Why does using u-substitution give a different answer?












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I am working on a problem that required finding $int (-7-y)^2dy$. Instead of foiling it out, I just used u-substitution to find the antiderivative. ($u=-7-y; u'=-1$). I got an antiderivative of $frac{-(-7-y)^3}{3}$. However, if you factor out this antiderivative you end up with a constant of $frac {343}{3}$ that's not there if you foil then integrate. Why would the +c be defined for u-sub but not for foiling then integrating?






calculus integration







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  • 2




    $begingroup$
    The first thing I would do is factor and kill those two minus signs...
    $endgroup$
    – Matthias
    Dec 15 '18 at 20:54
















2












$begingroup$


I am working on a problem that required finding $int (-7-y)^2dy$. Instead of foiling it out, I just used u-substitution to find the antiderivative. ($u=-7-y; u'=-1$). I got an antiderivative of $frac{-(-7-y)^3}{3}$. However, if you factor out this antiderivative you end up with a constant of $frac {343}{3}$ that's not there if you foil then integrate. Why would the +c be defined for u-sub but not for foiling then integrating?






calculus integration







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The first thing I would do is factor and kill those two minus signs...
    $endgroup$
    – Matthias
    Dec 15 '18 at 20:54














2












2








2





$begingroup$


I am working on a problem that required finding $int (-7-y)^2dy$. Instead of foiling it out, I just used u-substitution to find the antiderivative. ($u=-7-y; u'=-1$). I got an antiderivative of $frac{-(-7-y)^3}{3}$. However, if you factor out this antiderivative you end up with a constant of $frac {343}{3}$ that's not there if you foil then integrate. Why would the +c be defined for u-sub but not for foiling then integrating?






calculus integration







share|cite|improve this question









$endgroup$




I am working on a problem that required finding $int (-7-y)^2dy$. Instead of foiling it out, I just used u-substitution to find the antiderivative. ($u=-7-y; u'=-1$). I got an antiderivative of $frac{-(-7-y)^3}{3}$. However, if you factor out this antiderivative you end up with a constant of $frac {343}{3}$ that's not there if you foil then integrate. Why would the +c be defined for u-sub but not for foiling then integrating?






calculus integration




calculus integration






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asked Dec 15 '18 at 20:45









Null SparkNull Spark

1234




1234








  • 2




    $begingroup$
    The first thing I would do is factor and kill those two minus signs...
    $endgroup$
    – Matthias
    Dec 15 '18 at 20:54














  • 2




    $begingroup$
    The first thing I would do is factor and kill those two minus signs...
    $endgroup$
    – Matthias
    Dec 15 '18 at 20:54








2




2




$begingroup$
The first thing I would do is factor and kill those two minus signs...
$endgroup$
– Matthias
Dec 15 '18 at 20:54




$begingroup$
The first thing I would do is factor and kill those two minus signs...
$endgroup$
– Matthias
Dec 15 '18 at 20:54










2 Answers
2






active

oldest

votes


















1












$begingroup$

Notice that with $u$ substitution you get
$$frac{-(-7-y)^3}{3}+C=frac{y^3}{3}+7y^2+49y+frac{343}{3}+C$$
(don't forget the $+C$). If you foil it then integrate you would get
$$int (y^2+14y+49)dy=frac{y^3}{3}+7y^2+49y+C'$$
where $C'$ is some constant. Notice that these two anti-derivates still only vary by a constant, namely $frac{343}{3}+C-C'$. Should you wish, you could simply relabel $frac{343}{3}+C$ as $C''$ and the two answers would look symbolically the same.


At the end of the day we still have
$$left{frac{-(-7-y)^3}{3}+C; :; Cinmathbb{R}right}=left{frac{y^3}{3}+7y^2+49y+C'; :; C'inmathbb{R}right}$$
so these two ways of writing it give us the same set of solutions.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Remember: Indefinite integration yields a family of functions that differ by only a constant.



    $$int f(x) ,dx = F(x) + C$$



    What $C$ happens to be is unknown to us; we just know that if we differentiate $F(x)$, we get $f(x).$




    Let's see how this works in a simpler example:



    Suppose we have $$int (x-1), dx$$



    Method 1: Let $u = x-1$. Then $du = dx.$, and we have $$int u,du = frac{u^2}2 + C= frac{(x-1)^2}2 + c = frac 12(x^2-2x + 1) + c = frac 12 x -x + left(frac 12 + cright) tag 1$$



    Method 2: integrate directly: $$int (x-1), dx = frac{x^2}2 - x + Ctag 2$$



    $(1)$ and $(2)$ differ only by a constant: We can make them "equal" by setting $C = frac 12 + c$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The constant $dfrac {343}3$ is not the "constant of integration" from the u-substitution method you used. The constant of integration from that method is $dfrac{343}3 + c = C$, where $C$ is a place holder, which upon differentiation, disappears (and is the integrand).
      $endgroup$
      – Namaste
      Dec 15 '18 at 21:11













