<s>[INST] Integrals, arc length I found this on my test today, and i didn't manage to solve it.
My question is.
How to find arc length of $f(x) = x^3/3$ from $x=1$ to $x=2$?
If I use formula for arc length which is $\ell = \int_a^b \sqrt{1 + (f'(x))^2} \, dx$ where is $f'(x)=x^2$ I will get $\ell = \int_1^2 \sqrt{1+x^4} \, dx$
I used everything I could (know), partition and substitution, but nothing works...
How to solve this definite integral?
$$\ell =\int_1^2 \sqrt{1+x^4} \, dx$$
 [/INST] We have:
$$\ell = \int_{1}^{2}x\,\sqrt{x^2+\frac{1}{x^2}}\,dx=\int_{1}^{2}x\,\sqrt{\left(x-\frac{1}{x}\right)^2+2}\,dx$$
and by replacing $x-\frac{1}{x}$ with $t$ we have:
$$\ell = \frac{1}{4}\int_{0}^{3/2}\left(t+\sqrt{4+t^2}\right)^2\sqrt{\frac{2+t^2}{4+t^2}}\,dt$$
hence:
$$\ell = \frac{17\sqrt{17}}{48}-\frac{\sqrt{2}}{3}+\frac{1}{2}\int_{0}^{3/2}(2+t^2)^{3/2}(4+t^2)^{-1/2}\,dt $$
but:
$$\begin{eqnarray*} \frac{1}{2}\int_{0}^{3/2}(2+t^2)^{3/2}(4+t^2)^{-1/2}\,dt &=& \frac{1}{4}\int_{0}^{9/4}(2+t)^{3/2}(t(4+t))^{-1/2}\,dt\\&=&\sqrt{2}\int_{0}^{9/8}(1+t)^{3/2}((t+1)^2-1)^{-1/2}\\ &=& \frac{1}{4}\int_{0}^{9/4}(2+t)^{3/2}(t(4+t))^{-1/2}\,dt\\&=&\sqrt{2}\int_{1}^{17/8}t^{3/2}(t^2-1)^{-1/2}\,dt\end{eqnarray*}$$
that can be written as an incomplete beta function if we replace $t$ with $\frac{1}{u}$.
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