Answer
$2\displaystyle \ln 2+\frac{1}{2}\approx 1.8863$
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 =$\displaystyle \frac{1}{4-2}\int_{2}^{4}\frac{4(x+1)}{x^{2}}dx$
$\displaystyle \int\frac{4x+4}{x^{2}}dx=\int\frac{4x}{x^{2}}dx+\int\frac{4}{x^{2}}dx$
$=4(\displaystyle \int\frac{dx}{x}+\int x^{-2}dx) \qquad $Log rule, Power rule
$=4(\displaystyle \ln|x|+\frac{x^{-1}}{1})+C$
Average value =$ \displaystyle \frac{1}{2}\cdot 4[\ln|x|-\frac{1}{x}]_{2}^{4}$
$=2[(\displaystyle \ln 4-\frac{1}{4})-(\ln 2-\frac{1}{2})]$
$=2(2\displaystyle \ln 2-\frac{1}{4}-\ln 2+\frac{1}{2})$
$=2\displaystyle \ln 2+\frac{1}{2}\approx 1.8863$
