# How do you find the average value of #x^3# as x varies between 2 and 4?

The function is continuous in the interval [2 ,4] and the average value from x = 2 to x = 4 is the integral.

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To find the average value of (x^3) as (x) varies between 2 and 4, you use the formula for the average value of a function over an interval, which is given by:

[ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) , dx ]

Where (a) and (b) are the limits of integration (in this case, 2 and 4), and (f(x)) is the function (x^3).

[ \text{Average value} = \frac{1}{4 - 2} \int_{2}^{4} x^3 , dx ]

Integrate (x^3) with respect to (x) from 2 to 4:

[ \int_{2}^{4} x^3 , dx = \left[ \frac{x^4}{4} \right]_{2}^{4} ]

[ = \frac{4^4}{4} - \frac{2^4}{4} ]

[ = 64 - 4 = 60 ]

So,

[ \text{Average value} = \frac{1}{2} \times 60 = 30 ]

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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.

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.

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.

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