Antiderivative: does it represent any area?












2












$begingroup$


I was told that an antiderivative doesn't represent any area, it is just a family of functions, but today I noticed something quite interesting thing:



If we have $f(x)=2x$, then $F(x)=x^2+C$. The area under $f(x)$ from $0$ to $1$ is $1$, and $F(1)$ is one, the area under $f(x)$ from $0$ to $2$ is $4$, and $F(2)$ is $4$, and so on.



Then I started to think, that when $C=0$, $F(x)$ is the area under $f(x)$ from $0$ to $x$, and tried some others functions like $f(x)=x^2$, $x^3$, and so on.



Then I decided to try $f(x)=sin x$, and I failed in my guess. But then noticed, if I take $C=1$ it works! Finally, I noticed that it works in all cases when $F(0)=f(0)=0$.



So antiderivative is not only a bunch of functions but also some area?



How can I visualize it generally? How can I visualize $C$? Can this area be an endless area in both directions?



Now I study the fundamental theorem of calculus, and at first, I imagined $F(b)-F(a)$ as $(text{area from } -infty text{ or } 0 text{ to } b)-(text{ area from }-infty text{ or } 0 text{ to } a)$. Is it the right visualization?



I will be very grateful if someone could clarify my guesses. Thanks in advance



Upd: thank you all who helped me to solve my problem, i appreciate it so!
To one, who faced the same question: there`s a quite similar question, and some of answers are pretty meaningful, for example, this one:
https://math.stackexchange.com/a/2559329/595919






calculus integration area







share|cite|improve this question











$endgroup$












  • $begingroup$
    "I was told that an antiderivative doesn't represent any area": don't trust them.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 17:58










  • $begingroup$
    So it does represent some area?
    $endgroup$
    – Max Knyazeff
    Dec 18 '18 at 18:00










  • $begingroup$
    This is well-known.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 18:03










  • $begingroup$
    So could u tell me please what area does it represent?
    $endgroup$
    – Max Knyazeff
    Dec 18 '18 at 18:05










  • $begingroup$
    As usual, between the axis, curve and verticals.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 18:09
















2












$begingroup$


I was told that an antiderivative doesn't represent any area, it is just a family of functions, but today I noticed something quite interesting thing:



If we have $f(x)=2x$, then $F(x)=x^2+C$. The area under $f(x)$ from $0$ to $1$ is $1$, and $F(1)$ is one, the area under $f(x)$ from $0$ to $2$ is $4$, and $F(2)$ is $4$, and so on.



Then I started to think, that when $C=0$, $F(x)$ is the area under $f(x)$ from $0$ to $x$, and tried some others functions like $f(x)=x^2$, $x^3$, and so on.



Then I decided to try $f(x)=sin x$, and I failed in my guess. But then noticed, if I take $C=1$ it works! Finally, I noticed that it works in all cases when $F(0)=f(0)=0$.



So antiderivative is not only a bunch of functions but also some area?



How can I visualize it generally? How can I visualize $C$? Can this area be an endless area in both directions?



Now I study the fundamental theorem of calculus, and at first, I imagined $F(b)-F(a)$ as $(text{area from } -infty text{ or } 0 text{ to } b)-(text{ area from }-infty text{ or } 0 text{ to } a)$. Is it the right visualization?



I will be very grateful if someone could clarify my guesses. Thanks in advance



Upd: thank you all who helped me to solve my problem, i appreciate it so!
To one, who faced the same question: there`s a quite similar question, and some of answers are pretty meaningful, for example, this one:
https://math.stackexchange.com/a/2559329/595919






calculus integration area







share|cite|improve this question











$endgroup$












  • $begingroup$
    "I was told that an antiderivative doesn't represent any area": don't trust them.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 17:58










  • $begingroup$
    So it does represent some area?
    $endgroup$
    – Max Knyazeff
    Dec 18 '18 at 18:00










  • $begingroup$
    This is well-known.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 18:03










  • $begingroup$
    So could u tell me please what area does it represent?
    $endgroup$
    – Max Knyazeff
    Dec 18 '18 at 18:05










  • $begingroup$
    As usual, between the axis, curve and verticals.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 18:09














2












2








2





$begingroup$


I was told that an antiderivative doesn't represent any area, it is just a family of functions, but today I noticed something quite interesting thing:



If we have $f(x)=2x$, then $F(x)=x^2+C$. The area under $f(x)$ from $0$ to $1$ is $1$, and $F(1)$ is one, the area under $f(x)$ from $0$ to $2$ is $4$, and $F(2)$ is $4$, and so on.



Then I started to think, that when $C=0$, $F(x)$ is the area under $f(x)$ from $0$ to $x$, and tried some others functions like $f(x)=x^2$, $x^3$, and so on.



Then I decided to try $f(x)=sin x$, and I failed in my guess. But then noticed, if I take $C=1$ it works! Finally, I noticed that it works in all cases when $F(0)=f(0)=0$.



So antiderivative is not only a bunch of functions but also some area?



How can I visualize it generally? How can I visualize $C$? Can this area be an endless area in both directions?



