Calculus - How do you find the Average Value of a function over an specific interval?

Question

Example:

f(x) = 4 - x^2

Interval: [-2,2]

Answer 1

The equation for the average value of a function is:

1/(b-a) * integral[a,b] f(x) dx.

When given an interval its given in the form [a, b].

So for your function on the interval [-2, 2]...

1/(2-(-2)) * integral[-2,2] 4 - x^2 dx
= 1/4 * integral[-2,2] 4 - x^2 dx
= 1/4 * 4x - 1/3 x^3 evaluated from -2 to 2
= (1/4)(32/3)
= 32/12 = 8/3

Answer 2

You integrate the function over that interval, and then divide by the length of that interval. In this case

Int 4 - x^2 dx = 4x - (x^3)/3 + c,

so, evaluating this at x = 2 and x = -2 gives the area of the function over the interval as (8 - 8/3) - (-8 + 8/3) = 32/3

Since the length of the interval is 4, the average value is
-32/(3*4) = 8/3.

Answer 3

Integrate and divide by the length of the range.

On the interval [-2,2] integrate:

∫(4 - x²)dx = 4x - (1/3)x³ = (4*2 - 8/3) - (-4*2 + 8/3)
= 16/3 + 16/3 = 32/3

Average value over the interval is

(32/3)/(2 - (-2)) = (32/3)/4 = 8/3

Answer 4

integrate function over domain and then devide by the magnitude of the domain