# How do you integrate int 4^(x/2)dx from [-2,2]?

Dec 10, 2016

We can start by simplifying ${4}^{\frac{x}{2}}$

$\int {4}^{\frac{x}{2}} \mathrm{dx} = \int {2}^{x} \mathrm{dx}$

On the interval [-2, 2], we have:

int_-2^2 2^xdx = 2^x/ln2]_(-2)^2=2^2/ln2-2^(-2)/ln2

$= \frac{4}{\ln} 2 - \frac{1}{4 \ln 2}$

$= \frac{4}{\ln} 2 - \frac{1}{\ln} 16$