A simple integral with one question












5












$begingroup$


integral



Question is: For $x$ equals $4$ and $9$, why is $t$ not $pm2$ and $pm3$ but just $2$ and $3$ ?






integration







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  • $begingroup$
    $x$ varying from $4$ to $9$, $t = sqrt x$ varies from $2$ to $3$
    $endgroup$
    – Damien
    Jan 10 at 14:22








  • 1




    $begingroup$
    In integration we must take the positive roots due to convention and consistency: see here
    $endgroup$
    – TheSimpliFire
    Jan 10 at 14:24


















5












$begingroup$


integral



Question is: For $x$ equals $4$ and $9$, why is $t$ not $pm2$ and $pm3$ but just $2$ and $3$ ?






integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    $x$ varying from $4$ to $9$, $t = sqrt x$ varies from $2$ to $3$
    $endgroup$
    – Damien
    Jan 10 at 14:22








  • 1




    $begingroup$
    In integration we must take the positive roots due to convention and consistency: see here
    $endgroup$
    – TheSimpliFire
    Jan 10 at 14:24
















5












5








5





$begingroup$


integral



Question is: For $x$ equals $4$ and $9$, why is $t$ not $pm2$ and $pm3$ but just $2$ and $3$ ?






integration







share|cite|improve this question









$endgroup$




integral



Question is: For $x$ equals $4$ and $9$, why is $t$ not $pm2$ and $pm3$ but just $2$ and $3$ ?






integration




integration






share|cite|improve this question













share|cite|improve this question











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asked Jan 10 at 14:17









PetraPetra

566




566












  • $begingroup$
    $x$ varying from $4$ to $9$, $t = sqrt x$ varies from $2$ to $3$
    $endgroup$
    – Damien
    Jan 10 at 14:22








  • 1




    $begingroup$
    In integration we must take the positive roots due to convention and consistency: see here
    $endgroup$
    – TheSimpliFire
    Jan 10 at 14:24




















  • $begingroup$
    $x$ varying from $4$ to $9$, $t = sqrt x$ varies from $2$ to $3$
    $endgroup$
    – Damien
    Jan 10 at 14:22








  • 1




    $begingroup$
    In integration we must take the positive roots due to convention and consistency: see here
    $endgroup$
    – TheSimpliFire
    Jan 10 at 14:24


















$begingroup$
$x$ varying from $4$ to $9$, $t = sqrt x$ varies from $2$ to $3$
$endgroup$
– Damien
Jan 10 at 14:22






$begingroup$
$x$ varying from $4$ to $9$, $t = sqrt x$ varies from $2$ to $3$
$endgroup$
– Damien
Jan 10 at 14:22






1




1




$begingroup$
In integration we must take the positive roots due to convention and consistency: see here
$endgroup$
– TheSimpliFire
Jan 10 at 14:24






$begingroup$
In integration we must take the positive roots due to convention and consistency: see here
$endgroup$
– TheSimpliFire
Jan 10 at 14:24












3 Answers
3






active

oldest

votes


















5












$begingroup$

You can use the negative square root if you want; it all works out the same way in the end. If we denote the positive square root of $x$ by $sqrt{x}$, we can substitute $t = -sqrt{x}$ instead. We still have $t^2 = x$, so $2 t , dt = dx $
as before. So the integral becomes
begin{align*}
int_{-2}^{-3} frac{2 t , dt}{-t -1} &= 2 int_{-3}^{-2} frac{t}{t + 1} dt \
&= 2 left[ int_{-3}^{-2} dt - int_{-3}^{-2} frac{dt}{t+1} right] \
&= 2 Big[ t - ln |t+1| Big]_{-3}^{-2} \
&= 2 Big[ -2 - (-3) - left( ln(1) - ln(2) right)Big] \
&= 2 + 2 ln 2.
end{align*}
The steps look slightly different, but it works out to be exactly the same result. This is an important lesson in general: there is frequently more than one possible substitution that allows you to solve an integral.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I can see this answer as very valuable to someone just learning calc. Nice job (+1).
    $endgroup$
    – clathratus
    Jan 10 at 15:22



