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    2 Answers
    2






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    2 Answers
    2






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    active

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    active

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    1












    $begingroup$

    Notice that with $u$ substitution you get
    $$frac{-(-7-y)^3}{3}+C=frac{y^3}{3}+7y^2+49y+frac{343}{3}+C$$
    (don't forget the $+C$). If you foil it then integrate you would get
    $$int (y^2+14y+49)dy=frac{y^3}{3}+7y^2+49y+C'$$
    where $C'$ is some constant. Notice that these two anti-derivates still only vary by a constant, namely $frac{343}{3}+C-C'$. Should you wish, you could simply relabel $frac{343}{3}+C$ as $C''$ and the two answers would look symbolically the same.


    At the end of the day we still have
    $$left{frac{-(-7-y)^3}{3}+C; :; Cinmathbb{R}right}=left{frac{y^3}{3}+7y^2+49y+C'; :; C'inmathbb{R}right}$$
    so these two ways of writing it give us the same set of solutions.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Notice that with $u$ substitution you get
      $$frac{-(-7-y)^3}{3}+C=frac{y^3}{3}+7y^2+49y+frac{343}{3}+C$$
      (don't forget the $+C$). If you foil it then integrate you would get
      $$int (y^2+14y+49)dy=frac{y^3}{3}+7y^2+49y+C'$$
      where $C'$ is some constant. Notice that these two anti-derivates still only vary by a constant, namely $frac{343}{3}+C-C'$. Should you wish, you could simply relabel $frac{343}{3}+C$ as $C''$ and the two answers would look symbolically the same.


      At the end of the day we still have
      $$left{frac{-(-7-y)^3}{3}+C; :; Cinmathbb{R}right}=left{frac{y^3}{3}+7y^2+49y+C'; :; C'inmathbb{R}right}$$
      so these two ways of writing it give us the same set of solutions.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Notice that with $u$ substitution you get
        $$frac{-(-7-y)^3}{3}+C=frac{y^3}{3}+7y^2+49y+frac{343}{3}+C$$
        (don't forget the $+C$). If you foil it then integrate you would get
        $$int (y^2+14y+49)dy=frac{y^3}{3}+7y^2+49y+C'$$
        where $C'$ is some constant. Notice that these two anti-derivates still only vary by a constant, namely $frac{343}{3}+C-C'$. Should you wish, you could simply relabel $frac{343}{3}+C$ as $C''$ and the two answers would look symbolically the same.


        At the end of the day we still have
        $$left{frac{-(-7-y)^3}{3}+C; :; Cinmathbb{R}right}=left{frac{y^3}{3}+7y^2+49y+C'; :; C'inmathbb{R}right}$$
        so these two ways of writing it give us the same set of solutions.






        share|cite|improve this answer









        $endgroup$



        Notice that with $u$ substitution you get
        $$frac{-(-7-y)^3}{3}+C=frac{y^3}{3}+7y^2+49y+frac{343}{3}+C$$
        (don't forget the $+C$). If you foil it then integrate you would get
        $$int (y^2+14y+49)dy=frac{y^3}{3}+7y^2+49y+C'$$
        where $C'$ is some constant. Notice that these two anti-derivates still only vary by a constant, namely $frac{343}{3}+C-C'$. Should you wish, you could simply relabel $frac{343}{3}+C$ as $C''$ and the two answers would look symbolically the same.


        At the end of the day we still have
        $$left{frac{-(-7-y)^3}{3}+C; :; Cinmathbb{R}right}=left{frac{y^3}{3}+7y^2+49y+C'; :; C'inmathbb{R}right}$$
        so these two ways of writing it give us the same set of solutions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 20:52









        Will FisherWill Fisher

        4,04811032




        4,04811032























            1












            $begingroup$

            Remember: Indefinite integration yields a family of functions that differ by only a constant.



            $$int f(x) ,dx = F(x) + C$$



            What $C$ happens to be is unknown to us; we just know that if we differentiate $F(x)$, we get $f(x).$




            Let's see how this works in a simpler example:



            Suppose we have $$int (x-1), dx$$



            Method 1: Let $u = x-1$. Then $du = dx.$, and we have $$int u,du = frac{u^2}2 + C= frac{(x-1)^2}2 + c = frac 12(x^2-2x + 1) + c = frac 12 x -x + left(frac 12 + cright) tag 1$$



