Integral of (1/100-x)












1












$begingroup$


I don't understand how the integral below:



begin{align}
int{frac{1}{100-x}dx} = -log|x-100| +c
end{align}



When I integrate I get the answer:
begin{align}
-log|100-x| +c
end{align}



I understand that it probably has something to do with the absolute value function, would someone kindly explain it to me? Thanks!






calculus integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $|a|=|-a|$; $x-100=-(100-x)$.
    $endgroup$
    – David Mitra
    1 hour ago


















1












$begingroup$


I don't understand how the integral below:



begin{align}
int{frac{1}{100-x}dx} = -log|x-100| +c
end{align}



When I integrate I get the answer:
begin{align}
-log|100-x| +c
end{align}



I understand that it probably has something to do with the absolute value function, would someone kindly explain it to me? Thanks!






calculus integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $|a|=|-a|$; $x-100=-(100-x)$.
    $endgroup$
    – David Mitra
    1 hour ago
















1












1








1





$begingroup$


I don't understand how the integral below:



begin{align}
int{frac{1}{100-x}dx} = -log|x-100| +c
end{align}



When I integrate I get the answer:
begin{align}
-log|100-x| +c
end{align}



I understand that it probably has something to do with the absolute value function, would someone kindly explain it to me? Thanks!






calculus integration







share|cite|improve this question











$endgroup$




I don't understand how the integral below:



begin{align}
int{frac{1}{100-x}dx} = -log|x-100| +c
end{align}



When I integrate I get the answer:
begin{align}
-log|100-x| +c
end{align}



I understand that it probably has something to do with the absolute value function, would someone kindly explain it to me? Thanks!






calculus integration




calculus integration






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edited 55 mins ago









in_mathematica_we_trust

2,5301019




2,5301019










asked 1 hour ago









SmithHSmithH

215




215








  • 2




    $begingroup$
    $|a|=|-a|$; $x-100=-(100-x)$.
    $endgroup$
    – David Mitra
    1 hour ago
















  • 2




    $begingroup$
    $|a|=|-a|$; $x-100=-(100-x)$.
    $endgroup$
    – David Mitra
    1 hour ago










2




2




$begingroup$
$|a|=|-a|$; $x-100=-(100-x)$.
$endgroup$
– David Mitra
1 hour ago






$begingroup$
$|a|=|-a|$; $x-100=-(100-x)$.
$endgroup$
– David Mitra
1 hour ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

The absolute value function disregards sign, so |4| = |-4| for example.



That also means that $|x-100| = |100-x|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In the context of a differential equation, where x is the population of function x(t), t is time, and x(0) = 200. Is there any specific reason why I would have to use |x-100| instead of |100-x|?
    $endgroup$
    – SmithH
    50 mins ago










  • $begingroup$
    if you know that x>100, then use |x-100|
    $endgroup$
    – Saketh Malyala
    49 mins ago










  • $begingroup$
    In fact, if you know $x gt 100$, you can use $(x-100)$ without absolute value signs, making your integral $-(x-100)+c = 100-x+c$, noting that your integrand is negative
    $endgroup$
    – Henry
    48 mins ago












  • $begingroup$
    Thanks so much for the help so far. If the differential equation for this logistic population growth was 1000(dx/dt) = x(100-x), what would suggest that x has to be greater than 100?
    $endgroup$
    – SmithH
    43 mins ago










  • $begingroup$
    thank you for providing the equation
    $endgroup$
    – Saketh Malyala
    37 mins ago



















2












$begingroup$

Since the argument of the logarithm is in absolute value signs, it doesn't really matter the order in a case like this. As we can see,



$$|a-b| = |(-1)(b-a)| = |-1||b-a| = |b-a|$$






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    The absolute value function disregards sign, so |4| = |-4| for example.



    That also means that $|x-100| = |100-x|$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      In the context of a differential equation, where x is the population of function x(t), t is time, and x(0) = 200. Is there any specific reason why I would have to use |x-100| instead of |100-x|?
      $endgroup$
      – SmithH
      50 mins ago










    • $begingroup$
      if you know that x>100, then use |x-100|
      $endgroup$
      – Saketh Malyala
      49 mins ago










    • $begingroup$
      In fact, if you know $x gt 100$, you can use $(x-100)$ without absolute value signs, making your integral $-(x-100)+c = 100-x+c$, noting that your integrand is negative
      $endgroup$
      – Henry
      48 mins ago












    • $begingroup$
      Thanks so much for the help so far. If the differential equation for this logistic population growth was 1000(dx/dt) = x(100-x), what would suggest that x has to be greater than 100?
      $endgroup$
      – SmithH
      43 mins ago










    • $begingroup$
      thank you for providing the equation
      $endgroup$
      – Saketh Malyala
      37 mins ago
















    3












    $begingroup$

    The absolute value function disregards sign, so |4| = |-4| for example.



    That also means that $|x-100| = |100-x|$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      In the context of a differential equation, where x is the population of function x(t), t is time, and x(0) = 200. Is there any specific reason why I would have to use |x-100| instead of |100-x|?
      $endgroup$
      – SmithH
      50 mins ago










    • $begingroup$
      if you know that x>100, then use |x-100|
      $endgroup$
      – Saketh Malyala
      49 mins ago










    • $begingroup$
      In fact, if you know $x gt 100$, you can use $(x-100)$ without absolute value signs, making your integral $-(x-100)+c = 100-x+c$, noting that your integrand is negative
      $endgroup$
      – Henry
      48 mins ago












    • $begingroup$
      Thanks so much for the help so far. If the differential equation for this logistic population growth was 1000(dx/dt) = x(100-x), what would suggest that x has to be greater than 100?
      $endgroup$
      – SmithH
      43 mins ago










    • $begingroup$
      thank you for providing the equation
      $endgroup$
      – Saketh Malyala
      37 mins ago














    3












    3








    3





    $begingroup$

    The absolute value function disregards sign, so |4| = |-4| for example.



