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

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Answer 1 · 2016-12-10T23:40:40

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

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

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

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

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

$=4/ln2-1/ln16$

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