Integrals of Exponential Functions

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Primitives of Exponential Functions

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Key Questions

  • Since

    #(e^x)'=e^x#,

    we have

    #int e^x dx=e^x+C#.

    I hope that this was helpful.

  • By rewriting a bit,

    #int n^x dx=int e^{(lnn)x}dx=e^{(lnn)x}/{lnn}+C=n^x/{lnn}+C#

    I hope that this was helpful.

  • The answer is #I=-e^(-x)+C#.

    This integral can be solved by a substitution:

    #u=-x#
    #du=-dx#
    #-du=dx#

    So, now we can substitute:

    #int e^(-x)dx = int e^u (-du)#
    #=-int e^u du#
    #=-e^u + C#

    and substitute back:
    #=-e^(-x) + C#

    For simple looking integrands, you should try a quick check to see if substitution works before trying harder integration methods.

  • Answer:

    #int3^xdx=1/ln3 3^x+"c"#

    Explanation:

    We want to find #int3^xdx#.

    Make the natural substitution #u=3^x# so #du=3^xln3dx#.

    So

    #int3^xdx=1/ln3int1du=1/ln3 u+"c"=1/ln3 3^x+"c"#

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