Let `f(x)={{:(x^(2)+4,":",xlt0),(4-2x,":",xge0):}` then the area bounded by `y=f(x)` and the x - axis from `x=-1" to "x=3` is equal to

Options

A. 9 sq. units
B. `(28)/(3)` sq. units
C. `(29)/(3)` sq. units
D. 27 sq. units

Answer

To find the area bounded by the function \( f(x) \) and the x-axis from \( x = -1 \) to \( x = 3 \), we will break down the problem step by step.

### Step 1: Understand the function \( f(x) \)
The function is defined as:
\[ f(x) = \begin{cases} x^2 + 4 & \text{if } x < 0 \\ 4 - 2x & \text{if } x \geq 0 \end{cases} \]

### Step 2: Determine the area from \( x = -1 \) to \( x = 0 \)
For \( x \) in the interval \([-1, 0]\), we use the first part of the function \( f(x) = x^2 + 4 \).

To find the area under the curve from \( x = -1 \) to \( x = 0 \), we calculate the integral:
\[ \text{Area}_1 = \int_{-1}^{0} (x^2 + 4) \, dx \]

### Step 3: Calculate the integral
Calculating the integral:
\[ \int (x^2 + 4) \, dx = \frac{x^3}{3} + 4x \]
Now, we evaluate it from \(-1\) to \(0\):
\[ \text{Area}_1 = \left[ \frac{x^3}{3} + 4x \right]_{-1}^{0} = \left( \frac{0^3}{3} + 4 \cdot 0 \right) - \left( \frac{(-1)^3}{3} + 4 \cdot (-1) \right) \]
\[ = 0 - \left( -\frac{1}{3} - 4 \right) = 0 + \frac{1}{3} + 4 = \frac{1}{3} + 4 = \frac{1}{3} + \frac{12}{3} = \frac{13}{3} \]

### Step 4: Determine the area from \( x = 0 \) to \( x = 3 \)
For \( x \) in the interval \([0, 3]\), we use the second part of the function \( f(x) = 4 - 2x \).

To find the area under the curve from \( x = 0 \) to \( x = 3 \), we calculate the integral:
\[ \text{Area}_2 = \int_{0}^{3} (4 - 2x) \, dx \]

### Step 5: Calculate the integral
Calculating the integral:
\[ \int (4 - 2x) \, dx = 4x - x^2 \]
Now, we evaluate it from \(0\) to \(3\):
\[ \text{Area}_2 = \left[ 4x - x^2 \right]_{0}^{3} = \left( 4 \cdot 3 - 3^2 \right) - \left( 4 \cdot 0 - 0^2 \right) \]
\[ = (12 - 9) - 0 = 3 \]

### Step 6: Total area
Now, we add the two areas together:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{13}{3} + 3 = \frac{13}{3} + \frac{9}{3} = \frac{22}{3} \]

### Final Answer
The area bounded by \( y = f(x) \) and the x-axis from \( x = -1 \) to \( x = 3 \) is:
\[ \frac{22}{3} \text{ square units} \]