Show that two primitives/ antiderivatives are related via a constant












0












$begingroup$



Let $I subset R$ be an interval. A differentiable function $F: I rightarrow R$ is called a
primitive for the function $f : I rightarrow mathbb{R}$ if
$$F'
(x) = f(x)$$ for all $x in I$.



Show: If $F_ 1$ and $F_2$ are two primitives for $f$ on $I$ then there is a constant
$C in mathbb{R}$ such that $F_2 equiv F_1 + C$, i.e. $F_2(x) = F_1(x) + C$ for all $x in I$.




What we have covered so far is the formal definition of the derivative in terms of the limit:
$$ lim_{h rightarrow 0}frac{F_1(x+h)-F_1(x)}{h}= f(x)$$
$$ lim_{h rightarrow 0}frac{F_2(x+h)-F_2(x)}{h}= f(x)$$



I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant 'appears'.




We normally just define $F_2=F_1 +C$ and then it would follow immediately that these two functions have the same derivative.






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/1862231/…
    $endgroup$
    – Hans Lundmark
    Dec 11 '18 at 10:44






  • 1




    $begingroup$
    It is essential that the domain is an intervall.
    $endgroup$
    – gammatester
    Dec 11 '18 at 10:45










  • $begingroup$
    AH, the glorious $MVT$, cool!
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 10:46
















0












$begingroup$



Let $I subset R$ be an interval. A differentiable function $F: I rightarrow R$ is called a
primitive for the function $f : I rightarrow mathbb{R}$ if
$$F'
(x) = f(x)$$ for all $x in I$.



Show: If $F_ 1$ and $F_2$ are two primitives for $f$ on $I$ then there is a constant
$C in mathbb{R}$ such that $F_2 equiv F_1 + C$, i.e. $F_2(x) = F_1(x) + C$ for all $x in I$.




What we have covered so far is the formal definition of the derivative in terms of the limit:
$$ lim_{h rightarrow 0}frac{F_1(x+h)-F_1(x)}{h}= f(x)$$
$$ lim_{h rightarrow 0}frac{F_2(x+h)-F_2(x)}{h}= f(x)$$



I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant 'appears'.




We normally just define $F_2=F_1 +C$ and then it would follow immediately that these two functions have the same derivative.






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/1862231/…
    $endgroup$
    – Hans Lundmark
    Dec 11 '18 at 10:44






  • 1




    $begingroup$
    It is essential that the domain is an intervall.
    $endgroup$
    – gammatester
    Dec 11 '18 at 10:45










  • $begingroup$
    AH, the glorious $MVT$, cool!
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 10:46














0












0








0





$begingroup$



Let $I subset R$ be an interval. A differentiable function $F: I rightarrow R$ is called a
primitive for the function $f : I rightarrow mathbb{R}$ if
$$F'
(x) = f(x)$$ for all $x in I$.



Show: If $F_ 1$ and $F_2$ are two primitives for $f$ on $I$ then there is a constant
$C in mathbb{R}$ such that $F_2 equiv F_1 + C$, i.e. $F_2(x) = F_1(x) + C$ for all $x in I$.




What we have covered so far is the formal definition of the derivative in terms of the limit:
$$ lim_{h rightarrow 0}frac{F_1(x+h)-F_1(x)}{h}= f(x)$$
$$ lim_{h rightarrow 0}frac{F_2(x+h)-F_2(x)}{h}= f(x)$$



I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant 'appears'.




We normally just define $F_2=F_1 +C$ and then it would follow immediately that these two functions have the same derivative.






real-analysis







share|cite|improve this question









$endgroup$





Let $I subset R$ be an interval. A differentiable function $F: I rightarrow R$ is called a
primitive for the function $f : I rightarrow mathbb{R}$ if
$$F'
(x) = f(x)$$ for all $x in I$.



