The area bounded by the curve `y^(2) = 4x` and the line `2x-3y+4=0` is

Options

A. `1/3`
B. `2/3`
C. `4/3`
D. `5/3`

Answer

To find the area bounded by the curve \( y^2 = 4x \) and the line \( 2x - 3y + 4 = 0 \), we can follow these steps:

### Step 1: Find the points of intersection
First, we need to find the points where the curve and the line intersect.

1. Rearranging the line equation:
\[ 2x - 3y + 4 = 0 \implies y = \frac{2x + 4}{3} \]

2. Substitute \( y \) from the line equation into the curve equation:
\[ y^2 = 4x \implies \left(\frac{2x + 4}{3}\right)^2 = 4x \]
Expanding this gives:
\[ \frac{(2x + 4)^2}{9} = 4x \]
\[ (2x + 4)^2 = 36x \]
\[ 4x^2 + 16x + 16 = 36x \]
\[ 4x^2 - 20x + 16 = 0 \]
Dividing the entire equation by 4:
\[ x^2 - 5x + 4 = 0 \]

3. Factor the quadratic:
\[ (x - 4)(x - 1) = 0 \]
Thus, \( x = 4 \) and \( x = 1 \).

### Step 2: Find the corresponding \( y \) values
Now, we find the \( y \) values for these \( x \) values using the line equation:
- For \( x = 1 \):
\[ y = \frac{2(1) + 4}{3} = \frac{6}{3} = 2 \]
- For \( x = 4 \):
\[ y = \frac{2(4) + 4}{3} = \frac{12}{3} = 4 \]

Thus, the points of intersection are \( (1, 2) \) and \( (4, 4) \).

### Step 3: Set up the integral for the area
The area \( A \) between the curve and the line from \( x = 1 \) to \( x = 4 \) can be found using the integral:
\[ A = \int_{1}^{4} \left( \text{Upper function} - \text{Lower function} \right) \, dx \]
Here, the upper function is the line \( y = \frac{2x + 4}{3} \) and the lower function is the parabola \( y = 2\sqrt{x} \).

### Step 4: Calculate the area
The area is given by:
\[ A = \int_{1}^{4} \left( \frac{2x + 4}{3} - 2\sqrt{x} \right) \, dx \]

1. Break it down:
\[ A = \int_{1}^{4} \left( \frac{2x}{3} + \frac{4}{3} - 2\sqrt{x} \right) \, dx \]

2. Integrate term by term:
\[ A = \left[ \frac{2}{3} \cdot \frac{x^2}{2} + \frac{4}{3}x - \frac{4}{3}x^{3/2} \right]_{1}^{4} \]
\[ = \left[ \frac{1}{3}x^2 + \frac{4}{3}x - \frac{4}{3}x^{3/2} \right]_{1}^{4} \]

3. Evaluate at the limits:
- At \( x = 4 \):
\[ = \frac{1}{3}(4^2) + \frac{4}{3}(4) - \frac{4}{3}(4^{3/2}) = \frac{16}{3} + \frac{16}{3} - \frac{32}{3} = \frac{0}{3} = 0 \]
- At \( x = 1 \):
\[ = \frac{1}{3}(1^2) + \frac{4}{3}(1) - \frac{4}{3}(1^{3/2}) = \frac{1}{3} + \frac{4}{3} - \frac{4}{3} = \frac{1}{3} \]

4. Thus, the area is:
\[ A = 0 - \frac{1}{3} = -\frac{1}{3} \quad \text{(taking absolute value)} \]

### Final Result
The area bounded by the curve and the line is:
\[ \text{Area} = \frac{1}{3} \text{ square units.} \]