The area between the curve `y^(2)=9x` and the line y = 3x is

Options

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

Answer

To find the area between the curve \( y^2 = 9x \) and the line \( y = 3x \), we will follow these steps:

### Step 1: Find the points of intersection
We start by setting the two equations equal to each other to find their points of intersection.

1. The curve is given by \( y^2 = 9x \).
2. The line is given by \( y = 3x \).

Substituting \( y = 3x \) into the curve equation:

\[ (3x)^2 = 9x \]

This simplifies to:

\[ 9x^2 = 9x \]

Rearranging gives:

\[ 9x^2 - 9x = 0 \]

Factoring out \( 9x \):

\[ 9x(x - 1) = 0 \]

Thus, the solutions are:

\[ x = 0 \quad \text{and} \quad x = 1 \]

### Step 2: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be calculated using the integral:

\[ A = \int_{0}^{1} (y_{\text{upper}} - y_{\text{lower}}) \, dx \]

Here, \( y_{\text{upper}} = 3x \) (the line) and \( y_{\text{lower}} = \sqrt{9x} = 3\sqrt{x} \) (the curve).

Thus, the area becomes:

\[ A = \int_{0}^{1} (3x - 3\sqrt{x}) \, dx \]

### Step 3: Calculate the integral
Now we compute the integral:

\[ A = \int_{0}^{1} (3x - 3\sqrt{x}) \, dx = \int_{0}^{1} 3x \, dx - \int_{0}^{1} 3\sqrt{x} \, dx \]

Calculating each integral separately:

1. For \( \int 3x \, dx \):

\[ \int 3x \, dx = \frac{3x^2}{2} \bigg|_0^1 = \frac{3(1)^2}{2} - \frac{3(0)^2}{2} = \frac{3}{2} \]

2. For \( \int 3\sqrt{x} \, dx \):

\[ \int 3\sqrt{x} \, dx = 3 \cdot \frac{2}{3} x^{3/2} \bigg|_0^1 = 2x^{3/2} \bigg|_0^1 = 2(1) - 2(0) = 2 \]

Putting it all together:

\[ A = \frac{3}{2} - 2 = \frac{3}{2} - \frac{4}{2} = -\frac{1}{2} \]

Since we are looking for the area, we take the absolute value:

\[ A = \frac{1}{2} \]

### Final Answer
Thus, the area between the curve \( y^2 = 9x \) and the line \( y = 3x \) is:

\[ \boxed{\frac{1}{2}} \]