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

Answer 1

Isaiah Allen

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#

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Answer 2

Axel Alvarez

To integrate (\int 4^{\frac{x}{2}} , dx) from ([-2, 2]), first recognize that (4^{\frac{x}{2}}) can be expressed as ((2^2)^{\frac{x}{2}}), which simplifies to (2^x).

Then, integrate (2^x) with respect to (x), which gives (\frac{2^x}{\ln(2)}).

Evaluate the integral from -2 to 2 by substituting the upper and lower limits of integration into the antiderivative and subtracting the result of the lower limit from the result of the upper limit. This will give you the definite integral value.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.