Question

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

Answer 1

`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 ...`