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Thread: Antiderivative of 2^t

  1. #1
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    Antiderivative of 2^t

    $\displaystyle \int 2^t dt$

    I know the derivative of $\displaystyle 2^t$ is $\displaystyle 2^tln2$, but I can't figure out the anitderivative
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  2. #2
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    Quote Originally Posted by cammywhite View Post
    $\displaystyle \int 2^t dt$

    I know the derivative of $\displaystyle 2^t$ is $\displaystyle 2^tln2$, but I can't figure out the anitderivative
    $\displaystyle 2^t = e^{\ln(2^t)} = e^{t\ln(2)} $

    Hence

    $\displaystyle I = \int e^{t\ln(2)}dt $

    Can you handle that?
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  3. #3
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    Quote Originally Posted by Mush View Post
    $\displaystyle 2^t = e^{\ln(2^t)} = e^{t\ln(2)} $

    Hence

    $\displaystyle I = \int e^{t\ln(2)}dt $

    Can you handle that?

    I got

    $\displaystyle \frac {e^{t\ln(2)}} {ln2}$
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cammywhite View Post
    I got

    $\displaystyle \frac {e^{t\ln(2)}} {ln2} ~{\color{red} + C}$
    correct. but what is $\displaystyle e^{t \ln 2}$?
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  5. #5
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    Can I simplify the following function?

    what do you mean?

    Oh is it $\displaystyle 2^x$
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cammywhite View Post
    what do you mean?

    Oh is it $\displaystyle 2^x$
    where is $\displaystyle x$ coming from? it is $\displaystyle 2^t$.

    in general, $\displaystyle \int a^x~dx = \frac {a^x}{\ln a} + C$, for $\displaystyle a > 1$ a constant

    and we derive this in the same way you did here, by considering $\displaystyle \int e^{x \ln a}~dx$
    Last edited by Jhevon; Apr 5th 2009 at 06:33 PM. Reason: :) per post #9
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    where is $\displaystyle x$ coming from? it is $\displaystyle 2^t$.

    in general, $\displaystyle \int a^x~dx = \frac {a^x}{\ln a} + C$, for $\displaystyle a > 0$ a constant

    and we derive this in the same way you did here, by considering $\displaystyle \int e^{x \ln a}~dx$

    $\displaystyle
    \int 2^t dt = \frac {2^t}{ln2}+C
    $
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cammywhite View Post
    $\displaystyle
    \int 2^t dt = \frac {2^t}{ln2}+C
    $
    yes...i had thought we already came to that conclusion
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  9. #9
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    Quote Originally Posted by Jhevon View Post

    in general, $\displaystyle \int a^x~dx = \frac {a^x}{\ln a} + C$, for $\displaystyle a > 0$ a constant
    We're so mentalized on stating that but in this case it's actually $\displaystyle a>1.$
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    We're so mentalized on stating that but in this case it's actually $\displaystyle a>1.$
    yes, of course. ">0" is so fun to say
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  11. #11
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    Quote Originally Posted by Krizalid View Post
    We're so mentalized on stating that but in this case it's actually $\displaystyle a>1.$
    Of course, it is only necessary for $\displaystyle a>0$ and $\displaystyle a\neq1.$ That is, $\displaystyle a\in(0,1)\cup(1,\infty).$
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Reckoner View Post
    Of course, it is only necessary for $\displaystyle a>0$ and $\displaystyle a\neq1.$ That is, $\displaystyle a\in(0,1)\cup(1,\infty).$
    why, you're right! i guess ">1" should be added to the list of things that are fun to say but not think about. right behind "let $\displaystyle \epsilon > 0$"
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