The area (in sq. units) of the region {(x, y)ϵR|4x² ≤ y ≤ 8x + 12} is : (A) 125/3 (B) 128/3 (C) 124/

Q: Wait, why do we subtract the parabola from the line? Could it be the other way around?

Great question! The inequality $4x^2 \le y \le 8x + 12$ explicitly tells us that $y$ is greater than or equal to the parabola and less than or equal to the line. This means the line is 'above' the parabola in the region of interest. If you ever get confused, just pick a test point between your intersection values (like $x=0$) and see which function gives a larger $y$-value. The larger one is your upper boundary.

Q: Do I really need to factor the quadratic equation? Can I use the quadratic formula?

You absolutely can use the quadratic formula! Factoring is just a faster tool when the roots are clean integers. If you ever feel stuck or the numbers look messy, the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ will always lead you to the correct intersection points. Just be careful with your signs!

Q: Why do we integrate with respect to $x$ instead of $y$?

You could integrate with respect to $y$, but that would require expressing $x$ in terms of $y$ for both curves and splitting the integral into two parts because the left and right sides of the parabola are different functions ($x = \pm \sqrt{y}/2$). Integrating with respect to $x$ is much simpler here because the boundaries are defined as single functions of $x$ over the entire interval.

Question

The area (in sq. units) of the region $\{(x, y) \in R | 4x^2 \le y \le 8x + 12\}$ is :

Options:

(A) $\frac{125}{3}$

(B) $\frac{128}{3}$

(C) $\frac{124}{3}$

(D) $\frac{127}{3}$

Sigma Percentile · Accepted Answer

### The Geometry of the Trap

Imagine you are standing on a coordinate plane, looking at two distinct paths. One is a graceful, upward-opening curve, the parabola defined by $y = 4x^2$. The other is a sharp, straight line, $y = 8x + 12$.

The problem asks us to find the area of the region where $4x^2 \le y \le 8x + 12$. This is not just an abstract inequality; it is a physical trap. The parabola acts as the floor, and the line acts as the ceiling.

Our goal is to calculate the total space enclosed between these two boundaries. To do this, we must first understand where these two paths meet, as these intersection points will define the start and end of our journey.

### Finding the Boundaries

To find where the line and the parabola cross, we set their equations equal to each other:
$$4x^2 = 8x + 12$$

We simplify by dividing the entire equation by $4$, which gives us $x^2 = 2x + 3$. Rearranging this into a standard quadratic form, we get:
$$x^2 - 2x - 3 = 0$$

Now, we look for two numbers that multiply to $-3$ and add to $-2$. Those numbers are $-3$ and $1$. Thus, we can factor the equation as $(x - 3)(x + 1) = 0$.

This reveals our intersection points: $x = 3$ and $x = -1$. These are the boundaries of our region, meaning the area we seek is confined strictly between $x = -1$ and $x = 3$.

### The Engine of Calculation

With our limits defined, we turn to the power of calculus. The area $A$ between two curves is the integral of the upper curve minus the lower curve.

In our case, the line is the upper boundary, so we set up the integral:
$$A = \int_{-1}^{3} (8x + 12 - 4x^2) dx$$

This integral is the engine that will calculate the area for us. We integrate term by term:
1. The integral of $8x$ is $4x^2$.
2. The integral of $12$ is $12x$.
3. The integral of $-4x^2$ is $-\frac{4x^3}{3}$.

We are left with the expression evaluated from $-1$ to $3$:
$$\left[ 4x^2 + 12x - \frac{4x^3}{3} \right]_{-1}^{3}$$

### The Final Triumph

Now, we apply the Fundamental Theorem of Calculus. First, we substitute the upper limit, $x = 3$:
$$(4(3)^2 + 12(3) - \frac{4(3)^3}{3}) = (36 + 36 - 36) = 36$$

Next, we substitute the lower limit, $x = -1$:
$$(4(-1)^2 + 12(-1) - \frac{4(-1)^3}{3}) = (4 - 12 + \frac{4}{3}) = -8 + \frac{4}{3}$$

Finally, we subtract the lower limit result from the upper limit result:
$$A = 36 - (-8 + \frac{4}{3}) = 36 + 8 - \frac{4}{3} = 44 - \frac{4}{3}$$

Converting $44$ to a fraction with a denominator of $3$, we get:
$$\frac{132}{3} - \frac{4}{3} = \frac{128}{3}$$

And there it is! The area of the region is **$\frac{128}{3}$ square units**. You have successfully navigated the geometry, solved the quadratic, and mastered the integral.

Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region ${(x, y) \in R ∣4 x^{2} \leq y \leq 8 x + 12}$ is :

Select Answer:

Visualized Solution

Visualizing the Region

Finding Intersection Points

Solving for $x$

Identifying the Limits

Defining the Area Integral

Performing Integration

Applying Upper Limit

Applying Lower Limit

Final Calculation

The Sigma Insight: Area Bounded by Curves

The Geometry of the Trap

Finding the Boundaries

The Engine of Calculation

The Final Triumph

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area (in sq. units) of the region {(x,y)∈R∣4x2≤y≤8x+12} is :