What is the average value of the function f(x)=1x2 on the interval [1,3]?

1 Answer
Mar 23, 2016

13

Explanation:

The average value of the function f(x) on the continuous and differentiable interval [a,b] can be found through:

1babaf(x)dx

Applying this to the given situation gives:

=131311x2dx=1231x2dx

Using the integration rule

xndx=xn+1n+1+C

So, we want to evaluate

x2+12+1=x11=1x

From 1 to 3, all multiplied by 12.

=12[1x]31=12(13(11))=12(13+1)

=12(23)=13