To find the area of the region bounded by the graphs of $f(x) = x^2$ and $g(x) = x^3 + 9$ between $x = -1$ and $x = 1$, we need to calculate the integral of the absolute difference between these functions over this interval.

Steps:

1. Determine the points of intersection: Set $f(x) = g(x)$: $x^2 = x^3 + 9$ Rearrange to get: $x^3 - x^2 + 9 = 0$ There are no real roots in the interval $[-1, 1]$, so there are no intersections between these two curves in this interval.

2. Identify which function is on top: For the interval $[-1, 1]$, we check at a few points:

   • At $x = 0$: $f(0) = 0$ and $g(0) = 9$. Clearly, $g(x) > f(x)$.
   • At $x = 1$: $f(1) = 1$ and $g(1) = 10$. Again, $g(x) > f(x)$.

   Hence, $g(x) \geq f(x)$ over the entire interval $[-1, 1]$.

3. Calculate the area: The area $A$ is given by: [ A = \int_{-1}^{1} \left[ g(x) - f(x) \right] , dx = \int_{-1}^{1} \left[ (x^3 + 9) - x^2 \right] , dx ]

4. Evaluate the integral: $A = \int_{-1}^{1} \left( x^3 - x^2 + 9 \right) \, dx$ Calculate the integral separately for each term: $\int_{-1}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{-1}^{1} = 0$ $\int_{-1}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}$ $\int_{-1}^{1} 9 \, dx = 9 \left[ x \right]_{-1}^{1} = 9(1 - (-1)) = 18$

   Now, sum up the parts: $A = 0 - \frac{2}{3} + 18 = \frac{-2}{3} + \frac{54}{3} = \frac{52}{3} \approx 17.33$

Final Answer:

The area of the region is approximately $17.33$.

Would you like more details or have any questions?

Here are 5 related questions to further explore this topic:

1. How do you find the area between two curves with intersections at multiple points?
2. What if the functions intersect within the interval, how would that change the integral calculation?
3. Can you explain the geometric meaning of definite integrals in this context?
4. How does the sign of the integrand affect the result when calculating areas?
5. How would the area change if the interval were different?

Tip: Always sketch the curves if possible to get a visual understanding of which function is on top and how the region looks like.