1












$begingroup$

By convention, $sqrt{a}$ represents the non-negative square root, or the principal square root, of $x$. Hence, the only case in which there are two opposite solutions is when you have $pmsqrt{a}$. (Note the extra $pm$ sign.) Hence is important to note that $x = sqrt{a}$ (one non-negative solution). should not be confused with $x^2 = a iff vert xvert = sqrt{a} iff x = pmsqrt{a}$ (two solutions). So, for instance, $sqrt{4} = color{blue}{+}2$ and $sqrt{9} = color{blue}{+}3$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Because under standard and generally accepted notation,



    $$ forall xge0;sqrt x =|x^{frac12}|$$
    That is to say, the square root is used in this case to mean the positive square root.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      You can use the negative square root if you want; it all works out the same way in the end. If we denote the positive square root of $x$ by $sqrt{x}$, we can substitute $t = -sqrt{x}$ instead. We still have $t^2 = x$, so $2 t , dt = dx $
      as before. So the integral becomes
      begin{align*}
      int_{-2}^{-3} frac{2 t , dt}{-t -1} &= 2 int_{-3}^{-2} frac{t}{t + 1} dt \
      &= 2 left[ int_{-3}^{-2} dt - int_{-3}^{-2} frac{dt}{t+1} right] \
      &= 2 Big[ t - ln |t+1| Big]_{-3}^{-2} \
      &= 2 Big[ -2 - (-3) - left( ln(1) - ln(2) right)Big] \
      &= 2 + 2 ln 2.
      end{align*}
      The steps look slightly different, but it works out to be exactly the same result. This is an important lesson in general: there is frequently more than one possible substitution that allows you to solve an integral.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I can see this answer as very valuable to someone just learning calc. Nice job (+1).
        $endgroup$
        – clathratus
        Jan 10 at 15:22
















      5












      $begingroup$

      You can use the negative square root if you want; it all works out the same way in the end. If we denote the positive square root of $x$ by $sqrt{x}$, we can substitute $t = -sqrt{x}$ instead. We still have $t^2 = x$, so $2 t , dt = dx $
      as before. So the integral becomes
      begin{align*}
      int_{-2}^{-3} frac{2 t , dt}{-t -1} &= 2 int_{-3}^{-2} frac{t}{t + 1} dt \
      &= 2 left[ int_{-3}^{-2} dt - int_{-3}^{-2} frac{dt}{t+1} right] \
      &= 2 Big[ t - ln |t+1| Big]_{-3}^{-2} \
      &= 2 Big[ -2 - (-3) - left( ln(1) - ln(2) right)Big] \
      &= 2 + 2 ln 2.
      end{align*}
      The steps look slightly different, but it works out to be exactly the same result. This is an important lesson in general: there is frequently more than one possible substitution that allows you to solve an integral.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I can see this answer as very valuable to someone just learning calc. Nice job (+1).
        $endgroup$
        – clathratus
        Jan 10 at 15:22














      5












      5








      5





      $begingroup$

      You can use the negative square root if you want; it all works out the same way in the end. If we denote the positive square root of $x$ by $sqrt{x}$, we can substitute $t = -sqrt{x}$ instead. We still have $t^2 = x$, so $2 t , dt = dx $
      as before. So the integral becomes
      begin{align*}
      int_{-2}^{-3} frac{2 t , dt}{-t -1} &= 2 int_{-3}^{-2} frac{t}{t + 1} dt \
      &= 2 left[ int_{-3}^{-2} dt - int_{-3}^{-2} frac{dt}{t+1} right] \
      &= 2 Big[ t - ln |t+1| Big]_{-3}^{-2} \
      &= 2 Big[ -2 - (-3) - left( ln(1) - ln(2) right)Big] \
      &= 2 + 2 ln 2.
      end{align*}
      The steps look slightly different, but it works out to be exactly the same result. This is an important lesson in general: there is frequently more than one possible substitution that allows you to solve an integral.






      share|cite|improve this answer









      $endgroup$



      You can use the negative square root if you want; it all works out the same way in the end. If we denote the positive square root of $x$ by $sqrt{x}$, we can substitute $t = -sqrt{x}$ instead. We still have $t^2 = x$, so $2 t , dt = dx $
      as before. So the integral becomes
      begin{align*}
      int_{-2}^{-3} frac{2 t , dt}{-t -1} &= 2 int_{-3}^{-2} frac{t}{t + 1} dt \
      &= 2 left[ int_{-3}^{-2} dt - int_{-3}^{-2} frac{dt}{t+1} right] \
      &= 2 Big[ t - ln |t+1| Big]_{-3}^{-2} \
      &= 2 Big[ -2 - (-3) - left( ln(1) - ln(2) right)Big] \
      &= 2 + 2 ln 2.
      end{align*}
      The steps look slightly different, but it works out to be exactly the same result. This is an important lesson in general: there is frequently more than one possible substitution that allows you to solve an integral.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 10 at 14:36









