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Mathematics 18 Online
OpenStudy (anonymous):

What is the arc length of x^(3/2) on the interval [2,3]... having trouble any help ?

OpenStudy (anonymous):

You need to integrate to get the length... it's been a while since I did one of these. Lemme research it real quick.

OpenStudy (anonymous):

\begin{eqnarray*}s &=& \int_2^3 \sqrt{1 + f'(x)^2}dx \\&=&\int_2^3 \sqrt{1+\left(\frac{3}{2} \sqrt{x}\right)^2}dx \\&=&\int_2^3\sqrt{1+\frac{9}{4}x}dx \\&=&\frac{1}{27}\left(31\sqrt{31}-22\sqrt{22}\right).\end{eqnarray*}

OpenStudy (anonymous):

Yeah. That's the equation I found for how to do this. The integral from 2 to 3 of sqrt(1 + f'(x)^2)

OpenStudy (anonymous):

f'(x) is (3/2)x^(1/2)

OpenStudy (anonymous):

So yeah. Follow what he put, that makes your equation \[\int\limits_{2}^{3}\sqrt{1 + (9/4)x}dx\]

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