Proof of exponential function integral












2














I fully apologize if this is a duplicate, but I really can’t find the answer to this online here. But a proof of the rule



$$int a^x dx= frac{a^x}{ln(a)}+C$$



would be really helpful. I need it for something I’m doing. This is all I need. I have no sources on the problem as I want the proof myself, but I use this rule so much that it struck me as second nature when I wanted to solve it! Any possible answers would be appreciated.






calculus integration proof-writing exponential-function







share|cite|improve this question




















  • 5




    It's quite simple. $(a^x)' = a^xcdot ln(a)$, and the rule follows.
    – Jakobian
    Dec 16 at 23:21










  • @Jakobian : The only real answer. +1.
    – MPW
    Dec 16 at 23:46
















2














I fully apologize if this is a duplicate, but I really can’t find the answer to this online here. But a proof of the rule



$$int a^x dx= frac{a^x}{ln(a)}+C$$



would be really helpful. I need it for something I’m doing. This is all I need. I have no sources on the problem as I want the proof myself, but I use this rule so much that it struck me as second nature when I wanted to solve it! Any possible answers would be appreciated.






calculus integration proof-writing exponential-function







share|cite|improve this question




















  • 5




    It's quite simple. $(a^x)' = a^xcdot ln(a)$, and the rule follows.
    – Jakobian
    Dec 16 at 23:21










  • @Jakobian : The only real answer. +1.
    – MPW
    Dec 16 at 23:46














2












2








2


0





I fully apologize if this is a duplicate, but I really can’t find the answer to this online here. But a proof of the rule



$$int a^x dx= frac{a^x}{ln(a)}+C$$



would be really helpful. I need it for something I’m doing. This is all I need. I have no sources on the problem as I want the proof myself, but I use this rule so much that it struck me as second nature when I wanted to solve it! Any possible answers would be appreciated.






calculus integration proof-writing exponential-function







share|cite|improve this question















I fully apologize if this is a duplicate, but I really can’t find the answer to this online here. But a proof of the rule



$$int a^x dx= frac{a^x}{ln(a)}+C$$



would be really helpful. I need it for something I’m doing. This is all I need. I have no sources on the problem as I want the proof myself, but I use this rule so much that it struck me as second nature when I wanted to solve it! Any possible answers would be appreciated.






calculus integration proof-writing exponential-function




calculus integration proof-writing exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 at 23:54

























asked Dec 16 at 23:16









Math Bob

138




138








  • 5




    It's quite simple. $(a^x)' = a^xcdot ln(a)$, and the rule follows.
    – Jakobian
    Dec 16 at 23:21










  • @Jakobian : The only real answer. +1.
    – MPW
    Dec 16 at 23:46














  • 5




    It's quite simple. $(a^x)' = a^xcdot ln(a)$, and the rule follows.
    – Jakobian
    Dec 16 at 23:21










  • @Jakobian : The only real answer. +1.
    – MPW
    Dec 16 at 23:46








5




5




It's quite simple. $(a^x)' = a^xcdot ln(a)$, and the rule follows.
– Jakobian
Dec 16 at 23:21




It's quite simple. $(a^x)' = a^xcdot ln(a)$, and the rule follows.
– Jakobian
Dec 16 at 23:21












@Jakobian : The only real answer. +1.
– MPW
Dec 16 at 23:46




@Jakobian : The only real answer. +1.
– MPW
Dec 16 at 23:46










3 Answers
3






active

oldest

votes


















4














I assume it's kosher to use the exponential integral with base $e$, i.e. $int e^x dx = e^x +C$? Or, more generally, for a constant $k$,



$$int e^{kx}dx = frac{1}{k}e^{kx}+C$$



If so, then note:



$$int a^x dx = int e^{ln(a^x)} dx = int e^{x ln(a)}dx = frac{1}{ln(a)}e^{x ln(a)}+C= frac{1}{ln(a)}e^{ln(a^x)}+C= frac{a^x}{ln(a)}+C$$



This is mostly just manipulation of various logarithm properties: namely,



$$e^{ln(x)} = x$$
$$ln(a^b) = b ln(a)$$



Also, a nitpick: the integral in your question, OP, needs a $+C$ after it, since indefinite integration introduces an arbitrary constant.






