`int_(-2)^(2) |x(x-1)|dx`=

Options

A. `(16)/(3)`
B. `(17)/(3)`
C. `(11)/(3)`
D. `(13)/(3)`

Answer

To solve the integral \( I = \int_{-2}^{2} |x(x-1)| \, dx \), we first need to analyze the expression inside the absolute value, \( |x(x-1)| \).

### Step 1: Identify the points where \( x(x-1) = 0 \)

The expression \( x(x-1) \) is zero at:
- \( x = 0 \)
- \( x = 1 \)

These points will help us determine the intervals for integration.

### Step 2: Determine the sign of \( x(x-1) \) in each interval

We will evaluate the sign of \( x(x-1) \) in the intervals:
1. \( (-\infty, 0) \)
2. \( (0, 1) \)
3. \( (1, \infty) \)

- For \( x < 0 \): \( x(x-1) > 0 \) (since both \( x \) and \( x-1 \) are negative)
- For \( 0 < x < 1 \): \( x(x-1) < 0 \) (since \( x \) is positive and \( x-1 \) is negative)
- For \( x > 1 \): \( x(x-1) > 0 \) (since both \( x \) and \( x-1 \) are positive)

### Step 3: Rewrite the integral using the absolute value

Now we can express the integral as:
\[ I = \int_{-2}^{0} -x(x-1) \, dx + \int_{0}^{1} -x(x-1) \, dx + \int_{1}^{2} x(x-1) \, dx \]

### Step 4: Calculate each integral

1. **For \( \int_{-2}^{0} -x(x-1) \, dx \)**:
\[ = \int_{-2}^{0} (-x^2 + x) \, dx \]
\[ = \left[-\frac{x^3}{3} + \frac{x^2}{2}\right]_{-2}^{0} \]
Evaluating the limits:
\[ = \left(0\right) - \left[-\frac{(-2)^3}{3} + \frac{(-2)^2}{2}\right] \]
\[ = 0 - \left[\frac{8}{3} - 2\right] = 0 - \left[\frac{8}{3} - \frac{6}{3}\right] = -\left[\frac{2}{3}\right] = \frac{2}{3} \]

2. **For \( \int_{0}^{1} -x(x-1) \, dx \)**:
\[ = \int_{0}^{1} (-x^2 + x) \, dx \]
\[ = \left[-\frac{x^3}{3} + \frac{x^2}{2}\right]_{0}^{1} \]
Evaluating the limits:
\[ = \left[-\frac{1^3}{3} + \frac{1^2}{2}\right] - \left(0\right) \]
\[ = -\frac{1}{3} + \frac{1}{2} = \frac{-2 + 3}{6} = \frac{1}{6} \]

3. **For \( \int_{1}^{2} x(x-1) \, dx \)**:
\[ = \int_{1}^{2} (x^2 - x) \, dx \]
\[ = \left[\frac{x^3}{3} - \frac{x^2}{2}\right]_{1}^{2} \]
Evaluating the limits:
\[ = \left[\frac{2^3}{3} - \frac{2^2}{2}\right] - \left[\frac{1^3}{3} - \frac{1^2}{2}\right] \]
\[ = \left[\frac{8}{3} - 2\right] - \left[\frac{1}{3} - \frac{1}{2}\right] \]
\[ = \left[\frac{8}{3} - \frac{6}{3}\right] - \left[\frac{1}{3} - \frac{3}{6}\right] \]
\[ = \frac{2}{3} - \left[\frac{1}{3} - \frac{1.5}{3}\right] = \frac{2}{3} - \frac{-0.5}{3} = \frac{2}{3} + \frac{0.5}{3} = \frac{2.5}{3} = \frac{5}{6} \]

### Step 5: Combine the results

Now we can combine the results from all three integrals:
\[ I = \frac{2}{3} + \frac{1}{6} + \frac{5}{6} \]

Finding a common denominator (which is 6):
\[ = \frac{4}{6} + \frac{1}{6} + \frac{5}{6} = \frac{10}{6} = \frac{5}{3} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-2}^{2} |x(x-1)| \, dx = \frac{5}{3} \]