Answer
$\displaystyle \frac{1}{e-1}\approx 0.58198$
Work Step by Step
see definition on p.281.
Average value =$\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)dx$
Using the given data,
Average value =$\displaystyle \frac{1}{e-1}\int_{1}^{e}\frac{2\ln x}{x}dx$
Find the indefinite integral,
$\displaystyle \int\frac{2\ln x}{x}dx= \left[\begin{array}{l}
u=\ln x\\
du=\frac{dx}{x}
\end{array}\right] = 2\int udu$
$=2\displaystyle \cdot\frac{u^{2}}{2}+C=u^{2}+C =(\ln x)^{2}+C$
Average value =$\displaystyle \frac{1}{e-1}[(\ln x)^{2}]_{1}^{e}$
$=\displaystyle \frac{1}{e-1}(1^{2}-0)$
$=\displaystyle \frac{1}{e-1}\approx 0.58198$
