The area in square units of the region bounded by the curve `x^(2)=4y`, the line x=2 and the x-axis, is

Options

A. 1
B. `2//3`
C. `4//3`
D. `8//3`

Answer

To find the area of the region bounded by the curve \( x^2 = 4y \), the line \( x = 2 \), and the x-axis, we can follow these steps:

### Step 1: Understand the Curve
The equation \( x^2 = 4y \) represents a parabola that opens upwards. We can rewrite it in terms of \( y \):
\[ y = \frac{x^2}{4} \]

### Step 2: Identify the Boundaries
We need to find the area bounded by:
- The parabola \( y = \frac{x^2}{4} \)
- The vertical line \( x = 2 \)
- The x-axis (which is \( y = 0 \))

### Step 3: Set Up the Integral
The area \( A \) can be calculated using the definite integral of the function \( y = \frac{x^2}{4} \) from \( x = 0 \) to \( x = 2 \):
\[ A = \int_{0}^{2} \frac{x^2}{4} \, dx \]

### Step 4: Calculate the Integral
First, we can factor out the constant \( \frac{1}{4} \):
\[ A = \frac{1}{4} \int_{0}^{2} x^2 \, dx \]

Now, we need to compute the integral \( \int x^2 \, dx \):
\[ \int x^2 \, dx = \frac{x^3}{3} \]

### Step 5: Evaluate the Integral from 0 to 2
Now, we substitute the limits into the integral:
\[ A = \frac{1}{4} \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{1}{4} \left( \frac{2^3}{3} - \frac{0^3}{3} \right) \]
\[ = \frac{1}{4} \left( \frac{8}{3} - 0 \right) = \frac{1}{4} \cdot \frac{8}{3} = \frac{8}{12} = \frac{2}{3} \]

### Final Answer
Thus, the area of the region bounded by the curve, the line, and the x-axis is:
\[ \text{Area} = \frac{2}{3} \text{ square units} \]

---