How do you set up an integral for the length of the curve y=sqrtx, 1<=x<=2?

Jul 27, 2017

arc length is:

$L = {\int}_{1}^{2} \setminus \sqrt{1 + \frac{1}{4 x}} \setminus \mathrm{dx}$

Explanation:

The arc length of a curve $y = f \left(x\right)$ over an interval $\left[a , b\right]$ is given by:

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

So for the given function:

$y = \sqrt{x}$

Then differentiating wrt $x$ we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}}$

So then the arc length is:

$L = {\int}_{1}^{2} \setminus \sqrt{1 + {\left(\frac{1}{2 \sqrt{x}}\right)}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus = {\int}_{1}^{2} \setminus \sqrt{1 + \frac{1}{4 x}} \setminus \mathrm{dx}$

NB:
If we evaluate this integral we get:

$L = 1.08306427952 \ldots$