Integral with respect to $x +$ constant












1












$begingroup$


Is this a valid expression:
$$int xd(x+5)$$



I am trying to calculate the value using a u-sub, of $u = x + 5$. So then $du = d(x+5)$ and so the result is:



$$int (u - 5)du= frac{u^2}{2} - 5u + C = frac{(x+5)^2}{2} - 5(x+5) +C = frac{x^2}{2} -12.5 + C $$



Or is it correct to just do $d(x+5) = dx$ from the beginning and calculate $$int xd(x+5) = int xdx$$






calculus algebra-precalculus







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Is this a valid expression:
    $$int xd(x+5)$$



    I am trying to calculate the value using a u-sub, of $u = x + 5$. So then $du = d(x+5)$ and so the result is:



    $$int (u - 5)du= frac{u^2}{2} - 5u + C = frac{(x+5)^2}{2} - 5(x+5) +C = frac{x^2}{2} -12.5 + C $$



    Or is it correct to just do $d(x+5) = dx$ from the beginning and calculate $$int xd(x+5) = int xdx$$






    calculus algebra-precalculus







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Is this a valid expression:
      $$int xd(x+5)$$



      I am trying to calculate the value using a u-sub, of $u = x + 5$. So then $du = d(x+5)$ and so the result is:



      $$int (u - 5)du= frac{u^2}{2} - 5u + C = frac{(x+5)^2}{2} - 5(x+5) +C = frac{x^2}{2} -12.5 + C $$



      Or is it correct to just do $d(x+5) = dx$ from the beginning and calculate $$int xd(x+5) = int xdx$$






      calculus algebra-precalculus







      share|cite|improve this question









      $endgroup$




      Is this a valid expression:
      $$int xd(x+5)$$



      I am trying to calculate the value using a u-sub, of $u = x + 5$. So then $du = d(x+5)$ and so the result is:



      $$int (u - 5)du= frac{u^2}{2} - 5u + C = frac{(x+5)^2}{2} - 5(x+5) +C = frac{x^2}{2} -12.5 + C $$



      Or is it correct to just do $d(x+5) = dx$ from the beginning and calculate $$int xd(x+5) = int xdx$$






      calculus algebra-precalculus




      calculus algebra-precalculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Feb 1 at 5:29









      SladeSlade

      78111




      78111






















          2 Answers
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          $begingroup$

          Notice that you have $-12.5+C$, which is just another constant.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh, that makes a lot of sense now. Thanks!
            $endgroup$
            – Slade
            Feb 1 at 5:59



















          1












          $begingroup$

          Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)



          $$begin{align}
          int xe^{-x^2}dx &= int frac{xe^{-x^2}dx}{1}\
          &= int frac{xe^{-x^2}dx}{1}cdotfrac{frac{d(-x^2)}{dx}}{frac{d(-x^2)}{dx}}\
          &= int frac{xe^{-x^2}dxcdotfrac{d(-x^2)}{dx}}{-2x}\
          &= int frac{xe^{-x^2}d(-x^2)}{-2x}\
          &= color{red}{-frac{1}{2}int e^{-x^2}d(-x^2)} &[1]\
          &= color{green}{-frac{1}{2}int e^udu} &[2]\
          &= -frac{1}{2}e^u + C\
          &= -frac{1}{2}e^{-x^2} + C\
          end{align}$$






          share|cite|improve this answer











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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Notice that you have $-12.5+C$, which is just another constant.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Oh, that makes a lot of sense now. Thanks!
              $endgroup$
              – Slade
              Feb 1 at 5:59
















            2












            $begingroup$

            Notice that you have $-12.5+C$, which is just another constant.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Oh, that makes a lot of sense now. Thanks!
              $endgroup$
              – Slade
              Feb 1 at 5:59














            2












            2








            2





            $begingroup$

            Notice that you have $-12.5+C$, which is just another constant.






            share|cite|improve this answer









            $endgroup$



            Notice that you have $-12.5+C$, which is just another constant.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 1 at 5:40









