<s>[INST] The length of curves. I can't see what I'm doing wrong here, it's very simple.
The length of the curve $x^2$ between $0$ and $4$ is $\int_0^4 \sqrt{(1+4x^2)}dx$ isn't it?
I don't know how to calculate this but wolframalpha tells me that it's about $16.8$ units.  That can't be right: maybe it's an optical illusion but I have it drawn up on my board and the curve doesn't look any longer than 8 units.
What am I doing wrong?
EDIT: You know what? I realised my mistake: my y-axis is scaled. Doh!  
Feel free to delete this question :)
 [/INST] Integrating by parts:
$$
  \int \sqrt{1+4x^2} \mathrm{d x} = x \sqrt{1+4x^2} - \int \frac{4 x^2}{\sqrt{1+4x^2}} \mathrm{d} x = x \sqrt{1+4x^2} - \int \sqrt{1+4x^2} \mathrm{d} x + \int \frac{\mathrm{d}x }{\sqrt{1+4x^2}}
$$
Solving for $\int \sqrt{1+4x^2} \mathrm{d} x$:
$$
   \int \sqrt{1+4x^2} \mathrm{d} x = \frac{x}{2} \sqrt{1+4x^2} + \frac{1}{2} \int \frac{\mathrm{d}x }{\sqrt{1+4x^2}} = \frac{x}{2} \sqrt{1+4x^2} + \frac{1}{4} \operatorname{arcsinh}(2x)
$$
This gives
$$
  \int_0^4 \sqrt{1+4x^2} \mathrm{d} x = 2 \sqrt{65} + \frac{1}{4} \operatorname{arcsinh}(8)
  \approx 16.8186
$$
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