Key Concepts and Formulas

• Area Between Curves: If $f(x) \ge g(x)$ on $[a, b]$, the area between the curves $y=f(x)$ and $y=g(x)$ is given by $\int_a^b [f(x) - g(x)] \, dx$.
• Finding Intersection Points: To find where two curves intersect, set their equations equal to each other and solve for $x$.
• Even Function Integration: If $f(x)$ is an even function (i.e., $f(-x) = f(x)$), then $\int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx$.

Step-by-Step Solution

Step 1: Identify the Upper and Lower Bounding Curves

We are given the region $R = \{ (x,y):5{x^2} \le y \le 2{x^2} + 9\}$. This tells us that $y$ is bounded below by $y = 5x^2$ and above by $y = 2x^2 + 9$. Therefore, the lower bounding curve is $g(x) = 5x^2$ and the upper bounding curve is $f(x) = 2x^2 + 9$.

Why: Identifying the upper and lower functions is essential for setting up the integral correctly. The area between curves formula relies on subtracting the lower function from the upper function.

Step 2: Find the Points of Intersection

To find the limits of integration, we need to find the $x$-values where the curves $y = 5x^2$ and $y = 2x^2 + 9$ intersect. We set the two equations equal to each other:

$5x^2 = 2x^2 + 9$

Why: The intersection points define the interval over which the region is bounded. These $x$-values become the limits of integration in our definite integral.

Step 3: Solve for the x-coordinates of the Intersection Points

Subtract $2x^2$ from both sides: $3x^2 = 9$ Divide by 3: $x^2 = 3$ Take the square root: $x = \pm \sqrt{3}$ So, the points of intersection occur at $x = -\sqrt{3}$ and $x = \sqrt{3}$.

Why: Solving for $x$ provides the precise boundaries of our region along the x-axis, crucial for accurate area calculation.

Step 4: Set up the Definite Integral

Now we set up the definite integral to calculate the area of the region:

$\text{Area} = \int_{-\sqrt{3}}^{\sqrt{3}} [f(x) - g(x)] \, dx = \int_{-\sqrt{3}}^{\sqrt{3}} [(2x^2 + 9) - (5x^2)] \, dx$

Why: This integral represents the accumulated difference between the upper and lower curves over the interval defined by the intersection points, giving us the area of the region.

Step 5: Simplify the Integrand

Simplify the expression inside the integral:

$\text{Area} = \int_{-\sqrt{3}}^{\sqrt{3}} (9 - 3x^2) \, dx$

Why: Simplifying the integrand makes the integration process easier.

Step 6: Use Symmetry to Simplify the Integral

Notice that the integrand, $9 - 3x^2$, is an even function because $(9 - 3(-x)^2) = (9 - 3x^2)$. Since we are integrating over a symmetric interval $[-\sqrt{3}, \sqrt{3}]$, we can use the property of even functions to simplify the integral:

$\text{Area} = 2 \int_{0}^{\sqrt{3}} (9 - 3x^2) \, dx$

Why: Using symmetry reduces the computational effort by allowing us to integrate over only half the interval. Since the other half is symmetrical, we just multiply the result by 2.

Step 7: Evaluate the Integral

Now, we evaluate the integral:

$\text{Area} = 2 \left[ 9x - x^3 \right]_0^{\sqrt{3}}$

Why: This step applies the fundamental theorem of calculus to find the antiderivative and evaluate it at the limits of integration.

Step 8: Calculate the Definite Integral

Plug in the limits of integration:

$\text{Area} = 2 \left[ (9\sqrt{3} - (\sqrt{3})^3) - (9(0) - 0^3) \right]$ $\text{Area} = 2 \left[ 9\sqrt{3} - 3\sqrt{3} \right]$ $\text{Area} = 2 \left[ 6\sqrt{3} \right]$ $\text{Area} = 12\sqrt{3}$

Why: This is the final calculation, yielding the area of the region.

Common Mistakes & Tips

• Incorrectly Identifying Upper/Lower Curves: Always sketch a graph or visualize the functions to ensure you correctly identify which curve is above the other within the region of interest.
• Forgetting the $\pm$ when taking the square root: Remember to consider both positive and negative roots when solving for intersection points.
• Using Symmetry: Look for opportunities to use symmetry to simplify the integral, especially when dealing with even or odd functions and symmetric intervals.

Summary

We found the area of the region bounded by $y = 5x^2$ and $y = 2x^2 + 9$ by first identifying the upper and lower curves and then finding their intersection points. These intersection points gave us the limits of integration. We then set up the definite integral, simplified it using the symmetry of the integrand, and evaluated the integral to find the area.

The final answer is $6\sqrt{3}$.

Final Answer

The final answer is \boxed{6\sqrt 3}, which corresponds to option (A).