`f(x)=min({x,x^(2)}` and `g(x)=max{x,x^(2)}`

Options

A. `int_(-1)^(1)f(x)=-(1)/(6)`
B. `int_(-2)^(1)g(x)=(19)/(6)`
C. `int_(-1)^(1)f(x)=(19)/(6)`
D. `int_(-1)^(1)g(x)=-(1)/(6)`

Explanation

To solve the problem, we need to analyze the functions \( f(x) = \min(x, x^2) \) and \( g(x) = \max(x, x^2) \). We will find the values of these functions over specific intervals.

### Step 1: Analyze \( f(x) = \min(x, x^2) \)

1. **Identify the points where \( x = x^2 \)**:
\[ x = x^2 \implies x^2 - x = 0 \implies x(x - 1) = 0 \]
This gives us the points \( x = 0 \) and \( x = 1 \).

2. **Determine the intervals**:
We will analyze the function on the intervals \( (-\infty, 0) \), \( [0, 1] \), and \( (1, \infty) \).

3. **Evaluate \( f(x) \) in each interval**:
- For \( x < 0 \): \( x^2 > x \) (since \( x^2 \) is positive and \( x \) is negative), thus \( f(x) = x^2 \).
- For \( 0 \leq x \leq 1 \): \( x^2 \leq x \), thus \( f(x) = x^2 \).
- For \( x > 1 \): \( x^2 > x \), thus \( f(x) = x \).

### Step 2: Set up the integral for \( f(x) \)

We need to find the integral of \( f(x) \) over the intervals:
- From \( -1 \) to \( 0 \): \( f(x) = x^2 \)
- From \( 0 \) to \( 1 \): \( f(x) = x^2 \)

The integral can be expressed as:
\[ \int_{-1}^{0} x^2 \, dx + \int_{0}^{1} x^2 \, dx \]

### Step 3: Calculate the integrals

1. **Calculate \( \int x^2 \, dx \)**:
\[ \int x^2 \, dx = \frac{x^3}{3} + C \]

2. **Evaluate the first integral**:
\[ \int_{-1}^{0} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{0} = \left(0 - \left(-\frac{1}{3}\right)\right) = \frac{1}{3} \]

3. **Evaluate the second integral**:
\[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \left(\frac{1}{3} - 0\right) = \frac{1}{3} \]

### Step 4: Combine the results for \( f(x) \)

Adding both integrals:
\[ \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]

### Step 5: Analyze \( g(x) = \max(x, x^2) \)

1. **Evaluate \( g(x) \) in each interval**:
- For \( x < 0 \): \( g(x) = x \) (since \( x \) is greater than \( x^2 \)).
- For \( 0 \leq x \leq 1 \): \( g(x) = x \).
- For \( x > 1 \): \( g(x) = x^2 \).

### Step 6: Set up the integral for \( g(x) \)

We need to find the integral of \( g(x) \) over the intervals:
- From \( -1 \) to \( 0 \): \( g(x) = x \)
- From \( 0 \) to \( 1 \): \( g(x) = x \)
- From \( 1 \) to \( 2 \): \( g(x) = x^2 \)

The integral can be expressed as:
\[ \int_{-1}^{0} x \, dx + \int_{0}^{1} x \, dx + \int_{1}^{2} x^2 \, dx \]

### Step 7: Calculate the integrals for \( g(x) \)

1. **Calculate \( \int x \, dx \)**:
\[ \int x \, dx = \frac{x^2}{2} + C \]

2. **Evaluate the first integral**:
\[ \int_{-1}^{0} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{0} = \left(0 - \frac{1}{2}\right) = -\frac{1}{2} \]

3. **Evaluate the second integral**:
\[ \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \left(\frac{1}{2} - 0\right) = \frac{1}{2} \]

4. **Evaluate the third integral**:
\[ \int_{1}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{1}^{2} = \left(\frac{8}{3} - \frac{1}{3}\right) = \frac{7}{3} \]

### Step 8: Combine the results for \( g(x) \)

Adding all integrals:
\[ \int_{-1}^{0} g(x) \, dx + \int_{0}^{1} g(x) \, dx + \int_{1}^{2} g(x) \, dx = -\frac{1}{2} + \frac{1}{2} + \frac{7}{3} = \frac{7}{3} \]

### Final Results

- The value of \( f(x) \) is \( \frac{2}{3} \).
- The value of \( g(x) \) is \( \frac{7}{3} \).