Derivative of indefinite integral vs. definite











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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




    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
    – Ethan Bolker
    6 hours ago










  • "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.
    – user587192
    4 hours ago















up vote
2
down vote

favorite












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




















  • 1




    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
    – Ethan Bolker
    6 hours ago










  • "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.
    – user587192
    4 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











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















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




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edited 6 hours ago









Eevee Trainer

3,115224




3,115224










asked 6 hours ago









Josh White

543




543








  • 1




    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
    – Ethan Bolker
    6 hours ago










  • "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.
    – user587192
    4 hours ago














  • 1




    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
    – Ethan Bolker
    6 hours ago










  • "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.
    – user587192
    4 hours ago








1




1




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
– Ethan Bolker
6 hours ago




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
– Ethan Bolker
6 hours ago












"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.
– user587192
4 hours ago




"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.
– user587192
4 hours ago










2 Answers
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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



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






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    0
    down vote













    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

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      2 Answers
      2






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      up vote
      4
      down vote













      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

























        up vote
        4
        down vote













        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























          up vote
          4
          down vote










          up vote
          4
          down vote









          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












          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 6 hours ago









          Eevee Trainer

          3,115224




          3,115224






















              up vote
              0
              down vote













              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

























                up vote
                0
                down vote













                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























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  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












                  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 6 hours ago









                  Michael Wang

                  188




                  188






























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