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Sagot :
The average value of the function f(x) = 3x^2 is 3 on the inetrval [-3, 3].
According to the given question.
We have a function.
F(x) = 3x^2
Since, we know that " the average value of a function is found by taking the integral of the function over the interval and dividing by the length of the interval".
Here, the given interval is [-3, 3]
Therefore, the length of the interval = 3 - (-3) = 3 + 3 = 6
Now, the average value of the given function f(x)
[tex]=\frac{1}{6} \int\limits^3_{-3} {3x^{2} } \, dx[/tex]
[tex]= \frac{1}{6} [\frac{x^{3} }{3} ]_{3} ^{-3}[/tex]
[tex]= \frac{1}{6} \frac{(3)^{3} -(-3)^{3} }{3}[/tex]
[tex]= \frac{1}{18} (27 + 27)[/tex]
= 2(27)/18
= 27/9
= 3
Hence, the average value of the function f(x) = 3x^2 is 3 on the inetrval [-3, 3].
Find out more information about average value of a function here:
https://brainly.com/question/22155666
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