Now I study the fundamental theorem of calculus, and at first, I imagined $F(b)-F(a)$ as $(text{area from } -infty text{ or } 0 text{ to } b)-(text{ area from }-infty text{ or } 0 text{ to } a)$. Is it the right visualization?



I will be very grateful if someone could clarify my guesses. Thanks in advance



Upd: thank you all who helped me to solve my problem, i appreciate it so!
To one, who faced the same question: there`s a quite similar question, and some of answers are pretty meaningful, for example, this one:
https://math.stackexchange.com/a/2559329/595919






calculus integration area







share|cite|improve this question











$endgroup$




I was told that an antiderivative doesn't represent any area, it is just a family of functions, but today I noticed something quite interesting thing:



If we have $f(x)=2x$, then $F(x)=x^2+C$. The area under $f(x)$ from $0$ to $1$ is $1$, and $F(1)$ is one, the area under $f(x)$ from $0$ to $2$ is $4$, and $F(2)$ is $4$, and so on.



Then I started to think, that when $C=0$, $F(x)$ is the area under $f(x)$ from $0$ to $x$, and tried some others functions like $f(x)=x^2$, $x^3$, and so on.



Then I decided to try $f(x)=sin x$, and I failed in my guess. But then noticed, if I take $C=1$ it works! Finally, I noticed that it works in all cases when $F(0)=f(0)=0$.



So antiderivative is not only a bunch of functions but also some area?



How can I visualize it generally? How can I visualize $C$? Can this area be an endless area in both directions?



Now I study the fundamental theorem of calculus, and at first, I imagined $F(b)-F(a)$ as $(text{area from } -infty text{ or } 0 text{ to } b)-(text{ area from }-infty text{ or } 0 text{ to } a)$. Is it the right visualization?



I will be very grateful if someone could clarify my guesses. Thanks in advance



Upd: thank you all who helped me to solve my problem, i appreciate it so!
To one, who faced the same question: there`s a quite similar question, and some of answers are pretty meaningful, for example, this one:
https://math.stackexchange.com/a/2559329/595919






calculus integration area




calculus integration area






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 15:30







Max Knyazeff

















asked Dec 18 '18 at 17:06









Max KnyazeffMax Knyazeff

134




134












  • $begingroup$
    "I was told that an antiderivative doesn't represent any area": don't trust them.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 17:58










  • $begingroup$
    So it does represent some area?
    $endgroup$
    – Max Knyazeff
    Dec 18 '18 at 18:00










  • $begingroup$
    This is well-known.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 18:03










  • $begingroup$
    So could u tell me please what area does it represent?
    $endgroup$
    – Max Knyazeff
    Dec 18 '18 at 18:05










  • $begingroup$
    As usual, between the axis, curve and verticals.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 18:09


















  • $begingroup$
    "I was told that an antiderivative doesn't represent any area": don't trust them.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 17:58










  • $begingroup$
    So it does represent some area?
    $endgroup$
    – Max Knyazeff
    Dec 18 '18 at 18:00










  • $begingroup$
    This is well-known.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 18:03










  • $begingroup$
    So could u tell me please what area does it represent?
    $endgroup$
    – Max Knyazeff
    Dec 18 '18 at 18:05










  • $begingroup$
    As usual, between the axis, curve and verticals.
    $endgroup$
    – Yves Daoust
    Dec 18 '18 at 18:09
















$begingroup$
"I was told that an antiderivative doesn't represent any area": don't trust them.
$endgroup$
– Yves Daoust
Dec 18 '18 at 17:58




$begingroup$
"I was told that an antiderivative doesn't represent any area": don't trust them.
$endgroup$
– Yves Daoust
Dec 18 '18 at 17:58












$begingroup$
So it does represent some area?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:00




$begingroup$
So it does represent some area?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:00












$begingroup$
This is well-known.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:03




$begingroup$
This is well-known.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:03












$begingroup$
So could u tell me please what area does it represent?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:05




$begingroup$
So could u tell me please what area does it represent?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:05












$begingroup$
As usual, between the axis, curve and verticals.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:09




$begingroup$
As usual, between the axis, curve and verticals.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:09










2 Answers
2






active

oldest

votes


















0












$begingroup$

Let $fcolon Bbb RtoBbb R$ be a function that has an anti-derivative $F$. Let $a<b$ be real numbers, then you can use $F$ to calculate the (oriented) area between the vertical lines $x=a$ and $x=b$ that sits between the $x$-axis and the graph of $f$:




oriented area




More precisely we have
$$
color{blue}{text{blue area}} - color{orange}{text{orange area}} = int_a^b f(x),mathrm dx = F(b) - F(a).
$$



Note that the choice of constant in the anti-derivative doesn't matter here, since it cancels when calculating the difference $F(b)-F(a)$.



Letting $a=0$ and choosing the constant such that $F(0)=0$ you get that $F(a)$ is the area under the curve between $x=0$ and $x=a$, as you figured out.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Firstly, you can visualize an anti-derivative as a curve itself, whose derivative yields the function with which you are concerned. This new curve can be translated up and down without affecting the slope, therefore you can visualize C as an ambiguous translation of that curve up and down.



