1. ## 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

2. 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?

3. 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

$\displaystyle \frac {e^{t\ln(2)}} {ln2}$

4. Originally Posted by cammywhite
I got

$\displaystyle \frac {e^{t\ln(2)}} {ln2} ~{\color{red} + C}$
correct. but what is $\displaystyle e^{t \ln 2}$?

5. ## Can I simplify the following function?

what do you mean?

Oh is it $\displaystyle 2^x$

6. 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 $\displaystyle \int e^{x \ln a}~dx$

7. 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 $\displaystyle \int e^{x \ln a}~dx$

$\displaystyle \int 2^t dt = \frac {2^t}{ln2}+C$

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

9. 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 $\displaystyle a>1.$

10. 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

11. 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, $\displaystyle a\in(0,1)\cup(1,\infty).$

12. 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 $\displaystyle \epsilon > 0$"