Mean value of 1/x on the interval [1,3]

Text

Compute the integral mean value of $f(x)=\dfrac1x$ on the interval $[1,3]$ and find the point $c$ predicted by the mean value theorem for integrals.

Solution

Mean value. By definition $\mu=\frac{1}{3-1}\int_1^3\frac{dx}{x}=\frac12\big[\ln x\big]_1^3=\frac12\ln 3\approx 0.5493.$

Point $c$. The mean value theorem guarantees a $c\in(1,3)$ with $f(c)=\mu$: $\frac1c=\frac12\ln 3 \implies c=\frac{2}{\ln 3}\approx 1.8205\in(1,3).$

$\boxed{\mu=\tfrac12\ln 3\approx 0.5493,\qquad c=\dfrac{2}{\ln 3}\approx 1.8205}$