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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 :)

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Well, it's clearly longer than the vertical line from $(4,16)$ down to $(4,0)$, so something longer than $16$ is good. Perhaps you thought that the endpoint was $(4,4)$, rather than the correct value of $(4,16)$? –  David Speyer Jan 26 '12 at 23:32

2 Answers 2

up vote 4 down vote accepted

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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Connect $(0, 0)$ and $(4, 16)$. The arch between these points must be longer than the straight line. Let's calculate the length of the straight line:

$\sqrt{4^2 + 16^2} = 4\sqrt{17} \approx 16.4924$

Thus, the arch length must be greater than $16.4924$.

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