The area bounded by the curve `y = (1)/(2)x^(2)`, the X-axis and the lines x = 2 is

Options

A. `(1)/(3)` sq unit
B. `(2)/(3)` sq unit
C. 1 sq unit
D. `(4)/(3)` sq units

Explanation

To find the area bounded by the curve \( y = \frac{1}{2}x^2 \), the X-axis, and the line \( x = 2 \), we will use definite integrals. Here are the steps to solve the problem:

### Step 1: Identify the area to be calculated
The area we need to find is under the curve \( y = \frac{1}{2}x^2 \) from \( x = 0 \) to \( x = 2 \), above the X-axis.

### Step 2: Set up the integral
The area \( A \) can be calculated using the definite integral:
\[ A = \int_{0}^{2} \left( \frac{1}{2}x^2 - 0 \right) \, dx \]
This simplifies to:
\[ A = \int_{0}^{2} \frac{1}{2}x^2 \, dx \]

### Step 3: Calculate the integral
Now we compute the integral:
\[ A = \frac{1}{2} \int_{0}^{2} x^2 \, dx \]
The integral of \( x^2 \) is:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Thus, we have:
\[ A = \frac{1}{2} \left[ \frac{x^3}{3} \right]_{0}^{2} \]

### Step 4: Evaluate the definite integral
Now we evaluate the limits:
\[ A = \frac{1}{2} \left( \frac{2^3}{3} - \frac{0^3}{3} \right) = \frac{1}{2} \left( \frac{8}{3} - 0 \right) = \frac{1}{2} \cdot \frac{8}{3} = \frac{8}{6} = \frac{4}{3} \]

### Step 5: Conclusion
The area bounded by the curve \( y = \frac{1}{2}x^2 \), the X-axis, and the line \( x = 2 \) is:
\[ \boxed{\frac{4}{3}} \text{ square units} \]

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