# Find the length of the curve $y = \frac{x^4}{16} + \frac{1}{2x^2}$

$$y = \frac{x^4}{16} + \frac{1}{2x^2}$$, $$1 \leq x \leq 2$$

$$\frac{dy}{dx} = \frac{x^3}{4} - \frac{1}{x^3}dx$$ \begin{align} L &= \int_{1}^{2} \sqrt{1 + \bigg[ \frac{x^3}{4} - \frac{1}{x^3} \bigg]^2}dx \\ &= \int_{1}^{2} \sqrt{1 + \bigg[ \frac{x^6-4}{4x^3}\bigg]^2}dx \\ &= \int_{1}^{2} \sqrt{\frac{(4x^3)^2 + (x^6-4)^2}{(4x^3)^2}}dx \end{align}

I tried splitting the $$\sqrt{~~}$$ into the numerator and denominator, but that didn't really get me anywhere and I am stuck. Can someone give me a hint in the right direction possibly?

edit: continuing on...

\begin{align} L &= \int_{1}^{2} \sqrt{\frac{(x^6+4)^2}{(4x^3)^2}}dx \\ &= \int_{1}^{2} \frac{(x^6+4)}{(4x^3)} \\ &= 4 \int_{1}^{2} \frac{x^6+4}{x^3}dx \\ &= 4\int_{1}^{2} \bigg(x^3 + \frac{4}{x^3}\bigg)dx \\ &= 4 \bigg[ \frac{1}{4}x^4 - \frac{2}{x^2} \bigg]_{1}^{2} \\ &= 4 \bigg[ (4-\frac{1}{2}) - (\frac{1}{4} - 2)\bigg] \\ &= 21 \end{align}

However wolphram alpha says the answer is actually $$\frac{21}{16}$$.

## 1 Answer

$$(4x^3)^2 + (x^6-4)^2$$ $$=16x^6 + x^{12} + 16 - 8x^6$$ $$=(x^6+4)^2$$

So that square root will open up straight away.

• ah, I factored $(x^6+4)^2$ wrong, did it too hastily. I see my error, thank you! – Evan Kim May 8 '19 at 21:22
• Yeah sure. No problem! – Vizag May 8 '19 at 21:23
• Yeah why did you send that $4$ in the numerator? – Vizag May 8 '19 at 22:00
• $= \int_{1}^{2} \frac{(x^6+4)}{(4x^3)}$... your error is after this line when you mistakenly take $4$ in the numerator – Vizag May 8 '19 at 22:02
• Yeah sure. Please mark the answer as accepted. :) – Vizag May 12 '19 at 22:58