<s>[INST] Find the arc length of a path I'm trying to solve the arc length for the following. $c(t) = (2, 6t^2, 4t^3)$ from $0\le t\le 1.$
I've checked on WolframAlpha and I get the answer $8\sqrt2 - 4$ but when I work it out I get $16\sqrt2$ where am I going wrong?
\begin{align*}
L&= \int_0^1 \sqrt{(12t)^2 + (12t^2)^2}\, dt\\
&= \int_0^1 \sqrt{144t^2 + 144t^4}\, dt\\
&= \int_0^1 12t\sqrt{1 + t^2}\,dt
\end{align*}
Then I do
$$ 8t\sqrt{(1+t^2)^3} = 16\sqrt2.$$ 
 [/INST] From $$\int_{0}^{1}12t\sqrt{1+t^2}dt$$
Use $u=1+t^2\implies \frac{1}{2}du=tdt$ to get $$\int_{0}^{1}12t\sqrt{1+t^2}dt=\int12\sqrt{u}\cdot\frac{1}{2}du=6u^{3/2}\cdot\frac{2}{3}=4(1+t^2)^{3/2}\bigg\vert_{0}^{1}=4\cdot2^{3/2}-4\cdot1$$ $$=4\cdot2\cdot2^{1/2}-4\cdot1=8\sqrt{2}-4$$
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