The area (in sq. units) bounded by `y=4x-x^2` and `y=x`is

Options

A. `2/3`
B. `7/2`
C. `9/2`
D. `5/2`

Answer

To find the area bounded by the curves \( y = 4x - x^2 \) and \( y = x \), we will follow these steps:

### Step 1: Find the points of intersection
We need to determine where the two curves intersect by setting them equal to each other:

\[ 4x - x^2 = x \]

Rearranging the equation gives:

\[ -x^2 + 4x - x = 0 \implies -x^2 + 3x = 0 \]

Factoring out \( x \):

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

This gives us the solutions:

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

### Step 2: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 3 \) can be found using the integral:

\[ A = \int_{0}^{3} \left( (4x - x^2) - x \right) \, dx \]

This simplifies to:

\[ A = \int_{0}^{3} (4x - x^2 - x) \, dx = \int_{0}^{3} (3x - x^2) \, dx \]

### Step 3: Calculate the integral
Now, we will calculate the integral:

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

Calculating the integral:

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

Evaluating at the bounds:

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

Calculating the values:

\[ = \left( \frac{3}{2} \cdot 9 - \frac{1}{3} \cdot 27 \right) \]

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

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

\[ \boxed{\frac{9}{2}} \text{ square units} \]

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