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

##### 1 Answer

Dec 30, 2016

#### Explanation:

The Arc Length of curve

# L = int_a^b sqrt(1+f'(x)^2) \ dx #

So with

# \ \ \ \ \ f'(x) = (x)(1/x) + (1)(lnx) #

# :. f'(x) = 1 + lnx #

And so the required Arc Length is given by:

# L = int_1^(e^2) sqrt(1+(1 + lnx)^2) \ dx #

# \ \ = int_1^(e^2) sqrt(1+(1 + 2lnx + (lnx)^2)) \ dx #

# \ \ = int_1^(e^2) sqrt(2 + 2lnx + (lnx)^2) \ dx #

Using Wolfram Alpha this integral evaluates to: