Split up a double integral












2












$begingroup$


Is it true that



$int_a^b int_c^d f(x)g(y)dydx = int_a^b f(x)dx cdot int_c^dg(y)dy $



My intuition says it is true, but I also have the feeling that I am missing something, but I cannot prove it. In my application $a,b,c,d$ are length variables. $f(x)$ and $g(y)$ are sinusoidal functions.



I would like to prove the general case, can somebody help me out?






integration multivariable-calculus







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$endgroup$












  • $begingroup$
    both functions are independent so it can be split up like this
    $endgroup$
    – Henry Lee
    Jan 9 at 11:34










  • $begingroup$
    in the same way $left(int_a^bf(x)dxright)^2=int_a^bint_a^bf(x)f(y)dxdy$
    $endgroup$
    – Henry Lee
    Jan 9 at 11:35
















2












$begingroup$


Is it true that



$int_a^b int_c^d f(x)g(y)dydx = int_a^b f(x)dx cdot int_c^dg(y)dy $



My intuition says it is true, but I also have the feeling that I am missing something, but I cannot prove it. In my application $a,b,c,d$ are length variables. $f(x)$ and $g(y)$ are sinusoidal functions.



I would like to prove the general case, can somebody help me out?






integration multivariable-calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    both functions are independent so it can be split up like this
    $endgroup$
    – Henry Lee
    Jan 9 at 11:34










  • $begingroup$
    in the same way $left(int_a^bf(x)dxright)^2=int_a^bint_a^bf(x)f(y)dxdy$
    $endgroup$
    – Henry Lee
    Jan 9 at 11:35














2












2








2





$begingroup$


Is it true that



$int_a^b int_c^d f(x)g(y)dydx = int_a^b f(x)dx cdot int_c^dg(y)dy $



My intuition says it is true, but I also have the feeling that I am missing something, but I cannot prove it. In my application $a,b,c,d$ are length variables. $f(x)$ and $g(y)$ are sinusoidal functions.



I would like to prove the general case, can somebody help me out?






integration multivariable-calculus







share|cite|improve this question









$endgroup$




Is it true that



$int_a^b int_c^d f(x)g(y)dydx = int_a^b f(x)dx cdot int_c^dg(y)dy $



My intuition says it is true, but I also have the feeling that I am missing something, but I cannot prove it. In my application $a,b,c,d$ are length variables. $f(x)$ and $g(y)$ are sinusoidal functions.



I would like to prove the general case, can somebody help me out?






integration multivariable-calculus




integration multivariable-calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 11:26









seaverseaver

132




132












  • $begingroup$
    both functions are independent so it can be split up like this
    $endgroup$
    – Henry Lee
    Jan 9 at 11:34










  • $begingroup$
    in the same way $left(int_a^bf(x)dxright)^2=int_a^bint_a^bf(x)f(y)dxdy$
    $endgroup$
    – Henry Lee
    Jan 9 at 11:35


















  • $begingroup$
    both functions are independent so it can be split up like this
    $endgroup$
    – Henry Lee
    Jan 9 at 11:34










  • $begingroup$
    in the same way $left(int_a^bf(x)dxright)^2=int_a^bint_a^bf(x)f(y)dxdy$
    $endgroup$
    – Henry Lee
    Jan 9 at 11:35
















$begingroup$
both functions are independent so it can be split up like this
$endgroup$
– Henry Lee
Jan 9 at 11:34




$begingroup$
both functions are independent so it can be split up like this
$endgroup$
– Henry Lee
Jan 9 at 11:34












$begingroup$
in the same way $left(int_a^bf(x)dxright)^2=int_a^bint_a^bf(x)f(y)dxdy$
$endgroup$
– Henry Lee
Jan 9 at 11:35




$begingroup$
in the same way $left(int_a^bf(x)dxright)^2=int_a^bint_a^bf(x)f(y)dxdy$
$endgroup$
– Henry Lee
Jan 9 at 11:35










3 Answers
3






active

oldest

votes


















1












$begingroup$

Note that for any integrable $f colon defR{mathbb R}R to R$ and any $zeta in R$ we have
$$ int_R zeta f(x), dx = zeta int_R f(x) , dx $$
Now note that with $zeta := int_R f(y), dy$ which depends on $f$, but since $f$ is a constant this gives
$$ int_R left(int_R f(y), dyright), f(x), dx = int_R f(y), dy cdot int_R f(x), dx $$
Now, for every fixed $x in R$, just apply the above result, now for $zeta = f(x)$ (not dependent on $y$):
$$ left(int_R f(y), dyright), f(x) = int_R f(x)f(y), dy $$
Thus
$$ int_R int_R f(y)f(x), dy, dx = int_R f(y), dy cdot int_R f(x), dx $$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you very much for the quick response, this (and the other answers) helped me out!
    $endgroup$
    – seaver
    Jan 9 at 11:48



















