What is the arclength of #f(x)=3x^2-x+4# on #x in [2,3]#?

Question

Christopher Allers · Accepted Answer

#L=1/12(17sqrt290-11sqrt122)+1/12ln((17+sqrt290)/(11+sqrt122))# units.

#f(x)=3x^2−x+4# #f'(x)=6x−1# Arclength is given by: #L=int_2^3sqrt(1+(6x-1)^2)dx# Apply the substitution #6x-1=tantheta#: #L=1/6intsec^3thetad theta# This is a known integral: #L=1/12[secthetatantheta+ln|sectheta+tantheta|]# Reverse the substitution: #L=1/12[(6x-1)sqrt(1+(6x-1)^2)+ln|(6x-1)+sqrt(1+(6x-1)^2)|]_2^3# Hence #L=1/12(17sqrt290-11sqrt122)+1/12ln((17+sqrt290)/(11+sqrt122))#

Aurora Alvarado · Answer

undefined To find the arc length of the function $ f(x) = 3x^2 - x + 4 $ on the interval $[2,3]$, follow these steps:

1. Find the derivative of the function, $ f'(x) $.
2. Use the formula for arc length: $ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx $, where $ a $ and $ b $ are the endpoints of the interval.
3. Evaluate the integral from step 2 over the interval $[2,3]$ to find the arc length.

First, find the derivative of $ f(x) $:
$$ f'(x) = 6x - 1 $$

Now, plug $ f'(x) $ into the arc length formula:
$$ L = \int_{2}^{3} \sqrt{1 + (6x - 1)^2} \, dx $$

Integrate $ \sqrt{1 + (6x - 1)^2} $ over the interval $[2,3]$ to find the arc length. This integration may require techniques such as trigonometric substitution or integration by parts.

After integrating and evaluating the integral, you will find the arc length $ L $ of the function $ f(x) = 3x^2 - x + 4 $ on the interval $[2,3]$.