      Michael SeifertMichael Seifert

      4,982625




      4,982625












      • $begingroup$
        I can see this answer as very valuable to someone just learning calc. Nice job (+1).
        $endgroup$
        – clathratus
        Jan 10 at 15:22


















      • $begingroup$
        I can see this answer as very valuable to someone just learning calc. Nice job (+1).
        $endgroup$
        – clathratus
        Jan 10 at 15:22
















      $begingroup$
      I can see this answer as very valuable to someone just learning calc. Nice job (+1).
      $endgroup$
      – clathratus
      Jan 10 at 15:22




      $begingroup$
      I can see this answer as very valuable to someone just learning calc. Nice job (+1).
      $endgroup$
      – clathratus
      Jan 10 at 15:22











      1












      $begingroup$

      By convention, $sqrt{a}$ represents the non-negative square root, or the principal square root, of $x$. Hence, the only case in which there are two opposite solutions is when you have $pmsqrt{a}$. (Note the extra $pm$ sign.) Hence is important to note that $x = sqrt{a}$ (one non-negative solution). should not be confused with $x^2 = a iff vert xvert = sqrt{a} iff x = pmsqrt{a}$ (two solutions). So, for instance, $sqrt{4} = color{blue}{+}2$ and $sqrt{9} = color{blue}{+}3$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        By convention, $sqrt{a}$ represents the non-negative square root, or the principal square root, of $x$. Hence, the only case in which there are two opposite solutions is when you have $pmsqrt{a}$. (Note the extra $pm$ sign.) Hence is important to note that $x = sqrt{a}$ (one non-negative solution). should not be confused with $x^2 = a iff vert xvert = sqrt{a} iff x = pmsqrt{a}$ (two solutions). So, for instance, $sqrt{4} = color{blue}{+}2$ and $sqrt{9} = color{blue}{+}3$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          By convention, $sqrt{a}$ represents the non-negative square root, or the principal square root, of $x$. Hence, the only case in which there are two opposite solutions is when you have $pmsqrt{a}$. (Note the extra $pm$ sign.) Hence is important to note that $x = sqrt{a}$ (one non-negative solution). should not be confused with $x^2 = a iff vert xvert = sqrt{a} iff x = pmsqrt{a}$ (two solutions). So, for instance, $sqrt{4} = color{blue}{+}2$ and $sqrt{9} = color{blue}{+}3$.






          share|cite|improve this answer









          $endgroup$



          By convention, $sqrt{a}$ represents the non-negative square root, or the principal square root, of $x$. Hence, the only case in which there are two opposite solutions is when you have $pmsqrt{a}$. (Note the extra $pm$ sign.) Hence is important to note that $x = sqrt{a}$ (one non-negative solution). should not be confused with $x^2 = a iff vert xvert = sqrt{a} iff x = pmsqrt{a}$ (two solutions). So, for instance, $sqrt{4} = color{blue}{+}2$ and $sqrt{9} = color{blue}{+}3$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 14:36









          KM101KM101

          5,9251524




          5,9251524























              0












              $begingroup$

              Because under standard and generally accepted notation,



              $$ forall xge0;sqrt x =|x^{frac12}|$$
              That is to say, the square root is used in this case to mean the positive square root.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Because under standard and generally accepted notation,



                $$ forall xge0;sqrt x =|x^{frac12}|$$
                That is to say, the square root is used in this case to mean the positive square root.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Because under standard and generally accepted notation,



                  $$ forall xge0;sqrt x =|x^{frac12}|$$
                  That is to say, the square root is used in this case to mean the positive square root.






                  share|cite|improve this answer









                  $endgroup$



                  Because under standard and generally accepted notation,



                  $$ forall xge0;sqrt x =|x^{frac12}|$$
                  That is to say, the square root is used in this case to mean the positive square root.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 at 14:28









                  Rhys HughesRhys Hughes

                  5,9631529




                  5,9631529






























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