I seem to be having a huge case of mind lock, i can't seem to remember how to solve for the original function to evaluate the integral:

$\displaystyle

\int_0^1 \frac{1}{2^x}dx$

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- Jan 31st 2010, 04:56 PMpenguinpwnMy mind is escaping me
I seem to be having a huge case of mind lock, i can't seem to remember how to solve for the original function to evaluate the integral:

$\displaystyle

\int_0^1 \frac{1}{2^x}dx$ - Jan 31st 2010, 05:02 PMzhupolongjoe
Note that 1/(2^x)=2^-x.

Also recall that d/dx(2^x)=log(2)*2^x

So our integral is -1/log(2) * 2^(-x)+C. We now this because when we take the derivative, we have a -log(2) to get rid of so that we will be left with 2^(-x) (evaluate between 0 and 1 of course). - Jan 31st 2010, 05:07 PMTWiX
- Jan 31st 2010, 05:58 PMPlato
BTW: It is impossible for "My mind is escaping me".

One's mind is simply one's brain. - Jan 31st 2010, 06:12 PMBruno J.
- Feb 1st 2010, 04:06 AMHallsofIvy
- Feb 2nd 2010, 02:12 AMmr fantastic
This thread can be continued in the Philosophy subforum (if so desired).

Thread closed.