1












$begingroup$

Here is a try:
$$I=int_a^bint_c^df(x)g(y)dxdy$$
where $int f(x)dx=F(x)$ and $int g(y)dy=G(y)$
we can start by saying:
$$I=int_a^bint_c^df(x)g(y)dxdy=int_c^dleft[F(x)right]_a^b g(y)dy=left[F(x)right]_a^b left[G(x)right]_c^d$$
as $left[F(x)right]_a^b$ is a constant. They are separable






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The integral is linear so for any constant $K$ you have:
    $$int_a^b K , f(x),mbox{d}x=Kint_a^b f(x),mbox{d}x$$



    and with respect to $y$, $f(x)=K_1$ is a constant and likewise the expression $int_c^d g(y),mbox{d}y = K_2$ is a constant:
    $$int_a^b int_c^d overbrace{f(x)}^{K_1} g(y),mbox{d}y,mbox{d}x =int_a^b left( overbrace{f(x)}^{K_1}underbrace{int_c^d g(y),mbox{d}y}_{K_2} right),mbox{d}x=underbrace{int_c^d g(y),mbox{d}y}_{K_2}int_a^b f(x),mbox{d}x$$






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Note that for any integrable $f colon defR{mathbb R}R to R$ and any $zeta in R$ we have
      $$ int_R zeta f(x), dx = zeta int_R f(x) , dx $$
      Now note that with $zeta := int_R f(y), dy$ which depends on $f$, but since $f$ is a constant this gives
      $$ int_R left(int_R f(y), dyright), f(x), dx = int_R f(y), dy cdot int_R f(x), dx $$
      Now, for every fixed $x in R$, just apply the above result, now for $zeta = f(x)$ (not dependent on $y$):
      $$ left(int_R f(y), dyright), f(x) = int_R f(x)f(y), dy $$
      Thus
      $$ int_R int_R f(y)f(x), dy, dx = int_R f(y), dy cdot int_R f(x), dx $$






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Thank you very much for the quick response, this (and the other answers) helped me out!
        $endgroup$
        – seaver
        Jan 9 at 11:48
















      1












      $begingroup$

      Note that for any integrable $f colon defR{mathbb R}R to R$ and any $zeta in R$ we have
      $$ int_R zeta f(x), dx = zeta int_R f(x) , dx $$
      Now note that with $zeta := int_R f(y), dy$ which depends on $f$, but since $f$ is a constant this gives
      $$ int_R left(int_R f(y), dyright), f(x), dx = int_R f(y), dy cdot int_R f(x), dx $$
      Now, for every fixed $x in R$, just apply the above result, now for $zeta = f(x)$ (not dependent on $y$):
      $$ left(int_R f(y), dyright), f(x) = int_R f(x)f(y), dy $$
      Thus
      $$ int_R int_R f(y)f(x), dy, dx = int_R f(y), dy cdot int_R f(x), dx $$






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Thank you very much for the quick response, this (and the other answers) helped me out!
        $endgroup$
        – seaver
        Jan 9 at 11:48














      1












      1








      1





      $begingroup$

      Note that for any integrable $f colon defR{mathbb R}R to R$ and any $zeta in R$ we have
      $$ int_R zeta f(x), dx = zeta int_R f(x) , dx $$
      Now note that with $zeta := int_R f(y), dy$ which depends on $f$, but since $f$ is a constant this gives
      $$ int_R left(int_R f(y), dyright), f(x), dx = int_R f(y), dy cdot int_R f(x), dx $$
      Now, for every fixed $x in R$, just apply the above result, now for $zeta = f(x)$ (not dependent on $y$):
      $$ left(int_R f(y), dyright), f(x) = int_R f(x)f(y), dy $$
      Thus
      $$ int_R int_R f(y)f(x), dy, dx = int_R f(y), dy cdot int_R f(x), dx $$






      share|cite|improve this answer









      $endgroup$



      Note that for any integrable $f colon defR{mathbb R}R to R$ and any $zeta in R$ we have
      $$ int_R zeta f(x), dx = zeta int_R f(x) , dx $$
      Now note that with $zeta := int_R f(y), dy$ which depends on $f$, but since $f$ is a constant this gives
      $$ int_R left(int_R f(y), dyright), f(x), dx = int_R f(y), dy cdot int_R f(x), dx $$
      Now, for every fixed $x in R$, just apply the above result, now for $zeta = f(x)$ (not dependent on $y$):
      $$ left(int_R f(y), dyright), f(x) = int_R f(x)f(y), dy $$
      Thus
      $$ int_R int_R f(y)f(x), dy, dx = int_R f(y), dy cdot int_R f(x), dx $$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 9 at 11:43









      KevinKevin

      5,746823




      5,746823








      • 1




        $begingroup$
        Thank you very much for the quick response, this (and the other answers) helped me out!
        $endgroup$
        – seaver
        Jan 9 at 11:48














      • 1




        $begingroup$
        Thank you very much for the quick response, this (and the other answers) helped me out!
        $endgroup$
        – seaver
        Jan 9 at 11:48








      1




      1




      $begingroup$
      Thank you very much for the quick response, this (and the other answers) helped me out!
      $endgroup$
      – seaver
      Jan 9 at 11:48




      $begingroup$
      Thank you very much for the quick response, this (and the other answers) helped me out!
      $endgroup$
      – seaver
      Jan 9 at 11:48











