# What is the arc length of the curve given by y = ln(x)/2 - x^2/4  in the interval x in [2,4]?

Mar 7, 2018

The arc length is $\frac{1}{2} \ln 2 + 3$ units.

#### Explanation:

$y = \frac{1}{2} \ln x - \frac{1}{4} {x}^{2}$

$y ' = \frac{1}{2} \left(\frac{1}{x} - x\right)$

Arc length is given by:

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

Factor out the constant and expand:

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

Simplify:

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

Factorize:

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

Simplify:

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

Integrate term by term:

$L = \frac{1}{2} {\left[\ln x + \frac{1}{2} {x}^{2}\right]}_{2}^{4}$

Insert the limits of integration:

$L = \frac{1}{2} \ln 2 + 3$