The area enclosed by the parabola `y=3(1-x^(2))` and the x-axis is

Options

A. 4
B. 3
C. 9
D. 2

Answer

To find the area enclosed by the parabola \( y = 3(1 - x^2) \) and the x-axis, we will follow these steps:

### Step 1: Identify the points of intersection
We need to find the points where the parabola intersects the x-axis. This occurs when \( y = 0 \).

Setting the equation of the parabola to zero:
\[ 3(1 - x^2) = 0 \]
Dividing both sides by 3:
\[ 1 - x^2 = 0 \]
Rearranging gives:
\[ x^2 = 1 \]
Taking the square root of both sides, we find:
\[ x = \pm 1 \]

### Step 2: Set up the integral for the area
The area \( A \) enclosed by the curve and the x-axis from \( x = -1 \) to \( x = 1 \) can be calculated using the integral:
\[ A = \int_{-1}^{1} y \, dx \]
Substituting \( y = 3(1 - x^2) \):
\[ A = \int_{-1}^{1} 3(1 - x^2) \, dx \]

### Step 3: Simplify the integral
We can factor out the constant:
\[ A = 3 \int_{-1}^{1} (1 - x^2) \, dx \]

### Step 4: Evaluate the integral
Now we will compute the integral:
\[ \int (1 - x^2) \, dx = x - \frac{x^3}{3} \]
Now we evaluate this from \( -1 \) to \( 1 \):
\[ A = 3 \left[ \left( x - \frac{x^3}{3} \right) \bigg|_{-1}^{1} \right] \]

Calculating at the upper limit \( x = 1 \):
\[ 1 - \frac{1^3}{3} = 1 - \frac{1}{3} = \frac{2}{3} \]

Calculating at the lower limit \( x = -1 \):
\[ -1 - \frac{(-1)^3}{3} = -1 + \frac{1}{3} = -\frac{2}{3} \]

Now substituting back into the area calculation:
\[ A = 3 \left( \frac{2}{3} - \left(-\frac{2}{3}\right) \right) = 3 \left( \frac{2}{3} + \frac{2}{3} \right) = 3 \cdot \frac{4}{3} = 4 \]

### Conclusion
The area enclosed by the parabola \( y = 3(1 - x^2) \) and the x-axis is:
\[ \boxed{4} \text{ square units} \]