# How do you find the arc length of the curve y=x^2/2 over the interval [0, 1]?

Sep 5, 2016

$\approx 2.96$ units

#### Explanation:

$s = {\int}_{0}^{2} \sqrt{1 + {\left(y '\right)}^{2}} \setminus \mathrm{dx}$

$= {\int}_{0}^{2} \sqrt{1 + {x}^{2}} \setminus \mathrm{dx}$

$= \frac{1}{2} {\left[\left(x \sqrt{{x}^{2} + 1} + {\sinh}^{- 1} \left(x\right)\right)\right]}_{0}^{2}$

$= \left(\sqrt{5} + \frac{1}{2} {\sinh}^{- 1} \left(2\right)\right)$ units

$\approx 2.96$ units

Computer used for integration and numerical solution

[The basic integral, $\int \sqrt{1 + {x}^{2}} \setminus \mathrm{dx}$...

... can be approached, using the identity ${\cosh}^{2} z - {\sinh}^{2} z = 1$...

....so by the sub $x = \sinh z , \mathrm{dx} = \cosh z \setminus \mathrm{dz}$

... and then maybe a hyperbolic double angle formula]