Show: If $F_ 1$ and $F_2$ are two primitives for $f$ on $I$ then there is a constant
$C in mathbb{R}$ such that $F_2 equiv F_1 + C$, i.e. $F_2(x) = F_1(x) + C$ for all $x in I$.




What we have covered so far is the formal definition of the derivative in terms of the limit:
$$ lim_{h rightarrow 0}frac{F_1(x+h)-F_1(x)}{h}= f(x)$$
$$ lim_{h rightarrow 0}frac{F_2(x+h)-F_2(x)}{h}= f(x)$$



I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant 'appears'.




We normally just define $F_2=F_1 +C$ and then it would follow immediately that these two functions have the same derivative.






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 10:37









Wesley StrikWesley Strik

2,057423




2,057423












  • $begingroup$
    math.stackexchange.com/questions/1862231/…
    $endgroup$
    – Hans Lundmark
    Dec 11 '18 at 10:44






  • 1




    $begingroup$
    It is essential that the domain is an intervall.
    $endgroup$
    – gammatester
    Dec 11 '18 at 10:45










  • $begingroup$
    AH, the glorious $MVT$, cool!
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 10:46


















  • $begingroup$
    math.stackexchange.com/questions/1862231/…
    $endgroup$
    – Hans Lundmark
    Dec 11 '18 at 10:44






  • 1




    $begingroup$
    It is essential that the domain is an intervall.
    $endgroup$
    – gammatester
    Dec 11 '18 at 10:45










  • $begingroup$
    AH, the glorious $MVT$, cool!
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 10:46
















$begingroup$
math.stackexchange.com/questions/1862231/…
$endgroup$
– Hans Lundmark
Dec 11 '18 at 10:44




$begingroup$
math.stackexchange.com/questions/1862231/…
$endgroup$
– Hans Lundmark
Dec 11 '18 at 10:44




1




1




$begingroup$
It is essential that the domain is an intervall.
$endgroup$
– gammatester
Dec 11 '18 at 10:45




$begingroup$
It is essential that the domain is an intervall.
$endgroup$
– gammatester
Dec 11 '18 at 10:45












$begingroup$
AH, the glorious $MVT$, cool!
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:46




$begingroup$
AH, the glorious $MVT$, cool!
$endgroup$
– Wesley Strik
Dec 11 '18 at 10:46










3 Answers
3






active

oldest

votes


















2












$begingroup$

Hint: What is the derivative of $F_1-F_2$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 10:49












  • $begingroup$
    And then apply the MVT to the interval.
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 10:49






  • 1




    $begingroup$
    @WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
    $endgroup$
    – Arthur
    Dec 11 '18 at 10:52












  • $begingroup$
    It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
    $endgroup$
    – Wesley Strik
    Dec 11 '18 at 12:20





















1












$begingroup$

Hint:



If $F_1,F_2$ are primitives of $f$, by linearity of differentiation,



$$(F_1(x)-F_2(x))'=F'_1(x)-F'_2(x)=f(x)-f(x)=0.$$



Remains to prove that the antiderivative of $0$ is a constant function.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Let $G(x)= F_1 - F_2.$



    We observe that $G'(x)=0$. We claim this means that $G(x)$ is constant.



    Suppose on the contrary the function is non-constant this means there must exist $a, b in I$ and where $a<b$ such that $G(a) neq G(b)$.



    We consider $[a,b]$. Now We apply the mean value theorem to this interval and the function $G(x)$. We get that there must exist a $c$ such that:
    Which is a contradiction.
    $$G'(c)= frac{G(b)-G(a)}{b-a} land G'(c)=0,$$
    $$ frac{G(b)-G(a)}{b-a}=0 implies G(a)=G(b). $$
    We conclude that $G(x)$ is constant. Thus, $F_2$ and $F_1$ differ by a constant $square$.






    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









      2












      $begingroup$

      Hint: What is the derivative of $F_1-F_2$?