            高田航高田航

            1,365418




            1,365418












            • $begingroup$
              Oh, that makes a lot of sense now. Thanks!
              $endgroup$
              – Slade
              Feb 1 at 5:59


















            • $begingroup$
              Oh, that makes a lot of sense now. Thanks!
              $endgroup$
              – Slade
              Feb 1 at 5:59
















            $begingroup$
            Oh, that makes a lot of sense now. Thanks!
            $endgroup$
            – Slade
            Feb 1 at 5:59




            $begingroup$
            Oh, that makes a lot of sense now. Thanks!
            $endgroup$
            – Slade
            Feb 1 at 5:59











            1












            $begingroup$

            Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)



            $$begin{align}
            int xe^{-x^2}dx &= int frac{xe^{-x^2}dx}{1}\
            &= int frac{xe^{-x^2}dx}{1}cdotfrac{frac{d(-x^2)}{dx}}{frac{d(-x^2)}{dx}}\
            &= int frac{xe^{-x^2}dxcdotfrac{d(-x^2)}{dx}}{-2x}\
            &= int frac{xe^{-x^2}d(-x^2)}{-2x}\
            &= color{red}{-frac{1}{2}int e^{-x^2}d(-x^2)} &[1]\
            &= color{green}{-frac{1}{2}int e^udu} &[2]\
            &= -frac{1}{2}e^u + C\
            &= -frac{1}{2}e^{-x^2} + C\
            end{align}$$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)



              $$begin{align}
              int xe^{-x^2}dx &= int frac{xe^{-x^2}dx}{1}\
              &= int frac{xe^{-x^2}dx}{1}cdotfrac{frac{d(-x^2)}{dx}}{frac{d(-x^2)}{dx}}\
              &= int frac{xe^{-x^2}dxcdotfrac{d(-x^2)}{dx}}{-2x}\
              &= int frac{xe^{-x^2}d(-x^2)}{-2x}\
              &= color{red}{-frac{1}{2}int e^{-x^2}d(-x^2)} &[1]\
              &= color{green}{-frac{1}{2}int e^udu} &[2]\
              &= -frac{1}{2}e^u + C\
              &= -frac{1}{2}e^{-x^2} + C\
              end{align}$$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)



                $$begin{align}
                int xe^{-x^2}dx &= int frac{xe^{-x^2}dx}{1}\
                &= int frac{xe^{-x^2}dx}{1}cdotfrac{frac{d(-x^2)}{dx}}{frac{d(-x^2)}{dx}}\
                &= int frac{xe^{-x^2}dxcdotfrac{d(-x^2)}{dx}}{-2x}\
                &= int frac{xe^{-x^2}d(-x^2)}{-2x}\
                &= color{red}{-frac{1}{2}int e^{-x^2}d(-x^2)} &[1]\
                &= color{green}{-frac{1}{2}int e^udu} &[2]\
                &= -frac{1}{2}e^u + C\
                &= -frac{1}{2}e^{-x^2} + C\
                end{align}$$






                share|cite|improve this answer











                $endgroup$



                Of course it's a valid expression. So is this. Note that [1] and [2] are the same expression. :-)



                $$begin{align}
                int xe^{-x^2}dx &= int frac{xe^{-x^2}dx}{1}\
                &= int frac{xe^{-x^2}dx}{1}cdotfrac{frac{d(-x^2)}{dx}}{frac{d(-x^2)}{dx}}\
                &= int frac{xe^{-x^2}dxcdotfrac{d(-x^2)}{dx}}{-2x}\
                &= int frac{xe^{-x^2}d(-x^2)}{-2x}\
                &= color{red}{-frac{1}{2}int e^{-x^2}d(-x^2)} &[1]\
                &= color{green}{-frac{1}{2}int e^udu} &[2]\
                &= -frac{1}{2}e^u + C\
                &= -frac{1}{2}e^{-x^2} + C\
                end{align}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 2 at 13:05

























                answered Feb 2 at 12:56









                John JoyJohn Joy

                6,29911827




                6,29911827






























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