# Finding Arc Length of a curve

Okay so just now getting into this. It's a rather straight forward topic, but it seems that a lot of questions have a quick and easy way of being solved. I found where that definite integral in the end just becomes the original, but where if it were something like $x + 1$, it would become $x-1$. Here is an example.

$$\,y= \frac{x^3}3 + \frac1{4x} \ \text{ where }\ 1 \leq x \leq \,2$$ taking the derivative gives you $$\,f(x)' = x^2 - \frac1{4x^2}$$ Then plugging that into the arc length formula you get $$\int_{1}^{2} \sqrt{1+(x^2-\frac1{4x^2})^2}\,dx$$ After finally doing the algebra and the integral, it comes out to this $$\frac{x^3}3-\frac1{4x}$$ from there we just plug in from $2$ to $1$ and we get our answer. Though the biggest thing to me was that it took a few minutes of work to get from the beginning to here even though the only thing that changes was we went from $$\frac{x^3}3 + \frac1{4x} \rightarrow \frac{x^3}3 - \frac1{4x}$$ Is this a common thing to happen in the simpler arc length calculations? Is it something I should look for or should I always do the work to make sure I get the correct answer. If it is something that has rules I could follow it could save me quite some time. Anyways, I couldn't find anywhere else where someone pointed this out (not saying it hasn't been answered here) so I figured I'd just ask. Thanks!

• i have under the squar root $$\left(x^2+\frac{1}{4x^2}\right)^2$$ Commented Oct 21, 2015 at 19:26
• ''doing the algebra and the integral'' is the hard work....and here you have some mistake. If you write what have you done we can help you. Anyway see: wolframalpha.com/input/… Commented Oct 21, 2015 at 19:29

It's all a matter of rewriting to get a very simple integral:\begin{align*}\int_{1}^{2} \sqrt{1+\left(x^2-\frac1{4x^2}\right)^2}\,dx&=\int_{1}^{2} \sqrt{1+x^4-\frac{1}{2}+\frac1{16x^4}}\,dx\\[1ex] &=\int_{1}^{2} \sqrt{x^4+\frac{1}{2}+\frac1{16x^4}}\,dx\\[1ex] &=\int_{1}^{2} \sqrt{\left(x^2+\frac{1}{4x^2}\right)^2}\,dx\\[1ex] &=\int_{1}^{2} \left|x^2+\frac{1}{4x^2}\right|\,dx\\[1ex] &=\int_{1}^{2} \left(x^2+\frac{1}{4x^2}\right)\,dx\end{align*}