If `f(x)={{:(x-2",","for "x ge0),((x^(2))/(4)-2","," for "x lt0):}`, then find area bounded by `y=|f(x)|` and x - axis.

Answer

To find the area bounded by \( y = |f(x)| \) and the x-axis, we first need to analyze the function \( f(x) \) given by:

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

### Step 1: Find the points where \( f(x) = 0 \)

1. For \( x \geq 0 \):
\[ x - 2 = 0 \implies x = 2 \]

2. For \( x < 0 \):
\[ \frac{x^2}{4} - 2 = 0 \implies \frac{x^2}{4} = 2 \implies x^2 = 8 \implies x = -2\sqrt{2} \text{ (since } x < 0\text{)} \]

### Step 2: Determine the intervals for integration

The function \( f(x) \) changes at \( x = -2\sqrt{2} \) and \( x = 2 \). We will calculate the area in two parts: from \( -2\sqrt{2} \) to \( 0 \) and from \( 0 \) to \( 2 \).

### Step 3: Calculate the area from \( -2\sqrt{2} \) to \( 0 \)

In this interval, \( f(x) = \frac{x^2}{4} - 2 \). The area under the curve is given by:

\[ \text{Area}_1 = \int_{-2\sqrt{2}}^{0} |f(x)| \, dx = \int_{-2\sqrt{2}}^{0} \left(2 - \frac{x^2}{4}\right) \, dx \]

### Step 4: Calculate the area from \( 0 \) to \( 2 \)

In this interval, \( f(x) = x - 2 \). The area under the curve is given by:

\[ \text{Area}_2 = \int_{0}^{2} |f(x)| \, dx = \int_{0}^{2} (2 - x) \, dx \]

### Step 5: Evaluate the integrals

1. **For Area 1:**
\[ \text{Area}_1 = \int_{-2\sqrt{2}}^{0} \left(2 - \frac{x^2}{4}\right) \, dx \]
\[ = \left[ 2x - \frac{x^3}{12} \right]_{-2\sqrt{2}}^{0} \]
\[ = \left( 0 - 0 \right) - \left( 2(-2\sqrt{2}) - \frac{(-2\sqrt{2})^3}{12} \right) \]
\[ = 4\sqrt{2} + \frac{8\sqrt{2}}{12} = 4\sqrt{2} + \frac{2\sqrt{2}}{3} = 4\sqrt{2} + \frac{2\sqrt{2}}{3} \]
\[ = \frac{12\sqrt{2}}{3} + \frac{2\sqrt{2}}{3} = \frac{14\sqrt{2}}{3} \]

2. **For Area 2:**
\[ \text{Area}_2 = \int_{0}^{2} (2 - x) \, dx \]
\[ = \left[ 2x - \frac{x^2}{2} \right]_{0}^{2} \]
\[ = \left( 4 - 2 \right) - (0) = 2 \]

### Step 6: Total Area

The total area \( A \) is the sum of the two areas:

\[ A = \text{Area}_1 + \text{Area}_2 = \frac{14\sqrt{2}}{3} + 2 \]

### Final Answer

Thus, the area bounded by \( y = |f(x)| \) and the x-axis is:

\[ A = \frac{14\sqrt{2}}{3} + 2 \]