}min(sin(x),sin(2x)),dx&=
    -frac{1}{2}int_0^{2pi}|sin(x)-sin(2x)|,dx\
    &=-int_0^{pi}|sin(x)-sin(2x)|,dx\
    &=int_0^{pi/3}(sin(x)-sin(2x)),dx\
    &quad-int_{pi/3}^pi(sin(x)-sin(2x)),dx
    end{align}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 27 '18 at 8:50

























    answered Dec 27 '18 at 8:35









    Robert ZRobert Z

    98.6k1068139




    98.6k1068139












    • $begingroup$
      That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
      $endgroup$
      – Caharpuka
      Dec 27 '18 at 8:41


















    • $begingroup$
      That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
      $endgroup$
      – Caharpuka
      Dec 27 '18 at 8:41
















    $begingroup$
    That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
    $endgroup$
    – Caharpuka
    Dec 27 '18 at 8:41




    $begingroup$
    That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
    $endgroup$
    – Caharpuka
    Dec 27 '18 at 8:41











    0












    $begingroup$

    $$f(x)=sin2x-sin x=2sindfrac x2cosdfrac{3x}2$$



    As the $f(x+2pi)=f(x)$



    we can safely focus on $0le xle2pi$ where $sindfrac x2not<0$



    $f(x)$ will be $>0$



    if $cosdfrac{3x}2>0$



    $(i)0<dfrac{3x}2<dfracpi2$ or $(ii)dfrac{3pi}2<dfrac{3x}2<2pi$



    $(i)0<x<dfracpi3$ or $(ii)pi<x<dfrac{4pi}3$



    Consequently, $displaystyleint_0^{2pi}$min$(sin x,sin2x) dx$



    $$=int_0^{pi/3}sin x dx+int_{pi/3}^pisin2x dx+int_pi^{4pi/3}sin x dx+int_{4pi/3}^{2pi }sin2x dx=?$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $$f(x)=sin2x-sin x=2sindfrac x2cosdfrac{3x}2$$



      As the $f(x+2pi)=f(x)$



      we can safely focus on $0le xle2pi$ where $sindfrac x2not<0$



      $f(x)$ will be $>0$



      if $cosdfrac{3x}2>0$



      $(i)0<dfrac{3x}2<dfracpi2$ or $(ii)dfrac{3pi}2<dfrac{3x}2<2pi$



      $(i)0<x<dfracpi3$ or $(ii)pi<x<dfrac{4pi}3$



      Consequently, $displaystyleint_0^{2pi}$min$(sin x,sin2x) dx$



      $$=int_0^{pi/3}sin x dx+int_{pi/3}^pisin2x dx+int_pi^{4pi/3}sin x dx+int_{4pi/3}^{2pi }sin2x dx=?$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $$f(x)=sin2x-sin x=2sindfrac x2cosdfrac{3x}2$$



        As the $f(x+2pi)=f(x)$



        we can safely focus on $0le xle2pi$ where $sindfrac x2not<0$



        $f(x)$ will be $>0$



        if $cosdfrac{3x}2>0$



        $(i)0<dfrac{3x}2<dfracpi2$ or $(ii)dfrac{3pi}2<dfrac{3x}2<2pi$



        $(i)0<x<dfracpi3$ or $(ii)pi<x<dfrac{4pi}3$



        Consequently, $displaystyleint_0^{2pi}$min$(sin x,sin2x) dx$



        $$=int_0^{pi/3}sin x dx+int_{pi/3}^pisin2x dx+int_pi^{4pi/3}sin x dx+int_{4pi/3}^{2pi }sin2x dx=?$$






        share|cite|improve this answer









        $endgroup$



        $$f(x)=sin2x-sin x=2sindfrac x2cosdfrac{3x}2$$



        As the $f(x+2pi)=f(x)$



        we can safely focus on $0le xle2pi$ where $sindfrac x2not<0$



        $f(x)$ will be $>0$



        if $cosdfrac{3x}2>0$



        $(i)0<dfrac{3x}2<dfracpi2$ or $(ii)dfrac{3pi}2<dfrac{3x}2<2pi$



        $(i)0<x<dfracpi3$ or $(ii)pi<x<dfrac{4pi}3$



        Consequently, $displaystyleint_0^{2pi}$min$(sin x,sin2x) dx$



        $$=int_0^{pi/3}sin x dx+int_{pi/3}^pisin2x dx+int_pi^{4pi/3}sin x dx+int_{4pi/3}^{2pi }sin2x dx=?$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 12:08









        lab bhattacharjeelab bhattacharjee

        226k15157275




        226k15157275






























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