share|cite|improve this answer





















  • OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
    – Math Bob
    Dec 16 at 23:51












  • Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
    – Eevee Trainer
    Dec 17 at 0:18










  • Okay. I have taken this in so thanks for the answer!
    – Math Bob
    Dec 17 at 4:04



















1














Well this one can be found within every good integration table $($e.g. take at look at this$)$. Anyway it is actually not that hard to compute therefore I will demonstrate it for you. First of all the basic property of any exponential function with a basis $a$ is that it can be rewritten in terms of $e$ in the following way



$$a^x=left(e^{ln(a)}right)^x=e^{xln(a)}$$



Now we know that $e^x$ remains $e^x$ after integration aswell as after differentiation. Adding a constant $c$ before the $x$ within the exponent yields to



$$frac d{dx}e^{cx}=ce^{cx}text{ and }int e^{cx}dx=frac1ce^{cx}+k$$



From hereon we are basically done since $ln(a)$ can be seens as a constant while integrating. So plugging this together leads to



$$int a^x dx=int e^{xln(a)}dx=frac1{ln(a)}e^{xln(a)}+k=frac{a^x}{ln(a)}+k$$




$$int a^xdx=frac{a^x}{ln(a)}+k$$







share|cite|improve this answer





























    1














    Just differentiate the right hand side and see what you get. Note that
    $$
    frac{d}{dx}left(frac{a^x}{ln a}right)=frac{1}{ln a}frac{d}{dx}(e^{xln a})=frac{1}{ln a}times e^{xln a}times ln a=a^x
    $$
    where we used the chain rule in the second equality.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      I assume it's kosher to use the exponential integral with base $e$, i.e. $int e^x dx = e^x +C$? Or, more generally, for a constant $k$,



      $$int e^{kx}dx = frac{1}{k}e^{kx}+C$$



      If so, then note:



      $$int a^x dx = int e^{ln(a^x)} dx = int e^{x ln(a)}dx = frac{1}{ln(a)}e^{x ln(a)}+C= frac{1}{ln(a)}e^{ln(a^x)}+C= frac{a^x}{ln(a)}+C$$



      This is mostly just manipulation of various logarithm properties: namely,



      $$e^{ln(x)} = x$$
      $$ln(a^b) = b ln(a)$$



      Also, a nitpick: the integral in your question, OP, needs a $+C$ after it, since indefinite integration introduces an arbitrary constant.






      share|cite|improve this answer





















      • OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
        – Math Bob
        Dec 16 at 23:51












      • Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
        – Eevee Trainer
        Dec 17 at 0:18










      • Okay. I have taken this in so thanks for the answer!
        – Math Bob
        Dec 17 at 4:04
















      4














      I assume it's kosher to use the exponential integral with base $e$, i.e. $int e^x dx = e^x +C$? Or, more generally, for a constant $k$,



      $$int e^{kx}dx = frac{1}{k}e^{kx}+C$$



      If so, then note:



      $$int a^x dx = int e^{ln(a^x)} dx = int e^{x ln(a)}dx = frac{1}{ln(a)}e^{x ln(a)}+C= frac{1}{ln(a)}e^{ln(a^x)}+C= frac{a^x}{ln(a)}+C$$



      This is mostly just manipulation of various logarithm properties: namely,



      $$e^{ln(x)} = x$$
      $$ln(a^b) = b ln(a)$$



      Also, a nitpick: the integral in your question, OP, needs a $+C$ after it, since indefinite integration introduces an arbitrary constant.






      share|cite|improve this answer





















      • OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
        – Math Bob
        Dec 16 at 23:51












      • Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
        – Eevee Trainer
        Dec 17 at 0:18










      • Okay. I have taken this in so thanks for the answer!
        – Math Bob
        Dec 17 at 4:04














      4












      4








      4






      I assume it's kosher to use the exponential integral with base $e$, i.e. $int e^x dx = e^x +C$? Or, more generally, for a constant $k$,



      $$int e^{kx}dx = frac{1}{k}e^{kx}+C$$



      If so, then note:



      $$int a^x dx = int e^{ln(a^x)} dx = int e^{x ln(a)}dx = frac{1}{ln(a)}e^{x ln(a)}+C= frac{1}{ln(a)}e^{ln(a^x)}+C= frac{a^x}{ln(a)}+C$$



