# How do you set up an integral from the length of the curve y=1/x, 1<=x<=5?

Feb 25, 2017

${\int}_{1}^{5} \sqrt{1 + \frac{1}{x} ^ 4}$

#### Explanation:

The length of the curve of $y$ on $a < = x < = b$ is equal to:

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

For $y = \frac{1}{x}$, we see that $y = {x}^{-} 1$ so $\frac{\mathrm{dy}}{\mathrm{dx}} = - {x}^{-} 2 = - \frac{1}{x} ^ 2$. Then:

$L = {\int}_{1}^{5} \sqrt{1 + {\left(- \frac{1}{x} ^ 2\right)}^{2}} = {\int}_{1}^{5} \sqrt{1 + \frac{1}{x} ^ 4}$

Which can be plugged into a calculator for a result of $4.15145$.