The area bounded by `y = x ^(2), x =0, x =2 and ` x-axis is `"________".`

Answer

To find the area bounded by the curve \( y = x^2 \), the lines \( x = 0 \), \( x = 2 \), and the x-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the curves and lines**:
- The curve is \( y = x^2 \).
- The vertical lines are \( x = 0 \) and \( x = 2 \).
- The x-axis is represented by \( y = 0 \).

2. **Set up the integral**:
- The area \( A \) can be found using the integral of the function from the left boundary to the right boundary.
- The area can be expressed as:
\[ A = \int_{0}^{2} (f(x) - g(x)) \, dx \]
- Here, \( f(x) = x^2 \) (the curve) and \( g(x) = 0 \) (the x-axis).

3. **Write the integral**:
- Substitute \( f(x) \) and \( g(x) \) into the integral:
\[ A = \int_{0}^{2} (x^2 - 0) \, dx = \int_{0}^{2} x^2 \, dx \]

4. **Calculate the integral**:
- The integral of \( x^2 \) is:
\[ \int x^2 \, dx = \frac{x^3}{3} + C \]
- Now, evaluate the definite integral from \( 0 \) to \( 2 \):
\[ A = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} - 0 = \frac{8}{3} \]

5. **Final answer**:
- The area bounded by the curve, the lines, and the x-axis is:
\[ A = \frac{8}{3} \text{ square units} \]