Consider the curves
`f(x)=x|x|-1andg(x)={{:((3x)/(2)","xgt0),(2x","xle0):}`
What is the area bounded by the curves?

Options

A. `(17)/(6)` square units
B. `(8)/(3)` square units
C. 2 square units
D. `(1)/(3)` square unit

Explanation

To find the area bounded by the curves \( f(x) = x|x| - 1 \) and \( g(x) = \begin{cases} \frac{3x}{2} & \text{if } x > 0 \\ 2x & \text{if } x \leq 0 \end{cases} \), we will follow these steps:

### Step 1: Determine the expressions for \( f(x) \)
The function \( f(x) = x|x| - 1 \) can be expressed as:
- For \( x \geq 0 \): \( f(x) = x^2 - 1 \)
- For \( x < 0 \): \( f(x) = -x^2 - 1 \)

### Step 2: Determine the expressions for \( g(x) \)
The function \( g(x) \) is already defined piecewise:
- For \( x > 0 \): \( g(x) = \frac{3x}{2} \)
- For \( x \leq 0 \): \( g(x) = 2x \)

### Step 3: Find the points of intersection
To find the area bounded by the curves, we need to find the points where \( f(x) = g(x) \).

1. **For \( x > 0 \)**:
\[ x^2 - 1 = \frac{3x}{2} \]
Rearranging gives:
\[ x^2 - \frac{3x}{2} - 1 = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ 2x^2 - 3x - 2 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4} \]
This gives us:
\[ x = 2 \quad \text{and} \quad x = -\frac{1}{2} \]

2. **For \( x \leq 0 \)**:
\[ -x^2 - 1 = 2x \]
Rearranging gives:
\[ -x^2 - 2x - 1 = 0 \quad \Rightarrow \quad x^2 + 2x + 1 = 0 \quad \Rightarrow \quad (x + 1)^2 = 0 \]
Thus, \( x = -1 \).

### Step 4: Identify the area between the curves
The points of intersection are \( x = -1 \) and \( x = 2 \).

### Step 5: Set up the integral for the area
The area \( A \) between the curves from \( x = -1 \) to \( x = 2 \) is given by:
\[ A = \int_{-1}^{0} (g(x) - f(x)) \, dx + \int_{0}^{2} (g(x) - f(x)) \, dx \]

Calculating each integral separately:

1. **For \( x \in [-1, 0] \)**:
\[ A_1 = \int_{-1}^{0} (2x - (-x^2 - 1)) \, dx = \int_{-1}^{0} (2x + x^2 + 1) \, dx \]
\[ = \left[ x^2 + \frac{x^3}{3} + x \right]_{-1}^{0} = \left[ 0 + 0 + 0 \right] - \left[ 1 - \frac{1}{3} - 1 \right] = 0 - \left[ \frac{1}{3} \right] = \frac{1}{3} \]

2. **For \( x \in [0, 2] \)**:
\[ A_2 = \int_{0}^{2} \left(\frac{3x}{2} - (x^2 - 1)\right) \, dx = \int_{0}^{2} \left(\frac{3x}{2} - x^2 + 1\right) \, dx \]
\[ = \left[\frac{3x^2}{4} - \frac{x^3}{3} + x \right]_{0}^{2} \]
\[ = \left[\frac{3(2^2)}{4} - \frac{(2^3)}{3} + 2\right] - \left[0\right] = \left[\frac{12}{4} - \frac{8}{3} + 2\right] = 3 - \frac{8}{3} + 2 = 5 - \frac{8}{3} = \frac{15 - 8}{3} = \frac{7}{3} \]

### Step 6: Combine the areas
The total area \( A \) is:
\[ A = A_1 + A_2 = \frac{1}{3} + \frac{7}{3} = \frac{8}{3} \]

Thus, the area bounded by the curves is \( \frac{8}{3} \) square units.