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

1 Answer
Jan 1, 2018

#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#