Question

Find the average value of the function ()=3 on the interval [−3,3] and determine a number in this interval for which () is equal to the average value

Answer 1

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)

$=\frac{1}{6} \int\limits^3_{-3} {3x^{2} } \, dx$

$= \frac{1}{6} [\frac{x^{3} }{3} ]_{3} ^{-3}$

$= \frac{1}{6} \frac{(3)^{3} -(-3)^{3} }{3}$

$= \frac{1}{18} (27 + 27)$

= 2(27)/18

= 27/9

= 3

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

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