# How do you find the arc length of the curve y=x^5/6+1/(10x^3) over the interval [1,2]?

Mar 21, 2018

$\frac{1219}{240}$

#### Explanation:

$y = {x}^{5} / 6 + \frac{1}{10 {x}^{3}} \implies$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {x}^{4}}{6} - \frac{3}{10 {x}^{4}} \implies$

$1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = 1 + {\left(\frac{5 {x}^{4}}{6} - \frac{3}{10 {x}^{4}}\right)}^{2}$

Now, since $1 = 4 \times \frac{5 {x}^{4}}{6} \times \frac{3}{10 {x}^{4}}$, the expression on the right above simplifies !

$1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = {\left(\frac{5 {x}^{4}}{6} + \frac{3}{10 {x}^{4}}\right)}^{2}$

Thus, the required arc length

$L = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx} = {\int}_{1}^{2} \left(\frac{5 {x}^{4}}{6} + \frac{3}{10 {x}^{4}}\right) \mathrm{dx} = {\left({x}^{5} / 6 - \frac{1}{10 {x}^{3}}\right)}_{1}^{2} = \left({2}^{5} / 6 - \frac{1}{10 \times {2}^{3}}\right) - \left(\frac{1}{6} - \frac{1}{10}\right) = \frac{1219}{240}$