Area bounded by region `0leyle4x-x^2-3` is

Answer

To find the area bounded by the region defined by the inequalities \(0 \leq y \leq 4x - x^2 - 3\), we will follow these steps:

### Step 1: Set up the equation
We start with the equation of the curve:
\[ y = 4x - x^2 - 3 \]
We need to find the points where this curve intersects the x-axis, which occurs when \(y = 0\).

### Step 2: Solve for x-intercepts
Setting \(y = 0\):
\[ 0 = 4x - x^2 - 3 \]
Rearranging gives:
\[ x^2 - 4x + 3 = 0 \]
This is a quadratic equation, which we can factor:
\[ (x - 1)(x - 3) = 0 \]
Thus, the solutions are:
\[ x = 1 \quad \text{and} \quad x = 3 \]
These are the x-coordinates of the points where the curve intersects the x-axis.

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

### Step 4: Compute the integral
We will now compute the integral:
\[ A = \int_{1}^{3} (4x - x^2 - 3) \, dx \]
Calculating the integral:
\[ = \left[ 2x^2 - \frac{x^3}{3} - 3x \right]_{1}^{3} \]

### Step 5: Evaluate the definite integral
Now, we evaluate the expression at the limits:
1. For \(x = 3\):
\[ 2(3^2) - \frac{(3^3)}{3} - 3(3) = 2(9) - \frac{27}{3} - 9 = 18 - 9 - 9 = 0 \]
2. For \(x = 1\):
\[ 2(1^2) - \frac{(1^3)}{3} - 3(1) = 2(1) - \frac{1}{3} - 3 = 2 - \frac{1}{3} - 3 = -1 - \frac{1}{3} = -\frac{4}{3} \]

Now, substituting back into the integral:
\[ A = 0 - \left(-\frac{4}{3}\right) = \frac{4}{3} \]

### Final Answer
Thus, the area bounded by the region is:
\[ \boxed{\frac{4}{3}} \text{ square units} \]

---