The area bounded by `4x^2 le y le 8x+12` is -

Options

A. 1. `(125)/(3)`
B. 2.`(124)/(3)`
C. 3.`(128)/(3)`
D. 4.`(127)/(3)`

Answer

To find the area bounded by the curves \(4x^2 \leq y \leq 8x + 12\), we will follow these steps:

### Step 1: Identify the curves
The given inequalities can be expressed as two equations:
1. \(y = 4x^2\) (a parabola)
2. \(y = 8x + 12\) (a linear equation)

### Step 2: Find the points of intersection
To find the area between the curves, we need to determine the points where they intersect. We set the equations equal to each other:
\[ 4x^2 = 8x + 12 \]
Rearranging gives:
\[ 4x^2 - 8x - 12 = 0 \]
Dividing the entire equation by 4 simplifies it to:
\[ x^2 - 2x - 3 = 0 \]
Now, we can factor this quadratic:
\[ (x + 1)(x - 3) = 0 \]
Thus, the solutions are:
\[ x = -1 \quad \text{and} \quad x = 3 \]

### Step 3: Set up the integral for the area
The area \(A\) between the curves from \(x = -1\) to \(x = 3\) can be found using the integral:
\[ A = \int_{-1}^{3} ((8x + 12) - (4x^2)) \, dx \]

### Step 4: Evaluate the integral
Now we compute the integral:
\[ A = \int_{-1}^{3} (8x + 12 - 4x^2) \, dx \]
We can break this down:
\[ A = \int_{-1}^{3} (8x) \, dx + \int_{-1}^{3} (12) \, dx - \int_{-1}^{3} (4x^2) \, dx \]

Calculating each integral:
1. \(\int (8x) \, dx = 4x^2\)
2. \(\int (12) \, dx = 12x\)
3. \(\int (4x^2) \, dx = \frac{4}{3}x^3\)

Now substituting the limits:
\[ A = \left[ 4x^2 + 12x - \frac{4}{3}x^3 \right]_{-1}^{3} \]

Calculating at \(x = 3\):
\[ = 4(3^2) + 12(3) - \frac{4}{3}(3^3) = 36 + 36 - 36 = 36 \]

Calculating at \(x = -1\):
\[ = 4(-1^2) + 12(-1) - \frac{4}{3}(-1^3) = 4 - 12 + \frac{4}{3} = -8 + \frac{4}{3} = -\frac{24}{3} + \frac{4}{3} = -\frac{20}{3} \]

Now, substituting back into the area formula:
\[ A = 36 - \left(-\frac{20}{3}\right) = 36 + \frac{20}{3} = \frac{108}{3} + \frac{20}{3} = \frac{128}{3} \]

### Final Answer
Thus, the area bounded by the curves is:
\[ \boxed{\frac{128}{3}} \]