# What is the arc length of f(x)=((4x^5)/5) + (1/(48x^3)) - 1  on x in [1,2]?

Apr 4, 2016

$\frac{47651}{1920} \cong 24.818$

#### Explanation:

$f \left(x\right) = \frac{4 {x}^{5}}{5} + \frac{1}{48 {x}^{3}} - 1$
$f \text{'} \left(x\right) = 4 {x}^{4} - \frac{1}{16 {x}^{4}}$

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

$L = {\int}_{1}^{2} \sqrt{1 + {\left(4 {x}^{4} - \frac{1}{16 {x}^{4}}\right)}^{2}} \cdot \mathrm{dx}$
$L = {\int}_{1}^{2} \sqrt{1 + 16 {x}^{8} - \frac{1}{2} + \frac{1}{256 {x}^{8}}} \cdot \mathrm{dx}$
$L = {\int}_{1}^{2} \sqrt{16 {x}^{8} + \frac{1}{2} + \frac{1}{256 {x}^{8}}} \cdot \mathrm{dx}$
$L = {\int}_{1}^{2} \sqrt{{\left(4 {x}^{4} + \frac{1}{16 {x}^{4}}\right)}^{2}} \cdot \mathrm{dx}$
$L = {\int}_{1}^{2} \left(4 {x}^{4} + \frac{1}{16 {x}^{4}}\right) \cdot \mathrm{dx}$
$L = \left(\frac{4}{5} {x}^{5} - \frac{1}{48 {x}^{3}}\right) {|}_{1}^{2}$
$L = \frac{4 \cdot 32}{5} - \frac{1}{48 \cdot 8} - \left(\frac{4}{5} - \frac{1}{48}\right)$
$L = \frac{128 - 4}{5} - \frac{1 - 8}{384} = \frac{124}{5} + \frac{7}{384} = \frac{47616 + 35}{1920}$
$L = \frac{47651}{1920} \cong 24.818$