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Integral problem. Unsure of the approach.


Calculate the integral $int_0^{x} (lfloor t+1 rfloor)^3dt$Cauchy integral formula problemWhy is are these expressions in Leibniz notation not equivalent?Double integral problemIs it neccessary to split the integral?Evaluate the following line integral.$f_n(x) = frac{x^n}{1 + x^n}$ does not converge uniformly, yet we can change the order of integralBizarre Definite IntegralWhat's the integral of $(df/dx)^2$?Problem with a Hypergeometric function integral













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 31 mins ago









      Eevee Trainer

      6,54811237




      6,54811237










      asked 40 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









            $endgroup$













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              2 Answers
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              2 Answers
              2






              active

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              active

              oldest

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              active

              oldest

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              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 21 mins ago









                  Jyrki Lahtonen

                  109k13169374




                  109k13169374










                  answered 37 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 32 mins ago









                          Eevee TrainerEevee Trainer

                          6,54811237




                          6,54811237






























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