Questions: Write an integral expression that will give the length of the path given by (f(x)=-8 sin (2 x^-5)-4) from (x=3) to (x=6).

Transcript text: Write an integral expression that will give the length of the path given by $f(x)=-8 \sin \left(2 x^{-5}\right)-4$ from $x=3$ to $x=6$.

Solution

Solution Steps

Step 1: Recall the formula for arc length

The length $ L $ of a curve defined by $ y = f(x) $ from $ x = a $ to $ x = b $ is given by: \[ L = \int_{a}^{b} \sqrt{1 + \left( f'(x) \right)^2} \, dx \]

Step 2: Compute the derivative of $ f(x) $

Given $ f(x) = -8 \sin \left(2 x^{-5}\right) - 4 $, compute its derivative $ f'(x) $: \[ f'(x) = -8 \cdot \cos \left(2 x^{-5}\right) \cdot \left( -10 x^{-6} \right) = 80 x^{-6} \cos \left(2 x^{-5}\right) \]

Step 3: Substitute into the arc length formula

Substitute $ f'(x) $ into the arc length formula: \[ L = \int_{3}^{6} \sqrt{1 + \left( 80 x^{-6} \cos \left(2 x^{-5}\right) \right)^2} \, dx \]

This is the integral expression for the length of the path.