3
$\begingroup$

$y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$

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

squaring this $=\frac{25}{36}x^{8}+\frac{9}{100x^{8}}$

Plugging into the formula $ds=\sqrt{1+\left( \frac{dy}{dx}\right) ^{2}}$

$\int_{1}^{2}\sqrt{1+\frac{25}{36}x^{8}+\frac{9}{100x^{8}}}$

Is this correct so far? And how would I go about evaluating this integral.

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    @Krysten: Yes. It should be $\int_{1}^{2}\sqrt{\frac{25}{36}x^{8}+\frac{1}{2}+\frac{9}{100x^{8}}}dx$2011-02-27

1 Answers 1

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As Américo Tavares pointed out, you are missing the term $-1/2$ due to the crossterm when squaring $d y/ dx$. The integral you want to solve is $\int_1^2 dx \,\sqrt{\frac{1}{2}+ \frac{25 x^8}{36} + \frac{9}{100 x^8}} = \int_1^2 dx \, \sqrt{\frac{(9+25 x^8)^2}{900 x^8}}$ $ = \int_1^2 dx \,\frac{9+25 x^8}{30 x^4} = \left[-\frac{1}{10 x^3} + \frac{x^4}{6} \right]_{x=1}^2 = \frac{1261}{240}. $

I hope every step is reproducible.

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    I corrected the typo...2011-02-27