Take the 2-minute tour ×

Find the arc length of $y=x^{2/3}$ from $x=-1$ to $x=8$.

I have tried applying the arc length formula but for some reason I keep getting $7.63$, but the answer is $10.51$.

share|improve this question
The modulus of the arc length is certainly greater than 9. I agree with the answer you provide, I got 10.5131. –  Gamma Function Jun 10 '13 at 10:56
You certainly shouldn't ever obtain negative arclength. –  Gamma Function Jun 10 '13 at 11:04

2 Answers 2

up vote 6 down vote accepted

The complex outcome of Maazul seems to be caused by the definition of $x^{1/3}$ for $x < 0$. Many computer algebra packages do not take it to be the "natural choice" $-|x|^{1/3}$ (but one of the other complex roots), while it is almost certain that this is what is intended.

Therefore, the following calculation (which should be reminiscent of what OP did) is tempting:

$$\begin{align} \int_{-1}^8 \sqrt{1+\frac49x^{-2/3}}\,\mathrm dx &=\int_{-1}^8\sqrt{\frac{x^{2/3}+\frac49}{x^{2/3}}}\,\mathrm dx\\{\text{tentative}}&= \int_{-1}^8 \frac{\sqrt{x^{2/3}+\frac49}}{x^{1/3}}\,\mathrm dx \\&= \left.\left(\frac49+x^{2/3}\right)^{3/2}\right|_{x=-1}^8 = \left(\frac{40}9\right)^{3/2}-\left(\frac{13}9\right)^{3/2} \approx 7.63 \end{align}$$

but $\sqrt{x^{2/3}} \ne x^{1/3}$ for $x < 0$, at which point a correcting minus sign is required.

Therefore, we have to split the integral into two parts: $\int_{-1}^0$ and $\int_0^8$ (which will have different primitives). Since this is a homework assignment, I'll stop here.

Addendum after seeing OP evaluates using substitution: Similar problems to above arise when using $u = x^{2/3}$ because this $u$ is always positive. Therefore, splitting the interval of integration is again necessary. Then both integrals can be solved using (suitable variants of) this substitution.

share|improve this answer
I've managed to solve this problem using the fact that y=x^(2/3) is symmetrical about y-axis.But even after the integral is divided into two parts ,the ∫0−1 + ∫8-0 yields 7.63. –  RhK Jun 10 '13 at 11:00
The first of these integrals has a different primitive from the second (with an additional minus sign, because for $x <0$, $\sqrt{x^{2/3}} = -x^{1/3}$). –  Lord_Farin Jun 10 '13 at 11:01

$\displaystyle y=x^{2/3}$

$\displaystyle y'=\frac{2}{3}x^{-1/3}$

$\displaystyle L=\int_{-1}^{8} ds$

$\displaystyle L=\int_{-1}^{8} \sqrt{dx^2+dy^2}$

$\displaystyle L=\int_{-1}^{8} \sqrt{1+y'^2}dx$

$\displaystyle L=\int_{-1}^{8} \sqrt{1+\frac{4}{9}x^{-2/3}}dx\approx10+0.51i$

share|improve this answer
I doubt that the arc length is a non-real complex number. –  egreg Jun 10 '13 at 10:52
As @Lord_Farin advises, this should be split at the origin. –  Gamma Function Jun 10 '13 at 11:01

Your Answer


