What is the average value of the function #f(x) = 1/x^2# on the interval #[1,3]#?

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Answer 1 · 2016-03-23T02:32:23

#1/3#

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

#1/(b-a)int_a^bf(x)dx#

Applying this to the given situation gives:

#=1/(3-1)int_1^3 1/x^2dx=1/2int_1^3 x^-2dx#

Using the integration rule

#intx^ndx=x^(n+1)/(n+1)+C#

So, we want to evaluate

#x^(-2+1)/(-2+1)=x^-1/(-1)=-1/x#

From #1# to #3#, all multiplied by #1/2#.

#=1/2[-1/x]_1^3=1/2(-1/3-(-1/1))=1/2(-1/3+1)#

#=1/2(2/3)=1/3#