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

1 Answer
Dec 30, 2016

# 16.168917 ... #

Explanation:

The Arc Length of curve #y=f(x)# is calculated using the formula:

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

So with #f(x)=xln(x)#, we can apply the product rule to get:

# \ \ \ \ \ 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:

# L = int_1^(e^2) sqrt(2 + 2lnx + (lnx)^2) \ dx = 16.168917 ... #