What is the average value of the function f(x)=1x2 on the interval [1,3]?
1 Answer
Mar 23, 2016
Explanation:
The average value of the function
1b−a∫baf(x)dx
Applying this to the given situation gives:
=13−1∫311x2dx=12∫31x−2dx
Using the integration rule
∫xndx=xn+1n+1+C
So, we want to evaluate
x−2+1−2+1=x−1−1=−1x
From
=12[−1x]31=12(−13−(−11))=12(−13+1)
=12(23)=13