Length between curve

**elitespart** · November 24th 2008, 05:56 PM

I need to find the length of the curve y=(1/2)x^2 - (1/4)lnx between x=1 and x=8.

I derived it and squared it and added one to it so now I'm left with the integral from 1 to 8 of sqrt (x^2 + (1/2) + 1/16x^2) dx. How do I integrate this?

**o_O** · November 24th 2008, 06:23 PM

$L = \int_1^8 \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx$

Now, as you found: $\frac{dy}{dx} = x - \frac{1}{4x}$

So:
$\begin{aligned} L & = \int_1^8 \sqrt{1 + \left(x - \frac{1}{4x}\right)^2} \ dx \\ & = \int_1^8 \sqrt{{\color{red}\left(\frac{x^2}{x^2}\right)}\le ft(x^2 + \frac{1}{16x^2} + \frac{1}{2}\right)} \ dx \\ & = \int_1^8 \sqrt{\frac{x^4 +\frac{1}{2}x^2 + \frac{1}{16}}{x^2}} \ dx \\ & = \int_1^8 \sqrt{\frac{\left(x + \frac{1}{4}\right)^2}{x^2}} \end{aligned}$

So, taking the square root and our final integral becomes: $L = \int_1^8 \frac{x + \frac{1}{4}}{x} \ dx \ = \ \int_1^8 \left(1 + \frac{1}{4x}\right) dx$

**Soroban** · November 24th 2008, 06:23 PM

Hello, elitespart!

Find the length of the curve: . $y\:=\:\tfrac{1}{2}x^2 - \tfrac{1}{4}\ln x$ between $x=1$ and $x=8.$

I derived it and squared it and added one to it.

So now I'm left with: . $L \;=\;\int^8_1\sqrt{x^2 + \tfrac{1}{2} + \tfrac{1}{16x^2}}\,dx$
How do I integrate this?

Believe it or not, there is a perfect square under the radical . . .

. . $\sqrt{x^2 + \frac{1}{2} + \frac{1}{16x^2} }\;=\;\sqrt{\left(x + \frac{1}{4x}\right)^2} \;=\;x + \frac{1}{4x} \quad\hdots$ see?

**elitespart** · November 24th 2008, 06:26 PM

Awesome! Thanks a lot guys.

**Krizalid** · November 24th 2008, 06:38 PM

It's actually $\sqrt{\left( x+\frac{1}{4} \right)^{2}\cdot \frac{1}{x^{2}}}=\left| \left( x+\frac{1}{4} \right)\cdot \frac{1}{x} \right|,$ but note that $\left( x+\frac{1}{4} \right)\cdot \frac{1}{x}>0$ for $1\le x\le8,$ hence $\left| \left( x+\frac{1}{4} \right)\cdot \frac{1}{x} \right|=\left( x+\frac{1}{4} \right)\cdot \frac{1}{x}.$

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