Lengths of Curves

**lilaziz1** · April 3rd 2010, 02:42 PM

Hello everyone. Much to my chagrin, I am having an algebraic problem. Here is the question:

Question:

Find the length of the curve analytically by anti-differentiation. You will need to simplify the integrand algebraically before finding an anti-derivative.

$y = \frac{x^3}{3} + x^2 + x + \frac{1}{4x + 4}, 0 \leq x \leq 2$

Solution:

$L = \int_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^2} dx$

$\frac{dy}{dx} = x^2 + 2x + 1 - \frac{1}{4(x + 1)^2} = (x+1)^2 - \frac{1}{4(x+1)^2}$

$L = \int_{0}^{2} \sqrt{1 + ( (x+1)^2 - \frac{1}{4(x+1)^2} )^2}$

I understand everything up to this point. In the answer, this is what it does after the prior step:

$= \int_{0}^{2} \sqrt{( (x+1)^2 + \frac{1}{4(x+1)^2} )^2}$

What happened to the 1? What happened to the negative sign? Any help is appreciated. Thanks in advance.

**zzzoak** · April 3rd 2010, 03:12 PM

Sorry I get

$\frac{dy}{dx}$ = $\frac{x^2}{2}+2x+1-\frac{1}{4{(x+1)}^2}$

**lilaziz1** · April 3rd 2010, 05:06 PM

Oh. Sorry. I mistyped. :s ment to write x^3/3

**chiph588@** · April 3rd 2010, 05:37 PM

Originally Posted by lilaziz1

$L = \int_{0}^{2} \sqrt{1 + ( (x+1)^2 - \frac{1}{4(x+1)^2} )^2}$

I understand everything up to this point. In the answer, this is what it does after the prior step:

$= \int_{0}^{2} \sqrt{( (x+1)^2 + \frac{1}{4(x+1)^2} )^2}$

What happened to the 1? What happened to the negative sign? Any help is appreciated. Thanks in advance.

$1 + ( (x+1)^2 - \frac{1}{4(x+1)^2} )^2 = 1+(x+1)^4-2\frac{(x+1)^2}{4(x+1)^2}+\frac{1}{16(x+1)^4}$

$= 1+(x+1)^4-\frac12+\frac{1}{16(x+1)^4} =(x+1)^4+\frac12+\frac{1}{16(x+1)^4}$

$= (x+1)^4+2\frac{(x+1)^2}{4(x+1)^2}+\frac{1}{16(x+1) ^4} = \left((x+1)^2+\frac{1}{4(x+1)^2}\right)^2$

**lilaziz1** · April 3rd 2010, 05:46 PM

:O thx!

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