How do I evaluate this double integral by changing the order of integration?

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One of my past papers has the following question (without a solution).



Evaluate the integral by changing the order of integration.



$int_0^1dx$$int_x^1/a^1e^y^a+1dy$



where $a$ is a constant and $a neq -1, 0$.



I would assume changing the order of integration would make it easier to do but I can't see this.




integration multivariable-calculus




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  • 1




    Try it for $a=2$ and get back to us with where you get stuck.
    – B. Goddard
    Jul 24 at 17:10














up vote
-1
down vote

favorite












One of my past papers has the following question (without a solution).



Evaluate the integral by changing the order of integration.



$int_0^1dx$$int_x^1/a^1e^y^a+1dy$



where $a$ is a constant and $a neq -1, 0$.



I would assume changing the order of integration would make it easier to do but I can't see this.




integration multivariable-calculus




share|cite|improve this question















  • 1




    Try it for $a=2$ and get back to us with where you get stuck.
    – B. Goddard
    Jul 24 at 17:10












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











One of my past papers has the following question (without a solution).



Evaluate the integral by changing the order of integration.



$int_0^1dx$$int_x^1/a^1e^y^a+1dy$



where $a$ is a constant and $a neq -1, 0$.



I would assume changing the order of integration would make it easier to do but I can't see this.




integration multivariable-calculus




share|cite|improve this question











One of my past papers has the following question (without a solution).



Evaluate the integral by changing the order of integration.



$int_0^1dx$$int_x^1/a^1e^y^a+1dy$



where $a$ is a constant and $a neq -1, 0$.



I would assume changing the order of integration would make it easier to do but I can't see this.





integration multivariable-calculus





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 24 at 17:03









user499701

777




777







  • 1




    Try it for $a=2$ and get back to us with where you get stuck.
    – B. Goddard
    Jul 24 at 17:10












  • 1




    Try it for $a=2$ and get back to us with where you get stuck.
    – B. Goddard
    Jul 24 at 17:10







1




1




Try it for $a=2$ and get back to us with where you get stuck.
– B. Goddard
Jul 24 at 17:10




Try it for $a=2$ and get back to us with where you get stuck.
– B. Goddard
Jul 24 at 17:10










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










Draw on a graph and convince yourself that the region bounded by $x = 0 , x = 1$ and $y$ varying from $x^1/a$ to $1$ is same as the region enclosed by $y =0, y = 1$, and $x$ varying from $0$ to $y^a$, and change the integral accordingly






share|cite|improve this answer






























    up vote
    1
    down vote













    From $x^frac1aleq yleq1$ we have $xleq y^aleq1$ and also $0leq yleq1$ then the integral is
    $$int_0^1,mathrmdyint_0^y^ae^y^a+1,mathrmdx=int_0^1,y^a,e^y^a+1,mathrmdy=colorbluedfrac1a+1(e-1)$$






    share|cite|improve this answer























    • How did you get $0leq yleq1$?
      – user499701
      Jul 24 at 17:26










    • Simply $0leq xleq y^aleq1$.
      – Nosrati
      Jul 24 at 17:34










    • Downvote means "the answer isn't useful.
      – Nosrati
      Jul 24 at 17:35











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Draw on a graph and convince yourself that the region bounded by $x = 0 , x = 1$ and $y$ varying from $x^1/a$ to $1$ is same as the region enclosed by $y =0, y = 1$, and $x$ varying from $0$ to $y^a$, and change the integral accordingly






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      Draw on a graph and convince yourself that the region bounded by $x = 0 , x = 1$ and $y$ varying from $x^1/a$ to $1$ is same as the region enclosed by $y =0, y = 1$, and $x$ varying from $0$ to $y^a$, and change the integral accordingly






