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Math Help - Antiderivative of 2^t

  1. #1
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    Antiderivative of 2^t

    \int 2^t dt

    I know the derivative of 2^t is 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
    \int 2^t dt

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

    Hence

    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
     2^t = e^{\ln(2^t)} = e^{t\ln(2)}

    Hence

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

    Can you handle that?

    I got

    \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

    \frac {e^{t\ln(2)}} {ln2} ~{\color{red} + C}
    correct. but what is 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 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 2^x
    where is x coming from? it is 2^t.

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

    and we derive this in the same way you did here, by considering \int e^{x \ln a}~dx
    Last edited by Jhevon; April 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 x coming from? it is 2^t.

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

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

     <br />
\int 2^t dt = \frac {2^t}{ln2}+C<br />
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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
     <br />
\int 2^t dt = \frac {2^t}{ln2}+C<br />
    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, \int a^x~dx = \frac {a^x}{\ln a} + C, for a > 0 a constant
    We're so mentalized on stating that but in this case it's actually 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 a>1.
    yes, of course. ">0" is so fun to say
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  11. #11
    MHF Contributor Reckoner'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 a>1.
    Of course, it is only necessary for a>0 and a\neq1. That is, 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 a>0 and a\neq1. That is, 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 \epsilon > 0"
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