            Method 2: integrate directly: $$int (x-1), dx = frac{x^2}2 - x + Ctag 2$$



            $(1)$ and $(2)$ differ only by a constant: We can make them "equal" by setting $C = frac 12 + c$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The constant $dfrac {343}3$ is not the "constant of integration" from the u-substitution method you used. The constant of integration from that method is $dfrac{343}3 + c = C$, where $C$ is a place holder, which upon differentiation, disappears (and is the integrand).
              $endgroup$
              – Namaste
              Dec 15 '18 at 21:11


















            1












            $begingroup$

            Remember: Indefinite integration yields a family of functions that differ by only a constant.



            $$int f(x) ,dx = F(x) + C$$



            What $C$ happens to be is unknown to us; we just know that if we differentiate $F(x)$, we get $f(x).$




            Let's see how this works in a simpler example:



            Suppose we have $$int (x-1), dx$$



            Method 1: Let $u = x-1$. Then $du = dx.$, and we have $$int u,du = frac{u^2}2 + C= frac{(x-1)^2}2 + c = frac 12(x^2-2x + 1) + c = frac 12 x -x + left(frac 12 + cright) tag 1$$



            Method 2: integrate directly: $$int (x-1), dx = frac{x^2}2 - x + Ctag 2$$



            $(1)$ and $(2)$ differ only by a constant: We can make them "equal" by setting $C = frac 12 + c$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The constant $dfrac {343}3$ is not the "constant of integration" from the u-substitution method you used. The constant of integration from that method is $dfrac{343}3 + c = C$, where $C$ is a place holder, which upon differentiation, disappears (and is the integrand).
              $endgroup$
              – Namaste
              Dec 15 '18 at 21:11
















            1












            1








            1





            $begingroup$

            Remember: Indefinite integration yields a family of functions that differ by only a constant.



            $$int f(x) ,dx = F(x) + C$$



            What $C$ happens to be is unknown to us; we just know that if we differentiate $F(x)$, we get $f(x).$




            Let's see how this works in a simpler example:



            Suppose we have $$int (x-1), dx$$



            Method 1: Let $u = x-1$. Then $du = dx.$, and we have $$int u,du = frac{u^2}2 + C= frac{(x-1)^2}2 + c = frac 12(x^2-2x + 1) + c = frac 12 x -x + left(frac 12 + cright) tag 1$$



            Method 2: integrate directly: $$int (x-1), dx = frac{x^2}2 - x + Ctag 2$$



            $(1)$ and $(2)$ differ only by a constant: We can make them "equal" by setting $C = frac 12 + c$.






            share|cite|improve this answer











            $endgroup$



            Remember: Indefinite integration yields a family of functions that differ by only a constant.



            $$int f(x) ,dx = F(x) + C$$



            What $C$ happens to be is unknown to us; we just know that if we differentiate $F(x)$, we get $f(x).$




            Let's see how this works in a simpler example:



            Suppose we have $$int (x-1), dx$$



            Method 1: Let $u = x-1$. Then $du = dx.$, and we have $$int u,du = frac{u^2}2 + C= frac{(x-1)^2}2 + c = frac 12(x^2-2x + 1) + c = frac 12 x -x + left(frac 12 + cright) tag 1$$



            Method 2: integrate directly: $$int (x-1), dx = frac{x^2}2 - x + Ctag 2$$



            $(1)$ and $(2)$ differ only by a constant: We can make them "equal" by setting $C = frac 12 + c$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 15 '18 at 21:24

























            answered Dec 15 '18 at 20:52









            NamasteNamaste

            1




            1












            • $begingroup$
              The constant $dfrac {343}3$ is not the "constant of integration" from the u-substitution method you used. The constant of integration from that method is $dfrac{343}3 + c = C$, where $C$ is a place holder, which upon differentiation, disappears (and is the integrand).
              $endgroup$
              – Namaste
              Dec 15 '18 at 21:11




















            • $begingroup$
              The constant $dfrac {343}3$ is not the "constant of integration" from the u-substitution method you used. The constant of integration from that method is $dfrac{343}3 + c = C$, where $C$ is a place holder, which upon differentiation, disappears (and is the integrand).
              $endgroup$
              – Namaste
              Dec 15 '18 at 21:11


















            $begingroup$
            The constant $dfrac {343}3$ is not the "constant of integration" from the u-substitution method you used. The constant of integration from that method is $dfrac{343}3 + c = C$, where $C$ is a place holder, which upon differentiation, disappears (and is the integrand).
            $endgroup$
            – Namaste
            Dec 15 '18 at 21:11






            $begingroup$
            The constant $dfrac {343}3$ is not the "constant of integration" from the u-substitution method you used. The constant of integration from that method is $dfrac{343}3 + c = C$, where $C$ is a place holder, which upon differentiation, disappears (and is the integrand).
            $endgroup$
            – Namaste
            Dec 15 '18 at 21:11




















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