Integral problem. Unsure of the approach.












3












$begingroup$


I have this integral:



$$int_0^1 frac{1 + 12t}{1 + 3t}dt$$



I can split this up into:



$$int_0^1 frac{1}{1+3t} dt + int_0^1frac{12t}{1+3t}dt$$



The left side:



$u = 1+3t$ and $du = 3dt$ and $frac{du}{3} = dt$



so $$frac{1}{3} int_0^1 frac{1}{u} du = frac{1}{3} ln |u| + C$$



But what about the right?






calculus integration







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I have this integral:



    $$int_0^1 frac{1 + 12t}{1 + 3t}dt$$



    I can split this up into:



    $$int_0^1 frac{1}{1+3t} dt + int_0^1frac{12t}{1+3t}dt$$



    The left side:



    $u = 1+3t$ and $du = 3dt$ and $frac{du}{3} = dt$



    so $$frac{1}{3} int_0^1 frac{1}{u} du = frac{1}{3} ln |u| + C$$



    But what about the right?






    calculus integration







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I have this integral:



      $$int_0^1 frac{1 + 12t}{1 + 3t}dt$$



      I can split this up into:



      $$int_0^1 frac{1}{1+3t} dt + int_0^1frac{12t}{1+3t}dt$$



      The left side:



      $u = 1+3t$ and $du = 3dt$ and $frac{du}{3} = dt$



      so $$frac{1}{3} int_0^1 frac{1}{u} du = frac{1}{3} ln |u| + C$$



      But what about the right?






      calculus integration







      share|cite|improve this question











      $endgroup$




      I have this integral:



      $$int_0^1 frac{1 + 12t}{1 + 3t}dt$$



      I can split this up into:



      $$int_0^1 frac{1}{1+3t} dt + int_0^1frac{12t}{1+3t}dt$$



      The left side:



      $u = 1+3t$ and $du = 3dt$ and $frac{du}{3} = dt$



      so $$frac{1}{3} int_0^1 frac{1}{u} du = frac{1}{3} ln |u| + C$$



      But what about the right?






      calculus integration




      calculus integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 24 mins ago









      Eevee Trainer

      6,54811237




      6,54811237










      asked 32 mins ago









      Jwan622Jwan622

      2,16211530




      2,16211530






















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          Hint:



          $$frac{12t}{1+3t} = 4 - frac{4}{1+3t}.$$ Now use the same approach as you did for the first part.
          In fact you should’ve simplified the fraction before splitting into parts.



          Remember to evaluate the integral at t=0 and t=1! ;)






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            At least on an initial glance, I'd make the substitution



            $$u = 1+3t implies du = 3dt implies dt = frac{du}{3}$$



            Note that $t = (u-1)/3$ and thus $12t = 4(u-1)$.



            Then the integral becomes



            $$int_0^1 frac{12t}{1 + 3t} dt = int_ast^ast frac{4(u-1)}{u} frac{dt}{3} = frac 4 3 int_ast^ast frac{u-1}{u}du = frac 4 3 int_ast^ast 1 - frac{1}{u}du$$



            But this seems a bit excessive and unnecessary, even if it's the first thing to come to mind. (That's why I left the bounds as $ast$'s: in cases like these, overcomplication suggests an alternative approach.)




            Another trick is to "add and subtract the same thing from the top"; this can be handy in a lot of math classes for manipulations if you add and subtract the right thing. In this case, we notice:



            $$12t = 12t + 4 - 4 = 4(3t+1) - 4$$



            Thus,



            $$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$






            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%2f3122196%2fintegral-problem-unsure-of-the-approach%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









              6












              $begingroup$

              Hint:



              $$frac{12t}{1+3t} = 4 - frac{4}{1+3t}.$$ Now use the same approach as you did for the first part.
              In fact you should’ve simplified the fraction before splitting into parts.



              Remember to evaluate the integral at t=0 and t=1! ;)






              share|cite|improve this answer











              $endgroup$


















                6












                $begingroup$

                Hint:



                $$frac{12t}{1+3t} = 4 - frac{4}{1+3t}.$$ Now use the same approach as you did for the first part.
                In fact you should’ve simplified the fraction before splitting into parts.



                Remember to evaluate the integral at t=0 and t=1! ;)






                share|cite|improve this answer











                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  Hint:



                  $$frac{12t}{1+3t} = 4 - frac{4}{1+3t}.$$ Now use the same approach as you did for the first part.
                  In fact you should’ve simplified the fraction before splitting into parts.



                  Remember to evaluate the integral at t=0 and t=1! ;)






                  share|cite|improve this answer











                  $endgroup$



                  Hint:



                  $$frac{12t}{1+3t} = 4 - frac{4}{1+3t}.$$ Now use the same approach as you did for the first part.
                  In fact you should’ve simplified the fraction before splitting into parts.



                  Remember to evaluate the integral at t=0 and t=1! ;)







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 13 mins ago









                  Jyrki Lahtonen

                  109k13169374




                  109k13169374










                  answered 29 mins ago









                  Gareth MaGareth Ma

                  416213




                  416213























                      1












                      $begingroup$

                      At least on an initial glance, I'd make the substitution



                      $$u = 1+3t implies du = 3dt implies dt = frac{du}{3}$$



                      Note that $t = (u-1)/3$ and thus $12t = 4(u-1)$.



                      Then the integral becomes



                      $$int_0^1 frac{12t}{1 + 3t} dt = int_ast^ast frac{4(u-1)}{u} frac{dt}{3} = frac 4 3 int_ast^ast frac{u-1}{u}du = frac 4 3 int_ast^ast 1 - frac{1}{u}du$$



                      But this seems a bit excessive and unnecessary, even if it's the first thing to come to mind. (That's why I left the bounds as $ast$'s: in cases like these, overcomplication suggests an alternative approach.)




