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

Question

17358 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-21T15:03:10

#1/3#

The average value of the function #f(x)# on the interval #[a,b]# is

#"average value"=1/(b-a)int_a^bf(x)dx#

Thus, the average value here is

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

To integrate this, use the rule

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

Hence we see that

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

So, when we evaluate #1/2int_1^3x^-2dx#, we get

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

#=1/3#