# How do you find the arc length of the curve y=x^3 over the interval [0,2]?

Mar 9, 2018

A first-order approximation to the arc length gives $8 + {\tan}^{- 1} \left(2 \sqrt{3}\right)$ units.

#### Explanation:

$y = {x}^{3}$

$y ' = 3 {x}^{2}$

Arc length is given by:

$L = {\int}_{0}^{2} \sqrt{1 + 9 {x}^{4}} \mathrm{dx}$

Apply the substitution $\sqrt{3} x = u$:

$L = \frac{1}{\sqrt{3}} {\int}_{0}^{2 \sqrt{3}} \sqrt{1 + {u}^{4}} \mathrm{du}$

Complete the square in the square root:

$L = \frac{1}{\sqrt{3}} {\int}_{0}^{2 \sqrt{3}} \sqrt{{\left({u}^{2} + 1\right)}^{2} - 2 {u}^{2}} \mathrm{du}$

Factor out the larger piece:

$L = \frac{1}{\sqrt{3}} {\int}_{0}^{2 \sqrt{3}} \left({u}^{2} + 1\right) \sqrt{1 - \frac{2 {u}^{2}}{{u}^{2} + 1} ^ 2} \mathrm{du}$

For $x \in \left[0 , 2 \sqrt{3}\right]$, $\frac{2 {u}^{2}}{{u}^{2} + 1} ^ 2 < 1$. Take the series expansion of the square root:

$L = \frac{1}{\sqrt{3}} {\int}_{0}^{2 \sqrt{3}} \left({u}^{2} + 1\right) \left\{{\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{2 {u}^{2}}{{u}^{2} + 1} ^ 2\right)}^{n}\right\} \mathrm{du}$

Isolate the $n = 0$ and $n = 1$ cases:

$L = \frac{1}{\sqrt{3}} {\int}_{0}^{2 \sqrt{3}} \left\{\left({u}^{2} + 1\right) - {u}^{2} / \left({u}^{2} + 1\right)\right\} \mathrm{du} + \frac{1}{\sqrt{3}} {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- 2\right)}^{n} {\int}_{0}^{2 \sqrt{3}} {u}^{2 n} / {\left({u}^{2} + 1\right)}^{2 n - 1} \mathrm{du}$

Hence

$L = \frac{1}{\sqrt{3}} {\int}_{0}^{2 \sqrt{3}} \left({u}^{2} + \frac{1}{{u}^{2} + 1}\right) \mathrm{du} + \frac{1}{\sqrt{3}} {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- 2\right)}^{n} {\int}_{0}^{2 \sqrt{3}} {u}^{2 n} / {\left({u}^{2} + 1\right)}^{2 n - 1} \mathrm{du}$

Ignoring the higher-order terms, a first-order approximation gives:

$L \approx \frac{1}{\sqrt{3}} {\left[\frac{1}{3} {u}^{3} + {\tan}^{- 1} u\right]}_{0}^{2 \sqrt{3}}$

Insert the limits of integration:

$L \approx 8 + {\tan}^{- 1} \left(2 \sqrt{3}\right)$