# How do you find the arc length of the curve f(x)=coshx over the interval [0, 1]?

Oct 28, 2016

I used WolframAlpha
s = int_0^1 sqrt(1 + sinh^2(x))dx ≈ 1.1752

#### Explanation:

From the reference on Arc Length we write the equation:

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

Given: $a = 0 , b = 1 , \mathmr{and} y = \cosh \left(x\right)$

We know that $\frac{\mathrm{dy}}{\mathrm{dx}} = \sinh \left(x\right)$

The arc length integral is:

s = int_0^1 sqrt(1 + sinh^2(x))dx ≈ 1.1752