Question

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

Answer 1

We can start by simplifying `4^(x/2)`

`int4^(x/2)dx=int2^xdx`

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

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

`=4/ln2-1/(4ln2)`

`=4/ln2-1/ln16`