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 will follow these steps:

### Step 1: Find the points of intersection
We start by solving the equations of the curve and the line simultaneously.

1. The equation of the curve is:
\[ y^2 = 4x \]
We can express \( x \) in terms of \( y \):
\[ x = \frac{y^2}{4} \]

2. The equation of the line can be rearranged to express \( y \):
\[ 2x - 3y + 4 = 0 \implies 3y = 2x + 4 \implies y = \frac{2x + 4}{3} \]

3. Substitute \( x = \frac{y^2}{4} \) into the line equation:
\[ y = \frac{2\left(\frac{y^2}{4}\right) + 4}{3} \implies y = \frac{\frac{y^2}{2} + 4}{3} \]
Multiply through by 3:
\[ 3y = \frac{y^2}{2} + 4 \implies 3y - \frac{y^2}{2} - 4 = 0 \]
Multiply through by 2 to eliminate the fraction:
\[ 6y - y^2 - 8 = 0 \implies y^2 - 6y + 8 = 0 \]

4. Factor the quadratic:
\[ (y - 2)(y - 4) = 0 \]
Thus, \( y = 2 \) and \( y = 4 \).

5. Now, find the corresponding \( x \) values:
- For \( y = 2 \):
\[ x = \frac{2^2}{4} = 1 \]
- For \( y = 4 \):
\[ x = \frac{4^2}{4} = 4 \]

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

### Step 2: Set up the integral for the area
The area \( A \) between the curve and the line from \( x = 1 \) to \( x = 4 \) can be calculated using the formula:
\[ A = \int_{1}^{4} (y_{\text{curve}} - y_{\text{line}}) \, dx \]

6. The curve \( y = 2\sqrt{x} \) and the line \( y = \frac{2x + 4}{3} \).

### Step 3: Find the area using integration
Now we can set up the integral:
\[ A = \int_{1}^{4} \left( 2\sqrt{x} - \frac{2x + 4}{3} \right) dx \]

7. Simplifying the integrand:
\[ A = \int_{1}^{4} \left( 2\sqrt{x} - \frac{2}{3}x - \frac{4}{3} \right) dx \]

### Step 4: Calculate the integral
8. We can integrate term by term:
\[ A = \int_{1}^{4} 2\sqrt{x} \, dx - \int_{1}^{4} \frac{2}{3}x \, dx - \int_{1}^{4} \frac{4}{3} \, dx \]

9. The integrals are:
- \( \int 2\sqrt{x} \, dx = \frac{4}{3}x^{3/2} \)
- \( \int \frac{2}{3}x \, dx = \frac{1}{3}x^2 \)
- \( \int \frac{4}{3} \, dx = \frac{4}{3}x \)

10. Evaluating from 1 to 4:
\[ A = \left[ \frac{4}{3}x^{3/2} \right]_{1}^{4} - \left[ \frac{1}{3}x^2 \right]_{1}^{4} - \left[ \frac{4}{3}x \right]_{1}^{4} \]

Calculating each part:
- For \( \frac{4}{3}x^{3/2} \):
\[ \frac{4}{3}(4^{3/2}) - \frac{4}{3}(1^{3/2}) = \frac{4}{3}(8) - \frac{4}{3}(1) = \frac{32}{3} - \frac{4}{3} = \frac{28}{3} \]
- For \( \frac{1}{3}x^2 \):
\[ \frac{1}{3}(4^2) - \frac{1}{3}(1^2) = \frac{16}{3} - \frac{1}{3} = \frac{15}{3} = 5 \]
- For \( \frac{4}{3}x \):
\[ \frac{4}{3}(4) - \frac{4}{3}(1) = \frac{16}{3} - \frac{4}{3} = \frac{12}{3} = 4 \]

### Step 5: Combine results
Putting it all together:
\[ A = \frac{28}{3} - 5 - 4 = \frac{28}{3} - \frac{15}{3} = \frac{13}{3} \]

Thus, the area bounded by the curve \( y^2 = 4x \) and the line \( 2x - 3y + 4 = 0 \) is:
\[ \boxed{\frac{13}{3}} \]