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. `(22)/(3)` sq. units
C. `(29)/(3)` sq. units
D. 27 sq. units

Answer

To find the area bounded by the curve \( y = f(x) \) and the x-axis from \( x = -1 \) to \( x = 3 \), we first need to define the function \( f(x) \) based on the given piecewise definition:

\[ f(x) = \begin{cases} x^2 + 4 & \text{if } x < 0 \\ 4 - 2x & \text{if } x \geq 0 \end{cases} \]

### Step 1: Set Up the Integral
Since the function is defined piecewise, we will break the integral into two parts: one from \( x = -1 \) to \( x = 0 \) and the other from \( x = 0 \) to \( x = 3 \).

The area \( A \) can be calculated as:

\[ A = \int_{-1}^{0} f(x) \, dx + \int_{0}^{3} f(x) \, dx \]

### Step 2: Calculate the First Integral
For \( x < 0 \), we have \( f(x) = x^2 + 4 \). Thus, we calculate:

\[ \int_{-1}^{0} (x^2 + 4) \, dx \]

Calculating this integral:

\[ \int (x^2 + 4) \, dx = \frac{x^3}{3} + 4x \]

Now we evaluate from \( -1 \) to \( 0 \):

\[ \left[ \frac{0^3}{3} + 4(0) \right] - \left[ \frac{(-1)^3}{3} + 4(-1) \right] = 0 - \left[ -\frac{1}{3} - 4 \right] \]

This simplifies to:

\[ 0 + \frac{1}{3} + 4 = \frac{1}{3} + \frac{12}{3} = \frac{13}{3} \]

### Step 3: Calculate the Second Integral
For \( x \geq 0 \), we have \( f(x) = 4 - 2x \). Thus, we calculate:

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

Calculating this integral:

\[ \int (4 - 2x) \, dx = 4x - x^2 \]

Now we evaluate from \( 0 \) to \( 3 \):

\[ \left[ 4(3) - (3)^2 \right] - \left[ 4(0) - (0)^2 \right] = (12 - 9) - 0 = 3 \]

### Step 4: Combine the Areas
Now we combine the areas from both integrals:

\[ A = \frac{13}{3} + 3 = \frac{13}{3} + \frac{9}{3} = \frac{22}{3} \]

### Final Answer
Thus, the area bounded by \( y = f(x) \) and the x-axis from \( x = -1 \) to \( x = 3 \) is:

\[ \boxed{\frac{22}{3}} \]