# Math Help - Average value of function

1. ## Average value of function

Hi, I was wondering if anyone can help:

Find a positive value of k such that the average value of f(x) = 1/ (kx^2) over the interval [k,1] is 2.

Thanks,

2. Originally Posted by Wintaker99
Hi, I was wondering if anyone can help:

Find a positive value of k such that the average value of f(x) = 1/ (kx^2) over the interval [k,1] is 2.

Thanks,
The average of f(x) over [k,1] is:

Av(f) = [integral_{x=k to 1} f(x) dx]/(1-k) = [(1-k)/k^2]/(1-k) = 1/k^2,

so if Av(f) = 2, we have 1/k^2=2, and so k=1/sqrt(2).

RonL