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Integrate 2^x

  1. Dec 27, 2009 #1
    Intergrate [tex]2^x[/tex]

    ....

    lol?

    How would I even go about this? This is not a homework question, but I'm practising for exams.
     
  2. jcsd
  3. Dec 27, 2009 #2
    Re: Intergration

    What exponential function -DO- you know how to integrate?

    Hint:
    [tex]\frac{d}{dx}e^x=e^x[/tex]

    You can use that relationship to find the integral of any base exponential.
     
  4. Dec 27, 2009 #3
    Re: Intergration

    Can I use Ln2? Because I don't see how e^x will help when it can't integrate anyway
     
  5. Dec 27, 2009 #4
    Re: Intergration

    Actually that's bollocks, we have not learnt how to integrate ln2 yet, I'm stumped lol
     
  6. Dec 27, 2009 #5
    Re: Intergration

    Ok I'm going to use e^xln2 since that = 2^x
    Ok?

    How to integrate this? Because it in terms of dx, but what to do with ln2
     
  7. Dec 27, 2009 #6
    Re: Intergration

    [tex]\ln{2}[/tex] is a constant.
    It is a number.

    How would you rewrite [tex]2^x[/tex] to look more like [tex]e^t[/tex] which you do know how to integrate?
     
  8. Dec 27, 2009 #7
    Re: Intergration

    [tex]2^x = e^{(ln2^x)}[/tex]
    =
    [tex]2^x = e^{(xln2)}[/tex]
     
  9. Dec 27, 2009 #8
    Re: Intergration

    That's right!

    Now all you need to know how to do, is integrate [tex]e^{ax}[/tex]
    Knowing that [tex]\frac{d}{dx}e^x=e^x[/tex], what can you say about [tex]e^{ax}[/tex] and what can you infer about its integral?
     
    Last edited: Dec 27, 2009
  10. Dec 27, 2009 #9
    Re: Intergration

    What is a?
     
  11. Dec 27, 2009 #10
    Re: Intergration

    The integral is the inverse
     
  12. Dec 27, 2009 #11
    Re: Intergration

    [tex]a[/tex] is just any number.
    If you can solve for any number, you can solve for the very specific number, [tex]\ln{2}[/tex]

    And you're right about having to multiply by the reciprocal of [tex]a[/tex]!

    When you differentiate [tex]e^{ax}[/tex] you get [tex]e^{ax}[/tex] times [tex]\frac{d}{dx} ax[/tex] which is just [tex]a\cdot e^{ax}[/tex] because of the chain rule.

    So to get rid of the [tex]a[/tex] you get when differentiating, you multiply the original function by a constant to "cancel it out," [tex]\frac{1}{a}[/tex]
     
  13. Dec 28, 2009 #12
    Re: Intergration

    [tex]y=2^{x} \Rightarrow lny = xln2 \Rightarrow \frac{1}{y}\frac{dy}{dx} = ln2 \Rightarrow dy= yln2dx \Rightarrow \int dy = ln2\int 2^{x}dx\Rightarrow \int 2^{x}dx=\frac{y}{ln2} = \frac{2^{x}}{ln2}[/tex]

    Therefore [tex] \int 2^{x}dx = \frac{2^{x}}{ln2}+C[/tex]

    What do you think?
     
    Last edited: Dec 28, 2009
  14. Dec 28, 2009 #13
    Re: Intergration

    That's true, but it's a bit of a backwards way of approaching it (Though it has its beauty! I mean, what's simpler than taking the integral [tex]\int 1\cdot dy[/tex] ?).
    I think mine is the simplest approach in this case:

    [tex]2^x=e^{x\ln{2}}[/tex]

    [tex]\int 2^x = \int e^{x\ln{2}}=\frac{1}{ln{2}} e^{x\ln{2}} + C[/tex]
     
    Last edited by a moderator: Dec 28, 2009
  15. Dec 28, 2009 #14

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Intergration

    Which equals, of course,
    [tex]\frac{1}{ln 2} 2^x+ C[/tex]

    The derivative of [itex]a^x[/itex] is [itex]ln(a)a^x[/itex] and the integral is [itex](1/ln(a))a^x+ C[/itex]
     
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