Find the area bounded by the curve `y=2x-x^(2)`, and the line `y=x`

Answer

To find the area bounded by the curve \( y = 2x - x^2 \) and the line \( y = x \), we can follow these steps:

### Step 1: Find the points of intersection
To find the points where the curve and the line intersect, we set the equations equal to each other:

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

Rearranging gives:

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

Factoring out \( x \):

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

Thus, the solutions are:

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

### Step 2: Set up the integral for the area
The area \( A \) between the curve and the line from \( x = 0 \) to \( x = 1 \) can be found using the integral:

\[ A = \int_{0}^{1} (y_{\text{upper}} - y_{\text{lower}}) \, dx \]

Here, \( y_{\text{upper}} = 2x - x^2 \) (the curve) and \( y_{\text{lower}} = x \) (the line). Therefore, we have:

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

Simplifying the integrand:

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

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

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

Calculating the integral:

\[ A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} \]

Substituting the limits:

\[ A = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) \]

This simplifies to:

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

### Step 4: Simplify the result
Finding a common denominator (which is 6):

\[ A = \left( \frac{3}{6} - \frac{2}{6} \right) = \frac{1}{6} \]

### Conclusion
The area bounded by the curve \( y = 2x - x^2 \) and the line \( y = x \) is:

\[ \boxed{\frac{1}{6}} \text{ square units} \]