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How do you find the average value of #x^3# as x varies between -1 and 2?

Answer 1

Emily Ammerman

#= 5/4#

Average value of #f(x)# over interval #[a,b]# is #bar (f) (x) = (int_a^b f(x) dx)/(b-a)#

Here that means that:

#bar (f) (x) = (int_(-1)^2 x^3 dx)/(2 - (-1))#

#= ([ x^4/4]_(-1)^2)/3 = 5/4#

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Answer 2

Elijah Abram

The average value of ( x^3 ) as ( x ) varies between -1 and 2 is found by integrating ( x^3 ) over the interval ([-1, 2]) and then dividing by the length of the interval.

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

Substituting (a = -1) and (b = 2):

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

[ = \frac{1}{3} \left[ \frac{x^4}{4} \right]_{-1}^{2} ]

[ = \frac{1}{3} \left( \frac{2^4}{4} - \frac{(-1)^4}{4} \right) ]

[ = \frac{1}{3} \left( \frac{16}{4} + \frac{1}{4} \right) ]

[ = \frac{1}{3} \left( 4 + 1 \right) ]

[ = \frac{5}{3} ]

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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.