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Could you please help me with this problem? Find the arc length of the graph of $y = \frac{x^{3}}{3} + \frac{1}{4x}$ between $x = 1$ and $x = 2$. Note: It may be helpful to use identities like $$x^{2} + \frac{1}{4x^{2}} = x^4 + \frac{1}{2} + \frac{1}{16x^4}.$$

The answer is $59/24$, but I have no idea how to obtain that. I get stuck after trying to use the identity and having to integrate $$\left( \frac{17}{18} + \frac{x^{4}}{81} + \frac{1}{16x^4}\right)^{1/2}.$$ Is it even possible to integrate this...?

Thanks in advance!

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  • $\begingroup$ Could you please use LaTeX or at least parentheses ? this might be helpful : codecogs.com/eqnedit.php $\endgroup$ – Shabbeh Aug 11 '14 at 4:08
  • $\begingroup$ Thanks for the edit. Yep okay I think this is reason enough for me to start learning LaTeX.. haha. $\endgroup$ – inggumnator Aug 11 '14 at 4:21
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Observe that: \begin{align*} L &= \int_1^2 \sqrt{1 + \left[\frac{d}{dx}\left(\frac{1}{3}x^3 + \frac{1}{4x}\right)\right]^2} \, dx \\ &= \int_1^2 \sqrt{1 + \left[x^2 - \frac{1}{4x^2}\right]^2} \, dx \\ &= \int_1^2 \sqrt{1 + \left[x^4 - \frac{1}{2} + \frac{1}{16x^4}\right]} \, dx \\ &= \int_1^2 \sqrt{x^4 + \frac{1}{2} + \frac{1}{16x^4}} \, dx \\ &= \int_1^2 \sqrt{\left[x^2 + \frac{1}{4x^2}\right]^2} \, dx \\ &= \int_1^2 \left[x^2 + \frac{1}{4x^2}\right] \, dx \\ &= \left[\frac{x^3}{3} - \frac{1}{4x}\right]_1^2 \\ &= \left[\frac{8}{3} - \frac{1}{8}\right] - \left[\frac{1}{3} - \frac{1}{4}\right] \\ &= \frac{59}{24} \end{align*}

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  • $\begingroup$ Accomplishment! :) +1 $\endgroup$ – mrs Aug 11 '14 at 4:50
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Basically the arc length is given by : $$\int_a^b \sqrt{1+\left[f'(x)\right]^2} \ \mathrm{d}x $$ In your case $f'(x)=x^2-\frac{1}{4x^2}$, then $$\begin{aligned}\text{The arc length} &= \int_1^2 \sqrt{1+\left(x^2-\frac{1}{4x^2}\right)^2 } \ \mathrm{d}x\\ &= \int_1^2 \sqrt{\frac{(4x^4+1)^{2}}{16x^4} } \ \mathrm{d}x \\ &= \int_1^2x^2+\frac{1}{x^2} \ \mathrm{d}x \\ &= \frac{7}{3} +\frac{1}{8} = \frac{59}{24}. \end{aligned}$$

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I think it is easy to see that while $y=x^3/3+1/4x$ then $$\sqrt{1+y'^2}=\frac{1}4\sqrt{\frac{(4x^4+1)^2}{x^4}}=\frac{4x^4+1}{4x^2}, ~ (1<x<2)$$

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