Find $int_{0}^{1} int_{x}^{1}y^4e^{xy^2}dy dx$












3












$begingroup$


$$I:=int_{0}^{1} int_{x}^{1}y^4e^{xy^2}dy dx$$
Here the region of integration is the triangle with vertices $(0,0),(0,1)$ and $(1,1)$ and given as a type-1 region. We can convert it into a type-2 region which makes the integral easier.
$$I=int_{0}^{1}int_{0}^{y}y^4e^{xy^2}dx dy=int_{0}^{1} y^2(e^{y^3}-1)dy=frac {e-2}{3}$$
Is this correct? I'd like to add graphs but I'm still learning how to do that.






integration definite-integrals multiple-integral iterated-integrals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes, you have it. Well done.
    $endgroup$
    – Mark Viola
    Jan 30 at 4:37
















3












$begingroup$


$$I:=int_{0}^{1} int_{x}^{1}y^4e^{xy^2}dy dx$$
Here the region of integration is the triangle with vertices $(0,0),(0,1)$ and $(1,1)$ and given as a type-1 region. We can convert it into a type-2 region which makes the integral easier.
$$I=int_{0}^{1}int_{0}^{y}y^4e^{xy^2}dx dy=int_{0}^{1} y^2(e^{y^3}-1)dy=frac {e-2}{3}$$
Is this correct? I'd like to add graphs but I'm still learning how to do that.






integration definite-integrals multiple-integral iterated-integrals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes, you have it. Well done.
    $endgroup$
    – Mark Viola
    Jan 30 at 4:37














3












3








3





$begingroup$


$$I:=int_{0}^{1} int_{x}^{1}y^4e^{xy^2}dy dx$$
Here the region of integration is the triangle with vertices $(0,0),(0,1)$ and $(1,1)$ and given as a type-1 region. We can convert it into a type-2 region which makes the integral easier.
$$I=int_{0}^{1}int_{0}^{y}y^4e^{xy^2}dx dy=int_{0}^{1} y^2(e^{y^3}-1)dy=frac {e-2}{3}$$
Is this correct? I'd like to add graphs but I'm still learning how to do that.






integration definite-integrals multiple-integral iterated-integrals







share|cite|improve this question









$endgroup$




$$I:=int_{0}^{1} int_{x}^{1}y^4e^{xy^2}dy dx$$
Here the region of integration is the triangle with vertices $(0,0),(0,1)$ and $(1,1)$ and given as a type-1 region. We can convert it into a type-2 region which makes the integral easier.
$$I=int_{0}^{1}int_{0}^{y}y^4e^{xy^2}dx dy=int_{0}^{1} y^2(e^{y^3}-1)dy=frac {e-2}{3}$$
Is this correct? I'd like to add graphs but I'm still learning how to do that.






integration definite-integrals multiple-integral iterated-integrals




integration definite-integrals multiple-integral iterated-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 30 at 4:28









RhaldrynRhaldryn

352416




352416








  • 1




    $begingroup$
    Yes, you have it. Well done.
    $endgroup$
    – Mark Viola
    Jan 30 at 4:37














  • 1




    $begingroup$
    Yes, you have it. Well done.
    $endgroup$
    – Mark Viola
    Jan 30 at 4:37








1




1




$begingroup$
Yes, you have it. Well done.
$endgroup$
– Mark Viola
Jan 30 at 4:37




$begingroup$
Yes, you have it. Well done.
$endgroup$
– Mark Viola
Jan 30 at 4:37










2 Answers
2






active

oldest

votes


















1












$begingroup$

Basically , the integral means you have to sum the differential term (function ) for all the points in the given area .
Now in the 1st equation , the integral first summed it for all points with common x coordinate (for them , x becomes a constant) , and the value y takes is from x to 1( x is a constant). This creates sums for the strips in terms of x coordinate of the strip. Then when we integrate again , these strips are summed up over all x.



The second equation ,finds the sum in strips with common y coordinate (x goes from 0 to y in a strip). And then , similarly , strips are summed up for all y .
Both of them are equivalent and yes, the second one an easy one to evaluate.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Here's a graph, for your better understanding:



    enter image description here






    share|cite|improve this answer









    $endgroup$














      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3093102%2ffind-int-01-int-x1y4exy2dy-dx%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Basically , the integral means you have to sum the differential term (function ) for all the points in the given area .
      Now in the 1st equation , the integral first summed it for all points with common x coordinate (for them , x becomes a constant) , and the value y takes is from x to 1( x is a constant). This creates sums for the strips in terms of x coordinate of the strip. Then when we integrate again , these strips are summed up over all x.



      The second equation ,finds the sum in strips with common y coordinate (x goes from 0 to y in a strip). And then , similarly , strips are summed up for all y .
      Both of them are equivalent and yes, the second one an easy one to evaluate.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Basically , the integral means you have to sum the differential term (function ) for all the points in the given area .
        Now in the 1st equation , the integral first summed it for all points with common x coordinate (for them , x becomes a constant) , and the value y takes is from x to 1( x is a constant). This creates sums for the strips in terms of x coordinate of the strip. Then when we integrate again , these strips are summed up over all x.



        The second equation ,finds the sum in strips with common y coordinate (x goes from 0 to y in a strip). And then , similarly , strips are summed up for all y .
        Both of them are equivalent and yes, the second one an easy one to evaluate.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Basically , the integral means you have to sum the differential term (function ) for all the points in the given area .
          Now in the 1st equation , the integral first summed it for all points with common x coordinate (for them , x becomes a constant) , and the value y takes is from x to 1( x is a constant). This creates sums for the strips in terms of x coordinate of the strip. Then when we integrate again , these strips are summed up over all x.



          The second equation ,finds the sum in strips with common y coordinate (x goes from 0 to y in a strip). And then , similarly , strips are summed up for all y .
          Both of them are equivalent and yes, the second one an easy one to evaluate.






          share|cite|improve this answer









          $endgroup$



          Basically , the integral means you have to sum the differential term (function ) for all the points in the given area .
          Now in the 1st equation , the integral first summed it for all points with common x coordinate (for them , x becomes a constant) , and the value y takes is from x to 1( x is a constant). This creates sums for the strips in terms of x coordinate of the strip. Then when we integrate again , these strips are summed up over all x.



          The second equation ,finds the sum in strips with common y coordinate (x goes from 0 to y in a strip). And then , similarly , strips are summed up for all y .
          Both of them are equivalent and yes, the second one an easy one to evaluate.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 30 at 4:44









          Adarsha AmanAdarsha Aman

          113




          113























              1












              $begingroup$

              Here's a graph, for your better understanding:



              enter image description here






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Here's a graph, for your better understanding:



                enter image description here






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Here's a graph, for your better understanding:



                  enter image description here






                  share|cite|improve this answer









                  $endgroup$



                  Here's a graph, for your better understanding:



                  enter image description here







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 30 at 5:41









                  David G. StorkDavid G. Stork

                  11.5k41534




                  11.5k41534






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3093102%2ffind-int-01-int-x1y4exy2dy-dx%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      MongoDB - Not Authorized To Execute Command

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                      Read more

                      in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                      Image
                      1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                      Read more

                      How to fix TextFormField cause rebuild widget in Flutter

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                      Read more