How do you find the average value of #f(x)=x^5-4x^3+2x-1# as x varies between #[-2,2]#?

Question

Ezra Adams · Accepted Answer

#f_(ave)=-1#

The average value of a function is found using the following equation:

#f_(ave)=1/(b-a)*int_a^bf(x)dx#

on some interval #[a,b]#

Therefore:

#f_(ave)=1/4int_(-2)^2(x^5-4x^3+2x-1)dx#

This is a basic integral.

#=>f_(a v e)=1/4(1/6x^6-x^4+x^2-x)]_(-2)^2#

#=1/4*-4#

#=-1#

Evelyn Agard · Answer

Morgan has given a fine answer. I want to mention a fact that can simplify the integration needed for this question.

A function #f# is odd if and only if #f(-x) = -f(x)# for all #x# in the domain of #f#. If #f# is an odd function, and integrable on #[-a,a]# then #int_-a^a f(x) dx = 0# In this question the first four terms of the polynomial form an odd function that is integrable on any closed interval, so #int_-2^2 (x^5-4x^3+2x-1) dx = int_-2^2(-1) dx# # = {: -x]_-2^2 = -(2)-(-(-2)) = -4#

Elijah Abram · Answer

undefined To find the average value of the function $f(x) = x^5 - 4x^3 + 2x - 1$ over the interval $[-2, 2]$, you use the formula for the average value of a function:

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

Substituting the given function and interval into this formula:

$$
\text{Average value} = \frac{1}{2 - (-2)} \int_{-2}^{2} (x^5 - 4x^3 + 2x - 1) \, dx
$$

You then evaluate this integral to find the average value of the function over the given interval.