      This is mostly just manipulation of various logarithm properties: namely,



      $$e^{ln(x)} = x$$
      $$ln(a^b) = b ln(a)$$



      Also, a nitpick: the integral in your question, OP, needs a $+C$ after it, since indefinite integration introduces an arbitrary constant.






      share|cite|improve this answer












      I assume it's kosher to use the exponential integral with base $e$, i.e. $int e^x dx = e^x +C$? Or, more generally, for a constant $k$,



      $$int e^{kx}dx = frac{1}{k}e^{kx}+C$$



      If so, then note:



      $$int a^x dx = int e^{ln(a^x)} dx = int e^{x ln(a)}dx = frac{1}{ln(a)}e^{x ln(a)}+C= frac{1}{ln(a)}e^{ln(a^x)}+C= frac{a^x}{ln(a)}+C$$



      This is mostly just manipulation of various logarithm properties: namely,



      $$e^{ln(x)} = x$$
      $$ln(a^b) = b ln(a)$$



      Also, a nitpick: the integral in your question, OP, needs a $+C$ after it, since indefinite integration introduces an arbitrary constant.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 16 at 23:23









      Eevee Trainer

      3,370225




      3,370225












      • OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
        – Math Bob
        Dec 16 at 23:51












      • Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
        – Eevee Trainer
        Dec 17 at 0:18










      • Okay. I have taken this in so thanks for the answer!
        – Math Bob
        Dec 17 at 4:04


















      • OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
        – Math Bob
        Dec 16 at 23:51












      • Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
        – Eevee Trainer
        Dec 17 at 0:18










      • Okay. I have taken this in so thanks for the answer!
        – Math Bob
        Dec 17 at 4:04
















      OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
      – Math Bob
      Dec 16 at 23:51






      OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C?
      – Math Bob
      Dec 16 at 23:51














      Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
      – Eevee Trainer
      Dec 17 at 0:18




      Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = sum_{n=0}^infty frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$frac{d}{dx} frac{e^{kx}}{k} + C= k frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative."
      – Eevee Trainer
      Dec 17 at 0:18












      Okay. I have taken this in so thanks for the answer!
      – Math Bob
      Dec 17 at 4:04




      Okay. I have taken this in so thanks for the answer!
      – Math Bob
      Dec 17 at 4:04











      1














      Well this one can be found within every good integration table $($e.g. take at look at this$)$. Anyway it is actually not that hard to compute therefore I will demonstrate it for you. First of all the basic property of any exponential function with a basis $a$ is that it can be rewritten in terms of $e$ in the following way



      $$a^x=left(e^{ln(a)}right)^x=e^{xln(a)}$$



      Now we know that $e^x$ remains $e^x$ after integration aswell as after differentiation. Adding a constant $c$ before the $x$ within the exponent yields to



      $$frac d{dx}e^{cx}=ce^{cx}text{ and }int e^{cx}dx=frac1ce^{cx}+k$$



      From hereon we are basically done since $ln(a)$ can be seens as a constant while integrating. So plugging this together leads to



      $$int a^x dx=int e^{xln(a)}dx=frac1{ln(a)}e^{xln(a)}+k=frac{a^x}{ln(a)}+k$$




      $$int a^xdx=frac{a^x}{ln(a)}+k$$







      share|cite|improve this answer


























        1














        Well this one can be found within every good integration table $($e.g. take at look at this$)$. Anyway it is actually not that hard to compute therefore I will demonstrate it for you. First of all the basic property of any exponential function with a basis $a$ is that it can be rewritten in terms of $e$ in the following way



        $$a^x=left(e^{ln(a)}right)^x=e^{xln(a)}$$



        Now we know that $e^x$ remains $e^x$ after integration aswell as after differentiation. Adding a constant $c$ before the $x$ within the exponent yields to



        $$frac d{dx}e^{cx}=ce^{cx}text{ and }int e^{cx}dx=frac1ce^{cx}+k$$



        From hereon we are basically done since $ln(a)$ can be seens as a constant while integrating. So plugging this together leads to



