To find the area between the curves $f(x) = x^3 - 2$ and $g(x) = x^2 - 2$ between the lines $x = -2$ and $x = 2$, you need to calculate the definite integral of the difference between these two functions over the given interval.

Step 1: Set Up the Integral

The area $A$ between the two curves from $x = -2$ to $x = 2$ is given by: $A = \int_{-2}^{2} \left[g(x) - f(x)\right] \, dx$ where:

• $g(x) = x^2 - 2$
• $f(x) = x^3 - 2$

So, the integral becomes: $A = \int_{-2}^{2} \left[(x^2 - 2) - (x^3 - 2)\right] \, dx$ This simplifies to: $A = \int_{-2}^{2} \left(x^2 - x^3\right) \, dx$

Step 2: Compute the Integral

Now, integrate each term separately: $\int \left(x^2 - x^3\right) \, dx = \int x^2 \, dx - \int x^3 \, dx$

This results in: $\frac{x^3}{3} - \frac{x^4}{4}$

Step 3: Evaluate the Definite Integral

Now, evaluate the definite integral from $x = -2$ to $x = 2$: $A = \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_{-2}^{2}$

Substituting the limits:

• At $x = 2$: $\frac{2^3}{3} - \frac{2^4}{4} = \frac{8}{3} - 4 = \frac{8}{3} - \frac{12}{3} = -\frac{4}{3}$
• At $x = -2$: $\frac{(-2)^3}{3} - \frac{(-2)^4}{4} = \frac{-8}{3} - 4 = -\frac{8}{3} - \frac{12}{3} = -\frac{20}{3}$

Subtracting: $A = \left(-\frac{4}{3}\right) - \left(-\frac{20}{3}\right) = \frac{16}{3}$

Step 4: Result

The area between the curves is: $A = \frac{16}{3} \text{ square units}$

This corresponds to option C: $\frac{16}{3}$.

Would you like more details or have any questions about this problem?

Here are 5 related questions:

1. How do you determine which function is on top when calculating the area between curves?
2. What happens if the interval was from $x = -3$ to $x = 3$?
3. How can you use symmetry to simplify the calculation of areas between curves?
4. What is the significance of finding the points of intersection when calculating the area between curves?
5. How would the integral change if the curves were defined parametrically?

Tip: Always sketch the graphs of the functions to visually understand which function is above the other in the given interval. This helps avoid mistakes in setting up the integral.