The problem asks to set up an integral that represents the length of the part of the parametric curve shown in the graph. Given parametric equations $x = 9t^2 - 3t^3$ and $y = 4t^2 - 8t$, the arc length formula for a parametric curve is:

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

Step-by-step approach:

1. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

   • $\frac{dx}{dt} = \frac{d}{dt}(9t^2 - 3t^3) = 18t - 9t^2$
   • $\frac{dy}{dt} = \frac{d}{dt}(4t^2 - 8t) = 8t - 8$

2. Set up the arc length integral: The arc length integral becomes: $L = \int_{a}^{b} \sqrt{(18t - 9t^2)^2 + (8t - 8)^2} \, dt$

3. Determine the limits of integration: From the graph, it appears that the curve corresponds to $t$ ranging from $t = -1$ to $t = 3$. So, the limits of integration are $-1$ and $3$.

4. Write the final integral: The integral to calculate the arc length is: $L = \int_{-1}^{3} \sqrt{(18t - 9t^2)^2 + (8t - 8)^2} \, dt$

Now, you can use a calculator or computer to evaluate this integral numerically to four decimal places.

Would you like help calculating the final value or need further clarification?

Here are 5 related questions to expand your understanding:

1. How does the parametric arc length formula generalize to other parametric curves?
2. Can you graph the parametric curve using software to visualize the arc more clearly?
3. How do the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$ affect the curvature of the parametric curve?
4. What happens to the length if the limits of integration change?
5. How can this arc length problem be solved using polar or Cartesian coordinates if the curve was given differently?

Tip: When solving arc length problems, always check if simplifying the derivative expressions can help in solving the integral faster.