The area of region between `y=4-x^(2), 0 le x le 2` and the x-axis is

Options

A. `(4)/(3)`
B. `(7)/(3)`
C. `(16)/(3)`
D. `(5)/(3)`

Answer

To find the area of the region between the curve \( y = 4 - x^2 \) and the x-axis from \( x = 0 \) to \( x = 2 \), we can follow these steps:

### Step 1: Understand the Problem
We need to find the area between the curve \( y = 4 - x^2 \) and the x-axis within the limits \( x = 0 \) to \( x = 2 \).

### Step 2: Set Up the Integral
The area \( A \) can be calculated using the integral:
\[ A = \int_{0}^{2} (4 - x^2) \, dx \]
Here, \( 4 - x^2 \) is the function representing the height of the region above the x-axis.

### Step 3: Compute the Integral
Now, we will compute the integral:
\[ A = \int_{0}^{2} (4 - x^2) \, dx = \int_{0}^{2} 4 \, dx - \int_{0}^{2} x^2 \, dx \]

Calculating each part separately:
1. **Integral of 4**:
\[ \int_{0}^{2} 4 \, dx = 4x \bigg|_{0}^{2} = 4(2) - 4(0) = 8 \]

2. **Integral of \( x^2 \)**:
\[ \int_{0}^{2} x^2 \, dx = \frac{x^3}{3} \bigg|_{0}^{2} = \frac{(2)^3}{3} - \frac{(0)^3}{3} = \frac{8}{3} \]

### Step 4: Combine the Results
Now, substitute these results back into the area formula:
\[ A = 8 - \frac{8}{3} \]

To combine these, we can convert 8 into a fraction:
\[ 8 = \frac{24}{3} \]
Thus,
\[ A = \frac{24}{3} - \frac{8}{3} = \frac{24 - 8}{3} = \frac{16}{3} \]

### Step 5: Conclusion
The area of the region between the curve \( y = 4 - x^2 \) and the x-axis from \( x = 0 \) to \( x = 2 \) is:
\[ \boxed{\frac{16}{3}} \]