Answer
$\frac{92}{9}$
Work Step by Step
The formula for the arc length is
$s=\int_{a}^b \sqrt{1 + (y')^2} dy$
Differentiate $y = \frac{x^4}{8}+\frac{1}{4x^2}$ with respect to x.
$y' = \frac{4x^3}{8}- \frac{2}{4x^3}$
$=\frac{1}{2}(x^3 - x^{−3})$
Substitute the value of y' in $1 + (y')^2$
$1+(y')^2=1+[\frac{1}{2}(x^3 - x^{−3})]^2$
$=[\frac{1}{2}(x^3 + x^{−3})]^2$
Substitute the value of $1 + (y')^2$ in the formula for the arc length $s=\int_{a}^b \sqrt{1 + (y')^2} dy$
Solve for the distance s
$s=\int_{1}^3\sqrt{\frac{1}{2}(x^3 + x^{-3})}^{2} dx$
$=\frac{1}{2}\int_{1}^3 (x^3 + x^{-3})dx$
$=\frac{1}{2}[\frac{x^4}{4}-\frac{1}{2x^2}]_1^3\ $
$=\frac{92}{9}$
