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

Question

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

Scarlett Allison · Accepted Answer

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

Luke Alvarez · Answer

undefined To find the arc length of the curve $ f(x) = x - xe^{x^2} $ on the interval $ x \in [2, 4] $, you can use the formula for arc length:

$$ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$

First, find $ \frac{dy}{dx} $ which is the derivative of $ f(x) $:

$$ \frac{dy}{dx} = 1 - e^{x^2}(1 + 2x^2) $$

Then, substitute $ \frac{dy}{dx} $ into the arc length formula:

$$ L = \int_2^4 \sqrt{1 + \left(1 - e^{x^2}(1 + 2x^2)\right)^2} \, dx $$

After evaluating this integral, you'll get the arc length of the curve $ f(x) = x - xe^{x^2} $ on the interval $ x \in [2, 4] $.