Question

(a) What is the average value of f(x) on the interval from x=0 to x=2 ? average value = (b) If f(x) is even, how to find the average of f(x) on the interval x=−2 to x=2 ? (c) If f(x) is odd , how to find the average of f(x) on the interval x=−2

Answer 1

a. The average value of a function `f` on an interval `[a,b]` is

`1/(b-a) int_a^b f(x) dx`

So, for this function we have an average value on `[0,2]` of

`1/(2-0) int_0^2 f(x) dx = 1/2 * 6 =3`

b. If `f` is even

If `f` is even then the graph of `f` is symmetric with respect to the `y` axis.

Therefore, `int_-2^0 f(x) dx = int_0^2 f(x) dx=6`

So `int_-2^2 f(x) dx = 12` and the average value on `[-2,2]` is

`1/(2-(-2)) int_-2^2 f(x)dx = 1/4(12) =3`

(Note that there are other ways to arrive at the average value of an even function on `[-a,a]` is the same as the average value on `[0,a]`

c. If `f` is odd

If `f` is odd, then the graph of `f` is symmetric with respect to the origin.

Therefore, `int_-2^0 f(x) dx = -int_0^2 f(x) dx = -6`

So `int_-2^2 f(x) dx = 0` and the average value on `[-2,2]` is

`1/(2-(-2)) int_-2^2 f(x)dx = 1/4(0) = 0`

(Note that there are other ways to arrive at: the average value of an odd function on `[-a,a]` is `0`.)