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
        $endgroup$
        – Wesley Strik
        Dec 11 '18 at 10:49












      • $begingroup$
        And then apply the MVT to the interval.
        $endgroup$
        – Wesley Strik
        Dec 11 '18 at 10:49






      • 1




        $begingroup$
        @WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
        $endgroup$
        – Arthur
        Dec 11 '18 at 10:52












      • $begingroup$
        It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
        $endgroup$
        – Wesley Strik
        Dec 11 '18 at 12:20


















      2












      $begingroup$

      Hint: What is the derivative of $F_1-F_2$?






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
        $endgroup$
        – Wesley Strik
        Dec 11 '18 at 10:49












      • $begingroup$
        And then apply the MVT to the interval.
        $endgroup$
        – Wesley Strik
        Dec 11 '18 at 10:49






      • 1




        $begingroup$
        @WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
        $endgroup$
        – Arthur
        Dec 11 '18 at 10:52












      • $begingroup$
        It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
        $endgroup$
        – Wesley Strik
        Dec 11 '18 at 12:20
















      2












      2








      2





      $begingroup$

      Hint: What is the derivative of $F_1-F_2$?






      share|cite|improve this answer









      $endgroup$



      Hint: What is the derivative of $F_1-F_2$?







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 11 '18 at 10:41









      ArthurArthur

      116k7116198




      116k7116198












      • $begingroup$
        Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
        $endgroup$
        – Wesley Strik
        Dec 11 '18 at 10:49












      • $begingroup$
        And then apply the MVT to the interval.
        $endgroup$
        – Wesley Strik
        Dec 11 '18 at 10:49






      • 1




        $begingroup$
        @WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
        $endgroup$
        – Arthur
        Dec 11 '18 at 10:52












      • $begingroup$
        It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
        $endgroup$
        – Wesley Strik
        Dec 11 '18 at 12:20




















      • $begingroup$
        Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
        $endgroup$
        – Wesley Strik
        Dec 11 '18 at 10:49












      • $begingroup$
        And then apply the MVT to the interval.
        $endgroup$
        – Wesley Strik
        Dec 11 '18 at 10:49






      • 1




        $begingroup$
        @WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
        $endgroup$
        – Arthur
        Dec 11 '18 at 10:52












      • $begingroup$
        It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
        $endgroup$
        – Wesley Strik
        Dec 11 '18 at 12:20


















      $begingroup$
      Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 10:49






      $begingroup$
      Zero of course, so we can define a new function $H(x)=F_2 -F_1$ such that $H'(x) =0$
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 10:49














      $begingroup$
      And then apply the MVT to the interval.
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 10:49




      $begingroup$
      And then apply the MVT to the interval.
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 10:49




      1




      1




      $begingroup$
      @WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
      $endgroup$
      – Arthur
      Dec 11 '18 at 10:52






      $begingroup$
      @WesleyStrik I think you're thinking too complicated. If $H'(x)=0$, then $H(x)=C$ for some constant $C$. What does this say about the relationship between $F_1$ and $F_2$?
      $endgroup$
      – Arthur
      Dec 11 '18 at 10:52














      $begingroup$
      It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 12:20






      $begingroup$
      It means the function $F_1 - F_2$ is constant, see my proof below using the MVT.
      $endgroup$
      – Wesley Strik
      Dec 11 '18 at 12:20













      1












      $begingroup$

      Hint:



      If $F_1,F_2$ are primitives of $f$, by linearity of differentiation,



      $$(F_1(x)-F_2(x))'=F'_1(x)-F'_2(x)=f(x)-f(x)=0.$$



      Remains to prove that the antiderivative of $0$ is a constant function.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Hint:



        If $F_1,F_2$ are primitives of $f$, by linearity of differentiation,



        $$(F_1(x)-F_2(x))'=F'_1(x)-F'_2(x)=f(x)-f(x)=0.$$



        Remains to prove that the antiderivative of $0$ is a constant function.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint:



