Derivative of indefinite integral vs. definite












2












$begingroup$


So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$.



But if I have $frac{d}{dx} int f(x)dx$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?



And if I have the $int frac{d}{dx} f(x)dx$ - it is just the function as well, correct?



Thanks






integration definite-integrals indefinite-integrals







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  • 1




    $begingroup$
    You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
    $endgroup$
    – Ethan Bolker
    Dec 16 '18 at 1:20










  • $begingroup$
    "So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
    $endgroup$
    – user587192
    Dec 16 '18 at 2:35
















2












$begingroup$


So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$.



But if I have $frac{d}{dx} int f(x)dx$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?



And if I have the $int frac{d}{dx} f(x)dx$ - it is just the function as well, correct?



Thanks






integration definite-integrals indefinite-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
    $endgroup$
    – Ethan Bolker
    Dec 16 '18 at 1:20










  • $begingroup$
    "So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
    $endgroup$
    – user587192
    Dec 16 '18 at 2:35














2












2








2





$begingroup$


So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$.



But if I have $frac{d}{dx} int f(x)dx$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?



And if I have the $int frac{d}{dx} f(x)dx$ - it is just the function as well, correct?



Thanks






integration definite-integrals indefinite-integrals







share|cite|improve this question











$endgroup$




So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$.



But if I have $frac{d}{dx} int f(x)dx$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?



And if I have the $int frac{d}{dx} f(x)dx$ - it is just the function as well, correct?



Thanks






integration definite-integrals indefinite-integrals




integration definite-integrals indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 1:20









Eevee Trainer

6,0071936




6,0071936










asked Dec 16 '18 at 1:13









J.W.J.W.

544




544








  • 1




    $begingroup$
    You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
    $endgroup$
    – Ethan Bolker
    Dec 16 '18 at 1:20










  • $begingroup$
    "So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
    $endgroup$
    – user587192
    Dec 16 '18 at 2:35














  • 1




    $begingroup$
    You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
    $endgroup$
    – Ethan Bolker
    Dec 16 '18 at 1:20










  • $begingroup$
    "So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
    $endgroup$
    – user587192
    Dec 16 '18 at 2:35








1




1




$begingroup$
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
$endgroup$
– Ethan Bolker
Dec 16 '18 at 1:20




$begingroup$
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
$endgroup$
– Ethan Bolker
Dec 16 '18 at 1:20












$begingroup$
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
$endgroup$
– user587192
Dec 16 '18 at 2:35




$begingroup$
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
$endgroup$
– user587192
Dec 16 '18 at 2:35










2 Answers
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$begingroup$

Not quite. The arbitrary constant from integration changes it up. Explicitly,



$$frac{d}{dx} int f(x)dx = f(x)$$



$$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$




An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.



Then



$$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$



but



$$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.



    For the second question, you have to add the Constant of Integration to the function because of the rules of integration.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Not quite. The arbitrary constant from integration changes it up. Explicitly,



      $$frac{d}{dx} int f(x)dx = f(x)$$



      $$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$




      An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.



      Then



      $$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$



      but



      $$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Not quite. The arbitrary constant from integration changes it up. Explicitly,



        $$frac{d}{dx} int f(x)dx = f(x)$$



        $$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$




        An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.



        Then



        $$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$



        but



        $$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Not quite. The arbitrary constant from integration changes it up. Explicitly,



          $$frac{d}{dx} int f(x)dx = f(x)$$



          $$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$




          An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.



          Then



          $$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$



          but



          $$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$






          share|cite|improve this answer









          $endgroup$



          Not quite. The arbitrary constant from integration changes it up. Explicitly,



          $$frac{d}{dx} int f(x)dx = f(x)$$



          $$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$




          An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.



          Then



          $$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$



          but



          $$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 1:18









          Eevee TrainerEevee Trainer

          6,0071936




          6,0071936























              0












              $begingroup$

              For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.



              For the second question, you have to add the Constant of Integration to the function because of the rules of integration.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.



                For the second question, you have to add the Constant of Integration to the function because of the rules of integration.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.



                  For the second question, you have to add the Constant of Integration to the function because of the rules of integration.






                  share|cite|improve this answer









                  $endgroup$



                  For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.



                  For the second question, you have to add the Constant of Integration to the function because of the rules of integration.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 16 '18 at 1:19









                  Michael WangMichael Wang

                  186115




                  186115






























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