Key Concepts and Formulas

• Area Between Curves: The area between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$ on $[a,b]$, is given by $\int_a^b [f(x) - g(x)] \, dx$.
• Finding Intersection Points: To find the points of intersection between two curves, set their equations equal to each other and solve for $x$. These $x$ values will be your limits of integration ($a$ and $b$).
• Definite Integration: Evaluating a definite integral $\int_a^b f(x) \, dx$ gives the signed area under the curve $f(x)$ from $x=a$ to $x=b$.

Step-by-Step Solution

Step 1: Rewrite the inequalities to express y as a function of x

We are given the region $S = \{ (x,y):3{x^2} \le 4y \le 6x + 24\}$. We want to find the area of this region. First, we rewrite the inequalities to isolate $y$: $\frac{3}{4}x^2 \le y \le \frac{6}{4}x + \frac{24}{4}$ $\frac{3}{4}x^2 \le y \le \frac{3}{2}x + 6$ This means $g(x) = \frac{3}{4}x^2$ and $f(x) = \frac{3}{2}x + 6$.

Step 2: Find the intersection points of the two curves

To determine the limits of integration, we need to find where the two curves intersect. We set the two functions equal to each other: $\frac{3}{4}x^2 = \frac{3}{2}x + 6$ Multiply both sides by 4/3 to simplify: $x^2 = 2x + 8$ $x^2 - 2x - 8 = 0$ Factoring the quadratic, we have: $(x - 4)(x + 2) = 0$ So the intersection points occur at $x = 4$ and $x = -2$. These will be our limits of integration.

Step 3: Set up the definite integral

Now we set up the definite integral to calculate the area between the curves. Since $\frac{3}{2}x + 6 \ge \frac{3}{4}x^2$ on the interval $[-2, 4]$, we have: $\text{Area} = \int_{-2}^4 \left[ \left( \frac{3}{2}x + 6 \right) - \left( \frac{3}{4}x^2 \right) \right] \, dx$

Step 4: Evaluate the definite integral

We evaluate the integral: $\text{Area} = \int_{-2}^4 \left( \frac{3}{2}x + 6 - \frac{3}{4}x^2 \right) \, dx$ $\text{Area} = \left[ \frac{3}{4}x^2 + 6x - \frac{1}{4}x^3 \right]_{-2}^4$ $\text{Area} = \left( \frac{3}{4}(4)^2 + 6(4) - \frac{1}{4}(4)^3 \right) - \left( \frac{3}{4}(-2)^2 + 6(-2) - \frac{1}{4}(-2)^3 \right)$ $\text{Area} = \left( \frac{3}{4}(16) + 24 - \frac{1}{4}(64) \right) - \left( \frac{3}{4}(4) - 12 - \frac{1}{4}(-8) \right)$ $\text{Area} = (12 + 24 - 16) - (3 - 12 + 2)$ $\text{Area} = 20 - (-7)$ $\text{Area} = 20 + 7$ $\text{Area} = 27/3 = 9-6 = 3$ Area = 27

Step 5: Recalculating Definite Integral I noticed an error in previous calculation. Let's recalculate. $\text{Area} = \left[ \frac{3}{4}x^2 + 6x - \frac{1}{4}x^3 \right]_{-2}^4$ $\text{Area} = \left( \frac{3}{4}(4)^2 + 6(4) - \frac{1}{4}(4)^3 \right) - \left( \frac{3}{4}(-2)^2 + 6(-2) - \frac{1}{4}(-2)^3 \right)$ $\text{Area} = \left( 12 + 24 - 16 \right) - \left( 3 - 12 + 2 \right)$ $\text{Area} = \left( 20 \right) - \left( -7 \right)$ $\text{Area} = 20+7=27$ The area of S is 27

Common Mistakes & Tips

• Incorrect Limits of Integration: Make sure you correctly find the intersection points of the curves. A sketch can be helpful.
• Incorrect Order of Subtraction: Ensure you are subtracting the lower curve from the upper curve. If you get a negative area, you've likely subtracted in the wrong order. Take the absolute value.
• Algebra Errors: Be careful with algebraic manipulations and arithmetic when evaluating the definite integral.

Summary

To find the area of the region $S$, we first rewrote the inequalities to express $y$ as a function of $x$, identifying the upper and lower bounding curves. Then, we found the points of intersection to determine the limits of integration. Finally, we set up and evaluated the definite integral to find the area between the curves. The area of the region $S$ is 27.

Final Answer

The final answer is $\boxed{27}$.