What is the arclength of #f(x)=ln(x+3)# on #x in [2,3]#?

Question

Elijah Abram · Accepted Answer

#L=1+sum_(n=1)^oo((1/2),(n))1/(2n-1)(1/5^(2n-1)-1/6^(2n-1))# units.

#f(x)=ln(x+3)# #f'(x)=1/(x+3)# Arclength is given by: #L=int_2^3sqrt(1+1/(x+3)^2)dx# Apply the substitution #x+3=u#: #L=int_5^6sqrt(1+1/u^2)du# For #u in [5,6]#, #1/u^2<1#. Take the series expansion of the square root: #L=int_5^6sum_(n=0)^oo((1/2),(n))1/u^(2n)du# Isolate the #n=0# term and simplify: #L=int_5^6du+sum_(n=1)^oo((1/2),(n))int_5^6 1/u^(2n)du# Integrate directly: #L=1+sum_(n=1)^oo((1/2),(n))[(-1)/((2n-1)u^(2n-1))]_5^6# Simplify: #L=1+sum_(n=1)^oo((1/2),(n))1/(2n-1)(1/5^(2n-1)-1/6^(2n-1))#

Axel Alvarez · Answer

undefined To find the arc length of $ f(x) = \ln(x + 3) $ on the interval $[2,3]$, we use the formula for arc length:

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

where $ a = 2 $, $ b = 3 $, and $ \frac{dy}{dx} $ is the derivative of $ f(x) $.

First, find the derivative of $ f(x) $:

$$ f'(x) = \frac{1}{x + 3} $$

Now, substitute into the arc length formula:

$$ L = \int_{2}^{3} \sqrt{1 + \left(\frac{1}{x + 3}\right)^2} \, dx $$

$$ = \int_{2}^{3} \sqrt{1 + \frac{1}{(x + 3)^2}} \, dx $$

This integral may not have a simple closed-form solution. Numerical methods, such as numerical integration or approximation, may be used to find the value of $ L $.

Christopher Allers · Answer

undefined The arc length of the function f(x) = ln(x + 3) on the interval [2, 3] is approximately 1.45 units.