Find the area enclosed between the curves
`y=x^(2),y=2x-x^(2)`

Answer

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

### Step 1: Find the points of intersection
To find the area between the curves, we first need to determine where they intersect. We set the two equations equal to each other:

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

Rearranging gives:

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

Factoring out \( 2x \):

\[ 2x(x - 1) = 0 \]

This gives us the solutions:

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

### Step 2: Calculate the area between the curves
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be found using the integral:

\[ A = \int_{0}^{1} \left( (2x - x^2) - x^2 \right) \, dx \]

This simplifies to:

\[ A = \int_{0}^{1} (2x - 2x^2) \, dx \]

### Step 3: Evaluate the integral
Now, we will evaluate the integral:

\[ A = \int_{0}^{1} (2x - 2x^2) \, dx = 2 \int_{0}^{1} x \, dx - 2 \int_{0}^{1} x^2 \, dx \]

Calculating the first integral:

\[ \int x \, dx = \frac{x^2}{2} \quad \text{from } 0 \text{ to } 1 \]

So,

\[ \int_{0}^{1} x \, dx = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]

Calculating the second integral:

\[ \int x^2 \, dx = \frac{x^3}{3} \quad \text{from } 0 \text{ to } 1 \]

So,

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

Putting it all together:

\[ A = 2 \left(\frac{1}{2}\right) - 2 \left(\frac{1}{3}\right) = 1 - \frac{2}{3} = \frac{1}{3} \]

### Final Answer
The area enclosed between the curves \( y = x^2 \) and \( y = 2x - x^2 \) is:

\[ \boxed{\frac{1}{3}} \text{ square units} \]