# What is the arc length of f(x)=xlnx  in the interval [1,e^2]?

Dec 30, 2016

$16.168917 \ldots$

#### Explanation:

The Arc Length of curve $y = f \left(x\right)$ is calculated using the formula:

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

So with $f \left(x\right) = x \ln \left(x\right)$, we can apply the product rule to get:

$\setminus \setminus \setminus \setminus \setminus f ' \left(x\right) = \left(x\right) \left(\frac{1}{x}\right) + \left(1\right) \left(\ln x\right)$
$\therefore f ' \left(x\right) = 1 + \ln x$

And so the required Arc Length is given by:

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

Using Wolfram Alpha this integral evaluates to:

$L = {\int}_{1}^{{e}^{2}} \sqrt{2 + 2 \ln x + {\left(\ln x\right)}^{2}} \setminus \mathrm{dx} = 16.168917 \ldots$