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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    It is 1/(10x^3)2011-02-27
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    $\left( \dfrac{5}{6}x^{4}-\dfrac{3}{10x^{4}}\right) ^{2}=\dfrac{25}{36}x^{8}-\dfrac{1}{2}+\dfrac{9}{100x^{8}}$2011-02-27
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    @Tavares where does the 1/2 come from?2011-02-27
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    @Krysten from the identity $(a-b)^2=a^2-2ab+b^2$2011-02-27
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    ok now i see. so then the integral would be sqr[(25x^8)/36 + 1/2 + 9/(100x^8)]2011-02-27
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    @Krysten: Is the text of your question "$y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}$, $1\leq x\leq 2$ $\frac{dy}{dx}=\frac{5}{6}x^{4}-\frac{3}{10x^{4}}$ 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."?2011-02-27
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    @Tavares, yes it is2011-02-27
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    @Krysten: Yes, you get $$\int_{1}^{2}\left( \frac{25}{36}x^{8}+\frac{1}{2}+\frac{9}{100x^{8}}\right) dx$$ instead.2011-02-27
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    is there supposed to be a sqr root?2011-02-27
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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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    +1 for evaluating the radical. How did you recognize it?2011-02-27
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    how did you get the second qauntity exactly?2011-02-27
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    @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 $b=9$.2011-02-27
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    @Américo Tavares: Mathematica helps ;-)2011-02-27
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    Thanks. I finally understand, but shouldn't the evaluated integral be: -1/(10x^3) + x^5/6?2011-02-27
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    @Krysten: That's right!2011-02-27
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    great. thanks for your help!2011-02-27
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    I corrected the typo...2011-02-27