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Draw on a graph and convince yourself that the region bounded by $x = 0 , x = 1$ and $y$ varying from $x^1/a$ to $1$ is same as the region enclosed by $y =0, y = 1$, and $x$ varying from $0$ to $y^a$, and change the integral accordingly






        share|cite|improve this answer















        Draw on a graph and convince yourself that the region bounded by $x = 0 , x = 1$ and $y$ varying from $x^1/a$ to $1$ is same as the region enclosed by $y =0, y = 1$, and $x$ varying from $0$ to $y^a$, and change the integral accordingly







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 24 at 17:50


























        answered Jul 24 at 17:28









        ab123

        1,344319




        1,344319




















            up vote
            1
            down vote













            From $x^frac1aleq yleq1$ we have $xleq y^aleq1$ and also $0leq yleq1$ then the integral is
            $$int_0^1,mathrmdyint_0^y^ae^y^a+1,mathrmdx=int_0^1,y^a,e^y^a+1,mathrmdy=colorbluedfrac1a+1(e-1)$$






            share|cite|improve this answer























            • How did you get $0leq yleq1$?
              – user499701
              Jul 24 at 17:26










            • Simply $0leq xleq y^aleq1$.
              – Nosrati
              Jul 24 at 17:34










            • Downvote means "the answer isn't useful.
              – Nosrati
              Jul 24 at 17:35















            up vote
            1
            down vote













            From $x^frac1aleq yleq1$ we have $xleq y^aleq1$ and also $0leq yleq1$ then the integral is
            $$int_0^1,mathrmdyint_0^y^ae^y^a+1,mathrmdx=int_0^1,y^a,e^y^a+1,mathrmdy=colorbluedfrac1a+1(e-1)$$






            share|cite|improve this answer























            • How did you get $0leq yleq1$?
              – user499701
              Jul 24 at 17:26










            • Simply $0leq xleq y^aleq1$.
              – Nosrati
              Jul 24 at 17:34










            • Downvote means "the answer isn't useful.
              – Nosrati
              Jul 24 at 17:35













            up vote
            1
            down vote










            up vote
            1
            down vote









            From $x^frac1aleq yleq1$ we have $xleq y^aleq1$ and also $0leq yleq1$ then the integral is
            $$int_0^1,mathrmdyint_0^y^ae^y^a+1,mathrmdx=int_0^1,y^a,e^y^a+1,mathrmdy=colorbluedfrac1a+1(e-1)$$






            share|cite|improve this answer















            From $x^frac1aleq yleq1$ we have $xleq y^aleq1$ and also $0leq yleq1$ then the integral is
            $$int_0^1,mathrmdyint_0^y^ae^y^a+1,mathrmdx=int_0^1,y^a,e^y^a+1,mathrmdy=colorbluedfrac1a+1(e-1)$$







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 24 at 17:39


























            answered Jul 24 at 17:15









            Nosrati

            19.3k41544




            19.3k41544











            • How did you get $0leq yleq1$?
              – user499701
              Jul 24 at 17:26










            • Simply $0leq xleq y^aleq1$.
              – Nosrati
              Jul 24 at 17:34










            • Downvote means "the answer isn't useful.
              – Nosrati
              Jul 24 at 17:35

















            • How did you get $0leq yleq1$?
              – user499701
              Jul 24 at 17:26










            • Simply $0leq xleq y^aleq1$.
              – Nosrati
              Jul 24 at 17:34










            • Downvote means "the answer isn't useful.
              – Nosrati
              Jul 24 at 17:35
















            How did you get $0leq yleq1$?
            – user499701
            Jul 24 at 17:26




            How did you get $0leq yleq1$?
            – user499701
            Jul 24 at 17:26












            Simply $0leq xleq y^aleq1$.
            – Nosrati
            Jul 24 at 17:34




            Simply $0leq xleq y^aleq1$.
            – Nosrati
            Jul 24 at 17:34












            Downvote means "the answer isn't useful.
            – Nosrati
            Jul 24 at 17:35





            Downvote means "the answer isn't useful.
            – Nosrati
            Jul 24 at 17:35













             

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