                      Another trick is to "add and subtract the same thing from the top"; this can be handy in a lot of math classes for manipulations if you add and subtract the right thing. In this case, we notice:



                      $$12t = 12t + 4 - 4 = 4(3t+1) - 4$$



                      Thus,



                      $$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        At least on an initial glance, I'd make the substitution



                        $$u = 1+3t implies du = 3dt implies dt = frac{du}{3}$$



                        Note that $t = (u-1)/3$ and thus $12t = 4(u-1)$.



                        Then the integral becomes



                        $$int_0^1 frac{12t}{1 + 3t} dt = int_ast^ast frac{4(u-1)}{u} frac{dt}{3} = frac 4 3 int_ast^ast frac{u-1}{u}du = frac 4 3 int_ast^ast 1 - frac{1}{u}du$$



                        But this seems a bit excessive and unnecessary, even if it's the first thing to come to mind. (That's why I left the bounds as $ast$'s: in cases like these, overcomplication suggests an alternative approach.)




                        Another trick is to "add and subtract the same thing from the top"; this can be handy in a lot of math classes for manipulations if you add and subtract the right thing. In this case, we notice:



                        $$12t = 12t + 4 - 4 = 4(3t+1) - 4$$



                        Thus,



                        $$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          At least on an initial glance, I'd make the substitution



                          $$u = 1+3t implies du = 3dt implies dt = frac{du}{3}$$



                          Note that $t = (u-1)/3$ and thus $12t = 4(u-1)$.



                          Then the integral becomes



                          $$int_0^1 frac{12t}{1 + 3t} dt = int_ast^ast frac{4(u-1)}{u} frac{dt}{3} = frac 4 3 int_ast^ast frac{u-1}{u}du = frac 4 3 int_ast^ast 1 - frac{1}{u}du$$



                          But this seems a bit excessive and unnecessary, even if it's the first thing to come to mind. (That's why I left the bounds as $ast$'s: in cases like these, overcomplication suggests an alternative approach.)




                          Another trick is to "add and subtract the same thing from the top"; this can be handy in a lot of math classes for manipulations if you add and subtract the right thing. In this case, we notice:



                          $$12t = 12t + 4 - 4 = 4(3t+1) - 4$$



                          Thus,



                          $$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$






                          share|cite|improve this answer









                          $endgroup$



                          At least on an initial glance, I'd make the substitution



                          $$u = 1+3t implies du = 3dt implies dt = frac{du}{3}$$



                          Note that $t = (u-1)/3$ and thus $12t = 4(u-1)$.



                          Then the integral becomes



                          $$int_0^1 frac{12t}{1 + 3t} dt = int_ast^ast frac{4(u-1)}{u} frac{dt}{3} = frac 4 3 int_ast^ast frac{u-1}{u}du = frac 4 3 int_ast^ast 1 - frac{1}{u}du$$



                          But this seems a bit excessive and unnecessary, even if it's the first thing to come to mind. (That's why I left the bounds as $ast$'s: in cases like these, overcomplication suggests an alternative approach.)




                          Another trick is to "add and subtract the same thing from the top"; this can be handy in a lot of math classes for manipulations if you add and subtract the right thing. In this case, we notice:



                          $$12t = 12t + 4 - 4 = 4(3t+1) - 4$$



                          Thus,



                          $$int_0^1 frac{12t}{1 + 3t} dt = int_0^1 frac{4(3t+1) - 4}{1 + 3t} dt = int_0^1 4 - frac{4}{1 + 3t} dt$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 25 mins ago









                          Eevee TrainerEevee Trainer

                          6,54811237




                          6,54811237






























                              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%2f3122196%2fintegral-problem-unsure-of-the-approach%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

                              Magento 2 controller redirect on button click in phtml file

                              Image
                              0 I want to redirect customer using controller when button is clicked in one of my phtml files. So far I got, controller: public function execute(){ return $this->addToCart(); } public function addToCart(){ // some logic $this->_redirect('checkout/cart/index'); } The above is working fine (redirecting) when you access the controller via URL (www.mydomain.com/batchorder/index/addtocart). And here is my .phtml file: $("#addToCart").click(function(){ let controllerUrl = "<?php echo $block-getUrl('batchorder/index/addtocart'); ?>" $.post(controllerUrl, { "products": JSON.stringify(sideCart.SideCartProducts) }) The post request returns my logic as it should, however, it is not redirecting. Anyone can help? ...
                              Read more

                              Polycentropodidae

                              Image
                              Polycentropodidae Plectrocnemia conspersa Siyentipiko nga pagklasipika Ginhadi-an: Animalia Phylum: Arthropoda Ubosphylum: Hexapoda Klase: Insecta Orden: Trichoptera Labawbanay: Hydropsychoidea Banay: Polycentropodidae Binomial nga ngaran Polycentropodidae An Polycentropodidae [1] in uska familia han Insecta. An Polycentropodidae in nahilalakip ha familia nga Hydropsychoidea, ordo nga Trichoptera, classis nga Insecta, punoan nga Arthropoda, ngan regnum nga Animalia. [1] An familia nga Polycentropodidae in naglalakip hin 660 ka mga species, sumala ha Catalogue of Life [1] . An kladograma hini sumala ha Catalogue of Life [1] : Hydropsychoidea   Polycentropodidae  Adectophylax Antillopsyche Cernotina Cyrnellus Cyrnodes Cyrnopsis Cyrnus Eodipseudopsis Eoneureclipsis Holocentropus Kambaitipsyche Neucentropus ...
                              Read more