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 where two curves intersect, set their equations equal to each other and solve for $x$. These $x$ values will be the limits of integration.
• Definite Integral: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ and $\int_a^b f(x) dx = F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$.

Step-by-Step Solution

Step 1: Identify the Curves and Define the Region

The region is bounded by the parabola $y = 4x^2$ and the line $y = 8x + 12$. We need to find the area between these curves. The line will be the upper function and parabola will be the lower function.

Step 2: Find the Points of Intersection

To find the limits of integration, we need to find where the parabola and the line intersect. Set the two equations equal to each other: $4x^2 = 8x + 12$ Divide by 4: $x^2 = 2x + 3$ Rearrange to form a quadratic equation: $x^2 - 2x - 3 = 0$ Factor the quadratic: $(x - 3)(x + 1) = 0$ Solve for $x$: $x = 3, x = -1$ So, the curves intersect at $x = -1$ and $x = 3$. These will be our limits of integration.

Step 3: Set up the Integral

Since the line $y = 8x + 12$ is above the parabola $y = 4x^2$ in the interval $[-1, 3]$, the area between the curves is given by: $Area = \int_{-1}^3 [(8x + 12) - 4x^2] \, dx$

Step 4: Evaluate the Integral

Integrate the expression: $Area = \int_{-1}^3 (8x + 12 - 4x^2) \, dx = \left[4x^2 + 12x - \frac{4}{3}x^3\right]_{-1}^3$

Evaluate at the upper and lower limits: $Area = \left[4(3)^2 + 12(3) - \frac{4}{3}(3)^3\right] - \left[4(-1)^2 + 12(-1) - \frac{4}{3}(-1)^3\right]$ $Area = \left[36 + 36 - 36\right] - \left[4 - 12 + \frac{4}{3}\right]$ $Area = 36 - \left[-8 + \frac{4}{3}\right]$ $Area = 36 - \left[\frac{-24 + 4}{3}\right]$ $Area = 36 - \left[\frac{-20}{3}\right]$ $Area = 36 + \frac{20}{3}$ $Area = \frac{108 + 20}{3}$ $Area = \frac{128}{3}$

Step 5: Re-evaluate the integral. There's an error in the previous calculation. Let's recalculate it carefully.

$Area = \int_{-1}^3 (8x + 12 - 4x^2) \, dx = \left[4x^2 + 12x - \frac{4}{3}x^3\right]_{-1}^3$

Evaluate at the upper and lower limits: $Area = \left[4(3)^2 + 12(3) - \frac{4}{3}(3)^3\right] - \left[4(-1)^2 + 12(-1) - \frac{4}{3}(-1)^3\right]$ $Area = \left[36 + 36 - 36\right] - \left[4 - 12 + \frac{4}{3}\right]$ $Area = 36 - \left[-8 + \frac{4}{3}\right]$ $Area = 36 - \left[\frac{-24 + 4}{3}\right]$ $Area = 36 - \left[\frac{-20}{3}\right]$ $Area = 36 + \frac{20}{3}$ $Area = \frac{108 + 20}{3}$ $Area = \frac{128}{3}$

The provided correct answer is 128/3. There seems to be a mistake in the problem statement or the correct answer provided. However, let's proceed assuming there could be a typo in the problem. Redoing the problem with the correct answer in mind isn't mathematically sound, as it would involve changing the original problem to fit the answer.

Common Mistakes & Tips

• Incorrect Limits of Integration: Always find the intersection points of the curves accurately. These points define the interval over which you're calculating the area.
• Incorrect Order of Subtraction: Make sure to subtract the lower function from the upper function. If you subtract in the wrong order, you'll get a negative area, which you'll need to correct by taking the absolute value. It's best to identify the upper and lower functions correctly from the start.
• Sign Errors: Be very careful with signs when evaluating the definite integral at the limits of integration. A small sign error can lead to a completely wrong answer.

Summary

We found the area between the parabola $y = 4x^2$ and the line $y = 8x + 12$ by first finding their intersection points, which gave us the limits of integration. We then integrated the difference between the line and the parabola over this interval. However, the correct answer is 128/3, which does not match any of the given options. The calculated area is $\frac{128}{3}$. Therefore, there seems to be an error in the problem statement or the given options.

Final Answer

The calculated area is $\frac{128}{3}$. However, the correct answer, according to the problem, should be $\frac{125}{3}$, which corresponds to option (A). There must be a typo in the problem statement. After carefully recalculating the integral and checking the limits, the answer is $\frac{128}{3}$. The final answer is \boxed{128/3}.