    That also means that $|x-100| = |100-x|$.






    share|cite|improve this answer









    $endgroup$



    The absolute value function disregards sign, so |4| = |-4| for example.



    That also means that $|x-100| = |100-x|$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    Saketh MalyalaSaketh Malyala

    7,8481535




    7,8481535












    • $begingroup$
      In the context of a differential equation, where x is the population of function x(t), t is time, and x(0) = 200. Is there any specific reason why I would have to use |x-100| instead of |100-x|?
      $endgroup$
      – SmithH
      50 mins ago










    • $begingroup$
      if you know that x>100, then use |x-100|
      $endgroup$
      – Saketh Malyala
      49 mins ago










    • $begingroup$
      In fact, if you know $x gt 100$, you can use $(x-100)$ without absolute value signs, making your integral $-(x-100)+c = 100-x+c$, noting that your integrand is negative
      $endgroup$
      – Henry
      48 mins ago












    • $begingroup$
      Thanks so much for the help so far. If the differential equation for this logistic population growth was 1000(dx/dt) = x(100-x), what would suggest that x has to be greater than 100?
      $endgroup$
      – SmithH
      43 mins ago










    • $begingroup$
      thank you for providing the equation
      $endgroup$
      – Saketh Malyala
      37 mins ago


















    • $begingroup$
      In the context of a differential equation, where x is the population of function x(t), t is time, and x(0) = 200. Is there any specific reason why I would have to use |x-100| instead of |100-x|?
      $endgroup$
      – SmithH
      50 mins ago










    • $begingroup$
      if you know that x>100, then use |x-100|
      $endgroup$
      – Saketh Malyala
      49 mins ago










    • $begingroup$
      In fact, if you know $x gt 100$, you can use $(x-100)$ without absolute value signs, making your integral $-(x-100)+c = 100-x+c$, noting that your integrand is negative
      $endgroup$
      – Henry
      48 mins ago












    • $begingroup$
      Thanks so much for the help so far. If the differential equation for this logistic population growth was 1000(dx/dt) = x(100-x), what would suggest that x has to be greater than 100?
      $endgroup$
      – SmithH
      43 mins ago










    • $begingroup$
      thank you for providing the equation
      $endgroup$
      – Saketh Malyala
      37 mins ago
















    $begingroup$
    In the context of a differential equation, where x is the population of function x(t), t is time, and x(0) = 200. Is there any specific reason why I would have to use |x-100| instead of |100-x|?
    $endgroup$
    – SmithH
    50 mins ago




    $begingroup$
    In the context of a differential equation, where x is the population of function x(t), t is time, and x(0) = 200. Is there any specific reason why I would have to use |x-100| instead of |100-x|?
    $endgroup$
    – SmithH
    50 mins ago












    $begingroup$
    if you know that x>100, then use |x-100|
    $endgroup$
    – Saketh Malyala
    49 mins ago




    $begingroup$
    if you know that x>100, then use |x-100|
    $endgroup$
    – Saketh Malyala
    49 mins ago












    $begingroup$
    In fact, if you know $x gt 100$, you can use $(x-100)$ without absolute value signs, making your integral $-(x-100)+c = 100-x+c$, noting that your integrand is negative
    $endgroup$
    – Henry
    48 mins ago






    $begingroup$
    In fact, if you know $x gt 100$, you can use $(x-100)$ without absolute value signs, making your integral $-(x-100)+c = 100-x+c$, noting that your integrand is negative
    $endgroup$
    – Henry
    48 mins ago














    $begingroup$
    Thanks so much for the help so far. If the differential equation for this logistic population growth was 1000(dx/dt) = x(100-x), what would suggest that x has to be greater than 100?
    $endgroup$
    – SmithH
    43 mins ago




    $begingroup$
    Thanks so much for the help so far. If the differential equation for this logistic population growth was 1000(dx/dt) = x(100-x), what would suggest that x has to be greater than 100?
    $endgroup$
    – SmithH
    43 mins ago












    $begingroup$
    thank you for providing the equation
    $endgroup$
    – Saketh Malyala
    37 mins ago




    $begingroup$
    thank you for providing the equation
    $endgroup$
    – Saketh Malyala
    37 mins ago











    2












    $begingroup$

    Since the argument of the logarithm is in absolute value signs, it doesn't really matter the order in a case like this. As we can see,



    $$|a-b| = |(-1)(b-a)| = |-1||b-a| = |b-a|$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Since the argument of the logarithm is in absolute value signs, it doesn't really matter the order in a case like this. As we can see,



      $$|a-b| = |(-1)(b-a)| = |-1||b-a| = |b-a|$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Since the argument of the logarithm is in absolute value signs, it doesn't really matter the order in a case like this. As we can see,



        $$|a-b| = |(-1)(b-a)| = |-1||b-a| = |b-a|$$






        share|cite|improve this answer









        $endgroup$



        Since the argument of the logarithm is in absolute value signs, it doesn't really matter the order in a case like this. As we can see,



        $$|a-b| = |(-1)(b-a)| = |-1||b-a| = |b-a|$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Eevee TrainerEevee Trainer

        10.8k31944




        10.8k31944






























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