Let `f(x)={{:(2^([x]),, x lt0),(2^([-x]),,x ge0):}`, then the area bounded by `y=f(x) and y=0` is

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To find the area bounded by the function \( f(x) \) and the x-axis, we need to analyze the function defined as:

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

### Step 1: Analyze the function for \( x < 0 \)

For \( x < 0 \):
- When \( -1 < x < 0 \), \( \lfloor x \rfloor = -1 \) so \( f(x) = 2^{-1} = \frac{1}{2} \).
- When \( -2 < x < -1 \), \( \lfloor x \rfloor = -2 \) so \( f(x) = 2^{-2} = \frac{1}{4} \).
- When \( -3 < x < -2 \), \( \lfloor x \rfloor = -3 \) so \( f(x) = 2^{-3} = \frac{1}{8} \).
- This pattern continues for \( x < -3 \).

### Step 2: Analyze the function for \( x \geq 0 \)

For \( x \geq 0 \):
- When \( 0 \leq x < 1 \), \( \lfloor x \rfloor = 0 \) so \( f(x) = 2^{0} = 1 \).
- When \( 1 \leq x < 2 \), \( \lfloor x \rfloor = 1 \) so \( f(x) = 2^{-1} = \frac{1}{2} \).
- When \( 2 \leq x < 3 \), \( \lfloor x \rfloor = 2 \) so \( f(x) = 2^{-2} = \frac{1}{4} \).
- This pattern continues for \( x \geq 3 \).

### Step 3: Calculate the area for \( x < 0 \)

The area under the curve for \( x < 0 \) can be calculated as follows:

- From \( -1 \) to \( 0 \): Area = \( \frac{1}{2} \times 1 = \frac{1}{2} \)
- From \( -2 \) to \( -1 \): Area = \( \frac{1}{4} \times 1 = \frac{1}{4} \)
- From \( -3 \) to \( -2 \): Area = \( \frac{1}{8} \times 1 = \frac{1}{8} \)

This continues indefinitely, forming a geometric series:

\[ \text{Area} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \]

### Step 4: Calculate the area for \( x \geq 0 \)

The area under the curve for \( x \geq 0 \) is similar:

- From \( 0 \) to \( 1 \): Area = \( 1 \times 1 = 1 \)
- From \( 1 \) to \( 2 \): Area = \( \frac{1}{2} \times 1 = \frac{1}{2} \)
- From \( 2 \) to \( 3 \): Area = \( \frac{1}{4} \times 1 = \frac{1}{4} \)

This also forms a geometric series:

\[ \text{Area} = 1 + \frac{1}{2} + \frac{1}{4} + \ldots \]

### Step 5: Sum the areas

The total area bounded by \( f(x) \) and the x-axis is:

\[ \text{Total Area} = 2 \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \right) + \left( 1 + \frac{1}{2} + \frac{1}{4} + \ldots \right) \]

Using the formula for the sum of an infinite geometric series \( S = \frac{a}{1 - r} \):

For the series \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \):
- First term \( a = \frac{1}{2} \), common ratio \( r = \frac{1}{2} \)
- Sum = \( \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1 \)

For the series \( 1 + \frac{1}{2} + \frac{1}{4} + \ldots \):
- First term \( a = 1 \), common ratio \( r = \frac{1}{2} \)
- Sum = \( \frac{1}{1 - \frac{1}{2}} = 2 \)

Thus, the total area is:

\[ \text{Total Area} = 2 \times 1 + 2 = 4 \]

### Final Answer

The area bounded by \( y = f(x) \) and \( y = 0 \) is \( 2 \).