      1












      $begingroup$

      Here is a try:
      $$I=int_a^bint_c^df(x)g(y)dxdy$$
      where $int f(x)dx=F(x)$ and $int g(y)dy=G(y)$
      we can start by saying:
      $$I=int_a^bint_c^df(x)g(y)dxdy=int_c^dleft[F(x)right]_a^b g(y)dy=left[F(x)right]_a^b left[G(x)right]_c^d$$
      as $left[F(x)right]_a^b$ is a constant. They are separable






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Here is a try:
        $$I=int_a^bint_c^df(x)g(y)dxdy$$
        where $int f(x)dx=F(x)$ and $int g(y)dy=G(y)$
        we can start by saying:
        $$I=int_a^bint_c^df(x)g(y)dxdy=int_c^dleft[F(x)right]_a^b g(y)dy=left[F(x)right]_a^b left[G(x)right]_c^d$$
        as $left[F(x)right]_a^b$ is a constant. They are separable






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Here is a try:
          $$I=int_a^bint_c^df(x)g(y)dxdy$$
          where $int f(x)dx=F(x)$ and $int g(y)dy=G(y)$
          we can start by saying:
          $$I=int_a^bint_c^df(x)g(y)dxdy=int_c^dleft[F(x)right]_a^b g(y)dy=left[F(x)right]_a^b left[G(x)right]_c^d$$
          as $left[F(x)right]_a^b$ is a constant. They are separable






          share|cite|improve this answer









          $endgroup$



          Here is a try:
          $$I=int_a^bint_c^df(x)g(y)dxdy$$
          where $int f(x)dx=F(x)$ and $int g(y)dy=G(y)$
          we can start by saying:
          $$I=int_a^bint_c^df(x)g(y)dxdy=int_c^dleft[F(x)right]_a^b g(y)dy=left[F(x)right]_a^b left[G(x)right]_c^d$$
          as $left[F(x)right]_a^b$ is a constant. They are separable







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 11:39









          Henry LeeHenry Lee

          2,158319




          2,158319























              1












              $begingroup$

              The integral is linear so for any constant $K$ you have:
              $$int_a^b K , f(x),mbox{d}x=Kint_a^b f(x),mbox{d}x$$



              and with respect to $y$, $f(x)=K_1$ is a constant and likewise the expression $int_c^d g(y),mbox{d}y = K_2$ is a constant:
              $$int_a^b int_c^d overbrace{f(x)}^{K_1} g(y),mbox{d}y,mbox{d}x =int_a^b left( overbrace{f(x)}^{K_1}underbrace{int_c^d g(y),mbox{d}y}_{K_2} right),mbox{d}x=underbrace{int_c^d g(y),mbox{d}y}_{K_2}int_a^b f(x),mbox{d}x$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                The integral is linear so for any constant $K$ you have:
                $$int_a^b K , f(x),mbox{d}x=Kint_a^b f(x),mbox{d}x$$



                and with respect to $y$, $f(x)=K_1$ is a constant and likewise the expression $int_c^d g(y),mbox{d}y = K_2$ is a constant:
                $$int_a^b int_c^d overbrace{f(x)}^{K_1} g(y),mbox{d}y,mbox{d}x =int_a^b left( overbrace{f(x)}^{K_1}underbrace{int_c^d g(y),mbox{d}y}_{K_2} right),mbox{d}x=underbrace{int_c^d g(y),mbox{d}y}_{K_2}int_a^b f(x),mbox{d}x$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The integral is linear so for any constant $K$ you have:
                  $$int_a^b K , f(x),mbox{d}x=Kint_a^b f(x),mbox{d}x$$



                  and with respect to $y$, $f(x)=K_1$ is a constant and likewise the expression $int_c^d g(y),mbox{d}y = K_2$ is a constant:
                  $$int_a^b int_c^d overbrace{f(x)}^{K_1} g(y),mbox{d}y,mbox{d}x =int_a^b left( overbrace{f(x)}^{K_1}underbrace{int_c^d g(y),mbox{d}y}_{K_2} right),mbox{d}x=underbrace{int_c^d g(y),mbox{d}y}_{K_2}int_a^b f(x),mbox{d}x$$






                  share|cite|improve this answer











                  $endgroup$



                  The integral is linear so for any constant $K$ you have:
                  $$int_a^b K , f(x),mbox{d}x=Kint_a^b f(x),mbox{d}x$$



                  and with respect to $y$, $f(x)=K_1$ is a constant and likewise the expression $int_c^d g(y),mbox{d}y = K_2$ is a constant:
                  $$int_a^b int_c^d overbrace{f(x)}^{K_1} g(y),mbox{d}y,mbox{d}x =int_a^b left( overbrace{f(x)}^{K_1}underbrace{int_c^d g(y),mbox{d}y}_{K_2} right),mbox{d}x=underbrace{int_c^d g(y),mbox{d}y}_{K_2}int_a^b f(x),mbox{d}x$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 9 at 11:44

























                  answered Jan 9 at 11:38









                  StackTDStackTD

                  24.3k2254




                  24.3k2254






























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