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` unit
C. 1 sq unit
D. `4/3` sq unit

Answer

To find the area bounded by the curve \( y = \frac{1}{2}x^2 \), the X-axis, and the line \( x = 2 \), we can follow these steps:

### Step 1: Set up the integral for the area
The area \( A \) under the curve from \( x = 0 \) to \( x = 2 \) can be found using the integral:
\[ A = \int_{0}^{2} y \, dx \]
Since \( y = \frac{1}{2}x^2 \), we can rewrite the integral as:
\[ A = \int_{0}^{2} \frac{1}{2}x^2 \, dx \]

### Step 2: Calculate the integral
Now we need to 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 can evaluate the definite integral:
\[ A = \frac{1}{2} \left[ \frac{x^3}{3} \right]_{0}^{2} \]

### Step 3: Evaluate the limits
Now we substitute the limits into the evaluated integral:
\[ A = \frac{1}{2} \left( \frac{2^3}{3} - \frac{0^3}{3} \right) = \frac{1}{2} \left( \frac{8}{3} - 0 \right) = \frac{1}{2} \cdot \frac{8}{3} \]
This simplifies to:
\[ A = \frac{8}{6} = \frac{4}{3} \]

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

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