How do you find the average value of #f(x)=x^5-4x^3+2x-1# as x varies between #[-2,2]#?
2 Answers
Explanation:
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
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#
Morgan has given a fine answer. I want to mention a fact that can simplify the integration needed for this question.
Explanation:
A function
If
In this question the first four terms of the polynomial form an odd function that is integrable on any closed interval, so
# = {: -x]_-2^2 = -(2)-(-(-2)) = -4#