Integration of some root functions.












0












$begingroup$


while trying to help the daughter of my cousin to understand Integrals, I see that I forget about some of the integral calculations. Can someone help me with that one, please?



$$int_0^1 Bigl(sqrt[3]{(1-x^7)}-sqrt[7]{(1-x^3)} Bigr) dx$$



(this is not a homework, I'm trying to understand and explain to her how things work.)



I have a clue like..



$$
sqrt[3]{(1-x^7)} = y
$$
$$
(1-x^7) = y^3
$$
$$
(1-y^3) = x^7
$$
$$
x = sqrt[7]{(1-y^3)}
$$



but that is all...






definite-integrals







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    while trying to help the daughter of my cousin to understand Integrals, I see that I forget about some of the integral calculations. Can someone help me with that one, please?



    $$int_0^1 Bigl(sqrt[3]{(1-x^7)}-sqrt[7]{(1-x^3)} Bigr) dx$$



    (this is not a homework, I'm trying to understand and explain to her how things work.)



    I have a clue like..



    $$
    sqrt[3]{(1-x^7)} = y
    $$
    $$
    (1-x^7) = y^3
    $$
    $$
    (1-y^3) = x^7
    $$
    $$
    x = sqrt[7]{(1-y^3)}
    $$



    but that is all...






    definite-integrals







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      while trying to help the daughter of my cousin to understand Integrals, I see that I forget about some of the integral calculations. Can someone help me with that one, please?



      $$int_0^1 Bigl(sqrt[3]{(1-x^7)}-sqrt[7]{(1-x^3)} Bigr) dx$$



      (this is not a homework, I'm trying to understand and explain to her how things work.)



      I have a clue like..



      $$
      sqrt[3]{(1-x^7)} = y
      $$
      $$
      (1-x^7) = y^3
      $$
      $$
      (1-y^3) = x^7
      $$
      $$
      x = sqrt[7]{(1-y^3)}
      $$



      but that is all...






      definite-integrals







      share|cite|improve this question









      $endgroup$




      while trying to help the daughter of my cousin to understand Integrals, I see that I forget about some of the integral calculations. Can someone help me with that one, please?



      $$int_0^1 Bigl(sqrt[3]{(1-x^7)}-sqrt[7]{(1-x^3)} Bigr) dx$$



      (this is not a homework, I'm trying to understand and explain to her how things work.)



      I have a clue like..



      $$
      sqrt[3]{(1-x^7)} = y
      $$
      $$
      (1-x^7) = y^3
      $$
      $$
      (1-y^3) = x^7
      $$
      $$
      x = sqrt[7]{(1-y^3)}
      $$



      but that is all...






      definite-integrals




      definite-integrals






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 14 '18 at 13:20









      tanaydintanaydin

      1136




      1136






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Hint:



          What's is the area of curve $$x^3+y^7=1$$ in $xin[0,1]$



          $$int_0^1 y dx=int_{x=0}^{x=1} x dy$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ah yes, we missed that it is equals to 1. Great, thanks!
            $endgroup$
            – tanaydin
            Dec 14 '18 at 13:44










          • $begingroup$
            @tanaydin, What is $=1?$
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 13:50










          • $begingroup$
            area of the curve is 0, (if I'm not wrong)
            $endgroup$
            – tanaydin
            Dec 14 '18 at 14:03












          • $begingroup$
            @tanaydin, You are correct
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 14:05










          • $begingroup$
            ok, we will study a bit more :)
            $endgroup$
            – tanaydin
            Dec 14 '18 at 14:39



















          1












          $begingroup$

          There is a nice insight to be gleaned here.



          First of all, note that the two integrands are inverse functions of each other, that is, if $f(x) = (1-x^7)^{frac 13}$, then $f^{-1}(x) = (1-x^3)^{frac 17}$.



          Also note that $f(0) = 1$ and $f(1) = 0$. If you represent a function and its inverse on a graph (the exact shapes don't matter), they will be superimposed on each other by reflection about the line of identity $y=x$. A simple curve sketch (and your job is easy because the shapes don't matter that much, just where they start and end and the fact that one is the mirror image of the other about $y=x$) should convince you and your student that the difference in area under the curve is zero for the interval $[0,1]$.



