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

Options

A. `9//2`
B. `43//6`
C. `35//6`
D. `10//3`

Answer

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

### Step 1: Find the Points of Intersection
We need to find the points where the two curves intersect. 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 solutions are:

\[ x = 0 \quad \text{and} \quad x = 3 \]

### Step 2: Determine the Area Between the Curves
The area between the curves from \( x = 0 \) to \( x = 3 \) can be calculated by integrating the difference of the functions:

\[ \text{Area} = \int_{0}^{3} ((2x - x^2) - (-x)) \, dx = \int_{0}^{3} (2x - x^2 + x) \, dx = \int_{0}^{3} (3x - x^2) \, dx \]

### Step 3: Compute the Integral
Now we compute the integral:

\[ \int (3x - x^2) \, dx = \frac{3x^2}{2} - \frac{x^3}{3} \]

Evaluating this from \( 0 \) to \( 3 \):

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

Calculating the first part:

\[ = \frac{3 \cdot 9}{2} - \frac{27}{3} = \frac{27}{2} - 9 = \frac{27}{2} - \frac{18}{2} = \frac{9}{2} \]

### Step 4: Final Area Calculation
Thus, the area bounded by the curves \( y = 2x - x^2 \) and \( y = -x \) is:

\[ \text{Area} = \frac{9}{2} \text{ square units} \]