          If $F_1,F_2$ are primitives of $f$, by linearity of differentiation,



          $$(F_1(x)-F_2(x))'=F'_1(x)-F'_2(x)=f(x)-f(x)=0.$$



          Remains to prove that the antiderivative of $0$ is a constant function.






          share|cite|improve this answer









          $endgroup$



          Hint:



          If $F_1,F_2$ are primitives of $f$, by linearity of differentiation,



          $$(F_1(x)-F_2(x))'=F'_1(x)-F'_2(x)=f(x)-f(x)=0.$$



          Remains to prove that the antiderivative of $0$ is a constant function.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 11:02









          Yves DaoustYves Daoust

          128k675227




          128k675227























              0












              $begingroup$

              Let $G(x)= F_1 - F_2.$



              We observe that $G'(x)=0$. We claim this means that $G(x)$ is constant.



              Suppose on the contrary the function is non-constant this means there must exist $a, b in I$ and where $a<b$ such that $G(a) neq G(b)$.



              We consider $[a,b]$. Now We apply the mean value theorem to this interval and the function $G(x)$. We get that there must exist a $c$ such that:
              Which is a contradiction.
              $$G'(c)= frac{G(b)-G(a)}{b-a} land G'(c)=0,$$
              $$ frac{G(b)-G(a)}{b-a}=0 implies G(a)=G(b). $$
              We conclude that $G(x)$ is constant. Thus, $F_2$ and $F_1$ differ by a constant $square$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Let $G(x)= F_1 - F_2.$



                We observe that $G'(x)=0$. We claim this means that $G(x)$ is constant.



                Suppose on the contrary the function is non-constant this means there must exist $a, b in I$ and where $a<b$ such that $G(a) neq G(b)$.



                We consider $[a,b]$. Now We apply the mean value theorem to this interval and the function $G(x)$. We get that there must exist a $c$ such that:
                Which is a contradiction.
                $$G'(c)= frac{G(b)-G(a)}{b-a} land G'(c)=0,$$
                $$ frac{G(b)-G(a)}{b-a}=0 implies G(a)=G(b). $$
                We conclude that $G(x)$ is constant. Thus, $F_2$ and $F_1$ differ by a constant $square$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let $G(x)= F_1 - F_2.$



                  We observe that $G'(x)=0$. We claim this means that $G(x)$ is constant.



                  Suppose on the contrary the function is non-constant this means there must exist $a, b in I$ and where $a<b$ such that $G(a) neq G(b)$.



                  We consider $[a,b]$. Now We apply the mean value theorem to this interval and the function $G(x)$. We get that there must exist a $c$ such that:
                  Which is a contradiction.
                  $$G'(c)= frac{G(b)-G(a)}{b-a} land G'(c)=0,$$
                  $$ frac{G(b)-G(a)}{b-a}=0 implies G(a)=G(b). $$
                  We conclude that $G(x)$ is constant. Thus, $F_2$ and $F_1$ differ by a constant $square$.






                  share|cite|improve this answer











                  $endgroup$



                  Let $G(x)= F_1 - F_2.$



                  We observe that $G'(x)=0$. We claim this means that $G(x)$ is constant.



                  Suppose on the contrary the function is non-constant this means there must exist $a, b in I$ and where $a<b$ such that $G(a) neq G(b)$.



                  We consider $[a,b]$. Now We apply the mean value theorem to this interval and the function $G(x)$. We get that there must exist a $c$ such that:
                  Which is a contradiction.
                  $$G'(c)= frac{G(b)-G(a)}{b-a} land G'(c)=0,$$
                  $$ frac{G(b)-G(a)}{b-a}=0 implies G(a)=G(b). $$
                  We conclude that $G(x)$ is constant. Thus, $F_2$ and $F_1$ differ by a constant $square$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 11 '18 at 12:04

























                  answered Dec 11 '18 at 11:58









                  Wesley StrikWesley Strik

                  2,057423




                  2,057423






























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                      Plaza Victoria

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                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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                      Puebla de Zaragoza

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                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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                      Musa

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                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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