          Need something more rigorous and "mathematical"? Well, there's this nice identity which can be proven with integration by parts:



          $$int_a^bf(x) dx = bf(b) - af(a) - int_{f(a)}^{f(b)}f^{-1}(x)dx$$



          In this problem, $a = 0, b = 1, f(a) = 1, f(b) = 0$. You can now prove directly that the required integral vanishes.






          share|cite|improve this answer









          $endgroup$














            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Hint:



            What's is the area of curve $$x^3+y^7=1$$ in $xin[0,1]$



            $$int_0^1 y dx=int_{x=0}^{x=1} x dy$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Ah yes, we missed that it is equals to 1. Great, thanks!
              $endgroup$
              – tanaydin
              Dec 14 '18 at 13:44










            • $begingroup$
              @tanaydin, What is $=1?$
              $endgroup$
              – lab bhattacharjee
              Dec 14 '18 at 13:50










            • $begingroup$
              area of the curve is 0, (if I'm not wrong)
              $endgroup$
              – tanaydin
              Dec 14 '18 at 14:03












            • $begingroup$
              @tanaydin, You are correct
              $endgroup$
              – lab bhattacharjee
              Dec 14 '18 at 14:05










            • $begingroup$
              ok, we will study a bit more :)
              $endgroup$
              – tanaydin
              Dec 14 '18 at 14:39
















            2












            $begingroup$

            Hint:



            What's is the area of curve $$x^3+y^7=1$$ in $xin[0,1]$



            $$int_0^1 y dx=int_{x=0}^{x=1} x dy$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Ah yes, we missed that it is equals to 1. Great, thanks!
              $endgroup$
              – tanaydin
              Dec 14 '18 at 13:44










            • $begingroup$
              @tanaydin, What is $=1?$
              $endgroup$
              – lab bhattacharjee
              Dec 14 '18 at 13:50










            • $begingroup$
              area of the curve is 0, (if I'm not wrong)
              $endgroup$
              – tanaydin
              Dec 14 '18 at 14:03












            • $begingroup$
              @tanaydin, You are correct
              $endgroup$
              – lab bhattacharjee
              Dec 14 '18 at 14:05










            • $begingroup$
              ok, we will study a bit more :)
              $endgroup$
              – tanaydin
              Dec 14 '18 at 14:39














            2












            2








            2





            $begingroup$

            Hint:



            What's is the area of curve $$x^3+y^7=1$$ in $xin[0,1]$



            $$int_0^1 y dx=int_{x=0}^{x=1} x dy$$






            share|cite|improve this answer











            $endgroup$



            Hint:



            What's is the area of curve $$x^3+y^7=1$$ in $xin[0,1]$



            $$int_0^1 y dx=int_{x=0}^{x=1} x dy$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 14 '18 at 13:34

























            answered Dec 14 '18 at 13:26









            lab bhattacharjeelab bhattacharjee

            228k15159279




            228k15159279












            • $begingroup$
              Ah yes, we missed that it is equals to 1. Great, thanks!
              $endgroup$
              – tanaydin
              Dec 14 '18 at 13:44










            • $begingroup$
              @tanaydin, What is $=1?$
              $endgroup$
              – lab bhattacharjee
              Dec 14 '18 at 13:50










            • $begingroup$
              area of the curve is 0, (if I'm not wrong)
              $endgroup$
              – tanaydin
              Dec 14 '18 at 14:03












            • $begingroup$
              @tanaydin, You are correct
              $endgroup$
              – lab bhattacharjee
              Dec 14 '18 at 14:05










            • $begingroup$
              ok, we will study a bit more :)
              $endgroup$
              – tanaydin
              Dec 14 '18 at 14:39


















            • $begingroup$
              Ah yes, we missed that it is equals to 1. Great, thanks!
              $endgroup$
              – tanaydin
              Dec 14 '18 at 13:44










            • $begingroup$
              @tanaydin, What is $=1?$
              $endgroup$
              – lab bhattacharjee
              Dec 14 '18 at 13:50










            • $begingroup$
              area of the curve is 0, (if I'm not wrong)
              $endgroup$
              – tanaydin
              Dec 14 '18 at 14:03












            • $begingroup$
              @tanaydin, You are correct
              $endgroup$
              – lab bhattacharjee
              Dec 14 '18 at 14:05










            • $begingroup$
              ok, we will study a bit more :)
              $endgroup$
              – tanaydin
              Dec 14 '18 at 14:39
















            $begingroup$
            Ah yes, we missed that it is equals to 1. Great, thanks!
            $endgroup$
            – tanaydin
            Dec 14 '18 at 13:44




            $begingroup$
            Ah yes, we missed that it is equals to 1. Great, thanks!
            $endgroup$
            – tanaydin
            Dec 14 '18 at 13:44












            $begingroup$
            @tanaydin, What is $=1?$
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 13:50




            $begingroup$
            @tanaydin, What is $=1?$
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 13:50












            $begingroup$
            area of the curve is 0, (if I'm not wrong)
            $endgroup$
            – tanaydin
            Dec 14 '18 at 14:03






            $begingroup$
            area of the curve is 0, (if I'm not wrong)
            $endgroup$
            – tanaydin
            Dec 14 '18 at 14:03














            $begingroup$
            @tanaydin, You are correct
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 14:05




            $begingroup$
            @tanaydin, You are correct
            $endgroup$
            – lab bhattacharjee
            Dec 14 '18 at 14:05












            $begingroup$
            ok, we will study a bit more :)
            $endgroup$
            – tanaydin
            Dec 14 '18 at 14:39




            $begingroup$
            ok, we will study a bit more :)
            $endgroup$
            – tanaydin
            Dec 14 '18 at 14:39











            1












            $begingroup$

            There is a nice insight to be gleaned here.



