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

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

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Answer 1 · 2016-12-10T23:40:40

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

$\int 4^{\frac{x}{2}} d x = \int 2^{x} d x$ $int4^(x/2)dx=int2^xdx$

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

$\int_{- 2}^{2} 2^{x} d x = \frac{2^{x}}{ln 2}]_{- 2}^{2} = \frac{2^{2}}{ln 2} - \frac{2^{- 2}}{ln 2}$ $int_-2^2 2^xdx = 2^x/ln2]_(-2)^2=2^2/ln2-2^(-2)/ln2$

$= \frac{4}{ln 2} - \frac{1}{4 ln 2}$ $=4/ln2-1/(4ln2)$

$= \frac{4}{ln 2} - \frac{1}{ln 16}$ $=4/ln2-1/ln16$

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

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