    As for your understanding of the Fundamental Theorem of Calculus, I would not interpret it as two areas, but rather as the area between a and b itself, which can be found using your anti-derivative function.



    I hope that helps.






    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









      0












      $begingroup$

      Let $fcolon Bbb RtoBbb R$ be a function that has an anti-derivative $F$. Let $a<b$ be real numbers, then you can use $F$ to calculate the (oriented) area between the vertical lines $x=a$ and $x=b$ that sits between the $x$-axis and the graph of $f$:




      oriented area




      More precisely we have
      $$
      color{blue}{text{blue area}} - color{orange}{text{orange area}} = int_a^b f(x),mathrm dx = F(b) - F(a).
      $$



      Note that the choice of constant in the anti-derivative doesn't matter here, since it cancels when calculating the difference $F(b)-F(a)$.



      Letting $a=0$ and choosing the constant such that $F(0)=0$ you get that $F(a)$ is the area under the curve between $x=0$ and $x=a$, as you figured out.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Let $fcolon Bbb RtoBbb R$ be a function that has an anti-derivative $F$. Let $a<b$ be real numbers, then you can use $F$ to calculate the (oriented) area between the vertical lines $x=a$ and $x=b$ that sits between the $x$-axis and the graph of $f$:




        oriented area




        More precisely we have
        $$
        color{blue}{text{blue area}} - color{orange}{text{orange area}} = int_a^b f(x),mathrm dx = F(b) - F(a).
        $$



        Note that the choice of constant in the anti-derivative doesn't matter here, since it cancels when calculating the difference $F(b)-F(a)$.



        Letting $a=0$ and choosing the constant such that $F(0)=0$ you get that $F(a)$ is the area under the curve between $x=0$ and $x=a$, as you figured out.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Let $fcolon Bbb RtoBbb R$ be a function that has an anti-derivative $F$. Let $a<b$ be real numbers, then you can use $F$ to calculate the (oriented) area between the vertical lines $x=a$ and $x=b$ that sits between the $x$-axis and the graph of $f$:




          oriented area




          More precisely we have
          $$
          color{blue}{text{blue area}} - color{orange}{text{orange area}} = int_a^b f(x),mathrm dx = F(b) - F(a).
          $$



          Note that the choice of constant in the anti-derivative doesn't matter here, since it cancels when calculating the difference $F(b)-F(a)$.



          Letting $a=0$ and choosing the constant such that $F(0)=0$ you get that $F(a)$ is the area under the curve between $x=0$ and $x=a$, as you figured out.






          share|cite|improve this answer









          $endgroup$



          Let $fcolon Bbb RtoBbb R$ be a function that has an anti-derivative $F$. Let $a<b$ be real numbers, then you can use $F$ to calculate the (oriented) area between the vertical lines $x=a$ and $x=b$ that sits between the $x$-axis and the graph of $f$:




          oriented area




          More precisely we have
          $$
          color{blue}{text{blue area}} - color{orange}{text{orange area}} = int_a^b f(x),mathrm dx = F(b) - F(a).
          $$



          Note that the choice of constant in the anti-derivative doesn't matter here, since it cancels when calculating the difference $F(b)-F(a)$.



          Letting $a=0$ and choosing the constant such that $F(0)=0$ you get that $F(a)$ is the area under the curve between $x=0$ and $x=a$, as you figured out.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 15:45









          ChristophChristoph

          12.5k1642




          12.5k1642























              0












              $begingroup$

              Firstly, you can visualize an anti-derivative as a curve itself, whose derivative yields the function with which you are concerned. This new curve can be translated up and down without affecting the slope, therefore you can visualize C as an ambiguous translation of that curve up and down.



              As for your understanding of the Fundamental Theorem of Calculus, I would not interpret it as two areas, but rather as the area between a and b itself, which can be found using your anti-derivative function.



              I hope that helps.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Firstly, you can visualize an anti-derivative as a curve itself, whose derivative yields the function with which you are concerned. This new curve can be translated up and down without affecting the slope, therefore you can visualize C as an ambiguous translation of that curve up and down.



                As for your understanding of the Fundamental Theorem of Calculus, I would not interpret it as two areas, but rather as the area between a and b itself, which can be found using your anti-derivative function.



                I hope that helps.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Firstly, you can visualize an anti-derivative as a curve itself, whose derivative yields the function with which you are concerned. This new curve can be translated up and down without affecting the slope, therefore you can visualize C as an ambiguous translation of that curve up and down.



                  As for your understanding of the Fundamental Theorem of Calculus, I would not interpret it as two areas, but rather as the area between a and b itself, which can be found using your anti-derivative function.



                  I hope that helps.






                  share|cite|improve this answer









                  $endgroup$



                  Firstly, you can visualize an anti-derivative as a curve itself, whose derivative yields the function with which you are concerned. This new curve can be translated up and down without affecting the slope, therefore you can visualize C as an ambiguous translation of that curve up and down.



                  As for your understanding of the Fundamental Theorem of Calculus, I would not interpret it as two areas, but rather as the area between a and b itself, which can be found using your anti-derivative function.



                  I hope that helps.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 18 '18 at 17:49









                  RJJBRJJB

                  132




                  132






























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