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


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

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


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

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It is 1/(10x^3) –  Krysten Feb 27 '11 at 17:05
$\left( \dfrac{5}{6}x^{4}-\dfrac{3}{10x^{4}}\right) ^{2}=\dfrac{25}{36}x^{8}-\dfrac{1}{2}+\dfrac{9}{100x^{8}}$ –  Américo Tavares Feb 27 '11 at 17:22
@Tavares where does the 1/2 come from? –  Krysten Feb 27 '11 at 17:27
@Krysten from the identity $(a-b)^2=a^2-2ab+b^2$ –  Américo Tavares Feb 27 '11 at 17:30
ok now i see. so then the integral would be sqr[(25x^8)/36 + 1/2 + 9/(100x^8)] –  Krysten Feb 27 '11 at 17:37

1 Answer 1

up vote 2 down vote accepted

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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+1 for evaluating the radical. How did you recognize it? –  Américo Tavares Feb 27 '11 at 18:16
how did you get the second qauntity exactly? –  Krysten Feb 27 '11 at 18:17
@Krysten: Put all three terms under the square-root together with the common denominator $900x^8$. You then get $(625 x^{16} + 450 x^{8} +81)/900 x^8$ under the square-root. Then you use Américo Tavares' formula (usually is called binomial) $(a+b)^2 = a^2 + 2ab + b^2$ backwards with $a=25 x^8$ and $b=9$. –  Fabian Feb 27 '11 at 18:27
@Américo Tavares: Mathematica helps ;-) –  Fabian Feb 27 '11 at 18:28
Thanks. I finally understand, but shouldn't the evaluated integral be: -1/(10x^3) + x^5/6? –  Krysten Feb 27 '11 at 18:49

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