(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

Question

Charles Arocha · Accepted Answer

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#.)

Emilia Aguilar · Answer

undefined (a) The average value of $f(x)$ on the interval from $x = 0$ to $x = 2$ can be found using the formula:

$$ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx $$

Substitute $a = 0$ and $b = 2$ into the formula and evaluate the integral.

(b) If $f(x)$ is even, meaning $f(x) = f(-x)$, then the average of $f(x)$ on the interval $x = -2$ to $x = 2$ can be found by considering the symmetry of the function about the y-axis. In this case, the average value will be the same as the average value on the interval $x = 0$ to $x = 2$. So, you can use the same formula as in part (a).

(c) If $f(x)$ is odd, meaning $f(-x) = -f(x)$, then the average of $f(x)$ on the interval $x = -2$ to $x = 2$ is zero. This is because the positive and negative areas of the function on this interval cancel each other out due to the symmetry about the origin. Therefore, the average value of $f(x)$ is zero on this interval.

(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