Answer
$\ln 2$
Work Step by Step
The fundamental theorem of calculus states that the average value of any function $f(x) $ on the interval $[m,n]$ is:
$\dfrac{1}{n-m}\int_m^n f'(x) dx=\dfrac{1}{n-m}[f'(x)]_m^n$
or, $=\dfrac{f(n)-f(m)}{n-m}$
This represents an average change of $f(x)$ on $[m,n]$
Here, $f(x)=\dfrac{1}{x}$
and $f'(x)=\ln x+C$
We are given the interval: $[1,2]$
Then, $\dfrac{f(n)-f(m)}{n-m}=\dfrac{\ln 2-\ln 1}{2-1}=\ln 2$
