# How do you find the length of the curve y=x^5/6+1/(10x^3) between 1<=x<=2 ?

Sep 11, 2014

We can find the arc length to be $\frac{1261}{240}$ by the integral
$L = {\int}_{1}^{2} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Let us look at some details.

By taking the derivative,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {x}^{4}}{6} - \frac{3}{10 {x}^{4}}$

So, the integrand looks like:
sqrt{1+({dy}/{dx})^2}=sqrt{({5x^4)/6)^2+1/2+(3/{10x^4})^2
by completing the square
$= \sqrt{{\left(\frac{5 {x}^{4}}{6} + \frac{3}{10 {x}^{4}}\right)}^{2}} = \frac{5 {x}^{4}}{6} + \frac{3}{10 {x}^{4}}$

Now, we can evaluate the integral.
$L = {\int}_{1}^{2} \left(\frac{5 {x}^{4}}{6} + \frac{3}{10 {x}^{4}}\right) \mathrm{dx} = {\left[{x}^{5} / 6 - \frac{1}{10 {x}^{3}}\right]}_{1}^{2} = \frac{1261}{240}$