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

Options

A. `(3)/(2)` sq units
B. `(9)/(3)` sq units
C. `(9)/(2)` sq units
D. None of these

Explanation

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

### Step 1: Find the points of intersection
We need to find the points where the curve intersects the line. This is done by setting the equations equal to each other:

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

Rearranging gives:

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

Factoring out \( x \):

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

Thus, the points of intersection are \( x = 0 \) and \( x = 3 \).

### Step 2: Determine the area above and below the x-axis
The area can be split into two parts:
1. Area \( A_1 \) from \( x = 0 \) to \( x = 2 \) (where the curve is above the line).
2. Area \( A_2 \) from \( x = 2 \) to \( x = 3 \) (where the line is below the curve).

### Step 3: Set up the integrals
The area \( A_1 \) can be calculated using the integral of the curve minus the line:

\[ A_1 = \int_0^2 ((2x - x^2) - (-x)) \, dx = \int_0^2 (2x - x^2 + x) \, dx = \int_0^2 (3x - x^2) \, dx \]

The area \( A_2 \) can be calculated using the integral of the line minus the curve:

\[ A_2 = \int_2^3 ((-x) - (2x - x^2)) \, dx = \int_2^3 (-x - 2x + x^2) \, dx = \int_2^3 (x^2 - 3x) \, dx \]

### Step 4: Calculate the integrals
Calculating \( A_1 \):

\[ A_1 = \int_0^2 (3x - x^2) \, dx = \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_0^2 \]

Evaluating at the limits:

\[ = \left( \frac{3(2)^2}{2} - \frac{(2)^3}{3} \right) - \left( 0 \right) = \left( \frac{12}{2} - \frac{8}{3} \right) = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3} \]

Calculating \( A_2 \):

\[ A_2 = \int_2^3 (x^2 - 3x) \, dx = \left[ \frac{x^3}{3} - \frac{3x^2}{2} \right]_2^3 \]

Evaluating at the limits:

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

Calculating each term:

\[ = \left( \frac{27}{3} - \frac{27}{2} \right) - \left( \frac{8}{3} - 6 \right) = (9 - 13.5) - \left( \frac{8}{3} - \frac{18}{3} \right) \]

This simplifies to:

\[ = -4.5 + \frac{10}{3} = -4.5 + 3.33 = -1.17 \]

Taking the absolute value for area:

\[ A_2 = \frac{19}{6} \]

### Step 5: Total area
The total area \( A \) is the sum of the two areas:

\[ A = A_1 + |A_2| = \frac{10}{3} + \frac{19}{6} \]

Finding a common denominator:

\[ = \frac{20}{6} + \frac{19}{6} = \frac{39}{6} = \frac{13}{2} \]

### Final Answer
Thus, the area bounded by the curve and the line is:

\[ \frac{13}{2} \text{ square units} \]