Find the area bounded by the curves `y=6x -x^(2) and y= x^(2)-2x`

Answer

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

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

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

Rearranging the equation gives:

\[ 6x - x^2 - x^2 + 2x = 0 \]

This simplifies to:

\[ -2x^2 + 8x = 0 \]

Factoring out \( -2x \):

\[ -2x(x - 4) = 0 \]

Setting each factor to zero gives us:

\[ x = 0 \quad \text{or} \quad x = 4 \]

### Step 2: Determine the Limits of Integration
The points of intersection are \( x = 0 \) and \( x = 4 \). These will serve as our limits of integration.

### Step 3: Identify the Upper and Lower Curves
To determine which curve is above the other between \( x = 0 \) and \( x = 4 \), we can evaluate both functions at a point in this interval, say \( x = 2 \):

- For \( y = 6x - x^2 \):
\[ y = 6(2) - (2)^2 = 12 - 4 = 8 \]

- For \( y = x^2 - 2x \):
\[ y = (2)^2 - 2(2) = 4 - 4 = 0 \]

Since \( 8 > 0 \), the curve \( y = 6x - x^2 \) is the upper curve and \( y = x^2 - 2x \) is the lower curve in this interval.

### Step 4: Set Up the Integral for Area
The area \( A \) bounded by the curves can be calculated using the integral:

\[ A = \int_{0}^{4} \left( (6x - x^2) - (x^2 - 2x) \right) \, dx \]

This simplifies to:

\[ A = \int_{0}^{4} \left( 6x - x^2 - x^2 + 2x \right) \, dx = \int_{0}^{4} \left( 8x - 2x^2 \right) \, dx \]

### Step 5: Integrate
Now we integrate:

\[ A = \int_{0}^{4} (8x - 2x^2) \, dx \]

Calculating the integral:

\[ A = \left[ 4x^2 - \frac{2}{3}x^3 \right]_{0}^{4} \]

### Step 6: Evaluate the Integral
Now we evaluate the definite integral:

1. Calculate at the upper limit \( x = 4 \):
\[ A = 4(4^2) - \frac{2}{3}(4^3) = 4(16) - \frac{2}{3}(64) = 64 - \frac{128}{3} \]

2. Calculate at the lower limit \( x = 0 \):
\[ A = 0 \]

Thus, the area becomes:

\[ A = 64 - \frac{128}{3} = \frac{192}{3} - \frac{128}{3} = \frac{64}{3} \]

### Final Answer
The area bounded by the curves is:

\[ \boxed{\frac{64}{3}} \text{ square units} \]