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What is the average value of the function #f(x) = 9 − x^2# on the interval #[-3,3]#?

Answer 1

Evelyn Agard

It is #6#

The average value of a function #f(x)# on the interval #[a, b]# is

#f_(avg)=1/(b-a)*int_a^b f(x)dx#

Hence

#f_(avg)=1/(3-(-3))int_-3^3 (9-x^2)dx=> f_(avg)=1/636=6#

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Answer 2

Aurora Alvarado

To find the average value of the function ( f(x) = 9 - x^2 ) on the interval ([-3,3]), follow these steps:

Calculate the definite integral of the function over the given interval.
Divide the result from step 1 by the width of the interval.

Applying these steps:

[ \int_{-3}^{3} (9 - x^2) , dx = \left[ 9x - \frac{x^3}{3} \right]_{-3}^{3} ] [ = \left(9 \cdot 3 - \frac{3^3}{3}\right) - \left(9 \cdot (-3) - \frac{(-3)^3}{3}\right) ] [ = (27 - 9) - (-27 - 9) ] [ = 18 - (-36) ] [ = 54 ]
Divide the result by the width of the interval: ( \frac{54}{3 - (-3)} = \frac{54}{6} = 9 ).

Therefore, the average value of the function ( f(x) = 9 - x^2 ) on the interval ([-3,3]) is ( 9 ).

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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.