The area of the region bounded by the curve `y=x^(2)` and the line `y=16` (in square units) is

Options

A. `(32)/(3)`
B. `(256)/(3)`
C. `(64)/(3)`
D. `(128)/(3)`

Answer

To find the area of the region bounded by the curve \( y = x^2 \) and the line \( y = 16 \), we can follow these steps:

### Step 1: Find the points of intersection
We first need to determine where the curve \( y = x^2 \) intersects the line \( y = 16 \). To find the points of intersection, we set the two equations equal to each other:

\[ x^2 = 16 \]

Taking the square root of both sides, we get:

\[ x = \pm 4 \]

Thus, the points of intersection are \( (-4, 16) \) and \( (4, 16) \).

### Step 2: Set up the area integral
The area between the curve and the line can be found by integrating the difference between the line and the curve from \( x = -4 \) to \( x = 4 \):

\[ \text{Area} = \int_{-4}^{4} (16 - x^2) \, dx \]

### Step 3: Calculate the integral
We can compute the integral:

\[ \text{Area} = \int_{-4}^{4} 16 \, dx - \int_{-4}^{4} x^2 \, dx \]

Calculating the first integral:

\[ \int_{-4}^{4} 16 \, dx = 16 \cdot (4 - (-4)) = 16 \cdot 8 = 128 \]

Now, calculating the second integral:

\[ \int_{-4}^{4} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-4}^{4} = \frac{4^3}{3} - \frac{(-4)^3}{3} = \frac{64}{3} - \left(-\frac{64}{3}\right) = \frac{64}{3} + \frac{64}{3} = \frac{128}{3} \]

### Step 4: Combine the results
Now, substituting back into the area formula:

\[ \text{Area} = 128 - \frac{128}{3} \]

To combine these, we convert 128 into a fraction:

\[ 128 = \frac{384}{3} \]

Now we can subtract:

\[ \text{Area} = \frac{384}{3} - \frac{128}{3} = \frac{256}{3} \]

### Final Result
Thus, the area of the region bounded by the curve \( y = x^2 \) and the line \( y = 16 \) is:

\[ \text{Area} = \frac{256}{3} \text{ square units} \]

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