Question

(a) Use a graphing utility to graph the function \(f(x)=x^{2 / 3}\). (b) Can you integrate with respect to \(x\) to find the arc length of the graph of \(f\) on the interval [-1,8]\(?\) Explain. (c) Find the arc length of the graph of \(f\) on the interval [-1,8] .

Short answer

Arc length of the graph of \(f(x)=x^{2/3}\) over interval [-1,8] is given by sum of \(L=\int_{-1}^{0} \sqrt{1+(f'(x))^2} dx + \int_{0}^{8} \sqrt{1+(f'(x))^2} dx\), after substitution of \(f'(x)\) and performing the integrations.

Step-by-step solution

Graph the function

Plot the function \(f(x)=x^{2/3}\) to understand its shape and behavior. This can be done by either generating a set of (x, f(x)) points and plotting them or using a graphing utility.

Assess the integration for arc length

Considering the equation for the arc length of the function \(f\) on the interval [a,b] is given by \(L = \int_{a}^{b} \sqrt{1+(f'(x))^2} dx\), where \(f'(x)\) is the derivative of the function. Deriving \(f(x)=x^{2/3}\) gets \(f'(x) = \frac{2}{3} x^{-1/3}\). Function \(f'(x)\) is not defined at x=0 , hence technically speaking, arc length can not be found by directly applying integral on the given interval [-1,8] without considering that.

Calculate the arc length

To avoid the undefined point at x=0, break the interval into two: [-1,0) and (0,8]. Compute the arc length for these two intervals and add them. This gives \(L=\int_{-1}^{0} \sqrt{1+(f'(x))^2} dx + \int_{0}^{8} \sqrt{1+(f'(x))^2} dx\). After substitution of \(f'(x)\), perform the integrations and sum the results.