Find the area enclosed by the curve y=3x and y=6x-`x^(2)`

Answer

To find the area enclosed by the curves \( y = 3x \) and \( y = 6x - x^2 \), we will follow these steps:

### Step 1: Find the points of intersection
To find the points where the two curves intersect, we set them equal to each other:

\[ 3x = 6x - x^2 \]

Rearranging gives:

\[ x^2 - 3x = 0 \]

Factoring out \( x \):

\[ x(x - 3) = 0 \]

Thus, the points of intersection are:

\[ x = 0 \quad \text{and} \quad x = 3 \]

### Step 2: Determine the area between the curves
The area \( A \) enclosed between the curves from \( x = 0 \) to \( x = 3 \) can be found using the integral:

\[ A = \int_{0}^{3} \left( (6x - x^2) - (3x) \right) \, dx \]

This simplifies to:

\[ A = \int_{0}^{3} (3x - x^2) \, dx \]

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

\[ A = \int_{0}^{3} (3x - x^2) \, dx \]

Calculating the integral:

\[ A = \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_{0}^{3} \]

### Step 4: Evaluate the integral at the limits
Substituting the upper limit \( x = 3 \):

\[ A = \left( \frac{3(3^2)}{2} - \frac{(3^3)}{3} \right) - \left( \frac{3(0^2)}{2} - \frac{(0^3)}{3} \right) \]

Calculating the first part:

\[ = \left( \frac{3 \cdot 9}{2} - \frac{27}{3} \right) \]

This simplifies to:

\[ = \left( \frac{27}{2} - 9 \right) \]

Converting \( 9 \) to a fraction with a denominator of 2:

\[ = \left( \frac{27}{2} - \frac{18}{2} \right) = \frac{9}{2} \]

### Final Result
Thus, the area enclosed by the curves \( y = 3x \) and \( y = 6x - x^2 \) is:

\[ \boxed{\frac{9}{2}} \]