What is the area bounded by the curve `y=4x-x^(2)-3` and the x-axis?

Options

A. 2/3 sq unit
B. 4/3 sq unit
C. 5/3 sq unit
D. 4/5 sq unit

Answer

To find the area bounded by the curve \( y = 4x - x^2 - 3 \) and the x-axis, we will follow these steps:

### Step 1: Find the points where the curve intersects the x-axis
To find the points where the curve intersects the x-axis, we set \( y = 0 \):
\[ 4x - x^2 - 3 = 0 \]
Rearranging gives:
\[ -x^2 + 4x - 3 = 0 \]
Multiplying through by -1:
\[ x^2 - 4x + 3 = 0 \]

### Step 2: Factor the quadratic equation
Now we will factor the quadratic:
\[ (x - 3)(x - 1) = 0 \]
Setting each factor to zero gives us the solutions:
\[ x = 1 \quad \text{and} \quad x = 3 \]

### Step 3: Set up the integral to find the area
The area \( A \) bounded by the curve and the x-axis from \( x = 1 \) to \( x = 3 \) can be found using the integral:
\[ A = \int_{1}^{3} (4x - x^2 - 3) \, dx \]

### Step 4: Calculate the integral
Now we will calculate the integral:
\[ A = \int_{1}^{3} (4x - x^2 - 3) \, dx \]
First, we find the antiderivative:
\[ \int (4x - x^2 - 3) \, dx = 2x^2 - \frac{x^3}{3} - 3x \]
Now we evaluate this from 1 to 3:
\[ A = \left[ 2(3)^2 - \frac{(3)^3}{3} - 3(3) \right] - \left[ 2(1)^2 - \frac{(1)^3}{3} - 3(1) \right] \]

### Step 5: Substitute the limits into the antiderivative
Calculating the upper limit:
\[ = 2(9) - \frac{27}{3} - 9 = 18 - 9 - 9 = 0 \]
Calculating the lower limit:
\[ = 2(1) - \frac{1}{3} - 3 = 2 - \frac{1}{3} - 3 = -1 + \frac{1}{3} = -\frac{2}{3} \]

### Step 6: Find the area
Now we find the area:
\[ A = 0 - \left(-\frac{2}{3}\right) = \frac{2}{3} \]

### Final Result
Thus, the area bounded by the curve and the x-axis is:
\[ \frac{4}{3} \text{ square units} \]