Answer
If f is integrable on the interval [a,b] then its average value is given by
$avg(f) =\frac{\int_{a}^{b} f(x)\,dx}{a-b}$
$\huge \int_{\Large\frac12}^{\normalsize2} \large{\frac{x^2}{2}}+\large{\frac{1}{x^2}} = \frac{x^3}{6} - \frac{1}{x}$
Substituting the limit of integration gives the value of integration as $\frac{5}{6}-(\frac{-95}{48}) = \frac{135}{48}$
Average value of the integral is ${\frac{138\div48}{3\div2}} = \frac{23}{12}$
Work Step by Step
If f is integrable on the interval [a,b] then its average value is given by
$avg(f) =\frac{\int_{a}^{b} f(x)\,dx}{a-b}$
$\huge \int_{\Large\frac12}^{\normalsize2} \large{\frac{x^2}{2}}+\large{\frac{1}{x^2}} = \frac{x^3}{6} - \frac{1}{x}$
Substituting the limit of integration gives the value of integration as $\frac{5}{6}-(\frac{-95}{48}) = \frac{135}{48}$
Average value of the integral is ${\frac{138\div48}{3\div2}} = \frac{23}{12}$
