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# How do you find the arc length of the curve x=y+y^3 over the interval [1,4]?

Nov 1, 2016

I used WolframAlpha to do the integration:
$L = {\int}_{1}^{4} \sqrt{1 + {\left(1 + 3 {y}^{2}\right)}^{2}} \mathrm{dy} = 66.1093$

#### Explanation:

From the reference Arc Length

$L = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}^{2}} \mathrm{dy}$

$\frac{\mathrm{dx}}{\mathrm{dy}} = 1 + 3 {y}^{2}$

$L = {\int}_{1}^{4} \sqrt{1 + {\left(1 + 3 {y}^{2}\right)}^{2}} \mathrm{dy}$

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