Find the area of the region included between `y^(2)=9x" and "y=x`.

Explanation

To find the area of the region included between the curves \(y^2 = 9x\) and \(y = x\), we will follow these steps:

### Step 1: Find the Points of Intersection
We need to find the points where the two curves intersect.

1. The first equation is \(y^2 = 9x\).
2. The second equation is \(y = x\).

Substituting \(y = x\) into the first equation:

\[ x^2 = 9x \]

Rearranging gives:

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

Factoring out \(x\):

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

Thus, \(x = 0\) or \(x = 9\).

Now, substituting these \(x\) values back into \(y = x\):

- For \(x = 0\), \(y = 0\) → Point (0, 0)
- For \(x = 9\), \(y = 9\) → Point (9, 9)

### Step 2: Set Up the Integral for Area
The area \(A\) between the curves from \(x = 0\) to \(x = 9\) can be calculated using the integral:

\[ A = \int_{0}^{9} (y_{\text{upper}} - y_{\text{lower}}) \, dx \]

Here, \(y_{\text{upper}} = \sqrt{9x}\) (from \(y^2 = 9x\)) and \(y_{\text{lower}} = x\).

Thus, the area becomes:

\[ A = \int_{0}^{9} (\sqrt{9x} - x) \, dx \]

### Step 3: Simplify the Integral
We can simplify \(\sqrt{9x}\):

\[ \sqrt{9x} = 3\sqrt{x} \]

So the integral becomes:

\[ A = \int_{0}^{9} (3\sqrt{x} - x) \, dx \]

### Step 4: Evaluate the Integral
Now we will evaluate the integral:

\[ A = \int_{0}^{9} 3x^{1/2} \, dx - \int_{0}^{9} x \, dx \]

Calculating each integral separately:

1. For \(\int 3x^{1/2} \, dx\):

\[ = 3 \cdot \frac{x^{3/2}}{3/2} = 2x^{3/2} \]

Evaluating from 0 to 9:

\[ = 2[9^{3/2} - 0^{3/2}] = 2[27 - 0] = 54 \]

2. For \(\int x \, dx\):

\[ = \frac{x^2}{2} \]

Evaluating from 0 to 9:

\[ = \frac{9^2}{2} - \frac{0^2}{2} = \frac{81}{2} \]

### Step 5: Combine the Results
Now, substituting back into the area formula:

\[ A = 54 - \frac{81}{2} \]

To combine these, convert 54 into halves:

\[ 54 = \frac{108}{2} \]

Thus,

\[ A = \frac{108}{2} - \frac{81}{2} = \frac{27}{2} \]

### Final Answer
The area of the region included between the curves \(y^2 = 9x\) and \(y = x\) is:

\[ \boxed{\frac{27}{2}} \text{ square units} \]

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