1. ## Lengths of Curves

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.

2. Sorry I get

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

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

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

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

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

5. :O thx!