Using integration, find the area enclosed between the curve `y=2x-x^(2)`, and the x-axis.

Answer

To find the area enclosed between the curve \( y = 2x - x^2 \) and the x-axis, we can follow these steps:

### Step 1: Identify the curves
The first curve is given by:
\[ y = 2x - x^2 \]
The second curve is the x-axis, which can be represented as:
\[ y = 0 \]

### Step 2: Find the points of intersection
To find the area between the curve and the x-axis, we need to determine the points where the curve intersects the x-axis. This occurs when \( y = 0 \):
\[ 2x - x^2 = 0 \]
Factoring the equation:
\[ x(2 - x) = 0 \]
This gives us the solutions:
\[ x = 0 \quad \text{or} \quad x = 2 \]
Thus, the points of intersection are \( (0, 0) \) and \( (2, 0) \).

### Step 3: Set up the integral for the area
The area \( A \) enclosed between the curve and the x-axis from \( x = 0 \) to \( x = 2 \) can be calculated using the definite integral:
\[ A = \int_{0}^{2} (2x - x^2) \, dx \]

### Step 4: Calculate the integral
Now we will compute the integral:
\[ A = \int_{0}^{2} (2x - x^2) \, dx \]
This can be split into two separate integrals:
\[ A = \int_{0}^{2} 2x \, dx - \int_{0}^{2} x^2 \, dx \]

Calculating each integral:
1. For \( \int 2x \, dx \):
\[ \int 2x \, dx = x^2 \]
Evaluating from 0 to 2:
\[ [x^2]_{0}^{2} = 2^2 - 0^2 = 4 \]

2. For \( \int x^2 \, dx \):
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating from 0 to 2:
\[ \left[\frac{x^3}{3}\right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \]

Now substituting back into the area formula:
\[ A = 4 - \frac{8}{3} \]
To combine these, we can express 4 as a fraction:
\[ 4 = \frac{12}{3} \]
Thus:
\[ A = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \]

### Step 5: Final answer
The area enclosed between the curve and the x-axis is:
\[ \text{Area} = \frac{4}{3} \text{ square units} \]