            First of all, note that the two integrands are inverse functions of each other, that is, if $f(x) = (1-x^7)^{frac 13}$, then $f^{-1}(x) = (1-x^3)^{frac 17}$.



            Also note that $f(0) = 1$ and $f(1) = 0$. If you represent a function and its inverse on a graph (the exact shapes don't matter), they will be superimposed on each other by reflection about the line of identity $y=x$. A simple curve sketch (and your job is easy because the shapes don't matter that much, just where they start and end and the fact that one is the mirror image of the other about $y=x$) should convince you and your student that the difference in area under the curve is zero for the interval $[0,1]$.



            Need something more rigorous and "mathematical"? Well, there's this nice identity which can be proven with integration by parts:



            $$int_a^bf(x) dx = bf(b) - af(a) - int_{f(a)}^{f(b)}f^{-1}(x)dx$$



            In this problem, $a = 0, b = 1, f(a) = 1, f(b) = 0$. You can now prove directly that the required integral vanishes.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              There is a nice insight to be gleaned here.



              First of all, note that the two integrands are inverse functions of each other, that is, if $f(x) = (1-x^7)^{frac 13}$, then $f^{-1}(x) = (1-x^3)^{frac 17}$.



              Also note that $f(0) = 1$ and $f(1) = 0$. If you represent a function and its inverse on a graph (the exact shapes don't matter), they will be superimposed on each other by reflection about the line of identity $y=x$. A simple curve sketch (and your job is easy because the shapes don't matter that much, just where they start and end and the fact that one is the mirror image of the other about $y=x$) should convince you and your student that the difference in area under the curve is zero for the interval $[0,1]$.



              Need something more rigorous and "mathematical"? Well, there's this nice identity which can be proven with integration by parts:



              $$int_a^bf(x) dx = bf(b) - af(a) - int_{f(a)}^{f(b)}f^{-1}(x)dx$$



              In this problem, $a = 0, b = 1, f(a) = 1, f(b) = 0$. You can now prove directly that the required integral vanishes.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                There is a nice insight to be gleaned here.



                First of all, note that the two integrands are inverse functions of each other, that is, if $f(x) = (1-x^7)^{frac 13}$, then $f^{-1}(x) = (1-x^3)^{frac 17}$.



                Also note that $f(0) = 1$ and $f(1) = 0$. If you represent a function and its inverse on a graph (the exact shapes don't matter), they will be superimposed on each other by reflection about the line of identity $y=x$. A simple curve sketch (and your job is easy because the shapes don't matter that much, just where they start and end and the fact that one is the mirror image of the other about $y=x$) should convince you and your student that the difference in area under the curve is zero for the interval $[0,1]$.



                Need something more rigorous and "mathematical"? Well, there's this nice identity which can be proven with integration by parts:



                $$int_a^bf(x) dx = bf(b) - af(a) - int_{f(a)}^{f(b)}f^{-1}(x)dx$$



                In this problem, $a = 0, b = 1, f(a) = 1, f(b) = 0$. You can now prove directly that the required integral vanishes.






                share|cite|improve this answer









                $endgroup$



                There is a nice insight to be gleaned here.



                First of all, note that the two integrands are inverse functions of each other, that is, if $f(x) = (1-x^7)^{frac 13}$, then $f^{-1}(x) = (1-x^3)^{frac 17}$.



                Also note that $f(0) = 1$ and $f(1) = 0$. If you represent a function and its inverse on a graph (the exact shapes don't matter), they will be superimposed on each other by reflection about the line of identity $y=x$. A simple curve sketch (and your job is easy because the shapes don't matter that much, just where they start and end and the fact that one is the mirror image of the other about $y=x$) should convince you and your student that the difference in area under the curve is zero for the interval $[0,1]$.



                Need something more rigorous and "mathematical"? Well, there's this nice identity which can be proven with integration by parts:



                $$int_a^bf(x) dx = bf(b) - af(a) - int_{f(a)}^{f(b)}f^{-1}(x)dx$$



                In this problem, $a = 0, b = 1, f(a) = 1, f(b) = 0$. You can now prove directly that the required integral vanishes.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 14 '18 at 13:57









                DeepakDeepak

                18k11640




                18k11640






























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