        $$int a^x dx=int e^{xln(a)}dx=frac1{ln(a)}e^{xln(a)}+k=frac{a^x}{ln(a)}+k$$




        $$int a^xdx=frac{a^x}{ln(a)}+k$$







        share|cite|improve this answer
























          1












          1








          1






          Well this one can be found within every good integration table $($e.g. take at look at this$)$. Anyway it is actually not that hard to compute therefore I will demonstrate it for you. First of all the basic property of any exponential function with a basis $a$ is that it can be rewritten in terms of $e$ in the following way



          $$a^x=left(e^{ln(a)}right)^x=e^{xln(a)}$$



          Now we know that $e^x$ remains $e^x$ after integration aswell as after differentiation. Adding a constant $c$ before the $x$ within the exponent yields to



          $$frac d{dx}e^{cx}=ce^{cx}text{ and }int e^{cx}dx=frac1ce^{cx}+k$$



          From hereon we are basically done since $ln(a)$ can be seens as a constant while integrating. So plugging this together leads to



          $$int a^x dx=int e^{xln(a)}dx=frac1{ln(a)}e^{xln(a)}+k=frac{a^x}{ln(a)}+k$$




          $$int a^xdx=frac{a^x}{ln(a)}+k$$







          share|cite|improve this answer












          Well this one can be found within every good integration table $($e.g. take at look at this$)$. Anyway it is actually not that hard to compute therefore I will demonstrate it for you. First of all the basic property of any exponential function with a basis $a$ is that it can be rewritten in terms of $e$ in the following way



          $$a^x=left(e^{ln(a)}right)^x=e^{xln(a)}$$



          Now we know that $e^x$ remains $e^x$ after integration aswell as after differentiation. Adding a constant $c$ before the $x$ within the exponent yields to



          $$frac d{dx}e^{cx}=ce^{cx}text{ and }int e^{cx}dx=frac1ce^{cx}+k$$



          From hereon we are basically done since $ln(a)$ can be seens as a constant while integrating. So plugging this together leads to



          $$int a^x dx=int e^{xln(a)}dx=frac1{ln(a)}e^{xln(a)}+k=frac{a^x}{ln(a)}+k$$




          $$int a^xdx=frac{a^x}{ln(a)}+k$$








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 at 23:25









          mrtaurho

          3,0901930




          3,0901930























              1














              Just differentiate the right hand side and see what you get. Note that
              $$
              frac{d}{dx}left(frac{a^x}{ln a}right)=frac{1}{ln a}frac{d}{dx}(e^{xln a})=frac{1}{ln a}times e^{xln a}times ln a=a^x
              $$
              where we used the chain rule in the second equality.






              share|cite|improve this answer


























                1














                Just differentiate the right hand side and see what you get. Note that
                $$
                frac{d}{dx}left(frac{a^x}{ln a}right)=frac{1}{ln a}frac{d}{dx}(e^{xln a})=frac{1}{ln a}times e^{xln a}times ln a=a^x
                $$
                where we used the chain rule in the second equality.






                share|cite|improve this answer
























                  1












                  1








                  1






                  Just differentiate the right hand side and see what you get. Note that
                  $$
                  frac{d}{dx}left(frac{a^x}{ln a}right)=frac{1}{ln a}frac{d}{dx}(e^{xln a})=frac{1}{ln a}times e^{xln a}times ln a=a^x
                  $$
                  where we used the chain rule in the second equality.






                  share|cite|improve this answer












                  Just differentiate the right hand side and see what you get. Note that
                  $$
                  frac{d}{dx}left(frac{a^x}{ln a}right)=frac{1}{ln a}frac{d}{dx}(e^{xln a})=frac{1}{ln a}times e^{xln a}times ln a=a^x
                  $$
                  where we used the chain rule in the second equality.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 16 at 23:44









                  Foobaz John

                  20.7k41250




                  20.7k41250






























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                      Alcedinidae

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                      Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout ...
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                      Origin of the phrase “under your belt”?

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                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 17 1 Today, I found myself discussing what students should have "under their belts" during a lecture, and I wondered to myself if there was some inappropriate undertone here that I might be unaware of. Question : What is the origin of the phrase "under your belt"? A google search revealed some webpages (e.g. [1]) that assert its origin is related to consumption of food -- once a meal has been eaten, it's under your belt (which is a relief). However, with such websites, I have no way to determine fact from "random guy on the internet making stuff up". etymology idioms ...
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