# What is the arclength of f(x)=[4x^2–2ln(x)] /8 in the interval [1,e^3]?

Apr 4, 2016

${e}^{6} / 2 + \frac{1}{4} \cong 201.964$

#### Explanation:

$f \left(x\right) = {x}^{2} / 2 - \ln \frac{x}{4}$
$f ' \left(x\right) = x - \frac{1}{4 x}$

$L = {\int}_{a}^{b} \sqrt{1 + {\left[f ' \left(x\right)\right]}^{2}} \cdot \mathrm{dx}$

$L = {\int}_{1}^{{e}^{3}} \sqrt{1 + {\left(x - \frac{1}{4 x}\right)}^{2}} \cdot \mathrm{dx}$
$L = {\int}_{1}^{{e}^{3}} \sqrt{1 + {x}^{2} - \frac{1}{2} + \frac{1}{16 {x}^{2}}} \cdot \mathrm{dx}$
$L = {\int}_{1}^{{e}^{3}} \sqrt{{x}^{2} + \frac{1}{2} + \frac{1}{16 {x}^{2}}} \cdot \mathrm{dx}$
$L = {\int}_{1}^{{e}^{3}} \sqrt{{\left(x + \frac{1}{4 x}\right)}^{2}} \cdot \mathrm{dx}$
$L = {\int}_{1}^{{e}^{3}} \left(x + \frac{1}{4 x}\right) \cdot \mathrm{dx}$
$L = \left({x}^{2} / 2 + \frac{\ln | x |}{4}\right) {|}_{1}^{{e}^{3}}$
$L = {e}^{6} / 2 + \frac{\ln {e}^{3}}{4} - \left(\frac{1}{2} + 0\right) = {e}^{6} / 2 + \frac{3}{4} - \frac{1}{2}$
$L = {e}^{6} / 2 + \frac{1}{4}$