The area bounded by the curve `y=4x-x^2` and x-axis is (A) `30/7` sq. units (B) `31/7` sq. units (C) `32/3` sq. units (D) `34/3` sq. units

Options

A. `(30)/(7)`
B. `(31)/(7)`
C. `(32)/(3)`
D. `(34)/(3)`

Answer

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

### Step 1: Identify the points of intersection with the x-axis
To find the points where the curve intersects the x-axis, we set \( y = 0 \):
\[ 4x - x^2 = 0 \]
Factoring gives:
\[ x(4 - x) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{and} \quad x = 4 \]

### Step 2: Set up the integral for the area
The area \( A \) bounded by the curve and the x-axis from \( x = 0 \) to \( x = 4 \) can be calculated using the definite integral:
\[ A = \int_{0}^{4} (4x - x^2) \, dx \]

### Step 3: Compute the integral
We will evaluate the integral:
\[ A = \int_{0}^{4} (4x - x^2) \, dx \]
This can be split into two separate integrals:
\[ A = \int_{0}^{4} 4x \, dx - \int_{0}^{4} x^2 \, dx \]

Calculating each integral separately:
1. For \( \int 4x \, dx \):
\[ \int 4x \, dx = 2x^2 \]
Evaluating from 0 to 4:
\[ 2(4^2) - 2(0^2) = 2(16) - 0 = 32 \]

2. For \( \int x^2 \, dx \):
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating from 0 to 4:
\[ \frac{4^3}{3} - \frac{0^3}{3} = \frac{64}{3} - 0 = \frac{64}{3} \]

### Step 4: Combine the results
Now, substituting back into the area formula:
\[ A = 32 - \frac{64}{3} \]
To combine these, convert 32 into a fraction with a denominator of 3:
\[ 32 = \frac{96}{3} \]
Thus,
\[ A = \frac{96}{3} - \frac{64}{3} = \frac{32}{3} \]

### Final Answer
The area bounded by the curve \( y = 4x - x^2 \) and the x-axis is:
\[ \boxed{\frac{32}{3}} \text{ sq. units} \]