Question

What is the arc length of `f(x) = x-xe^(x^2)` on `x in [ 2,4]`?

Answer 1

`int_2^4sqrt(1+(1-e^(x^2)-2x^2e^(x^2))^2)\ dx~~2.93`

Explanation:

The formula for arc length of `f(x)` on the interval `[a,b]` is:
`int_a^bsqrt(1-(f'(x))^2)\ dx`

First we'll work out the derivative of our function:
`d/dx(x-xe^(x^2))=1-d/dx(xe^(x^2))`

To figure out the second part, we'll use the product rule:
`d/dx(f(x)g(x))=f'(x)g(x)+f(x)g'(x)`

In our case, we get:
`1-(1*e^(x^2)+x*d/dx(e^(x^2)))`

We can use the chain rule, which says:
`f(g(x))=f'(g(x))*g'(x)`

This gives us:
`1-(e^(x^2)+x*2xe^(x^2))`

`1-e^(x^2)-2x^2e^(x^2)`

If we then plug this into our formula, we get:
`int_2^4sqrt(1+(1-e^(x^2)-2x^2e^(x^2))^2)\ dx`

This integral doesn't have an elementary answer that I've been able to find, but the value can be approximated to be roughly `2.93`