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:
$\displaystyle
\int_0^1 \frac{1}{2^x}dx$
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).