# Antiderivative of 2^t

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• Apr 5th 2009, 04:53 PM
cammywhite
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
• Apr 5th 2009, 04:54 PM
Mush
Quote:

Originally Posted by cammywhite
$\displaystyle \int 2^t dt$

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?
• Apr 5th 2009, 05:49 PM
cammywhite
Quote:

Originally Posted by Mush
$\displaystyle 2^t = e^{\ln(2^t)} = e^{t\ln(2)}$

Hence

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

Can you handle that?

I got

$\displaystyle \frac {e^{t\ln(2)}} {ln2}$
• Apr 5th 2009, 05:50 PM
Jhevon
Quote:

Originally Posted by cammywhite
I got

$\displaystyle \frac {e^{t\ln(2)}} {ln2} ~{\color{red} + C}$

correct. but what is $\displaystyle e^{t \ln 2}$?
• Apr 5th 2009, 05:58 PM
cammywhite
Can I simplify the following function?
what do you mean?

Oh is it $\displaystyle 2^x$
• Apr 5th 2009, 06:05 PM
Jhevon
Quote:

Originally Posted by cammywhite
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 $\displaystyle \int e^{x \ln a}~dx$
• Apr 5th 2009, 06:13 PM
cammywhite
Quote:

Originally Posted by Jhevon
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 $\displaystyle \int e^{x \ln a}~dx$

$\displaystyle \int 2^t dt = \frac {2^t}{ln2}+C$
• Apr 5th 2009, 06:18 PM
Jhevon
Quote:

Originally Posted by cammywhite
$\displaystyle \int 2^t dt = \frac {2^t}{ln2}+C$

yes...i had thought we already came to that conclusion
• Apr 5th 2009, 06:31 PM
Krizalid
Quote:

Originally Posted by Jhevon

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 $\displaystyle a>1.$ :)
• Apr 5th 2009, 06:32 PM
Jhevon
Quote:

Originally Posted by Krizalid
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 (Nod)
• Apr 5th 2009, 08:07 PM
Reckoner
Quote:

Originally Posted by Krizalid
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, $\displaystyle a\in(0,1)\cup(1,\infty).$
• Apr 6th 2009, 07:00 PM
Jhevon
Quote:

Originally Posted by Reckoner
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 $\displaystyle \epsilon > 0$" :)