Let f(x) be maximum and g(x) be minimum of `{x|x|, x^(2)|x|}` then `int_(-1)^(1)[f(x)-g(x)]dx=`

Options

A. `(1)/(12)`
B. `(1)/(3)`
C. `(2)/(3)`
D. `(7)/(12)`

Answer

To solve the problem, we need to find the integral of the difference between the maximum function \( f(x) \) and the minimum function \( g(x) \) over the interval \([-1, 1]\).

### Step 1: Define the functions \( f(x) \) and \( g(x) \)

Given:
- \( f(x) = \max\{|x|^3, x^2|x|\} \)
- \( g(x) = \min\{|x|^3, x^2|x|\} \)

We will analyze the two cases separately: when \( x \) is negative and when \( x \) is positive.

### Step 2: Analyze the case when \( x < 0 \)

For \( x < 0 \):
- \( |x| = -x \)
- \( |x|^3 = (-x)^3 = -x^3 \)
- \( x^2|x| = x^2(-x) = -x^3 \)

Thus, for \( x < 0 \):
- \( f(x) = \max\{-x^3, -x^3\} = -x^3 \)
- \( g(x) = \min\{-x^3, -x^3\} = -x^3 \)

### Step 3: Analyze the case when \( x > 0 \)

For \( x > 0 \):
- \( |x| = x \)
- \( |x|^3 = x^3 \)
- \( x^2|x| = x^2 \cdot x = x^3 \)

Thus, for \( x > 0 \):
- \( f(x) = \max\{x^3, x^3\} = x^3 \)
- \( g(x) = \min\{x^3, x^3\} = x^3 \)

### Step 4: Combine the results

From the analysis:
- For \( x < 0 \): \( f(x) = -x^3 \) and \( g(x) = -x^3 \)
- For \( x > 0 \): \( f(x) = x^3 \) and \( g(x) = x^3 \)

### Step 5: Set up the integral

Now we need to compute the integral:
\[ \int_{-1}^{1} [f(x) - g(x)] \, dx \]

Since \( f(x) = g(x) \) for both intervals, we can break the integral into two parts:
\[ \int_{-1}^{0} [f(x) - g(x)] \, dx + \int_{0}^{1} [f(x) - g(x)] \, dx \]

### Step 6: Evaluate the integrals

1. For \( x < 0 \):
\[ f(x) - g(x) = -x^3 - (-x^3) = 0 \]
Thus,
\[ \int_{-1}^{0} [f(x) - g(x)] \, dx = \int_{-1}^{0} 0 \, dx = 0 \]

2. For \( x > 0 \):
\[ f(x) - g(x) = x^3 - x^3 = 0 \]
Thus,
\[ \int_{0}^{1} [f(x) - g(x)] \, dx = \int_{0}^{1} 0 \, dx = 0 \]

### Step 7: Combine the results

Combining both parts:
\[ \int_{-1}^{1} [f(x) - g(x)] \, dx = 0 + 0 = 0 \]

### Final Answer

The value of the integral is:
\[ \int_{-1}^{1} [f(x) - g(x)] \, dx = 0 \]